forallx (UBC edition)An Introduction to Formal Logicthis copy compiled October 2, 2019P.D. MagnusUniversity at Albany, State University of New YorkJonathan Jenkins IchikawaUniversity of British ColumbiaUBC edition version 1.2 [191002]source material: fecundity.com/logic, v1.30This book is offered under a Creative Commons license.(Attribution-ShareAlike 3.0)The original author, P.D. Magnus, would like to thank the people who made thisproject possible. Notable among these are Cristyn Magnus, who read many earlydrafts; Aaron Schiller, who was an early adopter and provided considerable, helpfulfeedback; and Bin Kang, Craig Erb, Nathan Carter, Wes McMichael, Selva Samuel,Dave Krueger, Brandon Lee, Toan Tran, and the students of Introduction to Logic,who detected various errors in previous versions of the book.Jonathan Ichikawa, who modified the Magnus edition to create this version, would,first and foremost, like to thank P.D. Magnus. He is also grateful to Greg Restall,and his book Logic, which is the first text he taught from and which is the modelfor the discussion of trees in this book. He also thanks the many students in hisIntroduction to Logic course at UBC in 2017 and 2018, who test-drove the draftsof the book and caught many errors. He’s also grateful for helpful conversationswith Roberta Ballarin, Roy Cook, Josh Dever, Dave Gilbert, Thony Gillies, and JonShaheen about various matters, for careful proofreading to Cavell Chan and JoeyDeeth, and for technical help to Laura Greenstreet.c© 2005–2019 by P.D. Magnus and Jonathan Ichikawa. Some rights reserved.You are free to copy this book, to distribute it, to display it, and to make derivative works,under the following conditions: (a) Attribution. You must give the original author credit. (b)Share Alike. If you alter, transform, or build upon this work, you may distribute the resultingwork only under a license identical to this one. — For any reuse or distribution, you must makeclear to others the license terms of this work. Any of these conditions can be waived if you getpermission from the copyright holder. Your fair use and other rights are in no way affected bythe above. — This is a human-readable summary of the full license, which is available on-lineat http://creativecommons.org/licenses/by-sa/3.0/Typesetting was carried out entirely in LATEX2ε. The style for typesetting naturaldeduction proofs is based on fitch.sty (v0.4) by Peter Selinger, University ofOttawa. Tree proofs typesetting is from prooftrees (v0.6) by Clea F. Rees atCardiff University.This copy of forallx (UBC edition) is current as of October 2, 2019.Preface to the UBC EditionThis preface outlines my approach to teaching logic, and explains the way thisversion of forall x differs from Magnus’s original. The preface is intended morefor instructors than for students.I have been teaching logic at the University of British Columbia since 2011;starting in 2017, I decided to prepare this textbook, based on and incorporatingmuch of P. D. Magnus’s forall x, which has been freely available for use andmodification since 2005. Preparing this text had two main advantages for me: itallowed me to tailor the text precisely to my teaching preferences and emphasis,and, because it is available for free, it is saving money for students. (I encourageinstructors to take this latter consideration pretty seriously. If you have ahundred students a year, requiring them each to buy a $50 textbook takes $5,000out of students’ pockets each year. If you teach with this or another free bookinstead, you’ll save your students $50,000 over ten years. It can be sort ofannoying to switch textbooks if you’re used to something already. But is stayingthe course worth $50,000 of your students’ money?)This text was designed for a one-semester, thirteen-week course with no prereq-uisites, introducing formal logic. At UBC, the course has quite a mix of studentswith diverse academic backgrounds. For many it is their first philosophy course.As I teach Introduction to Formal Logic, the course has three central aims: (1) tohelp students think more clearly about arguments and argumentative structure,in a way applicable to informal arguments in philosophy and elsewhere; (2) toprovide some familiarity and comfort with formal proof systems, including prac-tice setting out formal proofs with each step justified by a syntactically-definedrule; and (3) to provide the conceptual groundwork for metatheoretical proofs,introducing the ideas of rigorous informal proofs about formal systems, preparingstudents for possible future courses focusing on metalogic and computability. Itry to give those three elements roughly equal focus in my course, and in thisbook.The book introduces two different kinds of formal proof systems — analytictableaux (‘trees’) and Fitch-style natural deduction. Unlike many logic texts,34it puts its greater emphasis on trees. There are two reasons I have found thisto be useful. One is that the algorithmic nature of tree proofs means thatone can be assured to achieve successful proofs on the basis of patience andcareful diligence, as opposed to requiring a difficult-to-quantify (and difficult-to-teach) ‘flash of insight’. The other is that the soundness and completenesstheorems for tree methods are simpler and more intuitive than they are fornatural deduction systems, and I find it valuable to expose students to proofsof significant metatheoretical results early in their logical studies. (I provesoundness and completeness for a sentential logic tree system in the fifth weekof the semester.) As presented here, the soundness and completeness proofsemphasize contrasting the systems students learn with hypothetical alternativesystems that modify the rules in various ways. A rule like this would underminethe soundness of the system, but not its completeness. If we changed the rulesin this way, it would still be both sound and complete. Etc. This helps giveintuitive substance to these theorems.I also include a Fitch-style natural deduction system, both for sentential andquantified logic, both because its premise-conclusion form is particularly helpfulfor thinking about informal arguments, and because it is important to recognizeand follow proofs laid out in that kind of format, for example in more advancedphilosophical material. While students do learn to do Fitch-style proofs, Iemphasize less of that puzzle-solving kind of skill here than in many textbooks.The book begins with a systematic engagement with sentential logic in conven-tional ways: translations, sentential connectives, models, truth tables, and bothproof systems, including soundness and completeness for the tree system. Stu-dents are thereby able to become familiar with all the central metalogical ideas,incorporating relatively simple logical symbolism, before introducing predicates,quantifiers, and identity. Once we enrich the language, we go through thoseprevious ideas again, using our more complex vocabulary.The first book I used for teaching was Greg Restall’s Logic (McGill–Queen’sUniversity Press, 2006), which I used for several years. My approach to teachinglogic is heavily informed by that book; its influence in this text is particularlyclear in the discussion of trees. (The natural deduction system I use is ratherdifferent from Restall’s.)In preparing this text, I began with Magnus’s original and edited freely. Thereare sections where Magnus’s prose has been retained entirely, and many of theexercises I have taken unchanged from the original. But I have also restructuredmany things and added quite a bit of new material. Unlike my version, whichfocuses on sentential logic before introducing predicates and quantification,Magnus’s version integrated the discussion of sentential and quantificationalsystems, e.g. covering translation for both before discussing models and proofsfor either. The original also did not include trees or soundness and completenessproofs. The two chapters on trees (5 and 10) and soundness and completeness (65and 11) were written from scratch; my chapter on identity (12) is also original.The other material in this edition incorporates Magnus’s original material, someparts more heavily edited than others. I have slightly modified Magnus’s naturaldeduction rules.In December 2018 I released the 1.0 version of this text. I’d had ‘beta’ versionsavailable online for a couple years while teaching with them, but the presentversion represents a relatively stable version. A few small corrections since thenhave been implemented in this 1.2 version, as of October 2019.Many thanks, first and foremost, to P.D. Magnus for providing this wonderfulresource under a Creative Commons license, which made it freely available andgave me the right to modify and distribute it under the same licensing agreement.I hope other instructors will also feel free to either teach directly from thisversion, or to modify it to develop their own. The typesetting for trees is viaClea F. Rees’s prooftrees package; thanks to her for making it available.I’m grateful to the students in my 2017–19 PHIL 220 courses at UBC, who hadan in-progress version of this book as their course textbook. They patiently andhelpfully found and pointed out mistakes as I wrote them (incentivized, perhaps,by an offer of extra credit); this version has many fewer errors than it otherwisewould have had. Thanks also to Cavell Chan and Joey Deeth, who did carefulproofreading, and generated many solutions to exercises for the answer key, andto Laura Greenstreet for LaTeX and other technical help. These three assistantswere supported by a UBC Library Open Access Grant in 2018–19.I am maintaining a list of known issues and errors for this book, to be correctedin future editions, under ‘issues’ at https://github.com/jonathanichikawa/for-all-x. If you see any mistakes, please feel free to add them there directly,or to email me with them. The most recent version of the book is also alwaysavailable for download there too.Jonathan IchikawaUniversity of British ColumbiaOctober 2019ichikawa@gmail.comContentsPreface to the UBC edition 31 What is logic? 101.1 Arguments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.2 Sentences and propositions . . . . . . . . . . . . . . . . . . . . . 121.3 Two ways that arguments can go wrong . . . . . . . . . . . . . 131.4 Validity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151.5 Impossibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161.6 Other logical notions . . . . . . . . . . . . . . . . . . . . . . . . 171.7 Formal languages . . . . . . . . . . . . . . . . . . . . . . . . . . 19Practice Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212 Sentential logic 232.1 Sentence letters . . . . . . . . . . . . . . . . . . . . . . . . . . . 232.2 Connectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252.3 Negation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262.4 Conjunction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282.5 Disjunction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 312.6 Conditional . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 332.7 Biconditional . . . . . . . . . . . . . . . . . . . . . . . . . . . . 352.8 Other symbolization . . . . . . . . . . . . . . . . . . . . . . . . 362.9 Sentences of SL . . . . . . . . . . . . . . . . . . . . . . . . . . . 37Practice Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 423 Truth tables 453.1 Truth-functional connectives . . . . . . . . . . . . . . . . . . . . 453.2 Complete truth tables . . . . . . . . . . . . . . . . . . . . . . . 463.3 Using truth tables . . . . . . . . . . . . . . . . . . . . . . . . . 493.4 Partial truth tables . . . . . . . . . . . . . . . . . . . . . . . . . 523.5 Evaluating English arguments via SL . . . . . . . . . . . . . . . 543.6 Common student mistakes . . . . . . . . . . . . . . . . . . . . . 57Practice Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 584 Entailment and Models for SL 624.1 Semantics for SL . . . . . . . . . . . . . . . . . . . . . . . . . . 636CONTENTS 74.2 Defining truth in SL . . . . . . . . . . . . . . . . . . . . . . . . 634.3 Semantic entailment . . . . . . . . . . . . . . . . . . . . . . . . 664.4 Entailment, validity, and informally good arguments . . . . . . 664.5 Tautologies are entailed by the empty set . . . . . . . . . . . . 674.6 Inconsistent sentences entail absurdities . . . . . . . . . . . . . 684.7 Defining concepts in terms of entailment . . . . . . . . . . . . . 70Practice Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 705 SL Trees 715.1 Satisfiability and entailment . . . . . . . . . . . . . . . . . . . . 725.2 An example: proving validity . . . . . . . . . . . . . . . . . . . 735.3 An example: proving invalidity . . . . . . . . . . . . . . . . . . 755.4 Resolution rules for SL trees . . . . . . . . . . . . . . . . . . . . 775.5 Branch closure rules . . . . . . . . . . . . . . . . . . . . . . . . 825.6 Branch completion rules . . . . . . . . . . . . . . . . . . . . . . 835.7 Resolution order . . . . . . . . . . . . . . . . . . . . . . . . . . 845.8 Choosing the right root . . . . . . . . . . . . . . . . . . . . . . 85Practice Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 866 Soundness and Completeness for SL Trees 886.1 Informal proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . 896.2 Soundness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 896.3 Recursive proofs . . . . . . . . . . . . . . . . . . . . . . . . . . 906.4 Proving soundness . . . . . . . . . . . . . . . . . . . . . . . . . 916.5 Completeness . . . . . . . . . . . . . . . . . . . . . . . . . . . . 976.6 Proving completeness . . . . . . . . . . . . . . . . . . . . . . . . 996.7 Testing alternate rules . . . . . . . . . . . . . . . . . . . . . . . 103Practice Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1067 Natural Deduction Proofs in SL 1087.1 Basic rules for SL . . . . . . . . . . . . . . . . . . . . . . . . . . 1107.2 Derived rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1197.3 Rules of replacement . . . . . . . . . . . . . . . . . . . . . . . . 1217.4 Proof strategy . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1237.5 Proof-theoretic concepts . . . . . . . . . . . . . . . . . . . . . . 1247.6 Proofs and models . . . . . . . . . . . . . . . . . . . . . . . . . 1267.7 Soundness and completeness . . . . . . . . . . . . . . . . . . . . 127Practice Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1298 Quantified logic 1328.1 From sentences to predicates . . . . . . . . . . . . . . . . . . . 1328.2 Building blocks of QL . . . . . . . . . . . . . . . . . . . . . . . 1348.3 Singular terms . . . . . . . . . . . . . . . . . . . . . . . . . . . 1358.4 Predicates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1368.5 Quantifiers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1388.6 Universe of discourse . . . . . . . . . . . . . . . . . . . . . . . . 1408 CONTENTS8.7 Translating to QL . . . . . . . . . . . . . . . . . . . . . . . . . . 1418.8 Empty predicates . . . . . . . . . . . . . . . . . . . . . . . . . . 1438.9 Picking a universe of discourse . . . . . . . . . . . . . . . . . . 1448.10 Translating pronouns . . . . . . . . . . . . . . . . . . . . . . . . 1468.11 Quantifiers and scope . . . . . . . . . . . . . . . . . . . . . . . . 1488.12 Ambiguous predicates . . . . . . . . . . . . . . . . . . . . . . . 1488.13 Multiple quantifiers . . . . . . . . . . . . . . . . . . . . . . . . . 1508.14 Grammaticality rules for QL . . . . . . . . . . . . . . . . . . . . 1528.15 Common student errors . . . . . . . . . . . . . . . . . . . . . . 155Practice Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1569 A formal semantics for QL 1629.1 Interpretations in QL . . . . . . . . . . . . . . . . . . . . . . . . 1629.2 Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1649.3 Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1659.4 Working with models . . . . . . . . . . . . . . . . . . . . . . . . 1679.5 Constructing models . . . . . . . . . . . . . . . . . . . . . . . . 1689.6 Reasoning about all models . . . . . . . . . . . . . . . . . . . . 1709.7 Truth in QL . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1729.8 Satisfaction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1739.9 Truth in QL . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1759.10 Reasoning about all models (reprise) . . . . . . . . . . . . . . . 176Practice Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17710 QL Trees 18110.1 Trees with fixed domains . . . . . . . . . . . . . . . . . . . . . . 18110.2 Generalizing the tree method . . . . . . . . . . . . . . . . . . . 18510.3 Existentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18610.4 Universals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18710.5 Negated existentials . . . . . . . . . . . . . . . . . . . . . . . . 18810.6 Negated universals . . . . . . . . . . . . . . . . . . . . . . . . . 18910.7 Whither branching? . . . . . . . . . . . . . . . . . . . . . . . . 18910.8 The other development rules . . . . . . . . . . . . . . . . . . . . 18910.9 Branch closure rules . . . . . . . . . . . . . . . . . . . . . . . . 19010.10 Tree completion rules . . . . . . . . . . . . . . . . . . . . . . . . 19010.11 Resolution order . . . . . . . . . . . . . . . . . . . . . . . . . . 19210.12 Infinite trees . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19310.13 Common student errors . . . . . . . . . . . . . . . . . . . . . . 198Practice Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19911 Soundness and Completeness for QL Trees 20111.1 Soundness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20211.2 Completeness . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207Practice Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20912 Identity 212CONTENTS 912.1 Motivating identity as a logical category . . . . . . . . . . . . . 21212.2 = . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21512.3 Identity and ‘no one else’ . . . . . . . . . . . . . . . . . . . . . 21612.4 Identical objects satisfy identical predicates . . . . . . . . . . . 21712.5 Quantity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21812.6 Definite descriptions . . . . . . . . . . . . . . . . . . . . . . . . 21912.7 Identity and trees . . . . . . . . . . . . . . . . . . . . . . . . . . 221Practice Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22513 Natural Deduction Proofs in QL 22913.1 Natural Deduction: the basics . . . . . . . . . . . . . . . . . . . 22913.2 Basic quantifier rules . . . . . . . . . . . . . . . . . . . . . . . . 23013.3 Identity Introduction . . . . . . . . . . . . . . . . . . . . . . . . 23613.4 Identity Elimination . . . . . . . . . . . . . . . . . . . . . . . . 23613.5 Translation and evaluation . . . . . . . . . . . . . . . . . . . . . 23713.6 Natural deduction strategy . . . . . . . . . . . . . . . . . . . . 23913.7 Soundness and completeness . . . . . . . . . . . . . . . . . . . . 239Practice Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 240A Other symbolic notation 244B Solutions to selected exercises 247Chapter 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247Chapter 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 248Chapter 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 249Chapter 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251Chapter 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255Chapter 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 257Chapter 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 264Chapter 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 266Chapter 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 269Chapter 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 275Chapter 12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 278Chapter 13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 287C Quick ReferenceChapter 1What is logic?Logic is the business of evaluating arguments, sorting good ones from badones. In everyday language, we sometimes use the word ‘argument’ to referto belligerent shouting matches. If you and a friend have an argument in thissense, things are not going well between the two of you. This is not the kind of‘argument’ that will concern us. Arguments in the logical sense aren’t eventsthat happen between people; a logical argument is structured to give someone areason to believe some conclusion. Here are two examples of such arguments:(1) It is raining heavily.(2) When it rains, everyone outside without an umbrella gets wet..˙. You should take an umbrella.(1) It is either raining or snowing.(2) If it is colder than -10 degrees, it is not raining.(3) It is -18 degrees..˙. It is snowing.The three dots on the last line of each argument mean ‘Therefore’ and theyindicate that the final sentence is the conclusion of the argument. The othersentences are premises of the argument. If you believe the premises, then theargument provides you with a reason to believe the conclusion.This chapter discusses some basic logical notions that apply to arguments in anatural language like English. It is important to begin with a clear understandingof what arguments are and of what it means for an argument to be valid. Laterwe will translate arguments from English into a formal language. We wantformal validity, as defined in the formal language, to have at least some of theimportant features of natural-language validity.10ch. 1 what is logic? 111.1 ArgumentsA crucial part of analyzing an argument is identifying its conclusion. Everyargument has a conclusion — the conclusion is the claim the argument is tryingto establish. Premises are starting-points, used to lend support to the conclusion.Often, the conclusion will be signified by words like ‘so’ or ‘therefore’. Premisesmight be marked by words like ‘because’. These words can give a clue as to justwhat the argument is supposed to be.premise indicators: since, because, given thatconclusion indicators: therefore, hence, thus, then, soIn a natural language like English, sometimes, arguments start with their premisesand end with their conclusions, but not always. For some purposes in this course,we will be working with idealizations of natural language, where we work as ifsome generally applicable rules of thumb held without exception. Let’s define (aslightly technical notion of) an argument as a series of sentences. The sentencesat the beginning of the series are premises. The final sentence in the series isthe conclusion.Here is an example of an argument:People get wet whenever it rains.It often rains in Vancouver..˙. People often get wet in Vancouver.The idea of an argument is that the premises are supposed to give you reason toaccept the conclusion. If I’m trying to convince you that people often get wet inVancouver — the conclusion of the argument above — convincing you of thetwo premises might be a good way to get you there.Notice that our definition of an argument is quite general. Consider this example:Vancouver has more churches than any other Canadian city.Two oboes are fighting a duel under the fireworks..˙. J. Edgar Hoover was an honest man.It may seem odd to call this an argument, but that is because it would be aterrible argument. The two premises have nothing at all to do with the conclusion.(Moreover, they aren’t very plausible.) Nevertheless, given our definition, it stillcounts as an argument — albeit a bad one. One of our central aims in formallogic is to provide rigorous, formal tests for evaluating arguments.12 forallx (UBC edition)1.2 Sentences and propositionsThe premises and conclusions of arguments are sentences. But not just anyEnglish sentence is suitable for figuring into an argument. For example, questionscount as grammatical sentences of English, but logical arguments never havequestions as premises or conclusions. We are interested especially in sentencesthat can be true or false. Think of these as the sentences that purport to describethe way things are. Such sentences are sometimes described as expressingpropositions.We will not be theorizing in terms of questions and other non-propositionalsentences in formal logic. Since we are only interested in sentences that canfigure as a premise or conclusion of an argument, we’ll define a technical notionof a sentence as a sentence expressing a proposition, i.e., a sentence that canbe true or false.Don’t confuse the idea of a sentence that can be true or false with the differencebetween fact and opinion. Often, sentences in logic will express things thatwould count as facts — such as ‘Kierkegaard was a hunchback’ or ‘Kierkegaardliked almonds.’ They can also express things that you might think of as mattersof opinion — such as, ‘Almonds are yummy’ or ‘the U.S. invasion of Iraq wasunjustified’. These are all examples of things that are either true or false.It is also important to keep clear the distinction between something’s being trueand something’s being known. A sentence is the kind of thing that can be trueor false; that doesn’t mean you’ll always be able to tell whether it is true orfalse. For example, ‘there are an even number of humans on Earth right now’ isa sentence. It is either true or false, even though it is pretty much impossible totell which. Similarly, there are controversial propositions, where people disagreeabout whether they are true or false, and where it seems very difficult to settlethe debate. (I think that ‘Freedom of expression requires restricting oppressivespeech’ is an example.) This is either true or false, even if debates about which itis aren’t likely to resolve the question. So it counts as a propositional sentence.What are some examples of grammatical English sentences that do not expresspropositions? We’ve discussed one category already:Questions In a grammar class, ‘Are you sleepy yet?’ would count as aninterrogative sentence. Although you might be sleepy or you might be alert, thequestion itself is neither true nor false. So ‘Are you sleepy yet?’ is not a sentencein our technical, propositional sense. Suppose you answer the question: ‘I amnot sleepy.’ This is either true or false, and so it is a sentence in the logical sense.Generally, questions will not count as sentences, but answers will.ch. 1 what is logic? 13‘What is this course about?’ is not a sentence. ‘No one knows what this courseis about’ is a sentence.Imperatives Commands are often phrased as imperatives like ‘Wake up!’, ‘Situp straight’, and so on. In a grammar class, these would count as imperativesentences. Although it might be good for you to sit up straight or it might not,the command is neither true nor false. Note, however, that commands are notalways phrased as imperatives. ‘You will respect my authority’ is either true orfalse — either you will or you will not — and so it counts as a sentence in thelogical sense.Exclamations Expressions like ‘Ouch!’ or ‘Boo, Yankees!’ are sometimesdescribed as exclamatory sentence, but they are neither true nor false. We willtreat ‘Ouch, I hurt my toe!’ as meaning the same thing as ‘I hurt my toe.’ The‘ouch’ does not add anything that could be true or false.To recap: sentences, in the technical sense we’re interested in, are claims thatcan be true or false. One pretty good test you can run to see whether somethingis a sentence is to ask whether it makes sense to insert ‘it is true that’ or ‘it isfalse that’ in front of it. It’s perfectly fine to say ‘it is true that Kierkegaardliked almonds’ or ‘it is true that the U.S. invasion of Iraq was unjustified’. Butit doesn’t make sense to say ‘it is true that are you sleepy yet’ or ‘it is true thatsit up straight’.We will call truth or falsity the truth-value of a sentence.1.3 Two ways that arguments can go wrongConsider the argument that you should take an umbrella (on p. 10, above).(1) It is raining heavily.(2) When it rains, everyone outside without an umbrella gets wet..˙. You should take an umbrella.If premise (1) is false — if it is sunny outside — then the argument fails. It doesnot establish that you should take an umbrella. Or suppose that premise (2) isfalse. Maybe not everyone without an umbrella gets wet. (Maybe some peopleare able to stay under the awnings.) In this case, too, the argument does notestablish its conclusion. Arguments only succeed when all their premises aretrue.14 forallx (UBC edition)Suppose both premises are true. It really is raining heavily, and we’re talkingabout an area where there’s no protection from the rain other than umbrellas, sothat anyone outside without an umbrella will get wet when it rains. Now doesthe argument show you that you should take an umbrella? It certainly seemsto lend some support in that direction. But notice that it’s still not conclusive.There are still possible ways the conclusion might be false. For example, supposeyou like getting wet. Then, even if it is raining heavily, and even if everyone whogoes outside without an umbrella when it rains gets wet, maybe you have noreason to take the umbrella.A good argument — or at least, an argument that is good in one particularlyinteresting way — compels its conclusion.So it seems that there are two ways an argument could be weak. First, one ormore of the premises might be false. An argument gives you a reason to believeits conclusion only if you believe its premises. Second, the premises might failto support the conclusion, or fail to support it sufficiently strongly. Even if thepremises were true, the form of the argument might be weak. The example wejust considered is weak in both ways.Once again, we are working with a particular kind of idealization regardingarguments and what makes them good. Consider another example:You are reading this book.This is a logic book..˙. You are a logic student.In a straightforward and ordinary sense, this is not a terrible argument. Thepremises are true. And they do offer some support for the conclusion. Mostpeople who read this book are logic students. Still, it is possible for someonebesides a logic student to read this book. If the deans carefully read this book inorder to find out whether I’m criticizing the UBC administration, that wouldn’tmake them logic students. So the premises of this argument, even though theyare true, do not guarantee the truth of the conclusion.By contrast, the sample argument about snow is valid. There’s just no possibleway for all three of its premises to be true without it snowing.In logic, we are interested in arguments whose premises guarantee their conclu-sions. We call such arguments ‘deductively valid’ or just ‘valid.’ Even though wemight count the rain argument, or the logic book argument, as a good argumentin some sense, they are not valid; that is, they are ‘invalid.’ One important taskof logic is to sort valid arguments from invalid arguments.ch. 1 what is logic? 151.4 ValidityAn argument is deductively valid if and only if it is impossible for all thepremises to be true while the conclusion is false. (We’ll say a bit more aboutjust what we mean by ‘impossible’ in §1.5 below.)That’s our official definition of validity. Once again (get used to this pattern), ourformal definition is related to, but not quite the same as, the ordinary colloquialnotion with the same name. In ordinary English, to say something is ‘valid’means, somewhat vaguely, that it is good. (Think of a ‘valid point’ or a ‘validperspective’.) In logic, a valid argument is one that has this very specific feature:it’s impossible for the premises to be true if the conclusion is false. (Side note:this is a good terminological distinction to keep in mind when writing essaysin philosophy courses. Philosophy professors tend to wince and get annoyedwhen students use the word ‘valid’ in its imprecise sense. You can help stay ontheir good sides by only saying something is ‘valid’ when you mean that it is anargument whose premises are inconsistent with the falsity of its conclusion.)Notice that validity does not require that the premises be true. Consider thisexample:(1) Oranges are either fruits or musical instruments.(2) Oranges are not fruits..˙. Oranges are musical instruments.This is a valid argument. It is impossible for the premises to be true if theconclusion is false. Since it has a false premise — premise (2) — it does notactually establish its conclusion, but it does have a valid logical form. If bothpremises were true, then the conclusion would necessarily be true.Since this is a valid argument that is, in some important sense, a bad argument,this shows that validity isn’t the only feature we care about in arguments. Anargument is sound if and only if it is valid and all of its premises are true. Forexample, this variant argument on the above argument is sound:(1) Oranges are either fruits or musical instruments.(2) Oranges are not musical instruments..˙. Oranges are fruits.For reasons that will emerge, logic tends to focus primarily on validity, ratherthan soundness.We’ve seen that a valid argument does not need to have true premises or a trueconclusion. Conversely, having true premises and a true conclusion is also notenough to make an argument valid. Consider this example:16 forallx (UBC edition)Donald Trump is a U.S. citizen.Justin Trudeau is a Canadian citizen..˙. UBC is the largest employer in Vancouver.The premises and conclusion of this argument are, as a matter of fact, all true.Nevertheless, this is quite a poor argument. This is related to the fact that thepremises have nothing to do with the conclusion. More precisely, the definitionof validity is not satisfied: it is possible for the premises to be true while theconclusion is false. Although the conclusion is actually true, it is possible forthe premises of this argument to be true while the conclusion is false. We mayimagine, for example, that Lululemon hired an additional 100,000 employees inVancouver, while UBC remained the same size, and without any world leaderschanging their citizenship statuses. In such a possible scenario, the premiseswould be true, while the conclusion is false.The important thing to remember is that validity is not about the actual truthor falsity of the sentences in the argument. Instead, it is about the form ofthe argument: The truth of the premises is incompatible with the falsity of theconclusion.Here are some more valid arguments. Can you see why each is valid?1. Socrates is a man.2. All men are carrots.3. .˙. Socrates is a carrot.1. Abe Lincoln was either born in Illinois or he was once president.2. Abe Lincoln was never president..˙. Abe Lincoln was born in Illinois.1. Justin Trudeau is either from France or from Luxemborg.2. Justin Trudeau is not from Luxemborg..˙. Justin Trudeau is from France.1. If the world were to end today, then I would not need to get up tomorrowmorning.2. I will need to get up tomorrow morning..˙. The world will not end today.1.5 ImpossibilityThe definition of validity centrally invokes the notion of impossibility. What itmeans for a sentence to be impossible is that there is no possible way for it toch. 1 what is logic? 17be true. Impossible sentences don’t just happen to be false; they must be false.Consider the sentence, ‘The Whitecaps lost their last game.’ As it happens, atthe time I’m writing, that’s false. They actually won. But it’s not impossible;they might have lost, if a few things had gone differently. So that sentence isfalse, but not impossible.By contrast, blatant contradictions like ‘the Whitecaps won the game and theWhitecaps lost the game’ are impossible. This sentence doesn’t just happen tobe false. It couldn’t possibly be true.Even falsehoods that are really obviously false, and for which it’s hard to imaginehow they could be true, might be possible. It’s not impossible that pigs couldfly. Someone might invent powerful wings for them. When we’re interested inimpossible sentences in logic, we really mean impossible. The best examples arecontradictions.So to apply the definition of validity, we consider whether there is any possibility,no matter how far-fetched, where the premises are true and the conclusion isfalse.1.6 Other logical notionsHere are a few more relevant terms we’ll be working with.1.6.1 Logical truthIn considering arguments formally, we care about what would be true if thepremises were true. Generally, we are not concerned with the actual truth valueof any particular sentences — whether they are actually true or false. (Asindicated above, we will be more interested in validity than in soundness.) Yetthere are some sentences that must be true, just as a matter of logic.Compare these sentences:1. It is raining.2. Either it is hot outside, or it is not hot outside.3. There is an earthquake happening here right now and there are neverearthquakes here.Sentence 1 could be true or it could be false. (As I’m typing these words on asunny Vancouver summer day, it is false.) Sentences that could be true, or couldbe false, are called contingent sentences.18 forallx (UBC edition)Sentence 2 is different. Even though I don’t know what the weather is likeas you’re reading this book, I still know that it’s true. Sentence 2 must betrue. This sentence is logically true; no matter how the weather or anythingelse happens to be, this sentence has to be true. We call a sentence like this alogical truth or a tautology.You do not need to check the weather to know about sentence 3, either. It mustbe false, simply as a matter of logic. There could be an earthquake now, butif so, it couldn’t be the case that there are never earthquakes here. Or theremight never be any earthquakes, but if so, there couldn’t be one now. The thirdsentence is logically false; it is false regardless of what the world is like. Alogically false sentence is called a contradiction.I said above that a contingent sentence could be true and it also could befalse. We can also define contingency in terms of tautologies and contradictionsthus: a contingent sentence is a sentence that is neither a tautology nor acontradiction.1.6.2 Logical equivalenceWe can also ask about the logical relations between two sentences. For example:Sunil went to the store after he washed the dishes.Sunil washed the dishes before he went to the store.These two sentences are both contingent. (Do you see why?) Yet they must havethe same truth-value. If either of the sentences is true, then they both are; ifeither of the sentences is false, then they both are. When two sentences necessarilyhave the same truth value, we say that they are logically equivalent.Notice that both of these arguments are valid:Sunil went to the store after he washed the dishes..˙. Sunil washed the dishes before he went to the store.Sunil washed the dishes before he went to the store..˙. Sunil went to the store after he washed the dishes.In general, if two sentences are equivalent, then an argument with either one asa premise, and the other as the conclusion, will be valid.ch. 1 what is logic? 191.6.3 ConsistencyConsider these two sentences:B1 My only brother is taller than I am.B2 My only brother is shorter than I am.Logic alone cannot tell us which, if either, of these sentences is true. Yet we cansay that if the first sentence (B1) is true, then the second sentence (B2) mustbe false. And if B2 is true, then B1 must be false. It cannot be the case thatboth of these sentences are true.If a set of sentences could not all be true at the same time, like B1–B2, they aresaid to be inconsistent. Otherwise, they are consistent.We can ask about the consistency of any number of sentences. For example,consider the following list of sentences:G1 There are at least four giraffes at the wild animal park.G2 There are exactly seven gorillas at the wild animal park.G3 There are not more than two Martians at the wild animal park.G4 Every giraffe at the wild animal park is a Martian.G1 and G4 together imply that there are at least four Martian giraffes at thepark. This conflicts with G3, which implies that there are no more than twoMartian giraffes there. So the set of sentences G1–G4 is inconsistent. (Noticethat in this example the inconsistency has nothing at all to do with G2. G2 justhappens to be part of an inconsistent set.)1.7 Formal languagesSo far, we have been working with arguments expressed in English. But thereare some general argumentative structures that transcend English or any otherparticular natural language. Consider, for example, the following simple argu-ment:Vancouver is in Canada.Moscow is in Russia..˙. Vancouver is in Canada and Moscow is in Russia.Hopefully by now it is obvious to you that this argument is valid. (Not just that— it is even sound.) That argument has something important in common withthis one:20 forallx (UBC edition)Vancouver is on Mars.Moscow is on Venus..˙. Vancouver is on Mars and Moscow is on Venus.Although this one has false premises, and so can’t be sound, it is also valid. It isimpossible for the premises to be true if the conclusion is false. Not only that— it seems that the argument is valid in, in a sense to be articulated, exactlythe same way. The validity of the first argument is very similar to that of thesecond. Even though they are expressed by different strings of English sentences,it is useful to develop a language in which they are, in an important sense, thesame kind of argument. Over the course of this textbook, we’ll be developingand working with such a language. We’ll begin with a simple version of ourlogical language, called SL, for sentential logic. Later on, we’ll develop a morecomplicated logical language, QL, for quantified logic. Both languages are formalin the sense that there are precise rules that govern how we should evaluatesentences within them.English is a natural language, not a formal one. Its rules are vague and messy,and constantly changing. We will spend some time translating between Englishand our formal languages, but the translations will not always be precise. Thereis a tension between wanting to capture as much of the structure of English aspossible and wanting a simple formal language with tractable rules — simplerformal languages will approximate natural languages less closely. There is noperfect formal language. Some will do a better job than others in translatingparticular English-language arguments.In this book, we make the assumption that true and false are the only possibletruth-values. Logical languages that make this assumption are called bivalent,which means two-valued. SL and QL are both bivalent, but some philosophershave emphasized limits to the power of bivalent logic. Some logics, beyond thescope of this book, allow for sentences that are neither true nor false. Othersallow for sentences that are both true and false. Our logical system, which isoften called classical logic, will give every sentence exactly one truth value: eithertrue or false.Summary of logical notions. An argument is a structured set of sentences, with premises intended tooffer support for a conclusion.. A sentence, in our logical terminology, is a sentence that expresses aproposition, and can be either true or false.ch. 1 what is logic? 21. An argument is (deductively) valid if it is impossible for the premises tobe true and the conclusion false; it is invalid otherwise.. A tautology is a sentence that must be true, as a matter of logic.. A contradiction is a sentence that must be false, as a matter of logic.. A contingent sentence is neither a tautology nor a contradiction.. Two sentences are logically equivalent if they necessarily have thesame truth value.. A set of sentences is consistent if it is logically possible for all themembers of the set to be true at the same time; it is inconsistentotherwise.Practice ExercisesAt the end of each chapter, you will find a series of practice problems that reviewand explore the material covered in the chapter. There is no substitute foractually working through some problems, because logic is more about a way ofthinking than it is about memorizing facts. The answers to some of the problemsare provided at the end of the book in appendix B; the problems that are solvedin the appendix are marked with a ?.? Part A Which of the following are ‘sentences’ in the logical sense?1. England is smaller than China.2. Greenland is south of Jerusalem.3. Is New Jersey east of Wisconsin?4. The atomic number of helium is 2.5. The atomic number of helium is pi.6. I hate overcooked noodles.7. Blech! Overcooked noodles!8. Overcooked noodles are disgusting.9. Take your time.10. This is the last question.Part B For each of the following: Is it a tautology, a contradiction, or acontingent sentence?1. Caesar crossed the Rubicon.2. Someone once crossed the Rubicon.3. No one has ever crossed the Rubicon.22 forallx (UBC edition)4. If Caesar crossed the Rubicon, then someone has.5. Even though Caesar crossed the Rubicon, no one has ever crossed theRubicon.6. If anyone has ever crossed the Rubicon, it was Caesar.? Part C Look back at the sentences G1–G4 on p. 19, and consider each of thefollowing sets of sentences. Which are consistent? Which are inconsistent?1. G2, G3, and G42. G1, G3, and G43. G1, G2, and G44. G1, G2, and G3? Part D Which of the following is possible? If it is possible, give an example.If it is not possible, explain why.1. A valid argument that has one false premise and one true premise2. A valid argument that has a false conclusion3. A valid argument, the conclusion of which is a contradiction4. An invalid argument, the conclusion of which is a tautology5. A tautology that is contingent6. Two logically equivalent sentences, both of which are tautologies7. Two logically equivalent sentences, one of which is a tautology and one ofwhich is contingent8. Two logically equivalent sentences that together are an inconsistent set9. A consistent set of sentences that contains a contradiction10. An inconsistent set of sentences that contains a tautologyPart E Is this argument valid? Why or why not? (Hint: here and elsewhere inlogic, read the definitions of our formal terms literally.)(1) PHIL 220 is a course with a final exam.(2) No courses ever have final exams..˙. Everyone is going to get an A+ in PHIL 220.Chapter 2Sentential logicThis chapter introduces a logical language called SL. It is a version of sententiallogic, because the basic units of the language will represent entire sentences.(Recall from §1.2 that we’re only considering propositional sentences.)2.1 Sentence lettersIn SL, capital Roman letters (A, B, C, etc.) are used to represent basic sentences.Considered only as a symbol of SL, the letter A could mean any sentence. Sowhen translating from English into SL, it is important to provide a symbolizationkey. The key provides an English language sentence for each sentence letter usedin the symbolization.For example, consider this argument:Today is New Year’s Day.If today is New Year’s Day, then people are swimming in English Bay..˙. People are swimming in English Bay.This is obviously a valid argument in English. In symbolizing it, we want topreserve the structure of the argument that makes it valid. What happens if wereplace each sentence with a letter? Our symbolization key would look like this:A: Today is New Year’s Day.B: If today is New Year’s Day, then people are swimming in English Bay.C: People are swimming in English Bay.2324 forallx (UBC edition)We could then symbolize the argument in this way:AB.˙. CThis is a possible way to symbolize this argument, but it’s not a very interestingone. There is no necessary connection between some sentence A, which could beany sentence, and some other sentences B and C, which could be any sentences.Something important about the argument has been lost in translation. Theoriginal argument was valid, but this translation of the argument does notreflect that validity. Given a different symbolization key, for example, the sameargument formAB.˙. Ccould equally well stand in for this invalid argument:Today is Christmas Day.Tiny Tim has difficulty walking without crutches..˙. We’re all going to die tomorrow.A more interesting translation of the valid New Year’s argument will show howit is different from the invalid Christmas argument. The relevant thing aboutthe New Year’s argument is that the second premise is not just any sentence.Notice that the second premise contains the first premise and the conclusion asparts. Our symbolization key for the argument only needs to include meaningsfor A and C, and we can build the second premise from those pieces. So wesymbolize the argument this way:AIf A, then C..˙. CThis preserves the structure of the argument that makes it valid, but it stillmakes use of the English expression ‘If. . . then. . ..’ Although for our formallanguage we ultimately want to replace all of the English expressions with logicalnotation, this is a good start.The sentences that can be symbolized with sentence letters are called atomicsentences, because they are the basic building blocks out of which more complexch. 2 sentential logic 25sentences can be built. Whatever logical structure a sentence might have is lostwhen it is translated as an atomic sentence. From the point of view of SL, thesentence is just a letter. It can be used to build more complex sentences, but itcannot be taken apart.We use capital Roman alphabet letters to represent SL sentences. There areonly twenty-six such letters. We don’t want to impose this artificial limit ontoour formal language; it’s better to work with a language that allows an arbitrarynumber of atomic sentences. To achieve this, we allow atomic sentences that havea capital letter with a numerical subscript. So we could have a symbolizationkey that looks like this:A1: Aang is from the Air Nation.A2: Aang is vegetarian.A3: Aang can bend water.T1: Toph is blind.T2: Toph likes badgers.T3: Toph invented metal bending.Keep in mind that each of these is a different atomic sentence. Although it isoften convenient, for making it easier to remember what each letter stands for,to use letters corresponding to the sentences’ subject matters, as in the exampleabove, no such requirement is built into the rules of SL. There is no specialrelationship between A1 and A2, as far as SL goes. It’s just for our conveniencethat we might choose to make all the A sentences about Aang.2.2 ConnectivesLogical connectives are used to build complex sentences from atomic components.There are five logical connectives in SL. This table summarizes them. They areexplained below.symbol what it is called rough translation¬ negation ‘It is not the case that. . .’& conjunction ‘Both. . . and . . .’∨ disjunction ‘Either. . . or . . .’⊃ conditional ‘If . . . then . . .’≡ biconditional ‘. . . if and only if . . .’Natural languages like English are vague and imprecise, and carry many complexsubtleties of meaning. Our formal language, SL, has none of these properties. Itis defined by precise rigid rules. Consequently, the ‘translation’ provided in thetable is only an approximate one. We’ll see some of the differences below.26 forallx (UBC edition)2.3 NegationConsider how we might symbolize these sentences:1. Logic is hard.2. It is false that logic is hard.3. Logic isn’t hard.In order to symbolize sentence 1, we will need one sentence letter. We canprovide a symbolization key:H: Logic is hardNow, sentence 1 is simply H.Since sentence 2 is obviously related to sentence 1, we do not want to introducea different sentence letter. To put it partly in English, the sentence means ‘It isfalse that H.’ In order to symbolize this, we need a symbol for logical negation.We will use ‘¬.’ Now we can translate ‘Not H’ to ¬H. In general, ‘¬’ means ‘itis false that’.What of Sentence 3? It looks like a more natural way of saying the same thingas sentence 2. It’s saying that logic isn’t hard, which is just another way ofdenying that logic is hard. So sentences 2 and 3 have the same truth conditions.In the terminology of Ch. 1, they are logically equivalent. So we can translateboth sentence 2 and sentence 3 as ¬H.When translating from English into SL, the word ‘not’ is usually a pretty goodclue that ‘¬’ will be an appropriate symbol to use, but it’s important to thinkabout the actual meaning of the sentence, and not rely too much on which wordsappear in it.For any sentence Φ, a sentence can be symbolized as ¬Φ if it canbe paraphrased in English as ‘It is not the case that Φ.’(For more on the ‘Φ’ notation, see the ‘note about notation’ on p. 30.)Consider these further examples:4. Rodrigo is mortal.5. Rodrigo is immortal.ch. 2 sentential logic 276. Rodrigo is not immortal.If we let R mean ‘Rodrigo is mortal’, then sentence 4 can be translated as R.What about sentence 5? Being immortal is pretty much the same as not beingmortal. So it makes sense to treat 5 as the negation of 4, symbolizing it as ¬R.Sentence 6 can be paraphrased as ‘It is not the case that Rodrigo is immortal.’Using negation twice, we translate this as ¬¬R. The two negations in a roweach work as negations, so the sentence means ‘It is not the case that. . . it is notthe case that. . . R.’ It is the negation of the negation of R. One can negate anysentence of SL by putting the ¬symbol in front of it. It’s not only for atomicsentences.Here is an example that illustrates some of the complexities of translation.7. Elliott is happy.8. Elliott is unhappy.If we let H mean ‘Elliot is happy’, then we can symbolize sentence 7 as H.We might be tempted to symbolize sentence 8 as ¬H. But is being unhappy thesame thing as not being happy? This is perhaps debatable. One might thinkthat it is possible to be neither happy nor unhappy. Maybe Elliot is in thisin-between zone. If so, then we shouldn’t treat 8 as the negation of 7. If we’reallowing that ‘unhappy’ means something different from ‘not happy’, then wewill need to use a different atomic sentence to translate 8.What of the truth conditions for negated sentences?For any sentence Φ: If Φ is true, then ¬Φ is false. If Φ is false, then ¬Φ is true.Using ‘1’ for true and ‘0’ for false, we can summarize this in a characteristictruth table for negation:Φ ¬Φ1 00 1The left column shows the possible truth values for a given sentence; the rightcolumn shows the truth value of the negation of that sentence.We will discuss truth tables at greater length in Chapter 3.28 forallx (UBC edition)2.4 ConjunctionConsider these sentences:9. Jessica is strong.10. Luke is strong.11. Jessica is strong, and Luke is also strong.We will need separate sentence letters for 9 and 10, so we define this symbolizationkey:J: Jessica is strong.L: Luke is strong.Sentence 9 can be symbolized as J .Sentence 10 can be symbolized as L.Sentence 11 can be paraphrased as ‘J and L.’ In order to fully symbolize thissentence, we need another symbol. We will use ‘ & .’ We translate ‘J and L’ as(J &L). The logical connective ‘ & ’ is called conjunction, and J and L areeach called conjuncts.Notice that we make no attempt to symbolize ‘also’ in sentence 11. Words like‘both’ and ‘also’ function to draw our attention to the fact that two things arebeing conjoined. They are not doing any further logical work, so we do not needto represent them in SL. Note that Sentence 11 would have meant the samething had it simply said ‘Jessica is strong, and Luke is strong.’Here are some more examples:12. Jessica is strong and grumpy.13. Jessica and Matt are both strong.14. Although Luke is strong, he is not grumpy.15. Matt is strong, but Jessica is stronger than Matt.Sentence 12 is obviously a conjunction. The sentence says two things aboutJessica, so in English it is permissible to use her name only once. It might betempting to try this when translating the argument: Since J means ‘Jessicais strong’, one might attempt to paraphrase sentence 12 as ‘J and grumpy.’But this would be a mistake. Once we translate part of a sentence as J , anyfurther structure within the original sentence is lost. J is an atomic sentence; SLdoesn’t keep track of the fact that it was intended to be about Jessica. Moreover,ch. 2 sentential logic 29‘grumpy’ is not a sentence; on its own it is neither true nor false. So instead,we paraphrase sentence 12 as ‘J and Jessica is grumpy.’ Now we need to add asentence letter to the symbolization key. Let G1 mean ‘Jessica is grumpy.’ Nowthe sentence can be translated as J &G1.A sentence can be symbolized as (Φ&Ψ) if it can be paraphrasedin English as ‘Both Φ, and Ψ .’ Each of the conjuncts must be asentence.Sentence 13 says one thing about two different subjects. It says of both Jessicaand Matt that they are strong, and in English we use the word ‘strong’ onlyonce. In translating to SL, we want to make sure each conjunct is a sentence onits own, so once again, we’ll paraphrase it by repeating the elements: ‘Jessica isstrong, and Matt is strong.’ Once we add a new atomic sentence M for ‘Matt isstrong’, this translates as J &M .Sentence 14 is a bit more complicated. The word ‘although’ tends to suggesta kind of contrast between the first part of the sentence and the second part.Nevertheless, the sentence is still telling us two things: Luke is strong, and he’snot grumpy. So we can paraphrase sentence 14 as, ‘Both Luke is strong, andLuke is not grumpy.’ The second conjunct contains a negation, so we paraphrasefurther: ‘Both Luke is strong and it is not the case that Luke is grumpy.’ Let’slet G2 stand for ‘Luke is grumpy’, and we can translate sentence 14 as L&¬G2.Once again, this is an imperfect translation of the English sentence 14. Thatsentence implicated that there was a contrast between Luke’s two properties.Our translation merely says that he has both of them. Still, it is a translationthat preserves some of the important features of the original. In particular, itsays that Luke is strong, and it also says that he’s not grumpy.Sentence 15’s use of the word ‘but’ indicates a similar contrastive structure. It isirrelevant for the purpose of translating to SL, so we can paraphrase the sentenceas ‘Both Matt is strong, and Jessica is stronger than Matt.’ How should wetranslate the second conjunct? We already have the sentence letters J and M ,which each say that one of Jessica and Matt is strong, but neither of these saysanything comparative. We need a new sentence letter. Let S mean ‘Jessica isstronger than Matt.’ Now the sentence translates as (M &S).Sentences that can be paraphrased ‘Φ, but Ψ ’ or ‘Although Φ, Ψ ’are best symbolized using conjunction: (Φ&Ψ).It is important to keep in mind that the sentence letters J , M , G1, G2, andS are atomic sentences. Considered as symbols of SL, they have no meaningbeyond being true or false. We used J and M to symbolize different English30 forallx (UBC edition)language sentences that are about people being strong, but this similarity iscompletely lost when we translate to SL. Nor does SL recognize any particularsimilarity between G1 and G2.For any sentences Φ and Ψ , (Φ&Ψ) is true if and only if both Φ and Ψ are true.We can summarize this in the characteristic truth table for conjunction:Φ Ψ (Φ&Ψ)1 1 11 0 00 1 00 0 0The two left columns indicate the truth values of the two conjuncts. Sincethere are four possible combinations of truth values, there are four rows. Theconjunction is true when both conjuncts are true, and false in all other cases.2.4.1 A Note About NotationIn speaking generally about connectives, we’ve used the ‘Φ’ and ‘Ψ ’ symbolsas variables to stand in for sentences of SL. We saw, for instance, that for anysentence Φ, ¬Φ is the negation of Φ. This means:. ¬A is the negation of A. ¬B1 is the negation of B1. ¬B2 is the negation of B2. ¬¬C is the negation of ¬C. ¬(A&B) is the negation of (A&B). etc.Note that Φ and Ψ are not sentences of SL. They are symbols we use to talkabout SL sentences, but they’re not themselves part of our formal language.(Compare the use of variables like x in algebra. An ‘x’ symbol is used to standin for any number, but x is not itself a number.) We’ll return to this distinctionin §2.9, and again later on when we discuss proving generalities about SL.ch. 2 sentential logic 312.5 DisjunctionConsider these sentences:16. Denison will golf with me or he will watch movies.17. Either Denison or Ellery will golf with me.For these sentences we can use this symbolization key:D: Denison will golf with me.E: Ellery will golf with me.M: Denison will watch movies.Sentence 16 is ‘Either D or M .’ To fully symbolize this, we introduce a new sym-bol. The sentence becomes (D ∨M). The ‘∨’ connective is called disjunction,and D and M are called disjuncts.Sentence 17 is only slightly more complicated. There are two subjects, but theEnglish sentence only gives the verb once. In translating, we can paraphrase itas ‘Either Denison will play golf with me, or Ellery will play golf with me.’ Nowit obviously translates as (D ∨ E).A sentence can be symbolized as (Φ ∨ Ψ) if it can be paraphrasedin English as ‘Either Φ, or Ψ .’ Each of the disjuncts must be asentence.What truth conditions should we offer for ‘∨’ sentences? If I say, ‘Denisonwill golf with me or he will watch movies’, under what circumstances will thatsentence be true? When will it be false? Well, suppose he doesn’t do eitheractivity. Suppose he goes swimming, and doesn’t watch movies, and doesn’t golfwith me. Then my sentence is false.Suppose Denison skips the movies and golfs with me instead. Then it seemspretty clear that my disjunctive claim was true. Likewise if he goes to the moviesand leaves me without a golf partner. It’s a little less clear what to think aboutthe English sentence if both disjuncts end up true. Suppose that Denison comesgolfing with me, and also stays up late afterward to come with me to the movies.Then is it true that ‘he golfed with me or went to the movies’? This is notentirely clear. Certainly it would be strange to assert such a sentence if youknow that both elements were true. On the other hand, it doesn’t exactly seemfalse that he’ll golf with me or watch movies, if in fact he’ll do both. In a studyof the semantics of English, it would be appropriate to pursue this question32 forallx (UBC edition)much further. In this introduction to formal logic, we’ll simply stipulate thefeatures of our formal symbol. ‘∨’ stands for an inclusive or, which means it istrue if and only if at least one disjunct is true.So (D∨E) is true if D is true, if E is true, or if both D and E are true. It is falseonly if both D and E are false. We can summarize this with the characteristictruth table for disjunction:Φ Ψ (Φ∨Ψ)1 1 11 0 10 1 10 0 0Note that both conjunction and disjunction are symmetrical. (Φ&Ψ) is logicallyequivalent to (Ψ &Φ), and (Φ∨Ψ) is logically equivalent to (Ψ∨Φ).These sentences are somewhat more complicated:18. Either you will not have soup, or you will not have salad.19. You will have neither soup nor salad.20. You get soup or salad, but not both.Here’s a symbolization key:S1: You will get soup.S2: You will get salad.Sentence 18 can be paraphrased in this way: ‘Either it is not the case that youget soup, or it is not the case that you get salad.’ Translating this requires bothdisjunction and negation. It becomes (¬S1 ∨ ¬S2).Sentence 19 also requires negation. It can be paraphrased as, ‘It is not thecase that either that you get soup or that you get salad.’ We need some way ofindicating that the negation does not just negate the right or left disjunct, butrather negates the entire disjunction. In order to do this, we put parenthesesaround the disjunction: ‘It is not the case that (S1 ∨ S2).’ This becomes simply¬(S1 ∨ S2). (A second, equivalent, way to translate this sentence is (¬S1 &¬S2).We’ll see why this is equivalent later on.)Notice that the parentheses are doing important work here. The sentence(¬S1 ∨ S2) would mean ‘Either you will not have soup, or you will have salad,’which is very different.ch. 2 sentential logic 33Sentence 20 has a more complex structure. We can break it into two parts. Thefirst part says that you get one or the other. We translate this as (S1 ∨ S2).The second part says that you do not get both. We can paraphrase this as,‘It is not the case both that you get soup and that you get salad.’ Using bothnegation and conjunction, we translate this as ¬(S1 &S2). Now we just need toput the two parts together. As we saw above, ‘but’ can usually be translated asa conjunction. Sentence 20 can thus be translated as ((S1 ∨ S2) &¬(S1 &S2)).2.6 ConditionalFor the following sentences, use this symbolization key:R: You will cut the red wire.B: The bomb will explode.21. If you cut the red wire, then the bomb will explode.22. The bomb will explode only if you cut the red wire.Sentence 21 can be translated partially as ‘If R, then B.’ We will use the symbol‘⊃’ to represent this conditional relationship. The sentence becomes (R ⊃ B).The connective is called a conditional. The sentence on the left-hand side ofthe conditional (R in this example) is called the antecedent. The sentence onthe right-hand side (B) is called the consequent.Sentence 22 is also a conditional. Since the word ‘if’ appears in the second half ofthe sentence, it might be tempting to symbolize this in the same way as sentence21. That would be a mistake.The conditional (R ⊃ B) says that if R were true, then B would also be true. Itdoes not say that your cutting the red wire is the only way that the bomb couldexplode. Someone else might cut the wire, or the bomb might be on a timer.The sentence (R ⊃ B) does not say anything about what to expect if R is false.Sentence 22 is different. It says that the only conditions under which the bombwill explode are ones where you cut the red wire; i.e., if the bomb explodes,then you must have cut the wire. As such, sentence 22 should be symbolized as(B ⊃ R).It is important to remember that the connective ‘⊃’ says only that, if theantecedent is true, then the consequent is true. It says nothing about the causalconnection between the two events. Translating sentence 22 as (B ⊃ R) doesnot mean that the bomb exploding would somehow have caused your cuttingthe wire. Both sentence 21 and 22 suggest that, if you cut the red wire, your34 forallx (UBC edition)cutting the red wire would be the cause of the bomb exploding. They differ onthe logical connection. If sentence 22 were true, then an explosion would tellus — those of us safely away from the bomb — that you had cut the red wire.Without an explosion, sentence 22 tells us nothing.The paraphrased sentence ‘Φ only if Ψ ’ is logically equivalent to ‘IfΦ, then Ψ .’‘If Φ then Ψ ’ means that if Φ is true then so is Ψ . So we know that if theantecedent Φ is true but the consequent Ψ is false, then the conditional ‘If Φ thenΨ ’ is false. What is the truth value of ‘If Φ then Ψ ’ under other circumstances?Suppose, for instance, that the antecedent Φ happened to be false. ‘If Φ then Ψ ’would then not tell us anything about the actual truth value of the consequentΨ , and it is unclear what the truth value of ‘If Φ then Ψ ’ would be.In English, the truth of conditionals often depends on what would be the case ifthe antecedent were true — even if, as a matter of fact, the antecedent is false.This poses a serious challenge for translating conditionals into SL. Consideredas sentences of SL, R and B in the above examples have nothing intrinsic to dowith each other. In order to consider what the world would be like if R were true,we would need to analyze what R says about the world. Since R is an atomicsymbol of SL, however, there is no further structure to be analyzed. When wereplace a sentence with a sentence letter, we consider it merely as some atomicsentence that might be true or false.In order to translate conditionals into SL, we will not try to capture all thesubtleties of English’s ‘if. . . then. . ..’ construction. Instead, the symbol ‘⊃’ willsignify a material conditional. This means that when Φ is false, the conditional(Φ⊃Ψ) is automatically true, regardless of the truth value of Ψ . If both Φ and Ψare true, then the conditional (Φ⊃Ψ) is true.In short, (Φ⊃Ψ) is false if and only if Φ is true and Ψ is false. We can summarizethis with a characteristic truth table for the conditional.Φ Ψ Φ⊃Ψ1 1 11 0 00 1 10 0 1More than any other connective, the SL translation of the conditional is arough approximation. It has some very counterintuitive consequences aboutthe truth-values of conditionals. You can see from the truth table, for example,that an SL conditional is true any time the consequent is true, no matter whatch. 2 sentential logic 35the antecedent is. (Look at lines 1 and 3 in the chart.) And it is also true anytime the antecedent is false, no matter what the consequent is. (Look at lines 3and 4.) This is an odd consequence. In English, some conditionals with trueconsequents and/or false antecedents seem clearly to be false. For example:23. If there are no philosophy courses at UBC, then PHIL 220 is a philosophycourse at UBC.24. If this book has fewer than thirty pages, then it will win the 2018 Pulitzerprize for poetry.Both 23 and 24 seem clearly false. But each of them, translated into SL, wouldcome out true. (If this isn’t obvious, it’s worth taking a moment to translate themand consider the truth table.) I told you before that English translations into SLare only approximate! Despite these odd results, the approach to conditionalsoffered here actually preserves many of the most important logical features ofconditionals. We’ll see this in more detail once we start working with proofs.For now, I’ll just ask you to go along with this approach to conditionals, eventhough it will seem strange.Note that unlike conjunction and disjunction, the conditional is asymmetrical.You cannot swap the antecedent and consequent without changing the meaningof the sentence, because (Φ⊃Ψ) and (Ψ⊃Φ) are not logically equivalent.2.7 BiconditionalConsider these sentences:25. The figure on the board is a triangle only if it has exactly three sides.26. The figure on the board is a triangle if it has exactly three sides.27. The figure on the board is a triangle if and only if it has exactly threesides.T: The figure is a triangle.S: The figure has three sides.Sentence 25, for reasons discussed above, can be translated as (T ⊃ S).Sentence 26 is importantly different. It can be paraphrased as, ‘If the figure hasthree sides, then it is a triangle.’ So it can be translated as (S ⊃ T ).Sentence 27 says that T is true if and only if S is true; we can infer S fromT , and we can infer T from S. This is called a biconditional, because it36 forallx (UBC edition)entails the two conditionals S ⊃ T and T ⊃ S. We will use ‘≡’ to represent thebiconditional; sentence 27 can be translated as (S ≡ T ).We could abide without a new symbol for the biconditional. Since sentence 27means ‘(T ⊃ S) and (S ⊃ T ),’ we could translate it as a conjunction of thosetwo conditionals — as ((T ⊃ S) & (S ⊃ T )). Notice how the parentheses work:we need to add a new set of parentheses for the conjunction, in addition to theones that were already there for the conditionals.Because we could always write ((Φ ⊃ Ψ) & (Ψ ⊃ Φ)) instead of (Φ ≡ Ψ), wedo not strictly speaking need to introduce a new symbol for the biconditional.Nevertheless, logical languages usually have such a symbol. SL will have one,which makes it easier to translate phrases like ‘if and only if.’(Φ ≡ Ψ) is true if and only if Φ and Ψ have the same truth value: they’re eitherboth true, or they’re both false. This is the characteristic truth table for thebiconditional:Φ Ψ Φ≡Ψ1 1 11 0 00 1 00 0 12.8 Other symbolizationWe have now introduced all of the connectives of SL. We can use them togetherto translate many kinds of sentences. Consider these examples of sentences thatuse the English-language connective ‘unless’, with an associated symbolizationkey:28. Unless you wear a jacket, you will catch a cold.29. You will catch a cold unless you wear a jacket.J: You will wear a jacket.D: You will catch a cold.We can paraphrase sentence 28 as ‘Unless J , D.’ This means that if you do notwear a jacket, then you will catch a cold; with this in mind, we might translateit as ¬J ⊃ D. It also means that if you do not catch a cold, then you must haveworn a jacket; with this in mind, we might translate it as ¬D ⊃ J .ch. 2 sentential logic 37Which of these is the correct translation of sentence 28? Both translations arecorrect, because the two translations are logically equivalent in SL.Sentence 29, in English, is logically equivalent to sentence 28. It can be translatedas either ¬J ⊃ D or ¬D ⊃ J .When symbolizing sentences like sentence 28 and sentence 29, it is easy to getturned around. Since the conditional is not symmetric, it would be wrong totranslate either sentence as J ⊃ ¬D. Fortunately, there are other logicallyequivalent expressions. Both sentences mean that you will wear a jacket or —if you do not wear a jacket — then you will catch a cold. So we can translatethem as J ∨D.If a sentence can be paraphrased as ‘Unless Φ, Ψ ,’ then it can besymbolized as (¬Φ ⊃ Ψ), (¬Ψ ⊃ Φ), or (Φ ∨ Ψ).Symbolization of standard sentence types is summarized on p. .2.9 Sentences of SLThe sentence ‘Apples are red, or berries are blue’ is a sentence of English, andthe sentence ‘(A∨B)’ is a sentence of SL. Although we can identify sentences ofEnglish when we encounter them, we do not have a formal definition of ‘sentenceof English’. In SL, it is possible to formally define what counts as a sentence.This is one respect in which a formal language like SL is more precise than anatural language like English.It is important to distinguish between the logical language SL, which we aredeveloping, and the language that we use to talk about SL. When we talkabout a language, the language that we are talking about is called the objectlanguage. The language that we use to talk about the object language is calledthe metalanguage.The object language in this chapter is SL. The metalanguage is English —not conversational English, but English supplemented with some logical andmathematical vocabulary (including the ‘Φ’ and ‘Ψ ’ symbols). The sentence‘(A ∨B)’ is a sentence in the object language, because it uses only symbols ofSL. The word ‘sentence’ is not itself part of SL, however, so the sentence ‘Thisexpression is a sentence of SL’ is not a sentence of SL. It is a sentence in themetalanguage, a sentence that we use to talk about SL.In this section, we will give a formal definition for ‘sentence of SL.’ The definitionitself will be given in mathematical English, the metalanguage.38 forallx (UBC edition)2.9.1 ExpressionsThere are three kinds of symbols in SL:sentence letters A,B,C, . . . , Zwith subscripts, as needed A1, B1, Z1, A2, A25, J375, . . .connectives ¬, & ,∨,⊃,≡parentheses ( , )We define an expression of sl as any string of symbols of SL. Take any of thesymbols of SL and write them down, in any order, and you have an expression.2.9.2 Well-formed formulaeSince any sequence of symbols is an expression, many expressions of SL will begobbledegook. For example, these expressions don’t mean anything:¬¬¬¬))≡A4 ∨None of these are sentences in SL. A meaningful expression is called a well-formed formula. It is common to use the acronym wff ; the plural is wffs.Individual sentence letters like A and G13 are certainly wffs. We can formfurther wffs out of these by using the various connectives. Using negation, wecan get ¬A and ¬G13. Using conjunction, we can get (A&G13), (G13 &A),(A&A), and (G13 &G13). We could also apply negation repeatedly to get wffslike ¬¬A or apply negation along with conjunction to get wffs like ¬(A&G13)and ¬(G13 &¬G13). The possible combinations are endless, even starting withjust these two sentence letters, and there are infinitely many sentence letters. Sothere is no point in trying to list all the wffs.Instead, we will describe the rules that govern how wffs can be constructed.Consider negation: Given any wff Φ of SL, ¬Φ is a wff of SL. Remember, Φ is notitself a sentence letter; it is a variable that stands in for any wff at all (atomicor not). Since the variable Φ is not a symbol of SL, ¬Φ is not an expressionof SL. Instead, it is an expression of the metalanguage that allows us to talkabout infinitely many expressions of SL: all of the expressions that start withthe negation symbol. Because Φ is part of the metalanguage, it is called ametavariable.ch. 2 sentential logic 39Another way of saying that given any wff Φ of SL, ¬Φ is a wff of SL is to saythat any time you have a wff of SL, you can make a new wff by adding a ‘¬’symbol to the front of it.We can say similar things for each of the other connectives. For instance, if Φand Ψ are wffs of SL, then (Φ&Ψ) is a wff of SL. (That is to say, given anytwo wffs in SL, you can put a ‘ & ’ symbol between them, a ‘(’ in front of thefirst, and a ‘)’ after the second, and the whole thing will add up to a new wff.)Providing clauses like this for all of the connectives, we arrive at the followingformal definition for a well-formed formula of SL:1. Every atomic sentence is a wff.2. For all expressions Φ and Ψ ,(a) If Φ is a wff, then ¬Φ is a wff.(b) If Φ and Ψ are wffs, then (Φ&Ψ) is a wff.(c) If Φ and Ψ are wffs, then (Φ ∨ Ψ) is a wff.(d) If Φ and Ψ are wffs, then (Φ ⊃ Ψ) is a wff.(e) If Φ and Ψ are wffs, then (Φ ≡ Ψ) is a wff.3. Nothing else is a wff.This is a recursive definition of a wff, illustrating how one can, starting with thesimplest cases (atomic sentences), build up more complicated sentences of SL. Ifyou have a wff, you can stick a ‘¬’ in front of it to make a new wff. If you havetwo wffs, you can put a ‘(’ in front of the first one, followed by a ‘ & ’, followedby the second one, then finally a ‘)’, and end up with a new wff. Etc.Note that these are purely syntactic rules. They tell you how to constructan admissible (‘grammatical’) sentence in SL. They do not tell you what thesentence will mean. (We’ve touched on that already, in introducing characteristictruth tables. We’ll return to this topic in much more detail in Ch. 3.)Suppose we want to know whether or not ¬¬¬D is a wff of SL. Looking at thesecond clause of the definition, we know that ¬¬¬D is a wff if ¬¬D is a wff. Sonow we need to ask whether or not ¬¬D is a wff. Again looking at the secondclause of the definition, ¬¬D is a wff if ¬D is. Again, ¬D is a wff if D is a wff.Now D is a sentence letter, an atomic sentence of SL, so we know that D is awff by the first clause of the definition. So for a compound formula like ¬¬¬D,we must apply the definition repeatedly. Eventually we arrive at the atomicsentences from which the wff is built up.The connective that you look to first in decomposing a sentence is called themain connective (or main logical operator) of that sentence. For example:40 forallx (UBC edition)The main connective of ¬(E ∨ (F ⊃ G)) is negation, ¬. The main connectiveof (¬E ∨ (F ⊃ G)) is disjunction, ∨. Conversely, if you’re building up a wfffrom simpler sentences, the connective introduced by the last rule you apply isthe main connective. It is the connective that governs the interpretation of theentire sentence.2.9.3 SentencesRecall that a sentence is a meaningful expression that can be true or false. Sincethe meaningful expressions of SL are the wffs and since every wff of SL is eithertrue or false, the definition for a sentence of SL is the same as the definition fora wff. Not every formal language will have this nice feature. In the language QL,which is developed later in the book, there are wffs which are not sentences.The recursive structure of sentences in SL will be important when we considerthe circumstances under which a particular sentence would be true or false.The sentence ¬¬¬D is true if and only if the sentence ¬¬D is false, and so onthrough the structure of the sentence until we arrive at the atomic components:¬¬¬D is true if and only if the atomic sentence D is false. We will return tothis point in much more detail in Chapters 3 and 4.2.9.4 Notational conventionsA wff like (Q&R) must be surrounded by parentheses, because we might applythe definition again to use this as part of a more complicated sentence. If wenegate (Q&R), we get ¬(Q&R). If we just had Q&R without the parenthesesand put a negation in front of it, we would have ¬Q&R. It is most natural toread this as meaning the same thing as (¬Q&R), something very different than¬(Q&R). The sentence ¬(Q&R) means that it is not the case that both Qand R are true; Q might be false or R might be false, but the sentence does nottell us which. The sentence (¬Q&R) means specifically that Q is false and thatR is true. So parentheses are crucial to the meaning of the sentence.Consequently, strictly speaking, Q&R without parentheses is not a sentenceof SL. When using SL, however, we will sometimes be able to relax the precisedefinition so as to make things easier for ourselves. We will do this in severalways.First, we understand that Q&R means the same thing as (Q&R). As a matterof convention, we can leave off parentheses that occur around the entire sentence.Think of this as a kind of shorthand; we don’t always write out the parenthesesthat we know are really there.ch. 2 sentential logic 41Second, it can sometimes be confusing to look at long sentences with nested setsof parentheses. We adopt the convention of using square brackets ‘[’ and ‘]’ inplace of parenthesis. There is no logical difference between (P ∨Q) and [P ∨Q],for example. The unwieldy sentence(((H ⊃ I) ∨ (I ⊃ H)) & (J ∨K))could be written in this way, omitting the outer parentheses and using squarebrackets to make the inner structure easier to see:[(H ⊃ I) ∨ (I ⊃ H)]& (J ∨K)Third, we will sometimes want to translate the conjunction of three or moresentences. For the sentence ‘Alice, Bob, and Candice all went to the party’,suppose we let A mean ‘Alice went’, B mean ‘Bob went’, and C mean ‘Candicewent.’ The definition only allows us to form a conjunction out of two sentences,so we can translate it as (A&B) &C or as A& (B&C). There is no reason todistinguish between these, since the two translations are logically equivalent.There is no logical difference between the first, in which (A&B) is conjoined withC, and the second, in which A is conjoined with (B&C). So we might as welljust write A&B&C. As a matter of convention, we can leave out parentheseswhen we conjoin three or more sentences.Fourth, a similar situation arises with multiple disjunctions. ‘Either Alice, Bob,or Candice went to the party’ can be translated as (A∨B)∨C or as A∨ (B∨C).Since these two translations are logically equivalent, we may write A ∨B ∨ C.These latter two conventions only apply to multiple conjunctions or multipledisjunctions. If a series of connectives includes both disjunctions and conjunctions,then the parentheses are essential; as with (A&B) ∨ C and A& (B ∨ C). Theparentheses are also required if there is a series of conditionals or biconditionals;as with (A ⊃ B) ⊃ C and A ≡ (B ≡ C).We have adopted these four rules as notational conventions, not as changes tothe definition of a sentence. Strictly speaking, A ∨B ∨ C is still not a sentence.Instead, it is a kind of shorthand. We write it for the sake of convenience, butwe really mean the sentence (A ∨ (B ∨ C)).Unless and until you are very confident about wffs and the use of parentheses, itis probably good advice to stick to the formal rules. These notational conventionsare a way to skip steps when writing things down; if you’re unsure about whetherit’s OK to take the shortcut, the safest thing is to go by the formal definition.If we had given a different definition for a wff, then these could count aswffs. We might have written rule 3 in this way: “If Φ, Ψ , . . . Z are wffs,then (Φ&Ψ & . . . &Z), is a wff.” This would make it easier to translate someEnglish sentences, but would have the cost of making our formal language morecomplicated. We would have to keep the complex definition in mind when we42 forallx (UBC edition)develop truth tables and a proof system. We want a logical language that isexpressively simple and allows us to translate easily from English, but we alsowant a formally simple language. (As we’ll see later, this is important if we wantto be able to prove things about our language.) Adopting notational conventionsis a compromise between these two desires.Practice Exercises? Part A Using the symbolization key given, translate each English-languagesentence into SL.M: Those creatures are men in suits.C: Those creatures are chimpanzees.G: Those creatures are gorillas.1. Those creatures are not men in suits.2. Those creatures are men in suits, or they are not.3. Those creatures are either gorillas or chimpanzees.4. Those creatures are neither gorillas nor chimpanzees.5. If those creatures are chimpanzees, then they are neither gorillas nor menin suits.6. Unless those creatures are men in suits, they are either chimpanzees orthey are gorillas.Part B Using the symbolization key given, translate each English-languagesentence into SL.A: Mister Ace was murdered.B: The butler did it.C: The cook did it.D: The Duchess is lying.E: Mister Edge was murdered.F: The murder weapon was a frying pan.1. Either Mister Ace or Mister Edge was murdered.2. If Mister Ace was murdered, then the cook did it.3. If Mister Edge was murdered, then the cook did not do it.4. Either the butler did it, or the Duchess is lying.5. The cook did it only if the Duchess is lying.6. If the murder weapon was a frying pan, then the culprit must have beenthe cook.ch. 2 sentential logic 437. If the murder weapon was not a frying pan, then the culprit was eitherthe cook or the butler.8. Mister Ace was murdered if and only if Mister Edge was not murdered.9. The Duchess is lying, unless it was Mister Edge who was murdered.10. If Mister Ace was murdered, he was done in with a frying pan.11. Since the cook did it, the butler did not.12. Of course the Duchess is lying!? Part C Using the symbolization key given, translate each English-languagesentence into SL.E1: Ava is an electrician.E2: Harrison is an electrician.F1: Ava is a firefighter.F2: Harrison is a firefighter.S1: Ava is satisfied with her career.S2: Harrison is satisfied with his career.1. Ava and Harrison are both electricians.2. If Ava is a firefighter, then she is satisfied with her career.3. Ava is a firefighter, unless she is an electrician.4. Harrison is an unsatisfied electrician.5. Neither Ava nor Harrison is an electrician.6. Both Ava and Harrison are electricians, but neither of them find it satisfy-ing.7. Harrison is satisfied only if he is a firefighter.8. If Ava is not an electrician, then neither is Harrison, but if she is, then heis too.9. Ava is satisfied with her career if and only if Harrison is not satisfied withhis.10. If Harrison is both an electrician and a firefighter, then he must be satisfiedwith his work.11. It cannot be that Harrison is both an electrician and a firefighter.12. Harrison and Ava are both firefighters if and only if neither of them is anelectrician.? Part D Give a symbolization key and symbolize the following sentences in SL.1. Alice and Bob are both spies.2. If either Alice or Bob is a spy, then the code has been broken.3. If neither Alice nor Bob is a spy, then the code remains unbroken.4. The German embassy will be in an uproar, unless someone has broken thecode.44 forallx (UBC edition)5. Either the code has been broken or it has not, but the German embassywill be in an uproar regardless.6. Either Alice or Bob is a spy, but not both.? Part E Give a symbolization key and symbolize the following sentences in SL.1. If Gregor plays first base, then the team will lose.2. The team will lose unless there is a miracle.3. The team will either lose or it won’t, but Gregor will play first baseregardless.4. Gregor’s mom will bake cookies if and only if Gregor plays first base.5. If there is a miracle, then Gregor’s mom will not bake cookies.Part F For each argument, write a symbolization key and translate the argumentas well as possible into SL.1. If Dorothy plays the piano in the morning, then Roger wakes up cranky.Dorothy plays piano in the morning unless she is distracted. So if Rogerdoes not wake up cranky, then Dorothy must be distracted.2. It will either rain or snow on Tuesday. If it rains, Neville will be sad. If itsnows, Neville will be cold. Therefore, Neville will either be sad or cold onTuesday.3. If Zoog remembered to do his chores, then things are clean but not neat. Ifhe forgot, then things are neat but not clean. Therefore, things are eitherneat or clean — but not both.? Part G For each of the following: (a) Is it, by the strictest formal standards,a sentence of SL? (b) Is it an acceptable way to write down a sentence of SL,allowing for our notational conventions?1. (A)2. J374 ∨ ¬J3743. ¬¬¬¬F4. ¬&S5. (G&¬G)6. Φ ⊃ Φ7. (A ⊃ (A&¬F )) ∨ (D ≡ E)8. [(Z ≡ S) ⊃W ] & [J ∨X]9. (F ≡ ¬D ⊃ J) ∨ (C &D)Part H1. Are there any wffs of SL that contain no sentence letters? Why or whynot?Chapter 3Truth tablesThis chapter introduces a way of evaluating sentences and arguments of SL. Thetruth table method is a purely mechanical procedure that requires no intuitionor special insight. Given the (albeit imprecise) translatability between SL andnatural languages, this also amounts to a formal way of evaluating some naturallanguage arguments.3.1 Truth-functional connectivesAny non-atomic sentence of SL is composed of atomic sentences with sententialconnectives. In Ch. 2, we offered characteristic truth tables for each connective.Although we didn’t emphasize it at the time, the fact that it is possible togive truth tables like this is very significant. It means that our connectives aretruth-functional. That is to say, the only thing that matters for determiningthe truth value of a given sentence of SL is the truth values of its constituentparts. To determine the truth value of a sentence ¬Φ, the only thing that mattersis the truth value of Φ. You don’t have to know what Φ means, or where it camefrom, or what evidence there is for it and what that evidence might depend on.The truth-value of ¬Φ is a function of the truth-value of Φ. And so likewise forthe other connectives.This is a very interesting feature of SL. It is not inevitable. The syntax of English,for example, permits one to make a new, more complex English declarativesentence by prefixing the phrase ‘Donald Trump cares whether’ in front of anydeclarative English sentence. In this respect this phrase is syntactically similarto ‘¬’ in SL. But it is impossible to give a truth-functional characterization ofthe ‘Donald Trump cares whether’ operator in English. If you want to know4546 forallx (UBC edition)Φ ¬Φ1 00 1Φ Ψ Φ&Ψ Φ∨Ψ Φ⊃Ψ Φ≡Ψ1 1 1 1 1 11 0 0 1 0 00 1 0 1 1 00 0 0 0 1 1Table 3.1: The characteristic truth tables for the connectives of SL.whether Donald Trump cares whether the Canadian dollar is getting stronger,it’s not enough to know whether the Canadian dollar is getting stronger. If itis, he might or might not care; if it isn’t, he might or might not care. ‘DonaldTrump cares whether’ is not truth-functional. But all the connectives of SL are.For this reason, we can construct truth tables to determine the logical featuresof SL sentences.3.2 Complete truth tablesThe truth-value of sentences which contain only one connective are given bythe characteristic truth table for that connective. Truth tables list out all thepossible combinations of truth values for the atomic wffs at play; each rowcorresponds to an interpretation — a possible way of assigning truth valuesto atomic sentence. ({P = 1, Q = 0} is an example of an interpretation.)In the previous chapter, we wrote the characteristic truth tables with ‘1’ for trueand ‘0’ for false. It is important to note, however, that this is not about truth inany deep or cosmic sense. Philosophical investigation into the nature of truth isa worthy project in its own right, but the truth functions in SL are just ruleswhich transform input values into output values. (This is part of the reason why,in this book, we usually write ‘1’ and ‘0’ instead of ‘T’ and ‘F’.) Even thoughwe interpret ‘1’ as meaning ‘true’ and ‘0’ as meaning ‘false’, computers can beprogrammed to fill out truth tables in a purely mechanical way. In a machine,‘1’ might mean that a register is switched on and ‘0’ that the register is switchedoff. Mathematically, they are just the two possible values that a sentence of SLcan have. The truth tables for the connectives of SL, written in terms of 1s and0s, are given in table 3.1.The characteristic truth table for conjunction, for example, gives the truthconditions for any sentence of the form (Φ&Ψ). Even if the conjuncts Φ and Ψare long, complicated sentences, the conjunction is true if and only if both Φand Ψ are true.Let’s construct a truth table for a more complicated sentence. Consider thech. 3 truth tables 47sentence (H & I) ⊃ H. We consider all the possible combinations of true andfalse for H and I, which gives us four rows. We then copy the truth-values forthe sentence letters and write them underneath the letters in the sentence.H I (H & I)⊃H1 1 1 1 11 0 1 0 10 1 0 1 00 0 0 0 0So far all we’ve done is duplicate the first two columns. We’ve written the ‘H’column twice — once under each ‘H’, and the ‘I’ column once, under the ‘I’.Now consider the subsentence H & I. This is a conjunction Φ&Ψ with H as Φand with I as Ψ . H and I are both true on the first row. Since a conjunctionis true when both conjuncts are true, we write a 1 underneath the conjunctionsymbol. In the other three rows, at least one of the conjuncts is false, so theconjunction (H & I) is false. So we write 0s under the conjunction symbol onthose rows:H I (H & I)⊃HΦ & Ψ1 1 1 1 1 11 0 1 0 0 10 1 0 0 1 00 0 0 0 0 0The entire sentence is a conditional Φ⊃Ψ with (H & I) as Φ and with H as Ψ . Onthe second row, for example, (H & I) is false and H is true. Since a conditionalis true when the antecedent is false, we write a 1 in the second row underneaththe conditional symbol. We continue for the other three rows and get this:H I (H & I)⊃HΦ ⊃ Ψ1 1 1 1 11 0 0 1 10 1 0 1 00 0 0 1 0The column of 1s underneath the conditional tells us that the sentence (H & I) ⊃ Iis true regardless of the truth-values of H and I. They can be true or false in anycombination, and the compound sentence still comes out true. It is crucial thatwe have considered all of the possible combinations. If we only had a two-line48 forallx (UBC edition)truth table, we could not be sure that the sentence was not false for some othercombination of truth-values.In this example, we have not repeated all of the entries in every successive table,so that it’s easier for you to see which parts are new. When actually writingtruth tables on paper, however, it is impractical to erase whole columns orrewrite the whole table for every step. Although it is more crowded, the truthtable can be written in this way:H I (H & I)⊃H1 1 1 1 1 1 11 0 1 0 0 1 10 1 0 0 1 1 00 0 0 0 0 1 0Most of the columns underneath the sentence are only there for bookkeepingpurposes. When you become more adept with truth tables, you will probably nolonger need to copy over the columns for each of the sentence letters. In anycase, the truth-value of the sentence on each row is just the column underneaththe main logical operator of the sentence; in this case, the column underneaththe conditional. We’ve marked it in bold.A complete truth table has a row for all the possible combinations of 1 and0 for all of the sentence letters. The size of the complete truth table depends onthe number of different sentence letters in the table. A sentence that containsonly one sentence letter requires only two rows, as in the characteristic truthtable for negation. This is true even if the same letter is repeated many times, asin the sentence [(C ≡ C) ⊃ C] &¬(C ⊃ C). The complete truth table requiresonly two lines because there are only two possibilities: C can be true or it canbe false. A single sentence letter can never be marked both 1 and 0 on the samerow. The truth table for this sentence looks like this:C [(C ≡C )⊃C ] & ¬ (C ⊃C )1 1 1 1 1 1 0 0 1 1 10 0 1 0 0 0 0 0 0 1 0Looking at the column underneath the main connective, we see that the sentenceis false on both rows of the table; i.e., it is false regardless of whether C is trueor false. So it is a contradiction. (See §1.6.1.)A sentence that contains two sentence letters requires four lines for a completetruth table, as in the other characteristic truth tables and the table for (H & I) ⊃I above.A sentence that contains three sentence letters requires eight lines. For example:ch. 3 truth tables 49M N P M & (N ∨ P )1 1 1 1 1 1 1 11 1 0 1 1 1 1 01 0 1 1 1 0 1 11 0 0 1 0 0 0 00 1 1 0 0 1 1 10 1 0 0 0 1 1 00 0 1 0 0 0 1 10 0 0 0 0 0 0 0From this table, we know that the sentence M & (N ∨ P ) might be true or false,depending on the truth-values of M , N , and P .A complete truth table for a sentence that contains four different sentence lettersrequires 16 lines. In general, if a complete truth table has n different sentenceletters, then it must have 2n rows.In order to fill in the columns of a complete truth table, begin with the right-mostsentence letter and alternate 1s and 0s. In the next column to the left, write two1s, write two 0s, and repeat. For the third sentence letter, write four 1s followedby four 0s. This yields an eight line truth table like the one above. For a 16 linetruth table, the next column of sentence letters should have eight 1s followed byeight 0s. For a 32 line table, the next column would have 16 1s followed by 160s. And so on.3.3 Using truth tables3.3.1 Tautologies, contradictions, and contingent sentencesRecall from §1.6.1 that an English sentence is a tautology if it must be true as amatter of logic. With a complete truth table, we consider all of the ways thatthe world might be. If the sentence is true on every line of a complete truthtable, then it is true as a matter of logic, regardless of what the world is like.So a sentence is a tautology in sl if the column under its main connective is1 on every row of a complete truth table.Conversely, a sentence is a contradiction in sl if the column under its mainconnective is 0 on every row of a complete truth table.A sentence is contingent in sl if it is neither a tautology nor a contradiction;i.e. if it is 1 on at least one row and 0 on at least one row.From the truth tables in the previous section, we know that (H & I) ⊃ H is50 forallx (UBC edition)a tautology, that [(C ≡ C) ⊃ C] &¬(C ⊃ C) is a contradiction, and thatM & (N ∨ P ) is contingent.3.3.2 Logical equivalenceTwo sentences are logically equivalent in English if they have the same truth valueas a matter of logic. Once again, truth tables allow us to define an analogousconcept for SL: Two sentences are logically equivalent in sl if they havethe same truth-value on every row of a complete truth table.Consider the sentences ¬(A ∨B) and ¬A&¬B. Are they logically equivalent?To find out, we construct a truth table.A B ¬ (A ∨B) ¬A & ¬B1 1 0 1 1 1 0 1 0 0 11 0 0 1 1 0 0 1 0 1 00 1 0 0 1 1 1 0 0 0 10 0 1 0 0 0 1 0 1 1 0Look at the columns for the main connectives; negation for the first sentence,conjunction for the second. On the first three rows, both are 0. On the finalrow, both are 1. Since they match on every row, the two sentences are logicallyequivalent.3.3.3 ConsistencyA set of sentences in English is consistent if it is logically possible for them allto be true at once. A set of sentences is logically consistent in sl if thereis at least one line of a complete truth table on which all of the sentences aretrue. It is inconsistent otherwise.Look again at the truth table above. We can see that ¬(A ∨B) and ¬A&¬Bare consistent, because there is at least one row, namely the last one, where bothare assigned true.Are A, B, and ¬A&¬B logically consistent? No. If they were, there would haveto be a row where all three sentences are assigned true. But examination of thetruth table reveals that there is no such row.ch. 3 truth tables 513.3.4 ValidityAn argument in English is valid if it is logically impossible for the premises to betrue and for the conclusion to be false at the same time. An argument is validin sl if there is no row of a complete truth table on which the premises are all 1and the conclusion is 0; an argument is invalid in sl if there is such a row.Consider this argument:¬L ⊃ (J ∨ L)¬L.˙. JIs it valid? To find out, we construct a truth table.J L ¬L⊃ (J ∨L) ¬ L J1 1 0 1 1 1 1 1 0 1 11 0 1 0 1 1 1 0 1 0 10 1 0 1 1 0 1 1 0 1 00 0 1 0 0 0 0 0 1 0 0To determine whether the argument is valid, check to see whether there are anyrows on which both premises are assigned 1, but where the conclusion is assigned0. There are no such rows. The only row on which both the premises are 1 isthe second row, and on that row the conclusion is also 1. So the argument formis valid in SL.Here is another example. Is this argument valid?P ⊃ Q¬P.˙. ¬QTo evaluate it, construct a truth table and see whether there is a row that assigns1 to the premises and 0 to the conclusion:P Q P ⊃ Q ¬ P ¬ Q1 1 1 1 1 0 1 0 11 0 1 0 0 0 1 1 00 1 0 1 1 1 0 0 10 0 0 1 0 1 0 1 0There is. On the third row, the premises are true and the conclusion is false. Sothe argument is invalid.52 forallx (UBC edition)3.4 Partial truth tablesIn order to show that a sentence is a tautology, we need to show that it isassigned 1 on every row. So we need a complete truth table. To show that asentence is not a tautology, however, we only need one line: a line on which thesentence is 0. Therefore, in order to show that something is not a tautology, itis enough to provide a one-line partial truth table — regardless of how manysentence letters the sentence might have in it.Consider, for example, the sentence (U &T ) ⊃ (S&W ). We want to show thatit is not a tautology by providing a partial truth table. We fill in 0 for the entiresentence. The main connective of the sentence is a conditional. In order for theconditional to be false, the antecedent must be true (1) and the consequent mustbe false (0). So we fill these in on the table:S T U W (U & T )⊃ (S & W )1 0 0In order for the (U &T ) to be true, both U and T must be true. So we put a 1under those letters:S T U W (U & T )⊃ (S & W )1 1 1 1 1 0 0Remember that each instance of a given sentence letter must be assigned thesame truth value in a given row of a truth table. You can’t assign 1 to oneinstance of U and 0 to another instance of U in the same row. So here we put a1 under each instance of U and T .Now we just need to make (S&W ) false. To do this, we need to make at leastone of S and W false. We can make both S and W false if we want. All thatmatters is that the whole sentence turns out false on this line. Making anarbitrary decision, we finish the table in this way:S T U W (U & T )⊃ (S & W )0 1 1 0 1 1 1 0 0 0 0Showing that something is a tautology requires a complete truth table. Showingthat something is not a tautology requires only a one-line partial truth table,where the sentence is false on that one line. That’s what we’ve just done. In thesame way, to show that something is a contradiction, you must show that it isfalse on every row; to show that it is not a contradiction, you need only find onerow where it is true.ch. 3 truth tables 53A sentence is contingent if it is neither a tautology nor a contradiction. Soshowing that a sentence is contingent requires a two-line partial truth table: Thesentence must be true on one line and false on the other. For example, we canshow that the sentence above is contingent with this truth table:S T U W (U & T )⊃ (S & W )0 1 1 0 1 1 1 0 0 0 00 1 0 0 0 0 1 1 0 0 0Note that there are many combinations of truth values that would have madethe sentence true, so there are many ways we could have written the second line.Showing that a sentence is not contingent requires providing a complete truthtable, because it requires showing that the sentence is a tautology or that it is acontradiction. If you do not know whether a particular sentence is contingent,then you do not know whether you will need a complete or partial truth table.You can always start working on a complete truth table. If you complete rowsthat show the sentence is contingent, then you can stop. If not, then complete thetruth table. Even though two carefully selected rows will show that a contingentsentence is contingent, there is nothing wrong with filling in more rows.Showing that two sentences are logically equivalent requires providing a completetruth table. Showing that two sentences are not logically equivalent requiresonly a one-line partial truth table: Make the table so that one sentence is trueand the other false.Showing that a set of sentences is consistent requires providing one row of a truthtable on which all of the sentences are true. The rest of the table is irrelevant,so a one-line partial truth table will do. Showing that a set of sentences isinconsistent, on the other hand, requires a complete truth table: You must showthat on every row of the table at least one of the sentences is false.Showing that an argument is valid requires a complete truth table. Showingthat an argument is invalid only requires providing a one-line truth table: If youcan produce a line on which the premises are all true and the conclusion is false,then the argument is invalid.Table 3.2 summarizes when a complete truth table is required and when a partialtruth table will do. If you are trying to remember whether you need a completetruth table or not, the general rule is, if you’re looking to establish a claim aboutevery interpretation, you need a complete table.54 forallx (UBC edition)YES NOtautology? complete truth table one-line partial truth tablecontradiction? complete truth table one-line partial truth tablecontingent? two-line partial truth table complete truth tableequivalent? complete truth table one-line partial truth tableconsistent? one-line partial truth table complete truth tablevalid? complete truth table one-line partial truth tableTable 3.2: Do you need a complete truth table or a partial truth table? Itdepends on what you are trying to show.3.5 Evaluating English arguments via SLRecall from §1.4 that a natural language argument is valid if and only if it isimpossible for the premises to be true while the conclusion is false. This notionis rather closely related to validity in SL, which obtains if and only if there isno assignment of truth values to atomic sentences on which the premises areassigned ‘1’ and the conclusion is assigned ‘0’. This is of course by design. Inmany cases, validity in SL is a good indication of validity in English. Considerfor example this English argument:1. Either the butler is the murderer or the gardener isn’t who he says he is.2. The gardener is who he says he is..˙. The butler is the murderer.This argument is valid. There’s no way for the conclusion to be false if bothpremises are true. Let’s translate this argument into SL, and evaluate theresulting formal argument for validity with a truth table. We begin with asymbolization key:B: The butler is the murderer.G: The gardener is who he says he is.With this key, the argument, rendered in SL, looks like this:1. (B ∨ ¬G)2. G.˙. BWe can evaluate this formal argument for validity using a truth table. We’llset up a table with two atomic sentences, and check to see whether there is ach. 3 truth tables 55row that assigns ‘1’ to both premises and assigns ‘0’ to the conclusion. Thecompleted truth table looks like this:B G (B ∨¬G) G B1 1 1 1 0 1 1 11 0 1 1 1 0 0 10 1 0 0 0 1 1 00 0 0 1 1 0 0 0If the argument form were invalid, there’d be a line on which the first two boldvalues are ‘1’ but the third is ‘0’. There is no such line, so the argument form isvalid. This helps explain why the English version of the argument is also valid:it is an argument that has a valid form in SL.There are at least two advantages to evaluating natural language argument forvalidity by translating it into SL. For one thing, using truth tables to evaluatearguments is a formal method that does not require any particular rationalinsight or intuition. It is often relatively easy to tell whether the informaldefinition of validity is met — most of us have a good sense of what is and isn’tpossible. But it is an advantage to have an operationalized set of rules that asimple computer could apply. A second advantage, alluded to in Chapter 2, isthat talking about valid SL forms provides a nice way to explain what variousvalid English arguments have in common with one another. The validity of theargument form1. (B ∨ ¬G)2. G.˙. Bhelps explain why the butler argument is valid, but the explanation does notdepend on any of the specifics of what the individual letters stand for. It wouldbe a perfectly good explanation for why any argument of this form would haveto be valid. For example, consider this argument:1. Either Barney is a purple dinosaur or I don’t have a really weird-lookingcat.2. I do have a really weird-looking cat..˙. Barney is a purple dinosaur.That argument can be translated into the same SL argument just evaluated,using this symbolization key:B: Barney is a purple dinosaur.56 forallx (UBC edition)G: I have a really weird-looking cat.Since it too has a valid form in SL, it too must be a valid argument. Anyargument with this form will be a valid argument.If you have an argument form that is valid in SL, then any Englishargument that is properly translatable into that argument formwill be valid.Note that this is even true for arguments that have a more complex internalstructure, like this strange argument:1. Either Canada is a democracy if and only if Poland is neither part of theEuropean Union nor majority Catholic, or my dog is not not lazy.2. My dog is not lazy..˙. Canada is a democracy if and only if Poland is neither part of the EuropeanUnion nor majority Catholic.One could translate this argument in to a relatively complex argument in SL.(If you did, the first premise would probably be a disjunction with a complexbiconditional as one disjunct, and a negated negation as the other; the conclusionwould probably be a biconditional with an atom on one side, and a negateddisjunction on the other.) In this instance, however, that extra structure isn’tneeded to explain the validity of the argument. For this argument too has thesame valid form as the previous ones, as can be seen via this symbolization key:B: Canada is a democracy if and only if Poland is neither part of theEuropean Union nor majority Catholic.G: My dog is not lazy.For many purposes, this would be a poor choice of symbolization keys. It ignoresthe internal structure of two central sentences. But if, as here, that structureis irrelevant, you can save yourself some work by engaging at a higher level ofabstraction.Note also that it’s an implication of what we’ve just illustrated that there’snot just one SL argument that is the argument form for a given argument inEnglish. One can formalize arguments in various ways. If an argument has avalid argument form in SL, that’s a guarantee that the argument is valid. But itis important not to invert this relationship — it does not follow from the factthat an argument has an invalid form, that the argument is invalid. One reasonthis is so is that it is possible for some arguments to have valid forms and invalidforms.ch. 3 truth tables 573.6 Common student mistakesHere are a couple of issues that often come up when I teach this material tostudents.3.6.1 Validity is not relative to particular interpretationsTruth in SL is relative to an interpretation. A given wff is true on someinterpretations, and false on others. For example, P ≡ Q is true when {P = 1,Q = 1}, and false when {P = 1, Q = 0}. But validity, unlike truth, is a claimabout all interpretations.Let’s use a truth table to evaluate whether P ∨ Q, ¬P .˙. P ≡ Q is a validargument form. To do this, we can draw the complete truth table for those threewffs:P Q P ∨ Q ¬ P P ≡ Q1 1 1 1 1 0 1 1 1 11 0 1 1 0 0 1 1 0 00 1 0 1 1 1 0 0 0 10 0 0 0 0 1 0 0 1 0We then check to see whether there is any row where the disjunction and thenegation are true and the biconditional is false. Row three, {P = 0, Q = 1}, issuch a row, so the argument is invalid.Here is a common mistake. Students sometimes draw the truth table accurately,and correctly note that line three is inconsistent with the argument’s validity,but misdescribe the situation, saying something like this: ‘The argument isinvalid when P = 0, Q = 1.’ This is a mistake because the argument isn’tonly invalid in that row — the argument in invalid, period. That interpretationexplains why the argument is invalid, but validity and invalidity are claims aboutall interpretations. So we shouldn’t say the argument is invalid ‘when’ P = 0,Q = 1; it’s invalid always, because of the interpretation where P = 0, Q = 1.This sounds like a small and fussy point about speaking precisely about logic. Itis. But careful language here can help lead to careful thought. It is important,to thinking clearly about validity, to remember that validity, unlike truth, is afeature that arguments have or lack intrinsically. It is not something that anargument has in some interpretations and lacks in others.The same goes for tautologies, contradictions, and contingent sentences. Asentence isn’t a tautology on some interpretations and not on others; it either58 forallx (UBC edition)simply is, or it simply is not, a tautology.3.6.2 Valid arguments with invalid formsHere is a question students sometimes get wrong: Provide (in English) a validargument of the form P .˙. P &Q, or explain why it is impossible to do so.Students often think this is impossible, because P .˙. P &Q is an invalid argumentform. That certainly is an invalid argument form. The premise is true, and theconclusion false, when P = 1, Q = 0. But that doesn’t mean there can be novalid argument with this form.A valid argument form is a form that guarantees that any argument of thatform is valid. Since this form is invalid, it is going to be possible to find invalidarguments of that form. For example, we could interpret P and Q thus:P: Captain Hook is angry.Q: Captain Hook has at least two daughters.And of course the English argument, Captain Hook is angry, therefore CaptainHook is angry and he has at least two daughters is obviously invalid. But thatdoesn’t mean there couldn’t be other arguments of this form that are valid. Theform doesn’t guarantee that they’ll be valid, but they may be valid nevertheless.Consider this symbolization key:P: Captain Hook has three daughters.Q: Captain Hook has at least two daughters.The resulting argument is valid: Captain Hook has three daughters. Therefore,Captain Hook has three daughters and he has at least two daughters. It isimpossible for the premise to be true without the conclusion also being true, soit satisfies the definition of validity given in Chapter 1. This argument is valid,even though it has an SL argument form that is invalid. We will be particularlyinterested in arguments whose validity can be explained by their forms, but thedefinition of validity in English is more general.Practice ExercisesIf you want additional practice, you can construct truth tables for any of thesentences and arguments in the exercises for the previous chapter.ch. 3 truth tables 59? Part A Determine whether each sentence is a tautology, a contradiction, or acontingent sentence. Justify your answer with a complete or partial truth tablewhere appropriate.1. A ⊃ A2. ¬B&B3. C ⊃ ¬C4. ¬D ∨D5. (A ≡ B) ≡ ¬(A ≡ ¬B)6. (A&B) ∨ (B&A)7. (A ⊃ B) ∨ (B ⊃ A)8. ¬[A ⊃ (B ⊃ A)]9. (A&B) ⊃ (B ∨A)10. A ≡ [A ⊃ (B&¬B)]11. ¬(A ∨B) ≡ (¬A&¬B)12. ¬(A&B) ≡ A13.[(A&B) &¬(A&B)]&C14. A ⊃ (B ∨ C)15. [(A&B) &C] ⊃ B16. (A&¬A) ⊃ (B ∨ C)17. ¬[(C ∨A) ∨B]18. (B&D) ≡ [A ≡ (A ∨ C)]? Part B Determine whether each pair of sentences is logically equivalent.Justify your answer with a complete or partial truth table where appropriate.1. A, ¬A2. A, A ∨A3. A ⊃ A, A ≡ A4. A ∨ ¬B, A ⊃ B5. A&¬A, ¬B ≡ B6. ¬(A&B), ¬A ∨ ¬B7. ¬(A ⊃ B), ¬A ⊃ ¬B8. (A ⊃ B), (¬B ⊃ ¬A)9. [(A ∨B) ∨ C], [A ∨ (B ∨ C)]10. [(A ∨B) &C], [A ∨ (B&C)]? Part C Determine whether each set of sentences is consistent or inconsistent.Justify your answer with a complete or partial truth table where appropriate.1. A ⊃ A, ¬A ⊃ ¬A, A&A, A ∨A2. A&B, C ⊃ ¬B, C3. A ∨B, A ⊃ C, B ⊃ C4. A ⊃ B, B ⊃ C, A, ¬C60 forallx (UBC edition)5. B& (C ∨A), A ⊃ B, ¬(B ∨ C)6. A ∨B, B ∨ C, C ⊃ ¬A7. A ≡ (B ∨ C), C ⊃ ¬A, A ⊃ ¬B8. A, B, C, ¬D, ¬E, F? Part D Determine whether each argument is valid or invalid. Justify youranswer with a complete or partial truth table where appropriate.1. A ⊃ A .˙. A2. A ∨ [A ⊃ (A ≡ A)] .˙. A3. A ⊃ (A&¬A) .˙. ¬A4. A ≡ ¬(B ≡ A) .˙. A5. A ∨ (B ⊃ A) .˙. ¬A ⊃ ¬B6. A ⊃ B, B .˙. A7. A ∨B, B ∨ C, ¬A .˙. B&C8. A ∨B, B ∨ C, ¬B .˙. A&C9. (B&A) ⊃ C, (C &A) ⊃ B .˙. (C &B) ⊃ A10. A ≡ B, B ≡ C .˙. A ≡ C? Part E Answer each of the questions below and justify your answer.1. Suppose that Φ and Ψ are logically equivalent. What can you say aboutΦ ≡ Ψ?2. Suppose that (Φ&Ψ) ⊃ Ω is contingent. What can you say about theargument “Φ, Ψ , .˙.Ω”?3. Suppose that {Φ, Ψ,Ω} is inconsistent. What can you say about (Φ&Ψ &Ω)?4. Suppose that Φ is a contradiction. What can you say about the argument“Φ, Ψ , .˙.Ω”?5. Suppose that Ω is a tautology. What can you say about the argument “Φ,Ψ , .˙.Ω”?6. Suppose that Φ and Ψ are logically equivalent. What can you say about(Φ ∨ Ψ)?7. Suppose that Φ and Ψ are not logically equivalent. What can you sayabout (Φ ∨ Ψ)?Part F We could leave the biconditional (≡) out of the language. If we didthat, we could still write ‘A ≡ B’ so as to make sentences easier to read, butthat would be shorthand for (A ⊃ B) & (B ⊃ A). The resulting language wouldbe formally equivalent to SL, since A ≡ B and (A ⊃ B) & (B ⊃ A) are logicallyequivalent in SL. If we valued formal simplicity over expressive richness, wecould replace more of the connectives with notational conventions and still havea language equivalent to SL.ch. 3 truth tables 61There are a number of equivalent languages with only two connectives. It wouldbe enough to have only negation and the material conditional. Show this bywriting sentences that are logically equivalent to each of the following using onlyparentheses, sentence letters, negation (¬), and the material conditional (⊃).1.? A ∨B2.? A&B3.? A ≡ BWe could have a language that is equivalent to SL with only negation anddisjunction as connectives. Show this: Using only parentheses, sentence letters,negation (¬), and disjunction (∨), write sentences that are logically equivalentto each of the following.4. A&B5. A ⊃ B6. A ≡ BThe Sheffer stroke is a logical connective with the following characteristictruthtable:Φ Ψ Φ|Ψ1 1 01 0 10 1 10 0 17. Write a sentence using the connectives of SL that is logically equivalent to(A|B).Every sentence written using a connective of SL can be rewritten as a logicallyequivalent sentence using one or more Sheffer strokes. Using no connectivesother than the Sheffer stroke, write sentences that are equivalent to each of thefollowing.8. ¬A9. (A&B)10. (A ∨B)11. (A ⊃ B)12. (A ≡ B)Chapter 4Entailment and Models forSLThis chapter offers a formal semantics for SL, allowing us to be more preciseabout the notion of truth in SL. We’ll also highlight some important features ofentailment, a key concept in formal logic.A formal, logical language is built from two kinds of elements: logical symbolsand non-logical symbols. Connectives like ‘ & ’ and ‘⊃’ are logical symbols,because their meaning is specified within the formal language. When writinga symbolization key, you are not allowed to change the meaning of the logicalsymbols. You cannot say, for instance, that the ‘¬’ symbol will mean ‘not’ inone argument and ‘perhaps’ in another. The ‘¬’ symbol always means logicalnegation. It is used to translate the English language word ‘not’, but it is asymbol of a formal language and is defined by its truth conditions.The sentence letters in SL are non-logical symbols, because their meaning is notdefined by the logical structure of SL. When we translate an argument fromEnglish to SL, for example, the sentence letter M does not have its meaningfixed in advance; instead, we provide a symbolization key that says how Mshould be interpreted in that argument.In translating from English to a formal language, we provided symbolizationkeys which were interpretations of all the non-logical symbols we used in thetranslation. An interpretation gives a meaning to all the non-logical elementsof the language. We’ll also use the term ‘model’ as another word for aninterpretation. In our simple formal language SL, meaning is simply a matter oftruth and falsity, relative to a given interpretation. These notions too need tobe characterized in a formal way.62ch. 4 entailment and models for sl 634.1 Semantics for SLThis section provides a rigorous, formal characterization of truth in SL whichbuilds on what we already know from doing truth tables. We were able to usetruth tables to reliably test whether a sentence was a tautology in SL, whethertwo sentences were equivalent, whether an argument was valid, and so on. Forinstance: Φ is a tautology in SL if and only if it is assigned ‘1’ on every line of acomplete truth table.This worked because each line of a truth table corresponds to a way the worldmight be. We considered all the possible combinations of 1 and 0 for the sentenceletters that made a difference to the sentences we cared about. The truth tableallowed us to determine what would happen given these different combinations.Each of those combinations of truth values is an interpretation in SL.Once we construct a truth table, the symbols ‘1’ and ‘0’ are divorced from theirmetalinguistic meaning of ‘true’ and ‘false’. We interpret ‘1’ as meaning ‘true’,but the formal properties of 1 are defined by the characteristic truth tables forthe various connectives. The symbols in a truth table have a formal meaning thatwe can specify entirely in terms of how the connectives operate. For example, ifA is value 1, then ¬A is value 0.To formally define truth in SL, then, we want a function that assigns, for eachmodel, a 1 or 0 to each of the sentences of SL. We can interpret this function asa definition of truth for SL if it assigns 1 to all of the true sentences of SL and0 to all of the false sentences of SL. Call this function ‘v’ (for ‘valuation’). Wewant v to be a function such that for any sentence Φ, v(Φ) = 1 if Φ is true andv(Φ) = 0 if Φ is false.Recall that the recursive definition of a wff for SL had two stages: The first stepsaid that atomic sentences (solitary sentence letters) are wffs. The second stageallowed for wffs to be constructed out of more basic wffs. There were clausesof the definition for all of the sentential connectives. For example, if Φ is a wff,then ¬Φ is a wff.Our strategy for defining the truth function, v, will also be in two steps. Thefirst step will handle truth for atomic sentences; the second step will handletruth for compound sentences.4.2 Defining truth in SLHow can we define truth for an atomic sentence of SL? Consider, for example,the sentence M . Without an interpretation, we cannot say whether M is true or64 forallx (UBC edition)false. It might mean anything. If we use M to symbolize ‘The moon orbits theEarth’, then M is true. If we use M to symbolize ‘The moon is a giant turnip’,then M is false.When we give a symbolization key for SL, we provide an interpretation of thesentence letters that we use. The key gives an English language sentence foreach sentence letter that we use. In this way, the interpretation specifies whateach of the sentence letters means. However, this is not enough to determinewhether or not that sentence is true. The sentences about the moon, for instance,require that you know some rudimentary astronomy. Imagine a small child whobecame convinced that the moon is a giant turnip. She could understand whatthe sentence ‘The moon is a giant turnip’ means, but mistakenly think that itwas true.So an interpretation alone does not determine whether a sentence is true orfalse. Truth or falsity depends also on what the world is like. If M meant ‘Themoon is a giant turnip’ and the real moon were a giant turnip, then M wouldbe true. To put the point in a general way, truth or falsity is determined by aninterpretation plus a way that the world is.INTERPRETATION + STATE OF THE WORLD =⇒ TRUTH/FALSITYIn providing a logical definition of truth, we will not be able to give an accountof how an atomic sentence is made true or false by the world. Instead, wewill introduce a truth value assignment. Formally, this will be a function thattells us the truth value of all the atomic sentences. Call this function ‘a’ (for‘assignment’). We define a for all sentence letters P , such thata(P ) ={1 if P is true,0 otherwise.This means that a takes any atomic sentence of SL and assigns it either a one ora zero; one if the sentence is true, zero if the sentence is false.You can think of a as being like a row of a truth table. Whereas a truth tablerow assigns a truth value to a few atomic sentences, the truth value assignmentassigns a value to every atomic sentence of SL. There are infinitely many sentenceletters, and the truth value assignment gives a value to each of them. Whenconstructing a truth table, we only care about sentence letters that affect thetruth value of sentences that interest us. As such, we ignore the rest. Strictlyspeaking, every row of a truth table gives a partial truth value assignment.It is important to note that the truth value assignment, a, is not part of thelanguage SL. Rather, it is part of the mathematical machinery that we are usingto describe SL. It encodes which atomic sentences are true and which are false.ch. 4 entailment and models for sl 65We now define the truth function, v, using the same recursive structure that weused to define a wff of SL.1. If Φ is a sentence letter, then v(Φ) = a(Φ).2. If Φ is ¬Ψ for some sentence Ψ , thenv(Φ) ={1 if v(Ψ) = 0,0 otherwise.3. If Φ is (Ψ &Ω) for some sentences Ψ , Ω, thenv(Φ) ={1 if v(Ψ) = 1 and v(Ω) = 1,0 otherwise.You may be tempted to worry that this definition is circular, because it uses theword ‘and’ in trying to define ‘and.’ But remember, we are not attempting togive a definition of the English word ‘and’; we are giving a definition of truth forsentences of SL containing the logical symbol ‘ & .’ We define truth for objectlanguage sentences containing the symbol ‘ & ’ using the metalanguage word‘and.’ There is nothing circular about that.4. If Φ is (Ψ ∨Ω) for some sentences Ψ , Ω, thenv(Φ) ={0 if v(Ψ) = 0 and v(Ω) = 0,1 otherwise.5. If Φ is (Ψ ⊃ Ω) for some sentences Ψ , Ω, thenv(Φ) ={0 if v(Ψ) = 1 and v(Ω) = 0,1 otherwise.6. If Φ is (Ψ ≡ Ω) for some sentences Ψ , Ω, thenv(Φ) ={1 if v(Ψ) = v(Ω),0 otherwise.Since the definition of v has the same structure as the definition of a wff, weknow that v assigns a value to every wff of SL. Since the sentences of SL andthe wffs of SL are the same, this means that v returns the truth value of everysentence of SL.Truth in SL is always truth relative to some interpretation, because the definitionof truth for SL does not say whether a given sentence is true or false. Rather, itsays how the truth of that sentence relates to a truth value assignment.66 forallx (UBC edition)4.3 Semantic entailmentWe are now in a position to give more precise definitions of terms like ‘tautology’,‘contradiction’, and so on. Truth tables provided a way to check whether asentence was a tautology in SL, but they did not define what it means to be atautology in SL. We will give definitions of these concepts for SL in terms ofentailment.The relation of semantic entailment is about satisfiability — that is, whetherthere is any possible interpretation that meets a certain set of conditions. ‘Φentails Ψ ’, means that there is no interpretation for which Φ is true and Ψ isfalse. In SL, an interpretation is just an assignment of truth values to atomicsentences; so Φ entails Ψ if and only if every assignment of truth values to atomicsentences that makes Φ true also makes Ψ true.We abbreviate entailment with a symbol called the double turnstile: Φ |= Ψmeans ‘Φ semantically entails Ψ .’The double turnstile, like ‘Φ’, etc., is part of the metalanguage we use to discussSL; it is not part of SL itself.4.4 Entailment, validity, and informally good ar-gumentsEntailment is a formal notion. It is connected in important ways to the informalnotion of a good argument, but when considering entailment in SL, it is importantto remember to apply the definitions rigorously and precisely, rather than relyingon your sense of whether the argument is a good one. The notions can comeapart in some surprising ways.Let’s start with a straightforward example. Consider whether this entailmentclaim is true:(P &Q) |= (P ∨Q)This is true if and only if every interpretation that satisfies (P &Q), also satisfies(P ∨Q). Hopefully it is obvious to you why this is. Only an interpretation thatassigns 1 to both P and Q will satisfy the left, and any such interpretation willcertainly satisfy the right as well. (You could draw the truth table to verify thisif you want the practice.) So, as one might naturally expect, (P &Q) |= (P ∨Q).Here is a less intuitive example. What should we make of this claim?(P &Q) |= (A ≡ ¬¬A)ch. 4 entailment and models for sl 67Notice that the sentence letters on the left-hand-side here are completely differentletters from those on the right-hand-side. So there is a straightforward sensein which the two sides of the turnstile have nothing to do with one another.Nevertheless, this is a true entailment claim. This is because it satisfies thedefinition: every interpretation that satisfies the left (i.e., every interpretationthat assigns ‘1’ to both P and Q), also satisfies (A ≡ ¬¬A). This for the simplereason that every interpretation whatsoever satisfies (A ≡ ¬¬A); it is a tautology.From this example we can see that a tautology will be entailed by anythingwhatsoever. If the right-hand-side of the entailment claim is a tautology, itdoesn’t matter what’s on the left — you know it’s going to be true.So these are all true:P |= (P ∨ ¬P )Q |= (P ∨ ¬P )P &¬P |= (P ∨ ¬P )4.5 Tautologies are entailed by the empty setIn the examples so far, we’ve been talking about one sentence entailing another.But we can also describe a set of several sentences as jointly entailing an SLsentence:Φ1, Φ2, Φ3, · · · |= Ψmeans that there is no truth value assignment for which all of the sentences inthe set {Φ1, Φ2, Φ3, · · · } are true and Ψ is false. It will sometimes be convenientto use a variable that can stand in for any set of sentences; just as ‘Φ’ can standfor any sentence of SL, we can let ‘X ’ stand in for any set of sentences in SL.(In the case of ‘Φ |= Ψ ’ above, ‘X ’ is the singleton set containing just Φ.)In general, we may say thatX |= Ψmeans that there’s no interpretation satisfying every member of set X withoutalso satisfying sentence Ψ .We saw in the previous section that a tautology — an SL sentence that is true inevery interpretation — is entailed by any sentence. The reasoning applies moregenerally: a tautology will be entailed by any set of sentences. This includesthe empty set, which can be written ‘∅’. In general ‘X |= Φ’ says that everyinterpretation that satisfies everything in X , satisfies Φ; in the special case whereX is ∅, every interpretation trivially satisfies everything in X . So ‘∅ |= Φ’ saysthat absolutely every interpretation satisfies Φ. That is, it says that Φ is atautology.68 forallx (UBC edition)By convention, we can leave the left side of an entailment claim blank, insteadof writing in ∅. In other words,|= Φis shorthand for∅ |= Φ.So these entailment claims are true, because in each case the sentence on theright is a tautology:|= (P ∨ ¬P )|= (P ≡ P )|= ((P &¬P ) ⊃ (A ∨B))And these entailment claims are not true, because each sentence on the right isnot a tautology:|= P|= (P &¬P )|= ¬(P ⊃ ¬P )(If you need help seeing why a given sentence is or is not a tautology, draw outthe truth table and check to see whether it has a 1 on every row. See §3.3.)4.6 Inconsistent sentences entail absurditiesConsider this entailment claim:(P &¬P ) |= QThis statement is true. It says that every interpretation satisfying (P &¬P ) alsosatisfies Q. But (P &¬P ) is a contradiction — no interpretations satisfy it. Sotrivially, ‘every’ interpretation that satisfies it satisfies Q.This explanation had nothing to do with the specifics of the sentence Q. Exactlythe same considerations would demonstrate that, for any sentence Ψ , (P &¬P ) |=Ψ . And the same goes for any other contradictory sentence, or any set of sentencesthat are mutually inconsistent. These are all true:(P &¬P ) |= (¬Q ⊃ R)P, (¬Q&¬P ) |= (Q ≡ P )ch. 4 entailment and models for sl 69(P &¬P ) |= ((A1 ∨A2) ⊃ ¬(A3 ≡ (A4 &¬A2)))You can think of evaluating an entailment claim as like checking to see whether arule is being violated or not. If there’s a rule that says every student with a doghas to have a permit, you check each student with a dog, and make sure theyhave a permit. If you find someone without a dog, it doesn’t matter whetherthey have a permit or not. (If you don’t have a dog, you’re not breaking therule.) Or if you find someone with a permit, it doesn’t matter whether they havea dog. (Nobody with a permit can be breaking the rule.) An entailment claim islike a rule that says every interpretation that satisfies the left-hand-side mustalso satisfy the right-hand-side. To verify it, you can ignore any interpretationthat falsifies the left, or that satisfies the right.In each of the cases above, you don’t even need to examine the sentence atthe right to confirm that the entailment claim is true, since you already knowthat no interpretations satisfy the sentences on the left. (This is analogous toknowing that no students have a dog — in this case you don’t have to check tosee whether anyone has a permit to confirm that the rule is being respected.)Any entailment claim with an unsatisfiable set of sentences on the left is true.Unsatisfiable sets of sentences entail every sentence.We have a special notation for describing this situation. We write:X |= ⊥to indicate that the set X entails an absurdity; think of ‘⊥’ as standing in for anarbitrary unsatisfiable SL wff. So you can think of ‘X |= ⊥’ as saying that everyinterpretation that satisfies X satisfies the unsatisfiable. That’s just another wayof saying that no interpretation satisfies X . (Compare: ‘the only Klingons whohate honour are the Klingons who aren’t Klingons at all.’)So these entailment claims are true, because the set of sentences on the left isnot satisfiable:(P &¬P ) |= ⊥P, (¬Q&¬P ) |= ⊥(P ≡ ¬P ) |= ⊥And these entailment claims are not true, because the sentences on the left canbe satisfied:(P ∨ ¬P ) |= ⊥P,¬Q, (R ∨Q) |= ⊥¬P,¬Q, (P ⊃ ¬¬Q) |= ⊥70 forallx (UBC edition)(If you need help seeing why a given set of sentences can be satisfied, draw outthe truth table and check to see if any row assigns a 1 to each sentence. See§3.3.)4.7 Defining concepts in terms of entailmentThe double turnstile symbol allows us to give concise definitions for variousconcepts in SL:A tautology in sl is a sentence Φ such that ∅ |= Φ.A contradiction in sl is a sentence Φ such that Φ |= ⊥.A sentence is contingent in sl if and only if it is neither a tautologynor a contradiction.An argument with premises X and conclusion Φ is valid in sl ifand only if X |= Φ.Two sentences Φ and Ψ are logically equivalent in sl if andonly if both Φ |= Ψ and Ψ |= Φ.A set of sentences X is inconsistent in sl if and only if X |= ⊥.Otherwise X is consistent in sl.Practice ExercisesPart A Determine whether each entailment claim is true. You may construct atruth table to test it if you like, but these examples are simple enough so thatyou may be able to just think it through and get the right answer.1. Q |= (P ∨Q)2. P,Q |= (P ∨ P )3. S |= (Q ⊃ Q)4. (P &¬P ) |= (Q&¬Q)5. (P ∨ ¬P ) |=6. |= (P ∨ ¬P )7. |= (P &¬P )8. (A ∨B) ⊃ ¬P |= (P ∨ ¬P )9. (P ≡ Q) |= ((P &Q) ∨ ¬(P ∨Q))Chapter 5SL TreesSo far we have learned one way to evaluate SL arguments for validity: anargument is valid just in case no interpretation satisfies the premises but falsifiesthe conclusion. We can check to see whether an argument is valid by constructingtruth tables, representing all of the possible interpretations, and checking tosee whether any of them do so. This method has two main advantages: one isthat it can be done in a formal and formulaic way — it requires no particularrational insight, just a straightforward application of the rules of SL. Another isthat, assuming you follow the rules correctly, it will always succeed in identifyingwhether the argument is valid or not.The truth-table method also has a significant disadvantage, however: it rapidlybecomes extremely cumbersome for evaluating arguments using more than twoor three atomic sentence-letters. To evaluate an argument like this one, you’dneed to consider four rows of a truth-table:P ≡ Q¬Q¬Q ∨ (P ⊃ ¬Q).˙. ¬P ⊃ QBut change just two of the atoms to new letters, and the required table willgrow exponentially. For this argument, you’d need sixteen rows.P ≡ Q¬A¬Q ∨ (A ⊃ ¬B).˙. ¬P ⊃ B7172 forallx (UBC edition)For this one, you’d need two hundred fifty-six!A ≡ B¬B ⊃ (C ∨D)E ⊃ ¬C(¬D&F ) ∨G¬A&E.˙. H ∨GSo it is useful to have alternate ways of proving entailments. In this chapter wewill introduce a proof system for SL. The proofs we will construct in this chapterare called analytic tableaux. Another name for them is trees, because ofthe way they ‘branch’ out to represent possibilities. In Ch. 7 we will examine adifferent proof system.5.1 Satisfiability and entailmentThe method of trees is most directly a way to test for satisfiability of a set of SLsentences. Since validity is definable in terms of satisfiability, it also gives a wayto test for validity.Recall from §4.1 the definition of entailment in SL: X |= Φ means that there isno interpretation that satisfies every sentence in X but falsifies Φ. (Remember,‘X ’ can stand for any set of SL sentences.) This in turn is equivalent to theclaim that there is no interpretation that satisfies {X ,¬Φ}. Unsatisfiability canbe represented with a double-turnstile with ⊥ on the right-hand side.In general, X |= Ψ is equivalent to X ,¬Ψ |= ⊥.So if you want to tell whether an argument in SL is valid, check to see whetherthe premises, along with the negation of the conclusion, are jointly satisfiable.If they are, the argument is invalid. If they’re not, the argument is valid. Thisis the basic idea of the tree method. We’ll write down the sentences we’reattempting to satisfy, and then we’ll work through their consequences, in orderto see whether it’s possible to complete the task.Let’s begin by working through an example that illustrates the general idea,then come back to give precise rules for the tree method.ch. 5 sl trees 735.2 An example: proving validityLet’s attempt to prove whether this is valid in SL:A&B¬(C ∨D)(¬B ∨ C) ∨ E.˙. ETo evaluate this argument, we consider whether it’s possible to satisfy the threepremises along with the negation of the conclusion. We begin by writing downthe sentences we’re going to attempt to satisfy.1.2.3.4.A&B¬(C ∨D)(¬B ∨ C) ∨ E¬ENotice that we’re putting the negation of the conclusion here in line (4), sincewe’re looking to see whether it’s possible to satisfy the premises and falsify theconclusion. The sentences we write down at the beginning of the tree, we callthe root. Trees are designed to show whether the root is satisfiable, and if so,how.With proof trees, the idea is to continue writing down that which our previouslines already commit us to. Consider the first line, A&B. For this conjunctionto be true, both conjuncts must be true. So any interpretation that satisfies thetop four lines must also satisfy both A and B; the truth of those sentences is acommitment of what we already have. Each follow from what is already writtendown. We represent this by continuing the tree with those sentences:1.2.3.4.5.6.A&B X¬(C ∨D)(¬B ∨ C) ∨ E¬EABIn addition to introducing lines (5) and (6), we also add a check mark on (1), toindicate that we’ve now considered it and represented its consequences. Thinkof the check mark as saying that we’ve extended the tree in a way that encodesthe information from this line.74 forallx (UBC edition)Now consider line (2). This is a negated disjunction. Disjunctions are true anytime either disjunct is true, so this negated disjunction can only be true if bothdisjuncts are false. So we include new lines for ¬C and ¬D, adding a check markon line (2):1.2.3.4.5.6.7.8.A&B X¬(C ∨D) X(¬B ∨ C) ∨ E¬EAB¬C¬DLine (3) is a disjunction. Unlike the previous two cases, it doesn’t tell us whatcategorically must be the case; it says that one of two possibilities must be met.We represent this in our tree by branching into two different columns:1.2.3.4.5.6.7.8.9.A&B X¬(C ∨D) X(¬B ∨ C) ∨ E X¬EAB¬C¬D¬B ∨ C EThink of these two branches as representing the idea that the root implies thatat least one of these ways of continuing the tree must be possible.So now we have two branches to consider. Start by examining the right branch.It gives an atomic sentence, E. Notice, however, that line (4) was ¬E. Soif we’re looking for a way to satisfy (1)-(4), this right branch isn’t a possibleinterpretation. It requires E to be true, and it also requires ¬E to be true. If abranch contains any sentence and also contains its negation, we know that branchdoesn’t correspond to a possible interpretation. We’ll consider this branch closed,and mark this with an ‘×’.The left branch is another disjunction. It too branches out into its two disjuncts:ch. 5 sl trees 751.2.3.4.5.6.7.8.9.10.A&B X¬(C ∨D) X(¬B ∨ C) ∨ E X¬EAB¬C¬D¬B ∨ C X¬B×C×E×Both of these disjuncts also result in closed branches. The left branch at (10)is the negation of (6), and the right branch is the negation of (7). Now everybranch in this tree is closed. This corresponds to the idea that every possibleway there could be to satisfy (1)-(4) ended up requiring a contradiction. Thereis no interpretation that satisfies (1)-(4). In other words, the argument we beganwith was valid.In the previous chapter we used the double turnstile, ‘|=’, to represent entailment.In this chapter we will use a different but related symbol, the single turnstile,which looks like this: ‘`’. Think of this symbol as representing provability. Soin proving that E follows from the premises above, we’re showing that {A&B,¬(C ∨D), (¬B ∨ C) ∨ E} ` E.Obviously provability and entailment are closely connected; we’ll consider thatconnection in more detail in Chapter 6.5.3 An example: proving invalidityLet’s work through another example to further illustrate the idea. In §5.4 we’lllearn the formal rules for trees in SL. Consider this argument form:(D ∨A) &¬NN ∨ ¬A.˙. ¬N &A76 forallx (UBC edition)As before, we’ll construct a tree that includes the premises and the negation ofthe conclusion at the top, and we’ll attempt to find an interpretation satisfyingthose three lines. By convention, we write the claim we’re attempting to proveat the top of the tree. We begin by processing line (1), which is a conjunction;its conjuncts are given in (4) and (5). Line (6) processes the disjunction in line(2), and the left branch closes.{(D ∨A) &¬N,N ∨ ¬A} ` ¬N &A1.2.3.4.5.6.(D ∨A) &¬N XN ∨ ¬A X¬(¬N &A)D ∨A¬NN×¬ALine (3) is a negated conjunction. A conjunction is true if and only if bothconjuncts are true, so it is false if at least one conjunct is false. So to resolve(3), line (7) branches into the negation of each conjunct, ¬¬N and ¬A. Theformer closes because it’s the negation of line (5). The final sentence requiringresolution is the disjunction on line (4); it branches, and one branch closes.{(D ∨A) &¬N,N ∨ ¬A} ` ¬N &A1.2.3.4.5.6.7.8.(D ∨A) &¬N XN ∨ ¬A X¬(¬N &A) XD ∨A X¬NN×¬A¬¬N×¬AD↑A×The ↑ indicates that the open branch ending in D is a completed branch. (We’llprecisely define ‘completion’ in §5.4.) What this means is that this branchch. 5 sl trees 77represents a way to satisfy all three sentences at the root of the tree. In otherwords, this argument form is not valid in SL. Furthermore, examining theopen branch demonstrates an interpretation that satisfies the root. The branchincludes three wffs that are either atoms or negated atoms: ¬A, D, and ¬N . Soit suggests this interpretation:I = A = 0D = 1N = 0You can verify this result by evaluating the premises and the conclusion of theargument we began with on this interpretation. Since there is an interpretationthat satisfies the premises and falsifies the conclusion, our tree system provesthe argument invalid.5.4 Resolution rules for SL treesHopefully the examples we’ve worked through have given you a decent intuitivesense of the tree method for SL. Now let’s get more precise about it, by givingformal rules for trees. You should be able to recognize the following rules as ageneralization of the steps of the proofs given above.We begin with resolution rules. These rules identify ways that tree branchescan be extended, given sentences that are already in that branch. The rulesdepend on the main connective of the sentence in question. (If the main connectiveis negation, then they also depend on the main connective of the negand.)5.4.1 ConjunctionThe rule for conjunction is that if you have a conjunction in a branch, you maymake a linear extension of the branch that includes each conjunct, adding acheck mark next to the conjunction. Using our Greek symbols once again asvariables to stand in for any sentence of SL, any time you have this,Φ&Ψyou may extend each open branch of the tree to this:78 forallx (UBC edition)Φ&Ψ XΦΨIt is important to remember once again that Φ and Ψ here can stand for anysentences of SL, including complex ones.If you have extended a tree branch using the conjunction rule, we say that youhave resolved the conjunction. When sentences are resolved, they are markedwith a check. In stating the resolution rules, we’ll typically omit the check mark;here is the conjunction rule:Φ&ΨΦΨ5.4.2 Negated conjunctionIf you have a negated conjunction, you may resolve it by branching into thenegation of each conjunct. This makes sense because there are two ways for anegated conjunct to be true — either conjunct can be false.¬(Φ&Ψ)¬Φ ¬Ψch. 5 sl trees 795.4.3 DisjunctionDisjunctions branch into each disjunct.Φ ∨ ΨΦ Ψ5.4.4 Negated disjunctionSince disjunctions are true any time either disjunct is true, they are only falseif both disjuncts are false. So negated disjunctions are resolved with a lineardevelopment containing the negation of each disjunct.¬(Φ ∨ Ψ)¬Φ¬Ψ5.4.5 ConditionalRecall the characteristic truth table for the conditional in SL:Φ Ψ Φ⊃Ψ1 1 11 0 00 1 10 0 1Conditionals are true any time the antecedent is false, and also any time theconsequent is true. We can represent this with a branching tree development,similar to a disjunction.80 forallx (UBC edition)Φ ⊃ Ψ¬Φ Ψ5.4.6 Negated conditionalAs in the case of disjunction, the negation of a conditional is a relatively strongclaim — the only way for a conditional to be false is for its antecedent to be trueand for its consequent to be false. We represent this with a linear development:¬(Φ ⊃ Ψ)Φ¬Ψ5.4.7 BiconditionalBiconditionals require a slightly different structure. A biconditional says thattwo sentences have the same truth value: either both are true, or both are false.Since these represent two different ways a biconditional can be true, our treerules will develop biconditionals into two new branches. But unlike in the caseof our other rules, each new branch will contain two sentences. One branch willhave both sides of the biconditional; the other will have both sides’ negations:Φ ≡ ΨΦΨ¬Φ¬Ψch. 5 sl trees 815.4.8 Negated biconditionalNegated biconditionals yield the same structure, but instead of new brancheswhere each subsentence has the same truth value, they yield branches in whichthe subsentences have opposite values.¬(Φ ≡ Ψ)Φ¬Ψ¬ΦΨ5.4.9 Double negationThere is one more resolution rule, for double-negated sentences. One resolves adouble negation by developing a linear branch where both negations are removed.¬¬ΦΦ5.4.10 Resolution rules summaryThese nine resolution rules describe, in purely formal terms, how to resolvemost sentences of SL. The only exceptions are atomic sentences, and atomicsentences with a single negation sign in front of them. In our formal proof system,atoms and negated atoms have a special role to play; they are not resolved. Indeveloping a tree proof, you may apply any of these rules to any unresolvedsentence (other than an atom or negated atom) in the branch. Mark resolvedsentences with a check.Note that it is not a rule that sentences must be resolved in the order in whichthey appear in the tree; one can save sentences and come back later to resolvethem. However, if you are resolving a sentence after the tree has already donesome branching, you must perform the resolution rule under each open branchthat includes that sentence. Here is a tree illustrating the issue.82 forallx (UBC edition)1.2.3.4.5.6.(D&¬R) ∨Q X¬Q ∨R XD&¬R X¬QD¬R↑RD¬R×4, 6Q¬Q×3, 4R↑1 ∨2 ∨3 &This tree has two disjunctions in the root, so it will involve branching multipletimes. We could start with either one, but this tree resolved line (1) first,branching into the two disjuncts at line (3). The ‘1 ∨’ to the right at that line isthe explanation for what is happening there: it indicates that the disjunctionrule was applied to line (1). When the second disjunction is processed at line(4), it needs to be done on both branches, since both branches include line (2).That’s why our two branches split into four at line (4). One branch closes atthat point. At line (5), the conjunction at line (3) is processed. This needs tobe done on both branches that include that disjunction, but not the right-mostbranch, which does not include it. Resolving a sentence affects everything belowit in the tree in its own descendent branches, but it doesn’t affect branches thatare off to the side.Besides the tree resolution rules, there are two more rules to our proof system.5.5 Branch closure rulesIn the tree examples above, we closed branches when they contained sentencesalong with their negations. Here is our general rule for tree closure:If any branch contains some formula Φ, and also contains ¬Φ, thatbranch is closed. Any branch that is not closed is open. Wemark closed branches with the ‘×’ symbol.A tree is closed if and only if every branch in that tree is closed.A tree is open if and only if it is not closed.If every branch is closed, the method proves that the root is unsatisfiable.Note that ‘Φ’ stands in for any sentence of SL; it is not part of the closure rulethat one must have an atomic sentence and its negation. Consider for examplech. 5 sl trees 83this tree:1.2.3.4.(¬¬S&T ) ⊃ (¬P ≡ (Q ∨R)) X¬¬S&T¬(¬P ≡ (Q ∨R))¬(¬¬S&T )×2, 4(¬P ≡ (Q ∨R))×3, 4In this tree, we only resolved the first sentence, using the conditional rule, andthe tree immediately closed. The line numbers under the closure symbols specifyexactly why the branches close: the left branch developed into the negation of(2), and the right branch developed into the negation of (3). Notice that we couldhave developed lines (2) and (3), using the conjunction or negated biconditionalrules, respectively, but this would have resulted in a more complicated tree thatultimately yielded the same result. Similarly, if we hadn’t noticed that bothbranches close at line (4), we could have continued processing the tree, developingthe left branch using the negated conjunction rule on (4), and developing theright branch using the biconditionals rule. But this too would have involvedneedless complication for no benefit. (Eventually, the tree would have closedwith simpler sentences too.) You can save yourself a lot of work by noticingwhen branches are ready to mark as closed, and by thinking a bit strategicallyabout which sentences to resolve first.5.6 Branch completion rulesHere is the final piece of our formal proof system. In our examples above wehave used ‘↑’ to indicate that an open branch is complete. We need a formalcharacterization of branch completion.In SL, a resolvable wff is a wff for which we have a resolution rule. That is,anything other than an atom or a negated atom.A branch is complete if and only if either it is closed, or everyresolvable sentence in that branch has been resolved.A tree is complete if and only if every branch in that tree iscomplete.If there is at least one open branch in a completed tree, the tree method indicatesthat that branch corresponds to an interpretation that satisfies the root.We use the ‘` ’ symbol to indicate tree closure. ‘X ` ⊥’ means that a tree with84 forallx (UBC edition)root X closes. We use ‘0 ⊥’ to indicate that a completed tree remains open.We will also define our single turnstile symbol so that X ` Φ is equivalent toX ,¬Φ ` ⊥. We’ve been using this implicit definition already, by testing validityby putting the negation of the conclusion along with the premises in the root.This completes the introduction of the formal rules for the SL tree proof system.In the remainder of this chapter, we’ll work through some more examples andprovide some advice for working through trees more efficiently. In Chapter 6we’ll explain why the tree method works, and prove why it’s a good one.5.7 Resolution orderThe resolution rules do not specify the order in which you should resolve sentencesin a tree; you are free to resolve in any order you like. But some ways of processingtrees are more efficient than others. Here is an example to illustrate. Suppose wewant to consider whether {¬(C &A), D ≡ C,A ∨B,¬B} ` ⊥. So we put thosesentences in the root of a tree. Here’s what happens if we resolve the sentencesin the order in which they’re given:1.2.3.4.5.6.7.8.¬(C &A) XD ≡ C XA ∨B X¬B¬CDC×¬D¬CA↑B×¬ADCA×B×¬D¬CA×B×1 ¬&2 ≡3 ∨Again, the justifications for particular steps are added to the right of the tree forclarity. For example, ‘1¬& ’ indicates that line (5) was produced by performingthe negated conjunction resolution rule on line (1). This tree remains open,offering this interpretation that satisfies the root:I =A = 1B = 0C = 0D = 0ch. 5 sl trees 85This tree reminds us again that if a sentence in a tree has multiple open branchesbelow it, then resolving that sentence requires that the resolution be performedin each open branch below. That’s why, for example, resolving line (3) at line (8)requires three different new branchings. (We do not resolve under the left-mostcolumn, because it is already closed.) So when there are more open branches,resolving sentences requires more work on the tree. Consequently, it is sometimesa good idea to think a bit strategically, and close off parts of the tree earlier, tosave yourself some work.The example above is a perfectly fine instance of a completed tree, and it doesyield the correct answer. However, it’s possible to get there with less work, bychoosing to resolve sentences that will close off branches right away. Here isanother way of developing a tree with the same root as in the previous example.This version begins by resolving line (3), because doing so will close off one ofits new branches immediately. It then continues to resolve in the most efficientorder:1.2.3.4.5.6.7.8.¬(C &A) XD ≡ C XA ∨B X¬BA¬CDC×¬D¬C↑¬A×B×3 ∨1 ¬&2 ≡This tree gets us to the same result much more quickly, pointing to the sameinterpretation that satisfies the root. As a general rule of thumb, it’s good adviceto look a step ahead, and resolve sentences that won’t lead to new open branches,before resolving ones that will. For just the same reason, it is usually moreefficient to resolve those sentences that have linear rules before those that havebranching rules.5.8 Choosing the right rootTrees are used to indicate whether the root is satisfiable. We test argumentsfor validity with a tree by putting their premises along with the negation oftheir conclusions in the root; if the tree remains open, indicating that the set is86 forallx (UBC edition)satisfiable, that means it’s possible to satisfy the premises while falsifying theconclusion, which means the argument is invalid. If the tree closes, that suggeststhe argument is invalid.But trees can be used to evaluate many other kinds of questions, aside fromvalidity. Any question that can be converted into a question about satisfiabilitycan be answered with a tree. For example, if you want to find out whether aset of sentences is consistent, put that set of sentences into the tree. A set ofsentences is consistent if and only if there is an interpretation satisfying it, soyou can use a tree to find out.What if you want to find out whether a sentence is a tautology? A tautology is asentence that is satisfied by every interpretation. To test this claim with a tree,we’ll attempt to find an interpretation that falsifies it, by putting its negationin the root. If this tree remains open, it shows us how to satisfy the negation,which means that the sentence is not a tautology. If it closes, then the negationis unsatisfiable, which means the sentence is a tautology.To find out whether a sentence is a contradiction, check and see whether it ispossible to satisfy it, by putting it in the root of the tree. If the tree remainsopen, the sentence is satisfiable, which means it’s not a contradiction. If the treecloses, then it is a contradiction.To use a tree to answer a question, the first step is always to convert the questioninto a question about whether some sentence, or set of sentences, is satisfiable.Practice ExercisesIf you want additional practice, you can construct trees for any of the SLarguments and entailment claims in the exercises for the previous two chapters.? Part A To evaluate each of the following claims with a tree, (a) what wouldyou put in the root of the tree?, and (b) if the tree closes, does that show thatthe claim is true or false?1. {P, P ⊃ Q,Q ⊃ ¬P} ` ⊥2. (P ⊃ Q) ≡ (Q ⊃ P ) is a tautology.3. The following argument is valid:P &Q¬R ⊃ ¬Q.˙. P &R4. There is no interpretation that satisfies A∨B, B ⊃ C, and A ≡ C withoutalso satisfying C.ch. 5 sl trees 875. A ≡ ¬A is a contradiction.6. Every interpretation satisfying P , P ⊃ Q, and ¬Q also satisfies A.7. There is at least one interpretation that satisfies P ⊃ Q, ¬P ∨ ¬Q, andQ ⊃ P .? Part B Evaluate the argument given on p. 72 by constructing a tree. If it isinvalid, give a model demonstrating it so.? Part C Evaluate each claim from Part A by constructing a tree. If applicable,give the interpretation that demonstrates the claim true or false.Part D Recall the discussion of the Sheffer stroke in Chapter 3, page 60. Thatconnective, if added to SL, would have this characteristic truth table:Φ Ψ Φ|Ψ1 1 01 0 10 1 10 0 1What would be appropriate tree resolution rules for the Sheffer stroke, or thenegated Sheffer stroke?Chapter 6Soundness andCompleteness for SL TreesIn Chapter 5 we introduced a proof system for SL. Trees provide a method foranswering the question of whether a given set of SL sentences is jointly satisfiable.Our focus last chapter was on using trees to answer those questions; this chapterwe turn to the study of the proof system itself. This chapter engages in a projectof metalogic. In particular, we aim to first precisify, then answer, this question:is the tree method a good method?It might not be obvious at first that there is even a genuine question here. Thetree method is a formal method, with precisely defined rules. You might betempted to think that, by definition, following the rules makes it a good method.There is one sense in which that is right — the rules laid out last chapter tell youto follow the rules laid out last chapter, so following the rules is doing what thesystem tells you to do. But there is also a deeper question to be asked. The ruleswere not selected at random. They were designed to do something in particular:namely, to tell whether a set of sentences is satisfiable. This isn’t a questionabout trees. (Note that we were considering this question back in Chapter 4,before we had even introduced the idea of trees.) So there is a question to beasked about whether the tree method actually does what it’s supposed to do.Recall the distinction between our two turnstiles: we use ‘X |= ⊥’ to mean thatno interpretation satisfies X ; ‘X ` ⊥’ means that a tree with root X closes. Weinterpret the latter as our proof system saying that nothing satisfies the root.Our question now is, can our proof system be trusted? Is it reliable? The mainproject of this chapter is to prove in a rigorous way that it can, and is.88ch. 6 soundness and completeness for sl trees 896.1 Informal proofIn Chapter 5 we learned a formal proof system. This chapter, we will proveimportant results about that system. It is important to emphasize, however,that the formal proof system isn’t the only way to ‘prove’ things. In particular,the tree method is not an appropriate methodology for the task of this chapter.When we say we wish to prove that the tree method is good, we don’t meanthat we will put the negation of that claim — i.e., that the tree method is notgood — in the root of a tree. That would get us nowhere.Instead, we will be engaging in an informal proof about the formal system. Aninformal proof needn’t be any less conclusive or compelling than a formal proofis, but evaluating it makes use of our general ability to recognize what followsfrom what, rather than working through a list of syntactically-defined rules.6.2 SoundnessIf a tree with root X closes, we interpret that as the system telling us thatX is unsatisfiable. If our system is a good one, then a tree will never misleadus in this respect. Tree closure should guarantee unsatisfiability. We call thisproperty soundness. The soundness of our SL tree system is the first importantmetalogical theorem we will prove in this chapter. (A ‘metalogical’ theorem is atheorem about logic.)soundness: If a tree closes, that guarantees that its root is unsat-isfiable. In other words:X ` ⊥ ⇒ X |= ⊥Here is a way to illustrate that soundness is a substantive result, and to clarifywhat it is we’re trying to prove. Recall the resolution rule for disjunction (see p.79):Φ ∨ ΨΦ ΨLet’s suppose for the purpose of argument that we had a different disjunctionrule instead of this one. Suppose, for example, that our rule for disjunction hadbeen this:90 forallx (UBC edition)Φ ∨ ΨΦΨIf this had been our disjunction rule, and all the other rules remained the same,our tree system would have been unsound. It would have been possible forsatisfiable roots to result in closed trees. That is to say, there would have beenpossible sets of sentences X such that X ` ⊥, even though X 6|= ⊥. Consider forexample these sentences:P¬P ∨QThese sentences are obviously jointly satisfiable, by an interpretation that assigns1 to P and 1 to Q. But if we used the tree system with the alternate disjunctionrule above, the tree would close:1.2.3.4.P¬P ∨Q X¬PQ×1, 32 alt.∨This would be a counterexample to the soundness of the SL tree system. Thisexplains what’s wrong with the alternate disjunction rule — it would allowtrees to ‘prove’ that a root is unsatisfiable, even if it really is satisfiable. Inconsidering the soundness of our system, we are investigating whether our actualproof system is defective in the same way this hypothetical modification of thesystem would have been. We will prove that it is not.6.3 Recursive proofsOur proof of soundness will be a recursive proof. A recursive proof is a proofthat proceeds stepwise: one first demonstrates that the claim to be proven holdsfor some simple case, and then shows that, if it holds for some case, then it alsoholds for some other, slightly more complicated case. More ways of complicatingcases may also be discussed, each along with the assurance that if the claim holdsfor a simpler case, then it will also hold for each more complex one. Finally, if onecan demonstrate that the ways of complicating cases considered are exhaustive —ch. 6 soundness and completeness for sl trees 91that is, if these represent the only possible cases — then this has been shown forevery possible case. You may be familiar with recursive proofs already in thecontext of mathematical induction. (This is a topic that comes up in many highschool algebra classes.)Here is an example illustrating proof by induction. Suppose that Sir RodericMurgatroyd has been cursed. The curse is subject to the following rules:1. The only way for someone to escape the curse is to transfer it to someoneelse.2. There are only three ways to transfer the curse to someone else:(a) One may transfer it to one’s parent.(b) One may transfer it to one’s child.(c) One may transfer it to one’s sibling.Given these rules, it’s not difficult to see that the Murgatroyd curse will neverleave the family. We know that Sir Roderic has the curse. He could transfer itto a parent, a child, or a sibling, but none of those actions would remove thecurse from the family, since one’s parents, one’s children, and one’s siblings areall family members. And any of those people, if they had the curse, can onlytransfer it to someone else within the family. No curse transfer can get the curseoutside the family. So someone in the family will remain cursed forever.Slightly more precisely: we’re attempting to prove that the curse will always bein the family. Roderic is in the family. And, for any person, if they are in thefamily, then they cannot get rid of the curse without transferring it to anothermember of the family. The proof is perfectly general; it applies to Roderic’sgreat-great-great-grandchildren just as well as it applies to Sir Roderic himself.This is a simple example of a recursive proof. The proof of the soundness of theSL tree method is more complex, but it has the same basic structure.6.4 Proving soundnessSoundness is the claim that any time a tree closes, the root must be unsatisfiable.This is equivalent to the claim that any time the root is satisfiable, the treewon’t close. (Compare: if every new citizen swears loyalty to the Queen, theneveryone who doesn’t swear loyalty to the Queen must not be a new citizen.) Soto prove soundness, we can assume that the root is satisfiable, and show that itfollows that the tree doesn’t close.Suppose, then, we have some satisfiable set of sentences X in the root of atree. If the root is satisfiable, then there is some interpretation that satisfies it.92 forallx (UBC edition)(Recall that an interpretation in SL is an assignment of truth values to atomicsentences.) Call this interpretation I. We will begin by proving that, if our treefollows the rules given in Chapter 5, then I doesn’t just satisfy the root — itsatisfies every sentence in some branch of the completed tree. Once we establishthat claim, it’s only a short step to demonstrate that this branch doesn’t close.Branches only close when they contain some formula and its negation. Butno interpretation can satisfy a formula and its negation; so if I satisfies everyformula in the branch, that means that branch must not contain any formulaalong with its negation. So the tree will remain open.This is the broad structure of our proof. The key step is in proving that I hasthe property mentioned above — that it satisfies every sentence in an openbranch. We will prove this recursively.6.4.1 RootStart with the root, X . This is trivial. We are assuming that our tree beginswith a satisfiable root, because we are trying to prove what follows from thatassumption. (Namely, that the tree won’t close.) I is just our name for one ofthe interpretations we are assuming must exist. So I satisfies everything in theroot. This is a reasonable thing to assume, when proving soundness, becausesoundness only tells us what happens when the root is satisfiable: namely, thatthe tree won’t close. Soundness doesn’t say anything about what happens if theroot is unsatisfiable.We want our proof to be perfectly general, so we don’t want to make anyparticular assumptions about what the tree does beyond the root. But, given theresolution rules outlined in §5.4, there are only nine possible ways the tree mightdevelop at each step. (Compare the three possible ways the curse might move inthe Murgatroyd example.) We will prove, for each of these nine resolution rules,the following: if I satisfies all the sentences in the branch above, then I alsosatisfies at least one branch of what comes below. In other words, we’ll prove,for each inference rule, that that rule cannot take you from a satisfiable branchto a tree with no satisfiable branches.6.4.2 ConjunctionSuppose that a tree develops via the conjunction rule:ch. 6 soundness and completeness for sl trees 93Φ&ΨΦΨWe assume that I satisfies the branch above the development. So in particular,I must satisfy Φ&Ψ . We may write this asI(Φ&Ψ) = 1We know from the definition of truth in SL that any interpretation that assigns1 to a conjunction must assign 1 to each conjunct. (See page 65.) So:I(Φ) = 1I(Ψ) = 1What we’ve just shown is that, if I satisfies the branch above this development,then it also satisfies everything in the new development. The conjunction rulewill never take us from a satisfiable branch to an unsatisfiable one. We need toprove that every possible way of developing the tree is like that.6.4.3 Negated conjunctionNegated conjunctions develop in our system with a branching rule:¬(Φ&Ψ)¬Φ ¬ΨOnce again, we are assuming for the purpose of argument that our interpretationI satisfies everything up until this development. So I(¬(Φ&Ψ)) = 1. Since Isatisfies that negation, I(Φ&Ψ) = 0. Given our definition of truth in SL, anyinterpretation that assigns 0 to this conjunction must assign 0 to at least one ofits conjuncts. So either I(Φ) = 0 or I(Ψ) = 0. (It might assign 0 to both, butwhat we know for sure is that it assigns 0 to at least one.) If I(Φ) = 0, thenI(¬Φ) = 1, and so the new left branch is satisfied. If I(Ψ) = 0, then I(¬Ψ) = 1,and so the new right branch is satisfied. Since (at least) one of these must be thecase, we know that I satisfies at least one branch of our extended tree, assumingit satisfied that which came before the extension. So the negated conjunctionrule will never take us from a satisfiable branch to an unsatisfiable one.94 forallx (UBC edition)6.4.4 DisjunctionDisjunctions branch according to this rule:Φ ∨ ΨΦ ΨAssume I(Φ ∨ Ψ) = 1. Then either I(Φ) = 1 or I(Ψ) = 1. If I(Φ) = 1, thenI satisfies the left branch. If I(Ψ) = 1, then I satisfies the right branch. So,assuming that I satisfies the sentences above this resolution rule, it must satisfyat least one branch below it. So the disjunction rule will never take us from asatisfiable branch to an unsatisfiable one.Hopefully the pattern is becoming clear by now. We’ve proven, for three of ournine rules, that they cannot take us from a satisfiable branch to an unsatisfiableone. Six resolution rules remain to be considered.Consider again the variant disjunction rule hypothesized above:Φ ∨ ΨΦΨIf we attempted to go through the same reasoning we’ve been going through,we’d fail. Assume I(Φ ∨ Ψ) = 1. By the definition of truth in SL, we know thatI assigns 1 to at least one of Φ and Ψ , but there is no guarantee that it willsatisfy both. So we have no assurance that, if one follows this rule, I will satisfythe development of the tree, even if we assume it satisfies the sentences above.We cannot prove that the use of this rule will never lead to an inappropriatetree closure.6.4.5 Negated DisjunctionHere is the rule for negated disjunctions:¬(Φ ∨ Ψ)¬Φ¬Ψch. 6 soundness and completeness for sl trees 95Suppose that I(¬(Φ∨Ψ)) = 1. Given the definition of negation, this means thatI(Φ ∨ Ψ) = 0. This in turn means, given the definition of disjunction, that Imust assign 0 to both Φ and Ψ . And so of course, given the definition of negationagain, we know that I assigns 1 to ¬Φ, and also assigns 1 to ¬Ψ . Therefore, ifwe began with a satisfiable tree branch, invoking this rule, like the other goodrules we’ve considered, will preserve satisfiability; the negated disjunction rulewill never take one from a satisfiable branch to an unsatisfiable one.6.4.6 ConditionalThe conditional rule is:Φ ⊃ Ψ¬Φ ΨAs before, assume that the material above the branch is satisfiable; so someinterpretation I satisfies it. Any interpretation that satisfies a conditional mustassign 0 to the antecedent, or 1 to the consequent (or both). If I assigns 0 tothe antecedent, then it satisfies the left development of the tree. If I assigns 1to the consequent, then it satisfies the right development of the tree. So, giventhat it satisfies the conditional, I is guaranteed to satisfy at least one branch ofthe tree as developed by the conditional rule.6.4.7 Negated ConditionalThe rule given in Chapter 5 for negated conditionals was this:¬(Φ ⊃ Ψ)Φ¬ΨI hope the procedure is feeling a bit tedious by now. As before, we assume thatthe negated conditional is satisfiable, and prove that the tree as developed by thisrule will remain satisfiable. Since we’re assuming that ¬(Φ⊃Ψ) is satisfiable, itfollows that some interpretation I satisfies it. But I(¬(Φ ⊃ Ψ)) only if I(Φ) = 1and I(Ψ) = 0. So I will satisfy Φ and ¬Ψ . That is to say, it will satisfy thecontinuation of the branch given this rule. The negated conditional rule cannever take one from a satisfiable root to an unsatisfiable tree.96 forallx (UBC edition)We have three more rules to consider.6.4.8 BiconditionalHere is the biconditional rule:Φ ≡ ΨΦΨ¬Φ¬ΨAssuming that I satisfies Φ≡Ψ , it must either assign 1 to both Φ and Ψ , or itmust assign 0 to both Φ and Ψ . If the former, I will satisfy the left branch ofthe new development from this rule. If the latter, it will satisfy the right branch,since any interpretation that assigns 0 to a sentence must assign 1 to its negation.So this rule too can never take one from a satisfiable root to an unsatisfiabletree.6.4.9 Negated biconditional¬(Φ ≡ Ψ)Φ¬Ψ¬ΦΨThe reasoning is much as before. If our interpretation satisfies the negatedbiconditional, then it must assign opposite values to each side; i.e., eitherI(Φ) = 1 and I(Ψ) = 0, or I(Φ) = 0 and I(Ψ) = 1. If the former, I satisfiesthe left branch; if the latter, I satisfies the right branch. So if the negatedbiconditional is satisfiable, this rule will never result in an unsatisfiable tree.6.4.10 Double negationHere is our final tree resolution rule:¬¬ΦΦch. 6 soundness and completeness for sl trees 97One last time, we assume that we begin with something satisfiable; so we allowthat some interpretation I assigns 1 to ¬¬Φ. If it assigns 1 to this negation,then it must assign 0 to its negand, ¬Φ. That is to say, I(¬Φ) = 0. And since itassigns 0 to this negation, it must assign 1 to its negand: I(Φ) = 1. But thisjust is the new branch development. So if we began with something satisfiable,this rule will result in something satisfiable.6.4.11 Taking stockWe’ve shown, for the nine resolution rules in our tree system, that they each havethe following important feature: if you begin with a satisfiable set of sentences,applying the rule will always result in at least one continuation of the tree thatis also satisfiable. And since these nine rules are the only ways one can develop atree, we’ve proven that there is no possible way, consistent with the tree rules, fora tree with a satisfiable root to develop into a tree with no satisfiable branches.Branches can only be closed if they contain a sentence and its negation, which asatisfiable branch will never have. So, assuming we started with a satisfiable root,the rules will never result in a tree with all branches closed. Satisfiable rootswill always result in open trees. The tree method will never erroneously “prove”that a root is unsatisfiable. Equivalently, tree closure guarantees unsatisfiabilityof the root. The tree method is sound.soundness: If a tree closes, that guarantees that its root is unsat-isfiable. In other words:X ` ⊥ ⇒ X |= ⊥6.5 CompletenessSoundness is the first of two important metalogical theorems considered inthis chapter. The second is completeness. One can think of soundness as aguarantee against a system proving too much; in proving soundness, we wereassuring ourselves that a tree would close only if the root was unsatisfiable.Completeness, as the name suggests, concerns whether our system proves enough.We want our system to be sure to close, if the root is unsatisfiable. Remember,we take open branches in completed trees as an indication that the root issatisfiable. Completeness is about ensuring that this is a warranted conclusion.98 forallx (UBC edition)completeness: If a root is unsatisfiable, that guarantees that thetree will close. In other words:X |= ⊥ ⇒ X ` ⊥Consider the unsatisfiable set of sentences, {¬Q,P &Q}. Given our conjunctionrule, a tree with this root will close:1.2.3.4.¬QP &Q XPQ×1, 42 &In proving completeness, we wish to demonstrate that this will always be the case:whenever we begin with an unsatisfiable root, the entire tree will eventually close.Notice that if we had a different conjunction rule that called for a branchingdevelopment instead of a linear one, completeness would fail. Suppose we hadthis rule:Φ&ΨΦ ΨUsing this rule, the tree with root {¬Q,P &Q} would remain open:1.2.3.¬QP &Q XP↑Q×1, 32 alt. &The right branch closes, but this tree has a left branch that remains open, eventhough the root is unsatisfiable. So if we modified our proof system by usingthis rule instead of the linear rule for conjunction, we would have a system thatfails completeness. We wish to prove that, given the actual rules, our system iscomplete.ch. 6 soundness and completeness for sl trees 996.6 Proving completenessCompleteness is the claim that any time a set of sentences is unsatisfiable, acompleted tree with that set as its root will close. This is equivalent to theclaim that if a completed tree has a branch that remains open, then the rootis satisfiable. To prove completeness, we will assume that a completed tree hasa branch that remains open, and prove that, on this assumption, the root issatisfiable. In fact, we will prove something stronger than that: we will provethat every formula in a completed open branch is satisfiable, by demonstratinga recipe for constructing an interpretation that satisfies it. Since the root is partof every branch of the tree, this will suffice for proving completeness.As in the case of our soundness proof, we will be giving an informal proof aboutour formal system.Here is the broad shape of the proof. Suppose that a completed tree has atleast one open branch. Then we can construct an interpretation, I, basedon that branch, as follows: if any atomic sentence Φ is in the branch, thenI(Φ) = 1. If any negated atomic sentence ¬Ψ is in the branch, then I(Ψ) = 0.Let these assignments exhaust I. We can guarantee that there will be a coherentinterpretation like this. This recipe for constructing interpretations will failonly if some atomic sentence and its negation are both in the branch. But ifa sentence and a negation are both in the branch, then that branch will close;by hypothesis, we’re considering a completed branch that remains open. So weknow it contains no explicit contradictions of this kind.Now we want to prove that I satisfies every formula in the branch (includingthe root). We know it satisfies every atomic formula and every negated atomicformula in the branch, given the way it was constructed. We’ll now show thatit must satisfy every other formula too. We’ll exploit the recursive rules for SLgrammaticality: there are only a certain number of ways that sentences can becreated from simpler sentences. We’ll show, for every sentence form, that if Isatisfies a branch downstream of a sentence of that form in a completed branch,then it satisfies that sentence too.We begin with conjunction.6.6.1 ConjunctionSuppose our completed, open branch contains a conjunction of the form Φ&Ψ .Since it is a completed branch, this means that the conjunction resolutionrule must have been applied to this conjunction. (Remember, a branch isn’tcompleted until every complex formula has a check mark next to it.) So, giventhe conjunction rule,100 forallx (UBC edition)Φ&ΨΦΨwe know that the branch must also contain Φ and Ψ . If we assume that Isatisfies both these sentences, then we know from the definition of the truth of aconjunction in SL that I satisfies (Φ&Ψ) too. In other words, there’s no way tosatisfy the simpler sentences that come after this resolution rule, without alsosatisfying the conjunction.6.6.2 Negated conjunctionSuppose a negated conjunction appears in the open branch. Since the branch iscomplete, you know that the negated conjunction rule has been applied:¬(Φ&Ψ)¬Φ ¬ΨWe are assuming that the negated conjunction is in an open branch. This isconsistent with either one of the branches below closing, but they cannot bothclose. If they did, the negated conjunction would not be in an open branch.So we know that at least one branch is open. So either I(Φ) = 0 (if the leftbranch is our open branch) or I(Ψ) = 0 (if the right branch is the open branch).Since at least one of these sentences is assigned 0, their conjunction must also beassigned 0, which means the negated conjunction we’re considering is assigned 1.So once again, if the material in at least one branch below the resolution rule issatisfied, then the negated conjunction is satisfied too.6.6.3 DisjunctionDisjunctions are very similar to negated conjunctions. Since the tree is complete,any disjunction Φ ∨ Ψ has a branch below it containing Φ, and one containing Ψ .Whichever of these disjuncts is in the open branch, I satisfies that disjunct, andso satisfies the disjunction too.ch. 6 soundness and completeness for sl trees 1016.6.4 Negated disjunctionNegated disjunctions are similar to conjunctions. If a negated disjunction isin an open branch, then the negation of each disjunct is also in that branch.So, suppose that I assigns 0 to each disjunct. Then it also assigns 0 to theirdisjunction. So once again, if the material below the negated disjunction issatisfied, then so is the negated disjunction itself.6.6.5 ConditionalIf a conditional is in a completed open branch, then it has been resolved by thisbranching rule:Φ ⊃ Ψ¬Φ ΨIf the left development is the open branch, then we suppose that I(Φ) = 0, whichmeans that I(Φ ⊃ Ψ) = 1. If this right development is the open branch, then wesuppose that I(Ψ) = 1, which also means that I(Φ ⊃ Ψ) = 1. So if the materialbelow in at least one branch is satisfied, then the conditional is satisfied too.6.6.6 Negated conditionalIf a negated conditional ¬(Φ⊃Ψ) is in the open branch, then so too are Φ and¬Ψ . So I(Φ) = 1 and I(Ψ) = 0. So I falsifies the conditional, satisfying thenegated conditional.There are three more kinds of sentences that exist in SL.6.6.7 BiconditionalSuppose a biconditional is in an open branch. If the branch is completed, thenthis rule has been performed:Φ ≡ ΨΦΨ¬Φ¬Ψ102 forallx (UBC edition)One of these developments is the open branch. If it’s the left branch, then,supposing that I assigns 1 to both Φ and Ψ , I must also assign 1 to thebiconditional Φ ≡ Ψ . If it’s the right branch, then, supposing that I assigns 0 toboth Φ and Ψ , this also means that I must also assign 1 to the biconditionalΦ ≡ Ψ . So whichever branch is satisfied by I, the biconditional is also satisfied.6.6.8 Negated biconditionalExactly the same reasoning as above applies to negated biconditionals, exceptthis time, the branches each assign opposite truth values to Φ and Ψ . So for ourinterpretation to satisfy either branch, it must falsify the biconditional, thussatisfying the negated biconditional.6.6.9 Double negationFinally, suppose there is a double-negated sentence in our completed open branch.Then this rule has been performed:¬¬ΦΦIf I(Φ) = 1, then, given the definition of truth in SL, I(¬Φ) = 0, and I(¬¬Φ) = 1.So once again, if our interpretation satisfies what comes below, then it satisfiesthe double-negation above.6.6.10 Summarizing the completeness proofWhat we’ve just shown is that, for any sentence of SL, if it has one of thenine structures just canvassed — if it’s a conjunction, a negated conjunction, adisjunction, etc. — then, if it is in a completed open branch where the sentencesbelow it are satisfied by I, then it too is satisfied by I. Given the way thatI was selected, we know that I must satisfy every atomic sentence, and everynegated atomic sentence, in the open branch. And since the nine structuresconsidered are the only ways to develop more complex sentences, this impliesthat every SL sentence in the open branch is satisfied by I. This includes theroot. Since interpretation I satisfies the root, this of course means that the rootis satisfiable. That is to say, if a completed branch remains open, this guaranteesthat the root is satisfiable. Equivalently, if the root is unsatisfiable, a completedtree is guaranteed to close. Completeness is proven.ch. 6 soundness and completeness for sl trees 1036.7 Testing alternate rulesWe can use the reasoning involved in the soundness and completeness proofsabove to consider various alternative tree rules. We saw one example of this onp. 89 above, when we observed that an alternate, linear rule for disjunctionswould result in an unsound tree system. Here again was the rule we considered:Φ ∨ ΨΦΨNote, however, that if we think through the completeness reasoning, we’ll find thatthe completeness proof would still hold. If we assume that some interpretationI satisfies both Φ and Ψ , we can be assured that it also satisfies the disjunctionΦ ∨ Ψ . So changing the disjunction rule to this linear one would not interferewith the completeness of our tree system. Our trees would still close any timethey began with unsatisfiable roots. But as we saw on p. 94, the soundness proofwould fail, which is why a system with this rule could start in a satisfiable root,and result in a closed tree.When either the soundness or the completeness proof fails, you know you areworking with an inappropriate rule. To conclusively demonstrate this, you canprovide a counterexample to the failed metalogical theorem. A counterexample tosoundness would be a tree with a satisfiable root that closes. A counterexampleto completeness would be a completed tree, with an unsatisfiable root, thatremains open. The rule above violates soundness, so using that rule we canconstruct a tree with a satisfiable root, that closes. Constructing the rightcounterexample takes a bit of thought. The rule puts both disjuncts into a singlebranch below, and we want it to close, despite having a satisfiable root. Soadding the negation of just one of the disjuncts to the root will close the tree,without making the root unsatisfiable:A ∨B X¬AAB×This tree is a counterexample to soundness, using the hypothetical rule mentioned104 forallx (UBC edition)above. Note that a counterexample is a tree that uses SL sentences, not theGreek letters Φ and Ψ that we use in the statements of the rules.Let’s work through one more example. Suppose we changed the conjunction ruleto this one:Φ&ΨΦΨΨWould our system still be sound? To answer this, we assume that the conjunctionΦ&Ψ is satisfiable, and ask whether this guarantees that at least one branchbelow is also satisfiable. It does. (In fact, both branches are guaranteed to besatisfiable.) So the system will still be sound.Would the system still be complete? To answer this, we ask whether each branchis such that, if we assume that an interpretation satisfies the developments below,it is guaranteed to satisfy the conjunction above. Begin with the left branch.If some interpretation satisfies both Φ and Ψ , then it will certainly satisfy theconjunction Φ&Ψ . So that branch looks fine. (Indeed, that branch is exactly thesame as the linear development of our actual conjunction rule, so this reasoningis the same as that on p. 99.)But what of the right branch? Assume that some interpretation satisfies Ψ ; doesthat guarantee that it satisfies Φ&Ψ? Certainly not. So if the right branch is ouropen branch, a completed tree using this rule may remain open, even if its rootis unsatisfiable. Completeness will be violated. Let’s construct a counterexamplethat form. We want a tree that includes a conjunction, whose root is unsatisfiable,but whose right branch remains open. Notice that the right branch ‘ignores’ thefirst conjunct; this is a clue that a good way to construct a counterexample willbe to locate the unsatisfiability within that first conjunct. Suppose, for example,that we let Φ itself stand for a contradiction. If so, any conjunction with Φ as aconjunct will be unsatisfiable. But if Ψ is not a contradiction, then a tree withroot Φ&Ψ will remain open. Let’s develop a tree with this root: (P ≡ ¬P ) &Q.Note that this sentence has a contradictory first conjunct, and a contingent,atomic second conjunct.ch. 6 soundness and completeness for sl trees 105(P ≡ ¬P ) &Q X(P ≡ ¬P ) XQP¬P×¬P¬¬P×Q↑The left branch closes after we perform the unchanged biconditional rule onthe contradictory first conjunct, but the right branch remains open. Sincethis is a completed tree with an open branch and an unsatisfiable root, it is acounterexample to completeness.Remember that we are considering a modification to the tree system that uses adifferent conjunction rule. In this example I used P ≡ ¬P as my Φ, which letme use the unchanged biconditional rule. But if I had used a contradiction thatused a conjunction, like P &¬P , I would have had to have used the revised rulewithin the left branch too.It is also possible to construct counterexamples using simpler sentences if youadd to the root. Instead of introducing a contradictory conjunct, we could havesimply added to the root, in a way that makes the root unsatisfiable, but leavesthe right branch open. Suppose for instance we put both P &Q and ¬P in theroot. Then we’d have an unsatisfiable root, but the right branch would remainopen:P &Q X¬PPQ×Q↑This too is a counterexample to completeness, given the rule in question. Sowe see here two different kinds of strategies for generating counterexamples tocompleteness.106 forallx (UBC edition)Practice Exercises? Part A Following are possible modifications to our SL tree system. For each,imagine a system that is like the system laid out in this chapter, except for theindicated change. Would the modified tree system be sound? If so, explain howthe proof given in this chapter would extend to a system with this rule; if not,give a tree that is a counterexample to the soundness of the modified system.1. Change the rule for conjunctions to this rule:Φ&ΨΦ Ψ2. Change the rule for conjunctions to this rule:Φ&ΨΦ3. Change the rule for conjunctions to this rule:Φ&ΨΦ¬Ψ4. Change the rule for disjunctions to this rule:Φ ∨ ΨΦΨ5. Change the rule for disjunctions to this rule:Φ ∨ ΨΦ Ψ Φ&Ψch. 6 soundness and completeness for sl trees 1076. Change the rule for conditionals to this rule:Φ ⊃ Ψ¬Φ Ψ ∨ Φ7. Change the rule for conditionals to this rule:Φ ⊃ Ψ¬Φ ∨ Ψ8. Change the rule for biconditionals to this rule:Φ ≡ ΨΦΨ9. Change the rule for disjunctions to this rule:Φ ∨ ΨΦ Ψ Ω(This would mean that one can put whatever SL sentence onelikes in the rightmost branch.)Part B For each of the rule modifications given in Part A, would the modifiedtree system be complete? If so, explain how the proof given in this chapter wouldextend to a system with this rule; if not, give a tree that is a counterexample tothe completeness of the modified system.Chapter 7Natural Deduction Proofsin SLThis chapter introduces a different proof system in SL, separate from the treemethod. The tree method has advantages and disadvantages. One advantage oftrees is that, for the most part, they can be produced in a purely mechanicalway; another is that, when a tree remains open, the tree method gives us a recipefor constructing an interpretation that satisfies the root. One disadvantage isthat they do not always emphasize in an intuitive way why a conclusion followsfrom a set of premises; they show that something must be the case, on pain ofcontradiction, but they don’t always demonstrate, in a way closely connected tonatural reasoning, why some things follow from other things.The natural deduction system of this chapter will differ from the tree methodin all of these respects. It is intended to model human reasoning in a closerway, illustrating the connections between various claims; consequently, workingthrough a natural deduction proof requires a bit more insight and inspirationthan a tree proof does.Natural deduction proofs can be used to prove that an argument is valid; if anargument is invalid, our natural deduction system will not necessarily make thatobvious. (Any valid argument in SL can be shown to be valid via a naturaldeduction proof — i.e., this method is complete too (and sound) — but that’sno guarantee that any individual will be able to come up with the proof.)Consider two arguments in SL:108ch. 7 natural deduction proofs in sl 109Argument AP ∨Q¬P.˙. QArgument BP ⊃ QP.˙. QClearly, these are valid arguments. You can confirm that they are valid byconstructing four-line truth tables, or by using trees. Argument A makes use ofan inference form that is always valid: Given a disjunction and the negation ofone of the disjuncts, the other disjunct follows as a valid consequence. This ruleis called disjunctive syllogism.Argument B makes use of a different valid form: Given a conditional and itsantecedent, the consequent follows as a valid consequence. This is called modusponens.When we construct truth tables, we do not need to give names to different infer-ence forms. There is no reason to distinguish modus ponens from a disjunctivesyllogism. For this same reason, however, the method of truth tables does notclearly show why an argument is valid. If you were to do a 1024-line truth tablefor an argument that contains ten sentence letters, then you could check to seeif there were any lines on which the premises were all true and the conclusionwere false. If you did not see such a line and provided you made no mistakesin constructing the table, then you would know that the argument was valid.Yet you would not be able to say anything further about why this particularargument was a valid argument form.We aim to show that particular arguments are valid in a way that allows usto understand the reasoning involved in the argument. We begin with basicargument forms, like disjunctive syllogism and modus ponens. These forms canthen be combined to make more complicated arguments, like this one:(1) ¬L ⊃ (J ∨ L)(2) ¬L.˙. JBy modus ponens, (1) and (2) entail J ∨ L. This is an intermediate conclusion.It follows logically from the premises, but it is not the conclusion we want. NowJ ∨ L and (2) entail J , by disjunctive syllogism. We do not need a new rulefor this argument. The proof of the argument shows that it is really just acombination of rules we have already introduced.Formally, a proof is a sequence of sentences. The first sentences of the sequenceare assumptions; these are the premises of the argument. Every sentence later110 forallx (UBC edition)in the sequence follows from earlier sentences by one of the rules of proof. Thefinal sentence of the sequence is the conclusion of the argument.7.1 Basic rules for SLIn designing a proof system, we could just start with disjunctive syllogism andmodus ponens. Whenever we discovered a valid argument which could not beproven with rules we already had, we could introduce new rules. Proceeding inthis way, we would have an unsystematic grab bag of rules. We might accidentallyadd some strange rules, and we would surely end up with more rules than weneed.Instead, we will develop what is called a natural deduction system. In anatural deduction system, there will be two rules for each logical operator: anintroduction rule that allows us to prove a sentence that has it as the mainlogical operator and an elimination rule that allows us to prove somethinggiven a sentence that has it as the main logical operator.In addition to the rules for each logical operator, we will also have a reiterationrule. If you already have shown something in the course of a proof, the reiterationrule allows you to repeat it on a new line. For instance:1 Φ2 Φ R 1The numbers on the left indicate line numbers; they are used for reference injustifying later steps in the proof. New lines in a proof must always be justifiedby rules and reference to previous lines. The ‘R 1’ to the right on line 2 is thejustification for that line — that line is permitted by the reiteration rule (R),applied to line 1.Obviously, the reiteration rule will not allow us to show anything new. For that,we will need more rules. The remainder of this section will give introductionand elimination rules for all of the sentential connectives. This will give us acomplete proof system for SL.All of the rules introduced in this chapter are summarized in the Quick Referenceguide at the end of this book.ch. 7 natural deduction proofs in sl 1117.1.1 ConjunctionWhat would you need to show in order to prove E&F? The natural answeris, you’d need to prove E, and you’d also need to prove F . In fact this holdsmuch more generally; one can prove any conjunction Φ&Ψ by proving Φ andalso proving Ψ , whether or not these conjuncts are atomic sentences. If you canprove [(A ∨ J) ⊃ V ] and [(V ⊃ L) ≡ (F ∨N)], then you have effectively proven[(A ∨ J) ⊃ V ] & [(V ⊃ L) ≡ (F ∨N)].So this will be our conjunction introduction rule, which we abbreviate ‘ & I’:m Φn ΨΦ&Ψ & I m, nAs always, the Φ and Ψ stand in for any arbitrary sentence in SL; the m andn here stand in for arbitrary line numbers. In an actual proof, the lines arenumbered 1, 2, 3, . . . and rules must be applied to specific line numbers. Whenwe define the rule, however, we use variables to underscore the point that therule may be applied to any two lines that are already in the proof. Note thatit is not a requirement that m and n be consecutive lines, or that they appearin the order listed here. We require only that each line has been establishedsomewhere above in the proof. If you have K on line 15 and L on line 8, youcan prove (K &L) at some later point in the proof with the justification ‘ & I 15,8.’ (By convention, we won’t always worry about which order to write the linenumbers down in. ‘ & I 8, 15’ is OK too.)Now, consider the elimination rule for conjunction. What are you entitled toconclude from a sentence like E&F? Surely, you are entitled to conclude E; ifE&F were true, then E would be true. Similarly, you are entitled to concludeF . This will be our conjunction elimination rule, which we abbreviate ‘ & E’:m Φ&ΨΦ & E mΨ & E mWhen you have a conjunction on some line of a proof, you can use & E to deriveeither of the conjuncts. This rule allows either of the two developments listedhere. You may also apply the rule twice, to get both. The & E rule requires onlyone sentence, so we write one line number as the justification for applying it.Even with just these two rules, we can provide some proofs. Consider thisargument.112 forallx (UBC edition)[(A ∨B) ⊃ (C ∨D)] & [(E ∨ F ) ⊃ (G ∨H)].˙. [(E ∨ F ) ⊃ (G ∨H)] & [(A ∨B) ⊃ (C ∨D)]The main logical operator in both the premise and conclusion is conjunction.Since conjunction is symmetric, the argument is obviously valid. In order toprovide a proof, we begin by writing down the premise. After the premises, wedraw a horizontal line — everything below this line must be justified by a ruleof proof. So the beginning of the proof looks like this:1 [(A ∨B) ⊃ (C ∨D)] & [(E ∨ F ) ⊃ (G ∨H)]From the premise, we can get each of the conjuncts by & E. The proof now lookslike this:1 [(A ∨B) ⊃ (C ∨D)] & [(E ∨ F ) ⊃ (G ∨H)]2 [(A ∨B) ⊃ (C ∨D)] & E 13 [(E ∨ F ) ⊃ (G ∨H)] & E 1The rule & I requires that we have each of the conjuncts available somewherein the proof. They can be separated from one another, and they can appear inany order. So by applying the & I rule to lines 3 and 2, we arrive at the desiredconclusion. The finished proof looks like this:1 [(A ∨B) ⊃ (C ∨D)] & [(E ∨ F ) ⊃ (G ∨H)]2 [(A ∨B) ⊃ (C ∨D)] & E 13 [(E ∨ F ) ⊃ (G ∨H)] & E 14 [(E ∨ F ) ⊃ (G ∨H)] & [(A ∨B) ⊃ (C ∨D)] & I 2, 3This proof is not terribly interesting, but it shows how we can use rules of prooftogether to demonstrate the validity of an argument form. Note also that using atruth table to show that this argument is valid would have required a staggering256 lines, since there are eight sentence letters in the argument. A proof via treeswould be less unwieldy than that, but it would be less simple and elegant thanthis one. (Constructing such a proof would be a good exercise for tree review.)7.1.2 DisjunctionIf M is true, then M ∨ N must also be true. In general, the disjunctionintroduction rule (∨I) allows us to derive a disjunction if we have one of the twodisjuncts:ch. 7 natural deduction proofs in sl 113m ΦΦ ∨ Ψ ∨I mΨ ∨ Φ ∨I mNotice that Ψ can be any sentence whatsoever. So the following is a legitimateproof:1 M2 M ∨ ([(A ≡ B) ⊃ (C &D)] ≡ [E&F ]) ∨I 1It may seem odd that just by knowing M we can derive a conclusion that includessentences like A, B, and the rest — sentences that have nothing to do withM . Yet the conclusion follows immediately by ∨I. This is as it should be: Thetruth conditions for the disjunction mean that, if Φ is true, then Φ ∨ Ψ is trueregardless of what Ψ is. So the conclusion could not be false if the premise weretrue; the argument is valid.Now consider the disjunction elimination rule. What can you conclude fromM ∨ N? You cannot conclude M . It might be M ’s truth that makes M ∨ Ntrue, as in the example above, but it might not. From M ∨N alone, you cannotconclude anything about either M or N specifically. If you also knew that Nwas false, however, then you would be able to conclude M .This is the disjunctive syllogism rule mentioned in the introduction. It will beour official disjunction elimination rule (∨E). If you have a disjunction and alsothe negation of one of its disjuncts, you may conclude the other disjunct.m Φ ∨ Ψn ¬ΨΦ ∨E m, nm Φ ∨ Ψn ¬ΦΨ ∨E m, nWe represent two different inference patterns here, because the rule allows youto conclude either disjunct from the negation of the other. If we’d only listed theleft version of the rule above, then ∨E would’ve only permited one to concludethe first disjunct from the negation of the second one, along with the disjunction.Our rule lets us work with either disjunction. (If you want to be very fussyabout it, you could think of these as two different rules with a strong conceptualsimilarity that happen to have the same name.)114 forallx (UBC edition)7.1.3 ConditionalConsider this argument:R ∨ F.˙. ¬R ⊃ FThe argument seems like it should be valid. (You can confirm this by examiningthe truth tables.) What should the conditional introduction rule be, such thatwe can draw this conclusion?We begin the proof by writing down the premise of the argument and drawing ahorizontal line, like this:1 R ∨ FIf we had ¬R as a further premise, we could derive F by the ∨E rule. We do nothave ¬R as a premise of this argument, nor can we derive it directly from thepremise we do have — so we cannot simply prove F . What we will do instead isstart a subproof, a proof within the main proof. When we start a subproof, wedraw another vertical line to indicate that we are no longer in the main proof.Then we write in an assumption for the subproof. This can be anything we want.Here, it will be helpful to assume ¬R. Our proof now looks like this:1 R ∨ F2 ¬RIt is important to notice that we are not claiming to have proven ¬R. We donot need to write in any justification for the assumption line of a subproof. Thevertical line indicates that an assumption is being made. You can think of thesubproof as posing the question: What could we show if ¬R were true? For onething, we can derive F . So we do:1 R ∨ F2 ¬R3 F ∨E 1, 2This has shown that if we had ¬R as a premise, then we could prove F . Ineffect, we have proven ¬R ⊃ F . So the conditional introduction rule (⊃I) willallow us to close the subproof and derive ¬R ⊃ F in the main proof. Our finalproof looks like this:ch. 7 natural deduction proofs in sl 1151 R ∨ F2 ¬R3 F ∨E 1, 24 ¬R ⊃ F ⊃I 2–3The ⊃I lets us discharge the assumption we’d been making, ending that verticalline. During lines (2) and (3), we were assuming that ¬R; by the time we get toline (4), we are no longer making that assumption.Notice that the justification for applying the ⊃I rule is the entire subproof. That’swhy we justify it by reference to a range of lines, instead of a comma-separatedlist. Usually that will be more than just two lines.It may seem as if the ability to assume anything at all in a subproof would leadto chaos: Does it allow you to prove any conclusion from any premises? Theanswer is no, it does not. Consider this proof:1 Φ2 Ψ3 Ψ R 2Does this show that one can prove any arbitrary sentence Ψ from any arbitrarypremise Φ? After all, we’ve written Ψ on a line of a proof that began withΦ, without violating any of the rules of our system. The reason this doesn’thave that implication is the vertical line that still extends into line 3. That lineindicates that the assumption made at line 2 is still in effect. When the verticalline for the subproof ends, the subproof is closed. In order to complete a proof,you must close all of the subproofs. The conclusion to be proved must not be‘blocked off’ by a vertical line; it should be aligned with the premises.In this example, there is no way to close the subproof and use the R rule again online 4 to derive Ψ in the main proof. Once you close a subproof, you cannot referback to individual lines inside it. One can only close a subproof via particularrules that allow you to do so; ⊃I is one such rule. One can’t just close a subproofwilly-nilly. Closing a subproof is called discharging the assumptions of thatsubproof. So we can put the point this way: You cannot complete a proof untilyou have discharged all of the assumptions besides the original premises of theargument.Of course, it is legitimate to do this:116 forallx (UBC edition)1 Φ2 Ψ3 Ψ R 24 Ψ ⊃ Ψ ⊃I 2–3This should not seem so strange, though. Since Ψ⊃Ψ is a tautology, no particularpremises should be required to validly derive it. (Indeed, as we will see, atautology follows from any premises.)Put in its general form, the ⊃I rule looks like this:m Φ (want Ψ)n ΨΦ ⊃ Ψ ⊃I m–nWhen we introduce a subproof, we typically write what we want to derive off tothe right. This is just so that we do not forget why we started the subproof if itgoes on for five or ten lines. There is no ‘want’ rule. It is a note to ourselves,and not formally part of the proof.Although it is always permissible to open a subproof with any assumption youplease, there is some strategy involved in picking a useful assumption. Startinga subproof with an arbitrary, wacky assumption is not a good strategy. It willjust waste lines of the proof. In order to derive a conditional by the ⊃I rule, forinstance, you must assume the antecedent of the conditional in a subproof.The ⊃I rule also requires that the consequent of the conditional be the last lineof the subproof. It is always permissible to close a subproof and discharge itsassumptions, but it will not be helpful to do so until you get what you want. Thisis an illustration of the observation made above, that unlike the tree method,the natural deduction method requires some strategy and thinking ahead.Now consider the conditional elimination rule. Nothing follows from M ⊃ Nalone, but if we have both M ⊃ N and M , then we can conclude N . This rule,modus ponens, will be the conditional elimination rule (⊃E).m Φ ⊃ Ψn ΦΨ ⊃E m, nNow that we have rules for the conditional, consider this argument:ch. 7 natural deduction proofs in sl 117P ⊃ QQ ⊃ R.˙. P ⊃ RWe begin the proof by writing the two premises as assumptions. Since the mainlogical operator in the conclusion is a conditional, we can expect to use the⊃I rule. For that, we need a subproof — so we write in the antecedent of theconditional as assumption of a subproof:1 P ⊃ Q2 Q ⊃ R3 PWe made P available by assuming it in a subproof, allowing us to use ⊃E on thefirst premise. This gives us Q, which allows us to use ⊃E on the second premise.Having derived R, we close the subproof. By assuming P we were able to proveR, so we apply the ⊃I rule and finish the proof.1 P ⊃ Q2 Q ⊃ R3 P (want R)4 Q ⊃E 1, 35 R ⊃E 2, 46 P ⊃ R ⊃I 3–57.1.4 BiconditionalBiconditionals indicate that the two sides have the same truth value. Oneestablishes a biconditional by establishing each direction of it as conditionals.To derive W ≡ X, for instance, you must establish both W ⊃ X and X ⊃ W .Those conditionals may occur in either order; they need not be on consecutivelines. (Compare the shape of the & I rule.) Schematically, the rule works likethis:m Φ ⊃ Ψn Ψ ⊃ ΦΦ ≡ Ψ ≡I m, n118 forallx (UBC edition)The biconditional elimination rule (≡E) is a generalized version of modus ponens(⊃E). If you have the left-hand subsentence of the biconditional, you can derivethe right-hand subsentence. If you have the right-hand subsentence, you canderive the left-hand subsentence. This is the rule:m Φ ≡ Ψn ΦΨ ≡E m, nm Φ ≡ Ψn ΨΦ ≡E m, n7.1.5 NegationHere is a simple mathematical argument in English:Assume there is some greatest natural number. Call it A.That number plus one is also a natural number.Obviously, A+ 1 > A.So there is a natural number greater than A.This is impossible, since A is assumed to be the greatest natural number..˙. There is no greatest natural number.This argument form is traditionally called a reductio. Its full Latin name isreductio ad absurdum, which means ‘reduction to absurdity.’ In a reductio,we assume something for the sake of argument — for example, that there isa greatest natural number. Then we show that the assumption leads to twocontradictory sentences — for example, that A is the greatest natural numberand that it is not. In this way, we show that the original assumption must havebeen false.The basic rules for negation will allow for arguments like this. If we assumesomething and show that it leads to contradictory sentences, then we have proventhe negation of the assumption. This is the negation introduction (¬I) rule:m Φ (for reductio)n Ψo ¬Ψp ¬Φ ¬I m–n, m–oThe ¬I rule discharges the assumption for reductio, concluding its negation, whenit’s shown that some sentence and its negation each follow from the assumption.It cites two (overlapping) ranges: a subproof from the assumption to somech. 7 natural deduction proofs in sl 119sentence Ψ , and a subproof from that same assumption to ¬Ψ . We write ‘forreductio’ to the right of the assumption, as a note to ourselves, a reminder ofwhy we started the subproof. It is not formally part of the proof, but it is helpfulfor thinking clearly about the proof.To see how the rule works, suppose we want to prove an instance of the lawof non-contradiction: ¬(G&¬G). We can prove this without any premises byimmediately starting a subproof. We want to apply ¬I to the subproof, so weassume (G&¬G). We then get an explicit contradiction by & E. The prooflooks like this:1 G&¬G for reductio2 G & E 13 ¬G & E 14 ¬(G&¬G) ¬I 1–2, 1–3The ¬E rule will work in much the same way. If we assume ¬Φ and show that itleads to a sentence and its negation, we have effectively proven Φ. So the rulelooks like this:m ¬Φ for reduction Ψo ¬Ψp Φ ¬E m–n, m–o7.2 Derived rulesThe rules of the natural deduction system are meant to be systematic. There isan introduction and an elimination rule for each logical operator, but why thesebasic rules rather than some others? Some natural deduction systems have adisjunction elimination rule that works like this:m Φ ∨ Ψn Φ ⊃ Ωo Ψ ⊃ ΩΩ DIL m, n, o120 forallx (UBC edition)Let’s call this rule Dilemma (DIL). It might seem as if there will be some proofsthat we cannot do with our proof system, because we do not have this as a basicrule. Yet this is not the case. Any proof that you can do using the Dilemmarule can be done with basic rules of our natural deduction system. Consider aproof of this form:1 Φ ∨ Ψ2 Φ ⊃ Ω3 Ψ ⊃ Ω want Ω4 ¬Ω for reductio5 Φ for reductio6 Ω ⊃E 2, 57 ¬Ω R 48 ¬Φ ¬I 5–6, 5–79 Ψ for reductio10 Ω ⊃E 3, 911 ¬Ω R 412 Ψ ∨E 1, 813 ¬Ψ ¬I 9–10, 9–1114 Ω ¬E 4–12, 4–13Remember once again that Φ, Ψ , and Ω are meta-variables. They are notsymbols of SL, but stand-ins for arbitrary sentences of SL. So this is not, strictlyspeaking, a proof in SL. It is more like a recipe. It provides a pattern that canprove anything that the Dilemma rule can prove, using only the basic rules ofSL. This means that the Dilemma rule is not really necessary. Adding it to thelist of basic rules would not allow us to derive anything that we could not derivewithout it.Nevertheless, the Dilemma rule would be convenient. It would allow us to do inone line what requires eleven lines and several nested subproofs with the basicrules. So we will add it to the proof system as a derived rule.A derived rule is a rule of proof that does not make any new proofs possible.Anything that can be proven with a derived rule can be proven without it. Youcan think of a short proof using a derived rule as shorthand for a longer proofthat uses only the basic rules. Anytime you use the Dilemma rule, you couldalways take ten extra lines and prove the same thing without it.ch. 7 natural deduction proofs in sl 121For the sake of convenience, we will add several other derived rules. One ismodus tollens (MT).m Φ ⊃ Ψn ¬Ψ¬Φ MT m, nWe leave the proof of this rule as an exercise. Note that if we had already proventhe MT rule, then the proof of the DIL rule could have been done in only fivelines.We also add hypothetical syllogism (HS) as a derived rule. We have alreadygiven a proof of it on p. 117.m Φ ⊃ Ψn Ψ ⊃ ΩΦ ⊃ Ω HS m, n7.3 Rules of replacementConsider how you would prove this argument valid: F ⊃ (G&H) .˙. F ⊃ GPerhaps it is tempting to write down the premise and apply the & E rule to theconjunction (G&H). This is impermissible, however, because the basic rules ofproof can only be applied to whole sentences. In order to use & E, we need toget the conjunction (G&H) on a line by itself. Here is a proof:1 F ⊃ (G&H)2 F want G3 G&H ⊃E 1, 24 G & E 35 F ⊃ G ⊃I 2–4The rules we have seen so far must apply to wffs that are on a proof line bythemselves. We will now introduce some derived rules that may be applied topart of a sentence. These are called rules of replacement, because theycan be used to replace part of a sentence with a logically equivalent expression.One simple rule of replacement is commutativity (abbreviated Comm), which122 forallx (UBC edition)says that we can swap the order of conjuncts in a conjunction or the order ofdisjuncts in a disjunction. We define the rule this way:(Φ&Ψ)⇐⇒ (Ψ &Φ)(Φ ∨ Ψ)⇐⇒ (Ψ ∨ Φ)(Φ ≡ Ψ)⇐⇒ (Ψ ≡ Φ) CommThe double arrow means that you can take a subformula on one side of the arrowand replace it with the subformula on the other side. The arrow is double-headedbecause rules of replacement work in both directions.Consider this argument: (M ∨ P ) ⊃ (P &M), .˙. (P ∨M) ⊃ (M &P )It is possible to give a proof of this using only the basic rules, but it will be longand inconvenient. With the Comm rule, we can provide a proof easily:1 (M ∨ P ) ⊃ (P &M)2 (P ∨M) ⊃ (P &M) Comm 13 (P ∨M) ⊃ (M &P ) Comm 2Another rule of replacement is double negation (DN). With the DN rule, youcan remove or insert a pair of negations for any wff in a line, even if it isn’t thewhole line. This is the rule:¬¬Φ⇐⇒ Φ DNTwo more replacement rules are called De Morgan’s Laws, named for the 19th-century British logician August De Morgan. (Although De Morgan did formalizeand publish these laws, many others discussed them before him.) The rulescapture useful relations between negation, conjunction, and disjunction. Hereare the rules, which we abbreviate DeM:¬(Φ ∨ Ψ)⇐⇒ (¬Φ&¬Ψ)¬(Φ&Ψ)⇐⇒ (¬Φ ∨ ¬Ψ) DeMAs we have seen, Φ ⊃ Ψ is equivalent to ¬Φ ∨ Ψ . A further replacement rulecaptures this equivalence. We abbreviate the rule MC, for ‘material conditional.’It takes two forms:(Φ ⊃ Ψ)⇐⇒ (¬Φ ∨ Ψ)(Φ ∨ Ψ)⇐⇒ (¬Φ ⊃ Ψ) MCch. 7 natural deduction proofs in sl 123Now consider this argument: ¬(P ⊃ Q), .˙. P &¬QAs always, we could prove this argument valid using only the basic rules. Withrules of replacement, though, the proof is much simpler:1 ¬(P ⊃ Q)2 ¬(¬P ∨Q) MC 13 ¬¬P &¬Q DeM 24 P &¬Q DN 3A final replacement rule captures the relation between conditionals and bi-conditionals. We will call this rule biconditional exchange and abbreviate it≡ex.[(Φ ⊃ Ψ) & (Ψ ⊃ Φ)]⇐⇒ (Φ ≡ Ψ) ≡ex7.4 Proof strategyThere is no simple recipe for proofs, and there is no substitute for practice. Here,though, are some rules of thumb and strategies to keep in mind.Work backwards from what you want. The ultimate goal is to derive theconclusion. Look at the conclusion and ask what the introduction rule is for itsmain logical operator. This gives you an idea of what should happen just beforethe last line of the proof. Then you can treat this line as if it were your goal.Ask what you could do to derive this new goal.For example: If your conclusion is a conditional Φ ⊃ Ψ , plan to use the ⊃I rule.This requires starting a subproof in which you assume Φ. In the subproof, youwant to derive Ψ .Work forwards from what you have. When you are starting a proof, lookat the premises; later, look at the sentences that you have derived so far. Thinkabout the elimination rules for the main operators of these sentences. These willtell you what your options are.For example: If you have a conditional Φ⊃Ψ , and you also have Φ, ⊃E is a prettynatural choice.124 forallx (UBC edition)For a short proof, you might be able to eliminate the premises and introducethe conclusion. A long proof is formally just a number of short proofs linkedtogether, so you can fill the gap by alternately working back from the conclusionand forward from the premises.Change what you are looking at. Replacement rules can often make yourlife easier. If a proof seems impossible, try out some different substitutions.For example: It is often difficult to prove a disjunction using the basic rules. Ifyou want to show Φ ∨ Ψ , it is often easier to show ¬Φ ⊃ Ψ and use the MC rule.Some replacement rules should become second nature. If you see a negateddisjunction, for instance, you should immediately think of DeMorgan’s rule.Do not forget indirect proof. If you cannot find a way to show somethingdirectly, try assuming its negation.Remember that most proofs can be done either indirectly or directly. One waymight be easier — or perhaps one sparks your imagination more than the other— but either one is formally legitimate.Repeat as necessary. Once you have decided how you might be able to getto the conclusion, ask what you might be able to do with the premises. Thenconsider the target sentences again and ask how you might reach them.Persist. Try different things. If one approach fails, then try something else.7.5 Proof-theoretic conceptsAs we did in our discussion of trees, we will again use the symbol ‘`’ to indicateprovability. Provability is relative to a proof system, so the meaning of the ‘`’symbol featured in this chapter should be distinguished from the one we used fortrees. When necessary, we can specify the single turnstile with reference to theproof system in question, letting ‘`T ’ stand for provability in the tree system,and ‘`ND’ stand for provability in this natural deduction system. For the mostpart in this chapter, though, we’ll be interested in natural deduction, so unlessit is specified otherwise, you can understand ‘`’ to mean ‘`ND’.The double turnstile symbol ‘|=’, remains unchanged. It stands for semanticentailment, as described in ch. 4.ch. 7 natural deduction proofs in sl 125When we write {Φ1, Φ2, . . .} `ND Ψ , this means that it is possible to give anatural deduction proof of Ψ with Φ1,Φ2,. . . as premises. With just one premise,we leave out the curly braces, so Φ ` Ψ means that there is a proof of Ψ withΦ as a premise. Naturally, ` Φ means that there is a proof of Φ that has nopremises. You can think of it as shorthand for ∅ ` Φ.For notational completeness, we can understand X ` ⊥ to mean that from X ,we could prove an arbitrary contradiction. In other words, X is an inconsistentset.Logical proofs are sometimes called derivations. So Φ ` Ψ can be read as ‘Ψ isderivable from Φ.’A theorem is a sentence that is derivable without any premises; i.e., Φ is atheorem if and only if ` Φ.It is not too hard to show that something is a theorem — you just have to give aproof of it. How could you show that something is not a theorem? If its negationis a theorem, then you could provide a proof. For example, it is easy to prove¬(P &¬P ), which shows that (P &¬P ) cannot be a theorem. For a sentencethat is neither a theorem nor the negation of a theorem, however, there is noeasy way to show this. You would have to demonstrate not just that certainproof strategies fail, but that no proof is possible. Even if you fail in tryingto prove a sentence in a thousand different ways, perhaps the proof is just toolong and complex for you to make out. As we’ve emphasized already, this is adifference between our natural deduction system and the tree method.Two sentences Φ and Ψ are provably equivalent if and only if each can bederived from the other; i.e., Φ ` Ψ and Ψ ` Φ.It is relatively easy to show that two sentences are provably equivalent — it justrequires a pair of proofs. Showing that sentences are not provably equivalentwould be much harder. It would be just as hard as showing that a sentence isnot a theorem. (In fact, these problems are interchangeable. Can you thinkof a sentence that would be a theorem if and only if Φ and Ψ were provablyequivalent?)The set of sentences {Φ1, Φ2, . . .} is provably inconsistent if and only if contra-dictory sentences are derivable from it; i.e., for some sentence Ψ , {Φ1, Φ2, . . .} ` Ψand {Φ1, Φ2, . . .} ` ¬Ψ . This is equivalent to {Φ1, Φ2, . . .} ` ⊥.It is easy to show that a set is provably inconsistent: You just need to assumethe sentences in the set and prove a contradiction. Showing that a set is notprovably inconsistent will be much harder. It would require more than justproviding a proof or two; it would require showing that proofs of a certain kindare impossible.126 forallx (UBC edition)7.6 Proofs and modelsAs you might already suspect, there is a connection between theorems andtautologies.There is a formal way of showing that a sentence is a theorem: Prove it. Foreach line, we can check to see if that line follows by the cited rule. It may behard to produce a twenty line proof, but it is not so hard to check each lineof the proof and confirm that it is legitimate — and if each line of the proofindividually is legitimate, then the whole proof is legitimate. Showing that asentence is a tautology, though, requires reasoning in English about all possiblemodels. There is no formal way of checking to see if the reasoning is sound.Given a choice between showing that a sentence is a theorem and showing thatit is a tautology, it would be easier to show that it is a theorem.By contrast, there is no formal way of showing that a sentence is not a theorem.We would need to reason in English about all possible proofs. Yet there is aformal method for showing that a sentence is not a tautology. We need onlyconstruct a model in which the sentence is false. Given a choice between showingthat a sentence is not a theorem and showing that it is not a tautology, it wouldbe easier to show that it is not a tautology.Fortunately, a sentence is a theorem if and only if it is a tautology. If we providea proof of ` Φ and thus show that it is a theorem, it follows that Φ is a tautology;i.e., |= Φ. Similarly, if we construct a model in which Φ is false and thus showthat it is not a tautology, it follows that Φ is not a theorem.In general, Φ ` Ψ if and only if Φ |= Ψ . As such:. An argument is valid if and only if the conclusion is derivable from thepremises.. Two sentences are logically equivalent if and only if they are provablyequivalent.. A set of sentences is consistent if and only if it is not provably inconsistent.You can pick and choose when to think in terms of proofs and when to think interms of models, doing whichever is easier for a given task. Table 7.1 summarizeswhen it is best to give proofs and when it is best to give models.In this way, proofs and models give us a versatile toolkit for working witharguments. If we can translate an argument into SL, then we can measure itslogical weight in a purely formal way. If it is deductively valid, we can give aformal proof; if it is invalid, we can provide a formal counterexample.ch. 7 natural deduction proofs in sl 127YES NOIs Φ a tautology? prove ` Φ give a model in whichΦ is falseIs Φ a contradiction? prove ` ¬Φ give a model in whichΦ is trueIs Φ contingent? give a model in which Φis true and another inwhich Φ is falseprove ` Φ or ` ¬ΦAre Φ and Ψ equiva-lent?prove Φ ` Ψ and Ψ ` Φ give a model in whichΦ and Ψ have differenttruth valuesIs the set A consistent? give a model in whichall the sentences in Aare truetaking the sentences inA, prove Ψ and ¬ΨIs the argument ‘Φ, Ψ ,. . . .˙. Ω’ valid?prove Φ, Ψ, . . . ` Ω give a model in which{Φ, Ψ , . . . } is satisfiedand Ω is falsifiedTable 7.1: Sometimes it is easier to show something by providing proofs thanit is by providing models. Sometimes it is the other way round. It depends onwhat you are trying to show.7.7 Soundness and completenessChapter 6 considered the soundness and completeness of the tree method atlength; it proved that this method was both sound (X `T Φ only if X |= Φ)and complete (X |= Φ only if X `T Φ). The natural deduction system of thischapter is also both sound and complete for SL. In other words, X `ND Φ if andonly if X |= Φ. Given the soundness and completeness of the tree method, thisalso means that our two proof systems are equivalent in the sense that anythingprovable in one is also provable in the other (X `T Φ) iff (X `ND Φ).How can we know that our natural deduction method is sound? A proofsystem is sound if there are no derivations corresponding to invalid arguments.Demonstrating that the proof system is sound would require showing that anypossible proof in our system is the proof of a valid argument. There is a fairlysimple way of approaching this in a step-wise fashion. If using the & E rule onthe last line of a proof could never change a valid argument into an invalid one,then using the rule many times could not make an argument invalid. Similarly,if using the & E and ∨E rules individually on the last line of a proof could neverchange a valid argument into an invalid one, then using them in combinationcould not either.The strategy is to show for every rule of inference that it alone could not make a128 forallx (UBC edition)valid argument into an invalid one. It follows that the rules used in combinationwould not make a valid argument invalid. Since a proof is just a series of lines,each justified by a rule of inference, this would show that every provable argumentis valid.Consider, for example, the & I rule. Suppose we use it to add Φ&Ψ to a validargument. In order for the rule to apply, Φ and Ψ must already be available inthe proof. Since the argument so far is valid, Φ and Ψ are either premises of theargument or valid consequences of the premises. As such, any model in whichthe premises are true must be a model in which Φ and Ψ are true. Accordingto the definition of truth in sl, this means that Φ&Ψ is also true in such amodel. Therefore, Φ&Ψ validly follows from the premises. This means thatusing the & I rule to extend a valid proof produces another valid proof.In order to show that the proof system is sound, we would need to show this forthe other inference rules. Since the derived rules are consequences of the basicrules, it would suffice to provide similar arguments for the 16 other basic rules.The reasoning is extremely similar to that given in the soundness proof for treesin the previous chapter. We will not go through it in detail here.Given a proof that the proof system is sound, it follows that every theorem is atautology.What of completeness? Why think that every valid argument is an argumentthat can be proven in our natural deduction system? That is, why think thatΦ |= Ψ implies Φ ` Ψ? Our system is also complete, but the completeness prooffor natural deduction is a bit more complex than the completeness proof fortrees. (In the case of trees, we had a mechanical method that was guaranteed tofind proofs if they exist; we have seen no such method here, which makes provingthese general results harder.) This proof is beyond the scope of this book.The important point is that, happily, the proof system for SL is both sound andcomplete. Consequently, we may freely use this natural deduction method todraw conclusions about models in SL.Summary of definitions. A sentence Φ is a theorem if and only if ` Φ.. Two sentences Φ and Ψ are provably equivalent if and only if Φ ` Ψand Ψ ` Φ.. {Φ1, Φ2, . . .} is provably inconsistent if and only if, for some sentenceΨ , {Φ1, Φ2, . . .} ` (Ψ &¬Ψ).ch. 7 natural deduction proofs in sl 129Practice Exercises? Part A Provide a justification (rule and line numbers) for each line of proofthat requires one.1 W ⊃ ¬B2 A&W3 B ∨ (J &K)4 W5 ¬B6 J &K7 K1 L ≡ ¬O2 L ∨ ¬O3 ¬L4 ¬O5 L6 ¬L7 L1 Z ⊃ (C &¬N)2 ¬Z ⊃ (N &¬C)3 ¬(N ∨ C)4 ¬N &¬C5 Z6 C &¬N7 C8 ¬C9 ¬Z10 N &¬C11 N12 ¬N13 N ∨ C? Part B Give a proof for each argument in SL.1. K &L, .˙.K ≡ L2. A ⊃ (B ⊃ C), .˙.(A&B) ⊃ C3. P & (Q ∨R), P ⊃ ¬R, .˙.Q ∨ E4. (C &D) ∨ E, .˙.E ∨D5. ¬F ⊃ G, F ⊃ H, .˙.G ∨H6. (X &Y ) ∨ (X &Z), ¬(X &D), D ∨M .˙.MPart C Give a proof for each argument in SL.1. Q ⊃ (Q&¬Q), .˙. ¬Q2. J ⊃ ¬J , .˙. ¬J130 forallx (UBC edition)3. E ∨ F , F ∨G, ¬F , .˙. E&G4. A ≡ B, B ≡ C, .˙. A ≡ C5. M ∨ (N ⊃M), .˙. ¬M ⊃ ¬N6. S ≡ T , .˙. S ≡ (T ∨ S)7. (M ∨N) & (O ∨ P ), N ⊃ P , ¬P , .˙. M &O8. (Z &K) ∨ (K &M), K ⊃ D, .˙. D? Part D Show that each of the following sentences is a theorem in SL.1. O ⊃ O2. N ∨ ¬N3. ¬(P &¬P )4. ¬(A ⊃ ¬C) ⊃ (A ⊃ C)5. J ≡ [J ∨ (L&¬L)]Part E Show that each of the following pairs of sentences are provably equivalentin SL.1. ¬¬¬¬G, G2. T ⊃ S, ¬S ⊃ ¬T3. R ≡ E, E ≡ R4. ¬G ≡ H, ¬(G ≡ H)5. U ⊃ I, ¬(U &¬I)? Part F Provide proofs to show each of the following.1. M & (¬N ⊃ ¬M) ` (N &M) ∨ ¬M2. {C ⊃ (E&G), ¬C ⊃ G} ` G3. {(Z &K) ≡ (Y &M), D& (D ⊃M)} ` Y ⊃ Z4. {(W ∨X) ∨ (Y ∨ Z), X ⊃ Y , ¬Z} ` W ∨ YPart G For the following, provide proofs using only the basic rules. The proofswill be longer than proofs of the same claims would be using the derived rules.1. Show that MT is a legitimate derived rule. Using only the basic rules,prove the following: Φ⊃Ψ , ¬Ψ , .˙. ¬Φ2. Show that Comm is a legitimate rule for the biconditional. Using only thebasic rules, prove that Φ ≡ Ψ and Ψ ≡ Φ are equivalent.3. Using only the basic rules, prove the following instance of DeMorgan’sLaws: (¬A&¬B), .˙. ¬(A ∨B)4. Show that ≡ex is a legitimate derived rule. Using only the basic rules,prove that D ≡ E and (D ⊃ E) & (E ⊃ D) are equivalent.ch. 7 natural deduction proofs in sl 131Part H1. If you know that Φ ` Ψ , what can you say about (Φ&Ω) ` Ψ? Explainyour answer.2. If you know that Φ ` Ψ , what can you say about (Φ ∨ Ω) ` Ψ? Explainyour answer.Chapter 8Quantified logicThis chapter introduces a logical language called QL. It is a version of quantifiedlogic, because it allows for quantifiers like all and some. Quantified logic is alsosometimes called predicate logic, because the basic units of the language arepredicates and terms.8.1 From sentences to predicatesConsider the following argument, which is valid in English:If everyone knows logic, then either no one will be confused oreveryone will. Everyone will be confused only if we try to believe acontradiction. Everyone knows logic..˙. If we don’t try to believe a contradiction, then no one will beconfused.In order to symbolize this in SL, we will need a symbolization key.L: Everyone knows logic.N: No one will be confused.E: Everyone will be confused.B: We try to believe a contradiction.Notice that N and E are both about people being confused, but they are twoseparate sentence letters. We can’t replace E with ¬N . Why not? ¬N means‘It is not the case that no one will be confused.’ This would be the case if even132ch. 8 quantified logic 133one person were confused, so it is a long way from saying that everyone will beconfused.Because we have separate sentence letters forN and E, however, our formalizationdoes not encode any connection between the two. They are just two atomicsentences which might be true or false independently. It is impossible for it tobe the case that both no one and everyone was confused. As sentences of SL,however, there is a truth-value assignment for which N and E are both true.This is a limitation of the descriptive power of SL. Some features of Englishsentences are not preserved in SL. Our new language, QL, will preserve more ofthis structure.Expressions like ‘no one’, ‘everyone’, and ‘anyone’ are called quantifiers. Bytranslating N and E as separate atomic sentences, we leave out the quantifierstructure of the sentences. In the example we’ve been discussing, the quantifierstructure is not terribly important. The argument is valid without reference toit. As such, we can safely ignore it. To see this, we translate the argument toSL:L ⊃ (N ∨ E)E ⊃ BL.˙. ¬B ⊃ NThis is a valid argument form in SL. (You can construct a truth table, a tree, ora natural deduction proof to confirm this.)Now consider another argument. This one is also valid in English.Willard is a logician. All logicians wear funny hats..˙. Willard wears a funny hat.To symbolize it in SL, we define a symbolization key:L: Willard is a logician.A: All logicians wear funny hats.F: Willard wears a funny hat.Now we symbolize the argument:LA.˙. F134 forallx (UBC edition)This is pretty obviously an invalid SL form. Nevertheless, this is clearly a validargument in English. It’s impossible for the premises to be true without theconclusion also being true. But the SL symbolization leaves out all quantifica-tional structure in virtue of which it is valid. The sentence ‘All logicians wearfunny hats’ says something specific about both logicians and hat-wearing. Bynot translating this structure, treating the whole sentence as an atom, we losethe connection between Willard’s being a logician and Willard’s wearing a hat.Some arguments with quantifier structure can be captured in SL, like the firstexample, even though SL ignores the quantifier structure. Other arguments’validity cannot be captured in SL, like the second example. Notice that theproblem is not that we have made a mistake while symbolizing the secondargument. We gave a perfectly appropriate translation; indeed, these are thebest symbolizations we can give for these arguments in SL.If an English argument is properly translated into a form that is valid in SL, thatargument is valid, even if it involves quantifiers. But if it does not have a validSL form, that doesn’t mean the English argument is invalid; valid argumentscan have invalid forms. It just means that SL doesn’t show that the argument isvalid. This will be the case when the argument’s quantifier structure plays animportant role in its validity.Similarly, if a sentence with quantifiers comes out as a tautology in SL, then theEnglish sentence is logically true. If it comes out as contingent in SL, then thismight be because of the structure of the quantifiers that gets removed when wetranslate into the formal language.In order to symbolize arguments that rely on quantifier structure, we need todevelop a different, richer, logical language. We will call this language ‘quantifiedlogic’, or QL.8.2 Building blocks of QLOur first key notion in QL will be predicates. A predicate is analogous toan English description — an expression like ‘is a dog’ or ‘has black fur’. Suchdescriptions are not sentences on their own. They are neither true nor false.They are descriptions that apply to some objects, but not to others. In orderto be true or false, we need to specify an object: Who or what is it that is saidto be a dog, or to have black fur? In SL, we have no way to give meaningfultranslations to English terms that are not full sentences. In QL, we will.The details of this will be explained in the rest of the chapter, but here is thebasic idea: In QL, we will represent predicates with capital letters. For instance,we might let D stand for ‘ is a dog.’ We will use lower-case letters asch. 8 quantified logic 135the names of specific things. For instance, we might let b stand for Bertie.The expression Db will be a sentence in QL. Given this symbolization, Db is atranslation of the sentence ‘Bertie is a dog.’In order to represent quantifier structure, we will also have symbols that representquantifiers. For instance, ‘∃’ will mean ‘There is some .’ So to say thatthere is a dog, we can write ∃xDx; that is: There is some x such that x is a dog.That will come later. We start by defining singular terms and predicates.8.3 Singular termsIn English, a singular term is a word or phrase that refers to a specific person,place, or thing. The word ‘dog’ is not a singular term, because it is a generalterm that could apply to many individual animals. The phrase ‘Jonathan’s dogMezzo’ is a singular term, because it refers to a specific little poodle mix. She isblack, and soft, and wonderful.A proper name is a singular term that picks out an individual directly. Thename ‘Emerson’ is a proper name, and the name alone does not tell you anythingabout Emerson. Of course, some names are traditionally given to boys, andothers are traditionally given to girls. If ‘Jack Hathaway’ is used as a singularterm, you might guess that it refers to a man. However, the name does notnecessarily mean that the person referred to is a man — or even that the creaturereferred to is a person. Jack might be a giraffe for all you could tell just fromthe name. There is a great deal of philosophical action surrounding this issue,but the important point here is that a name is a singular term because it picksout a single, specific individual.Other singular terms more obviously convey information about the thing towhich they refer. For instance, you can tell without being told anything furtherthat ‘Jonathan’s dog Mezzo’ is a singular term that refers to a dog. A definitedescription picks out an individual by means of a unique description. InEnglish, definite descriptions are often phrases of the form ‘the such-and-so.’They refer to the specific thing that matches the given description. For example,‘the tallest member of Monty Python’ and ‘the first emperor of China’ are definitedescriptions. A description that does not pick out a specific individual is not adefinite description. ‘A member of Monty Python’ and ‘an emperor of China’are not definite descriptions. We’ll discuss definite descriptions in more detail inChapter 12.In English, the specification of a singular term may depend on context; ‘Willard’means a specific person and not just someone named Willard; ‘P.D. Magnus’ asa logical singular term means the original author of this textbook, not the other136 forallx (UBC edition)person who has the same name. We live with this kind of ambiguity in English,but it is important to keep in mind that singular terms in QL must refer to justone specific thing.In QL, we will symbolize singular terms with lower-case letters a through w.We can add subscripts if we want to use some letter more than once. Soa, b, c, . . . w, a1, f32, j390, and m12 are all terms in QL.Singular terms are called constants because they pick out specific individuals.Note that x, y, and z are not constants in QL. They will be variables, letterswhich do not stand for any specific thing. We will need them when we introducequantifiers.8.4 PredicatesSimple one-place predicates are properties of individuals. They are things youcan say about an object. Here are some one-place predicates:‘ is a dog’‘ is a member of Monty Python’‘ ’s favourite ramen place is in Gastown’‘An anvil was dropped from a very high height onto ’s head’Predicates like these are called one-place or monadic, because there is onlyone blank to fill in. A one-place predicate and a singular term combine to makea sentence.Other predicates are about the relation between two things. For instance:‘ is bigger than ’‘ is to the left of ’‘ owes money to ’These are two-place or dyadic predicates, because they need to be filled inwith two terms in order to make a sentence.In general, you can think about predicates as schematic sentences that needto be filled out with some number of terms. Conversely, you can start withsentences and make predicates out of them by removing terms. Consider thesentence, ‘Buchanan Tower is North of Barber but South of Buchanan E.’ Byremoving a singular term, we can recognize this sentence as using any of threedifferent monadic predicates:ch. 8 quantified logic 137is North of Barber but South of Buchanan E.Buchanan Tower is North of but South of Buchanan E.Buchanan Tower is North of Barber but South of .By removing two singular terms, we can recognize three different dyadic predi-cates:Buchanan Tower is North of but South of .is North of Barber but South of .is North of but South of Buchanan E.By removing all three singular terms, we can recognize one three-place ortriadic predicate:is North of but South of .If we are translating this sentence into QL, should we translate it with a one-,two-, or three-place predicate? It depends on what we want to be able to say. Ifwe’re only interested in discussing the location of buildings relative to Barber,then the generality of the three-place predicate is unnecessary. If we only wantto discuss whether buildings are North of Barber and South of BUCH E, aone-place predicate will be enough.In general, we can have predicates with as many places as we need. Predicateswith more than one place are called polyadic. Predicates with n places, for somenumber n, are called n-place or n-adic. You can make an n-adic predicate byreplacing n names in any sentence with blanks. You can even have a 0-placepredicate — this would be the result of replacing 0 names in a sentence withblanks. In other words, a 0-place predicate is just a sentence.In QL, we symbolize predicates with capital letters A through Z, with orwithout subscripts. When we give a symbolization key for predicates, we willnot use blanks; instead, we will use variables. By convention, constants arelisted at the end of the key. So we might write a key that looks like this:138 forallx (UBC edition)Ax: x is angry.Hx: x is happy.T1xy: x is at least as tall as y.T2xy: x is at least as tough as y.Bxyz: y is between x and z.d: Donaldg: Gregorm: MarybethWe can symbolize sentences that use any combination of these predicates andterms. For example:1. Donald is angry.2. If Donald is angry, then so are Gregor and Marybeth.3. Marybeth is at least as tall and as tough as Gregor.4. Donald is shorter than Gregor.5. Gregor is between Donald and Marybeth.Sentence 1 is straightforward: Ad. The ‘x’ in the key entry ‘Ax’ is just aplaceholder; we replace it with other terms when translating.Sentence 2 can be paraphrased as, ‘If Ad, then Ag and Am.’ QL has all thetruth-functional connectives of SL, so we translate this as Ad ⊃ (Ag&Am).Sentence 3 can be translated as T1mg&T2mg.Sentence 4 might seem as if it requires a new predicate. If we only neededto symbolize this sentence, we could define a predicate like Sxy to mean ‘xis shorter than y.’ However, this would ignore the logical connection between‘shorter’ and ‘taller.’ Considered only as symbols of QL, there is no connectionbetween S and T1. They might mean anything at all. Instead of introducing anew predicate, we paraphrase sentence 4 using predicates already in our key: ‘Itis not the case that Donald is as tall or taller than Gregor.’ We can translate itas ¬T1dg.Sentence 5 requires that we pay careful attention to the order of terms in thekey. It becomes Bdgm.8.5 QuantifiersWe are now ready to introduce quantifiers. Quantifiers (unlike names andpredicates) are a new kind of logical connective; like conjunction, negation, etc.,they can govern the interpretation of a QL sentence. Consider these Englishsentences:ch. 8 quantified logic 1396. Everyone is happy.7. Everyone is at least as tough as Donald.8. Someone is angry.It might be tempting to translate sentence 6 as Hd&Hg&Hm. Yet this wouldonly say that Donald, Gregor, and Marybeth are happy. We want to say thateveryone is happy, even if we have not defined a constant to name them. Inorder to do this, we introduce the ‘∀’ symbol. This is called the universalquantifier.A quantifier is placed in front of the formula that it binds. It always comesalong with an associated variable. Typically that formula will include the samevariable. We can translate sentence 6 as ∀xHx. Paraphrased partially intoEnglish, this means ‘For all x, x is happy.’We call ∀x an x-quantifier. The formula that follows the quantifier is calledthe scope of the quantifier. We will give a formal definition of scope later, butintuitively it is the part of the sentence that the quantifier quantifies over. In∀xHx, the scope of the universal quantifier is Hx.Sentence 7 can be paraphrased as, ‘For all x, x is at least as tough as Donald.’This translates as ∀xT2xd.In these quantified sentences, the variable x is serving as a kind of placeholder.The expression ∀x means that you can pick anyone and put them in as x. Thereis no special reason to use x rather than some other variable. The sentence∀xHx means exactly the same thing as ∀yHy, ∀zHz, and ∀x5Hx5.To translate sentence 8, we introduce another new symbol: the existentialquantifier, ∃. Like the universal quantifier, the existential quantifier requiresa variable. Sentence 8 can be translated as ∃xAx. This means that there issome x which is angry. More precisely, it means that there is at least one angryperson. Once again, the variable is a kind of placeholder; we could just as easilyhave translated sentence 8 as ∃zAz.Consider these further sentences:9. No one is angry.10. There is someone who is not happy.11. Not everyone is happy.Sentence 9 can be paraphrased as, ‘It is not the case that someone is angry.’This can be translated using negation and an existential quantifier: ¬∃xAx. Yetsentence 9 could also be paraphrased as, ‘Everyone is not angry.’ With this inmind, it can be translated using negation and a universal quantifier: ∀x¬Ax.140 forallx (UBC edition)Both of these are acceptable translations, because they are logically equivalent.The critical thing is whether the negation comes before or after the quantifier.In general, ∀xΦ is logically equivalent to ¬∃x¬Φ. This means that any sentencewhich can be symbolized with a universal quantifier can be symbolized with anexistential quantifier, and vice versa. One translation might seem more naturalthan the other, but there is no logical difference in translating with one quantifierrather than the other. For some sentences, it will simply be a matter of taste.Sentence 10 is most naturally paraphrased as, ‘There is some x such that x isnot happy.’ This becomes ∃x¬Hx. Equivalently, we could write ¬∀xHx.Sentence 11 is most naturally translated as ¬∀xHx. This is logically equivalentto sentence 10 and so could also be translated as ∃x¬Hx.8.6 Universe of discourseGiven the symbolization key we have been using, ∀xHx means ‘Everyone ishappy.’ Who is included in this everyone? When we use sentences like this inEnglish, we usually do not mean everyone now alive on the Earth. We certainlydo not mean everyone who was ever alive or who will ever live. We meansomething more modest: everyone in the building, everyone in the class, oreveryone in the room.In order to eliminate this ambiguity, we will need to specify a universe ofdiscourse — abbreviated UD. The UD is the set of things that we are talkingabout. So if we want to talk about people in Chicago, we define the UD to bepeople in Chicago. We write this at the beginning of the symbolization key, likethis:UD: people in ChicagoThe quantifiers range over the universe of discourse. Given this UD, ∀x means‘Everyone in Chicago’ and ∃x means ‘Someone in Chicago.’ Each constant namessome member of the UD, so we can only use this UD with the symbolization keyabove if Donald, Gregor, and Marybeth are all in Chicago. If we want to talkabout people in places besides Chicago, then we need to include those people inthe UD.In QL, the UD must be non-empty; that is, it must include at least one thing.It is possible to construct formal languages that allow for empty UDs, but thisintroduces complications; such languages are beyond the scope of this book.Even allowing for a UD with just one member can produce some strange results.ch. 8 quantified logic 141Suppose we have this as a symbolization key:UD: the Eiffel TowerPx: x is in Paris.The sentence ∀xPx might be paraphrased in English as ‘Everything is in Paris.’Yet that would be misleading. It means that everything in the UD is in Paris.This UD contains only the Eiffel Tower, so with this symbolization key ∀xPxjust means that the Eiffel Tower is in Paris. We will rarely work with suchbizarre domains as this.It is a rule in QL that each constant will pick out exactly one member of theUD. (There is no rule prohibiting multiple different constants from referring tothe same member of the UD.)8.7 Translating to QLWe now have the basic pieces of QL. Translating more complicated sentenceswill only be a matter of knowing the right way to combine predicates, constants,quantifiers, variables, and sentential connectives. Consider these sentences:12. Every coin in my pocket is a loonie.13. Some coin on the table is a dime.14. Not all the coins on the table are loonies.15. None of the coins in my pocket are dimes.In providing a symbolization key, we need to specify a UD. Since we are talkingabout coins in my pocket and on the table, the UD must at least contain all ofthose coins. Since we are not talking about anything besides coins, we let theUD be all coins. Since we are not talking about any specific coins, we do notneed to define any constants. So we define this key:UD: all coinsPx: x is in my pocket.Tx: x is on the table.Lx: x is a loonie.Dx: x is a dime.Sentence 12 is most naturally translated with a universal quantifier. The universalquantifier says something about everything in the UD, not just about the coinsin my pocket. Sentence 12 means that, for any coin, if that coin is in my pocket,then it is a loonie. So we can translate it as ∀x(Px ⊃ Lx).142 forallx (UBC edition)Since sentence 12 is about coins that are both in my pocket and that are loonies,it might be tempting to translate it using a conjunction. However, the sentence∀x(Px&Lx) would mean that everything in the UD is both in my pocket and aloonie: All the coins that exist are loonies in my pocket. This would be nice,but it means something very different than sentence 12.Sentence 13 is most naturally translated with an existential quantifier. It saysthat there is some coin which is both on the table and which is a dime. So wecan translate it as ∃x(Tx&Dx).Notice that we needed to use a conditional with the universal quantifier, but weused a conjunction with the existential quantifier. This is a common pattern.What would it mean to write ∃x(Tx ⊃ Dx)? Probably not what you think. Itmeans that there is some member of the UD which would satisfy the subformula;roughly speaking, there is some name α such that (Tα ⊃ Dα) is true. In SL,Φ ⊃ Ψ is logically equivalent to ¬Φ ∨ Ψ , and this will also hold in QL. So∃x(Tx ⊃ Dx) is true if there is some α such that (¬Tα ∨Dα); i.e., it is true ifsome coin is either not on the table or is a dime. Of course there is a coin that isnot on the table — there are coins in lots of other places. So ∃x(Tx ⊃ Dx) makesan extremely weak claim. A conditional will usually be the natural connectiveto use with a universal quantifier, but a conditional within the scope of anexistential quantifier can do very strange things. It’s a pretty good rule of thumbthat you shouldn’t be putting conditionals in the scope of existential quantifiers.This is pretty much never a good translation of any natural English sentence.Sentence 14 can be paraphrased as, ‘It is not the case that every coin on thetable is a loonie.’ So we can translate it as ¬∀x(Tx ⊃ Lx). You might lookat sentence 14 and paraphrase it instead as, ‘Some coin on the table is not aloonie.’ You would then translate it as ∃x(Tx&¬Lx). Although it is probablynot obvious, these two translations are logically equivalent. (This is due to thelogical equivalence between ¬∀xΦ and ∃x¬Φ, along with the equivalence between¬(Φ ⊃ Ψ) and Φ&¬Ψ . We’ll be able to prove this later.)Sentence 15 can be paraphrased as, ‘It is not the case that there is some dimein my pocket.’ This can be translated as ¬∃x(Px&Dx). It might also beparaphrased as, ‘Everything in my pocket is a non-dime,’ and then could betranslated as ∀x(Px ⊃ ¬Dx). Again the two translations are logically equivalent.Both are correct translations of sentence 15.We can now translate the argument from p. 133, the one that motivated theneed for quantifiers:Willard is a logician. All logicians wear funny hats..˙. Willard wears a funny hat.UD: peoplech. 8 quantified logic 143Lx: x is a logician.Fx: x wears a funny hat.w: WillardTranslating, we get:Lw∀x(Lx ⊃ Fx).˙. FwThis captures the structure that was left out of the SL translation of thisargument, and this is a valid argument in QL.8.8 Empty predicatesA predicate need not apply to anything in the UD. A predicate that applies tonothing in the UD is called an empty predicate.Suppose we want to symbolize these two sentences:16. Every monkey knows sign language.17. Some monkey knows sign language.It is possible to write the symbolization key for these sentences in this way:UD: animalsMx: x is a monkey.Sx: x knows sign language.Sentence 16 can now be translated as ∀x(Mx ⊃ Sx).Sentence 17 becomes ∃x(Mx&Sx).It is tempting to say that sentence 16 entails sentence 17; that is: if every monkeyknows sign language, then it must be that some monkey knows sign language.However, the entailment does not hold in QL. It is possible for the sentence∀x(Mx ⊃ Sx) to be true even though the sentence ∃x(Mx&Sx) is false.How can this be? The answer comes from considering whether these sentenceswould be true or false if there were no monkeys.144 forallx (UBC edition). A UD must have at least one member.. A predicate may apply to some, all, or no members of theUD.. A constant must pick out exactly one member of the UD.A member of the UD may be picked out by one constant,many constants, or none at all.We have defined ∀ and ∃ in such a way that ∀Φ is equivalent to ¬∃¬Φ. Assuch, the universal quantifier doesn’t involve the existence of anything — onlynon-existence. If sentence 16 is true, then there are no monkeys who don’t knowsign language. If there were no monkeys, then ∀x(Mx ⊃ Sx) would be true and∃x(Mx&Sx) would be false.A second reason to allow empty predicates is that we want to be able to saythings like, ‘I do not know if there are any monkeys, but any monkeys that thereare know sign language.’ That is, we want to be able to have predicates that donot (or might not) refer to anything.Third, consider: ∀x(Px ⊃ Px). This should be a tautology. But if sentence16 implied sentence 17, then this would imply ∃x(Px&Px). It would becomea logical truth that for any predicate there is something that satisfies thatpredicate.What happens if we add an empty predicate R to the interpretation above? Forexample, we might define Rx to mean ‘x is a refrigerator.’ Now the sentence∀x(Rx ⊃ Mx) will be true. This is counterintuitive, since we do not want tosay that there are a whole bunch of refrigerator monkeys. It is important toremember, though, that ∀x(Rx ⊃Mx) means that any member of the UD whichis a refrigerator is a monkey. Since the UD is animals, there are no refrigeratorsin the UD and so the sentence is trivially true.If you were actually translating the sentence ‘All refrigerators are monkeys’, thenyou would want to include appliances in the UD. Then the predicate R wouldnot be empty and the sentence ∀x(Rx ⊃Mx) would be false.8.9 Picking a universe of discourseThe appropriate symbolization of an English language sentence in QL will dependon the symbolization key. In some ways, this is obvious: It matters whether Dxmeans ‘x is dainty’ or ‘x is dangerous.’ The meaning of sentences in QL alsoch. 8 quantified logic 145depends on the UD.Let Rx mean ‘x is a rose,’ let Tx mean ‘x has a thorn,’ and consider this sentence:18. Every rose has a thorn.It is tempting to say that sentence 18 should be translated as ∀x(Rx ⊃ Tx). Ifthe UD contains all roses, that would be correct. Yet if the UD is merely thingson my kitchen table, then ∀x(Rx ⊃ Tx) would only mean that every rose on mykitchen table has a thorn. If there are no roses on my kitchen table, the sentencewould be trivially true.The universal quantifier only ranges over members of the UD, so we need toinclude all roses in the UD in order to translate sentence 18. We have twooptions. First, we can restrict the UD to include all roses but only roses. Thensentence 18 becomes ∀xTx. This means that everything in the UD has a thorn;since the UD just is the set of roses, this means that every rose has a thorn.This option can save us trouble if every sentence that we want to translate usingthe symbolization key is about roses.Second, we can let the UD contain things besides roses: rhododendrons, rats,rifles, and whatall else. Then sentence 18 must be ∀x(Rx ⊃ Tx).If we wanted the universal quantifier to mean every thing, without restriction,then we might try to specify a UD that contains everything. But this notionis somewhat obscure. Does ‘everything’ include things that have only beenimagined, like fictional characters? On the one hand, we want to be able tosymbolize arguments about Hamlet or Sherlock Holmes. So we need to have theoption of including fictional characters in the UD. On the other hand, we neverneed to talk about every thing that does not exist. That might not even makesense. There are philosophical issues here that we will not try to address. Wecan avoid these difficulties by always specifying the UD. For example, if we meanto talk about plants, people, and cities, then the UD might be ‘living things andplaces.’Suppose that we want to translate sentence 18 and, with the same symbolizationkey, translate these sentences:19. Esmerelda has a rose in her hair.20. Everyone is cross with Esmerelda.We need a UD that includes roses (so that we can symbolize sentence 18) and aUD that includes people (so we can translate sentence 19–20.) Here is a suitablekey:146 forallx (UBC edition)UD: people and plantsPx: x is a person.Rx: x is a rose.Tx: x has a thorn.Cxy: x is cross with y.Hxy: x has y in their hair.e: EsmereldaSince we do not have a predicate that means ‘. . . has a rose in her hair’, translatingsentence 19 will require paraphrasing. The sentence says that there is a rosein Esmerelda’s hair; that is, there is something which is both a rose and is inEsmerelda’s hair. So we get: ∃x(Rx&Hex).It is tempting to translate sentence 20 as ∀xCxe. Unfortunately, this wouldmean that every member of the UD is cross with Esmerelda — both people andplants. It would mean, for instance, that the rose in Esmerelda’s hair is crosswith her. Of course, sentence 20 does not mean that.‘Everyone’ means every person, not every member of the UD. So we can para-phrase sentence 20 as, ‘Every person is cross with Esmerelda.’ We know how totranslate sentences like this: ∀x(Px ⊃ Cxe).In general, the universal quantifier can be used to mean ‘everyone’ if the UDcontains only people. If there are people and other things in the UD, then‘everyone’ must be treated as ‘every person’.8.10 Translating pronounsWhen translating to QL, it is important to understand the structure of thesentences you want to translate. What matters is the final translation in QL,and sometimes you will be able to move from an English language sentencedirectly to a sentence of QL. Other times, it helps to paraphrase the sentenceone or more times. Each successive paraphrase should move from the originalsentence closer to something that you can translate directly into QL.For the next several examples, we will use this symbolization key:UD: peopleGx: x can play guitar.Rx: x is a rock star.c: Chrisl: LemmyNow consider these sentences:ch. 8 quantified logic 14721. If Chris can play guitar, then they are a rock star.22. If a person can play guitar, then they are a rock star.Sentence 21 and sentence 22 have the same words in the consequent (‘. . . theyare a rock star’), but they cannot be translated in the same way. It helps toparaphrase the original sentences, replacing pronouns with explicit references.Sentence 21 can be paraphrased as, ‘If Chris can play guitar, then Chris is arockstar.’ The word ‘they’ in sentence 21 is being used to refer to a specificindividual, Chris. This can obviously be translated as Gc ⊃ Rc.Sentence 22 must be paraphrased differently: ‘If a person can play guitar, thenthat person is a rock star.’ The pronoun ‘they’ here is not about any particularperson, so we need a variable. Translating halfway, we can paraphrase thesentence as, ‘For any person x, if x can play guitar, then x is a rock star.’ Nowthis can be translated as ∀x(Gx ⊃ Rx). This is the same as, ‘Everyone who canplay guitar is a rock star.’Consider these further sentences:23. If anyone can play guitar, then Lemmy can.24. If anyone can play guitar, then they are a rock star.These two sentences have the same antecedent (‘If anyone can play guitar. . .’),but they have different logical structures.Sentence 23 can be paraphrased, ‘If someone can play guitar, then Lemmy canplay guitar.’ The antecedent and consequent are separate sentences, so it can besymbolized with a conditional as the main logical operator: ∃xGx ⊃ Gl.Sentence 24 can be paraphrased, ‘For anyone, if that one can play guitar, thenthat one is a rock star.’ It would be a mistake to symbolize this with an existentialquantifier, because it is talking about everybody. The sentence is equivalent to‘All guitar players are rock stars.’ It is best translated as ∀x(Gx ⊃ Rx).The English words ‘any’ and ‘anyone’ should typically be translated usingquantifiers. As these two examples show, they sometimes call for an existentialquantifier (as in sentence 23) and sometimes for a universal quantifier (as insentence 24). If you have a hard time determining which is required, paraphrasethe sentence with an English language sentence that uses words besides ‘any’ or‘anyone.’148 forallx (UBC edition)8.11 Quantifiers and scopeIn the sentence ∃xGx ⊃ Gl, the scope of the existential quantifier is the expressionGx. Would it matter if the scope of the quantifier were the whole sentence?That is, does the sentence ∃x(Gx ⊃ Gl) mean something different?With the key given above, ∃xGx ⊃ Gl means that if there is some guitarist, thenLemmy is a guitarist. ∃x(Gx ⊃ Gl) would mean that there is some person suchthat if that person were a guitarist, then Lemmy would be a guitarist. Recallthat the conditional here is a material conditional; the conditional is true anytime the antecedent is false. Let the constant p denote the author of this book,someone who is certainly not a guitarist. The sentence Gp ⊃ Gl is true becauseGp is false. Since someone (namely p) satisfies the sentence, then ∃x(Gx ⊃ Gl) istrue. The sentence is true because there is a non-guitarist, regardless of Lemmy’sskill with the guitar.Something strange happened when we changed the scope of the quantifier,because the conditional in QL is a material conditional. In order to keep themeaning the same, we would have to change the quantifier: ∃xGx ⊃ Gl meansthe same thing as ∀x(Gx ⊃ Gl), and ∃x(Gx ⊃ Gl) means the same thing as∀xGx ⊃ Gl.Note that quantifiers count as logical connectives, so one can sensibly askwhether the main connective of a given sentence is a quantifier or somethingelse. (Calling it a ‘connective’ can be slightly confusing, since, unlike connectiveslike conjunction and disjunction, it doesn’t literally connect two sentences.Quantifiers are like negations in this respect — each does count as a connective.)If the scope of the quantifier is the entire sentence, then that quantifier is themain connective, as in ∀x(Gx ⊃ Gl). If the scope of the quantifier is limited toa subsentence, then that quantifier is not the main connective. For example, in∃xGx ⊃ Gl, the main quantifier is the conditional; the existential is part of theantecedent.8.12 Ambiguous predicatesSuppose we just want to translate this sentence:25. Adina is a skilled surgeon.Let the UD be people, let Kx mean ‘x is a skilled surgeon’, and let a meanAdina. Sentence 25 is simply Ka.Suppose instead that we want to translate this argument:ch. 8 quantified logic 149The hospital will only hire a skilled surgeon. All surgeons are greedy.Billy is a surgeon, but is not skilled. Therefore, Billy is greedy, butthe hospital will not hire him.We need to distinguish being a skilled surgeon from merely being a surgeon. Sowe define this symbolization key:UD: peopleGx: x is greedy.Hx: The hospital will hire x.Rx: x is a surgeon.Kx: x is skilled.b: BillyNow the argument can be translated in this way:∀x[¬(Rx&Kx) ⊃ ¬Hx]∀x(Rx ⊃ Gx)Rb&¬Kb.˙. Gb&¬HbNext suppose that we want to translate this argument:Carol is a skilled surgeon and a tennis player. Therefore, Carol is askilled tennis player.If we start with the symbolization key we used for the previous argument, wecould add a predicate (let Tx mean ‘x is a tennis player’) and a constant (let cmean Carol). Then the argument becomes:(Rc&Kc) &Tc.˙. T c&KcThis translation is a disaster! It takes what in English is a terrible argument andtranslates it as a valid argument in QL. The problem is that there is a differencebetween being skilled as a surgeon and skilled as a tennis player. Translatingthis argument correctly requires two separate predicates, one for each type ofskill. If we let K1x mean ‘x is skilled as a surgeon’ and K2x mean ‘x is skilledas a tennis player,’ then we can symbolize the argument in this way:(Rc&K1c) &Tc.˙. T c&K2c150 forallx (UBC edition)Like the English language argument it translates, this is invalid.The moral of these examples is that you need to be careful of symbolizingpredicates in an ambiguous way. Similar problems can arise with predicates likegood, bad, big, and small. Just as skilled surgeons and skilled tennis players havedifferent skills, big dogs, big mice, and big problems are big in different ways.Is it enough to have a predicate that means ‘x is a skilled surgeon’, rather thantwo predicates ‘x is skilled’ and ‘x is a surgeon’? Sometimes. As sentence 25shows, sometimes we do not need to distinguish between skilled surgeons andother surgeons.Must we always distinguish between different ways of being skilled, good, bad,or big? No. As the argument about Billy shows, sometimes we only need to talkabout one kind of skill. If you are translating an argument that is just aboutdogs, it is fine to define a predicate that means ‘x is big.’ If the UD includesdogs and mice, however, it is probably best to make the predicate mean ‘x isbig for a dog.’8.13 Multiple quantifiersConsider this following symbolization key and the sentences that follow it:UD: People and dogsDx: x is a dog.Fxy: x is a friend of y.Oxy: x owns y.f: Fifig: Gerald26. Fifi is a dog.27. Gerald is a dog owner.28. Someone is a dog owner.29. All of Gerald’s friends are dog owners.30. Every dog owner is the friend of a dog owner.Sentence 26 is easy: Df .Sentence 27 can be paraphrased as, ‘There is a dog that Gerald owns.’ This canbe translated as ∃x(Dx&Ogx).Sentence 28 can be paraphrased as, ‘There is some y such that y is a dog owner.’The subsentence ‘y is a dog owner’ is just like sentence 27, except that it isch. 8 quantified logic 151about y rather than being about Gerald. So we can translate sentence 28 as∃y∃x(Dx&Oyx).Sentence 29 can be paraphrased as, ‘Every friend of Gerald is a dog owner.’Translating part of this sentence, we get ∀x(Fxg ⊃ ‘x is a dog owner’). Again, itis important to recognize that ‘x is a dog owner’ is structurally just like sentence27. Since we already have an x-quantifier, we will need a different variable forthe existential quantifier. Any other variable will do. Using z, sentence 29 canbe translated as ∀x[Fxg ⊃ ∃z(Dz&Oxz)].Sentence 30 can be paraphrased as ‘For any x that is a dog owner, there is adog owner who is x’s friend.’ Partially translated, this becomes∀x[x is a dog owner ⊃ ∃y(y is a dog owner &Fxy)].Completing the translation, sentence 30 becomes∀x[∃z(Dz&Oxz) ⊃ ∃y(∃z(Dz&Oyz) &Fxy)].Consider this symbolization key and these sentences:UD: peopleLxy: x likes y.i: Imre.k: Karl.31. Imre likes everyone that Karl likes.32. There is someone who likes everyone who likes everyone that he likes.Sentence 31 can be partially translated as ∀x(Karl likes x ⊃ Imre likes x). Thisbecomes ∀x(Lkx ⊃ Lix).Sentence 32 is complex. There is little hope of writing down the whole translationimmediately, but we can proceed by small steps. An initial, partial translationmight look like this:∃x everyone who likes everyone that x likes is liked by xThe part that remains in English is a universal sentence, so we translate further:∃x∀y(y likes everyone that x likes ⊃ x likes y).The antecedent of the conditional is structurally just like sentence 31, with yand x in place of Imre and Karl. So sentence 32 can be completely translated inthis way∃x∀y[∀z(Lxz ⊃ Lyz) ⊃ Lxy]152 forallx (UBC edition)When symbolizing sentences with multiple quantifiers, it is best to proceed bysmall steps. Paraphrase the English sentence so that the logical structure isreadily symbolized in QL. Then translate piecemeal, replacing the dauntingtask of translating a long sentence with the simpler task of translating shorterformulae.8.14 Grammaticality rules for QLIn this section, we provide a formal definition for a well-formed formula (wff)and sentence of QL.8.14.1 ExpressionsThere are six kinds of symbols in QL:predicates A,B,C, . . . , Zwith subscripts, as needed A1, B1, Z1, A2, A25, J375, . . .constants a, b, c, . . . , wwith subscripts, as needed a1, w4, h7,m32, . . .variables x, y, zwith subscripts, as needed x1, y1, z1, x2, . . .sentential connectives ¬, & ,∨,⊃,≡parentheses ( , )quantifiers ∀,∃We define an expression of ql as any string of symbols of QL. Take any of thesymbols of QL and write them down, in any order, and you have an expression.8.14.2 Well-formed formulaeBy definition, a term of ql is either a constant or a variable.An atomic formula of ql is an n-place predicate followed by n terms. nhere can be any non-negative integer, including 0. (A 0-place predicate is a anatomic formula of QL just on its own. It does not require the addition of a termfor meaningfulness or a truth value. In fact, you could think of atoms in SL as0-place QL predicates.)Just as we did for SL, we will give a recursive definition for a wff of QL. Infact, most of the definition will look like the definition of a wff of SL: Everych. 8 quantified logic 153atomic formula is a wff, and you can build new wffs by applying the sententialconnectives.We could just add a rule for each of the quantifiers and be done with it. Forinstance: If Φ is a wff, then ∀xΦ and ∃xΦ are wffs. However, this would allowfor some confusing sentences like ∀x∃xDx. What could these possibly mean?There are possible ways to give interpretations of such sentences, but instead wewill write the definition of a wff so that such abominations do not even count aswell-formed. QL will include the rule that in order for ∀xΦ or ∃xΦ to be a wff,Φ must not already contain an x-quantifier. So ∀x∃xDx will not count as a wffbecause ∃xDx already contains an x-quantifier.1. Every atomic formula is a wff.2. If Φ is a wff, then ¬Φ is a wff.3. If Φ and Ψ are wffs, then (Φ&Ψ), is a wff.4. If Φ and Ψ are wffs, (Φ ∨ Ψ) is a wff.5. If Φ and Ψ are wffs, then (Φ ⊃ Ψ) is a wff.6. If Φ and Ψ are wffs, then (Φ ≡ Ψ) is a wff.7. If Φ is a wff, x is a variable, and Φ contains no x -quantifiers, then ∀x Φ isa wff.8. If Φ is a wff, x is a variable, and Φ contains no x -quantifiers, then ∃x Φ isa wff.9. All and only wffs of QL can be generated by applications of these rules.Notice that the ‘x ’ that appears in the definition above is not the variable x. Itis a meta-variable that stands in for any variable of QL. So ∀xAx is a wff, butso are ∀yAy, ∀zAz, ∀x4Ax4, and ∀z9Az9.We can now give a formal definition for scope: The scope of a quantifier is thesubformula for which the quantifier is the main logical operator.8.14.3 SentencesA sentence is something that can be either true or false. In SL, every wff was asentence. This will not be the case in QL. Consider the following symbolizationkey:UD: people154 forallx (UBC edition)Lxy: x loves yb: BorisConsider the expression Lzz. It is an atomic formula: a two-place predicatefollowed by two terms. All atomic formula are wffs, so Lzz is a wff. Does itmean anything? You might think that it means that z loves himself, in thesame way that Lbb means that Boris loves himself. Yet z is a variable; it doesnot name some person the way a constant would. The wff Lzz does not tellus how to interpret z. Does it mean everyone? anyone? someone? If we hada z-quantifier, it would tell us how to interpret z. For instance, ∃zLzz wouldmean that someone loves themselves.Some formal languages treat a wff like Lzz as implicitly having a universalquantifier in front. We will not do this for QL. If you mean to say that everyoneloves themself, then you need to write the quantifier: ∀zLzzIn order to make sense of a variable, we need a quantifier to tell us how tointerpret that variable. The scope of an x-quantifier, for instance, is the part ofthe formula where the quantifier tells how to interpret x.In order to be precise about this, we define a bound variable to be anoccurrence of a variable x that is within the scope of an x -quantifier. A freevariable is an occurrence of a variable that is not bound.For example, consider the wff ∀x(Ex ∨ Dy) ⊃ ∃z(Rzx ⊃ Lzx). The scope ofthe universal quantifier ∀x is (Ex ∨Dy), so the first x is bound by the universalquantifier but the second and third xs are free. There is no y-quantifier, so they is free. The scope of the existential quantifier ∃z is (Rzx ⊃ Lzx), so bothoccurrences of z are bound by it.We define a sentence of QL as a wff of QL that contains no free variables.8.14.4 Notational conventionsWe will adopt the same notational conventions that we did for SL (p. 40).First, we may leave off the outermost parentheses of a formula. Second, we willsometimes use square brackets ‘[’ and ‘]’ in place of parentheses to increase thereadability of formulae. Third, we give ourselves permission to leave out paren-theses between each pair of conjuncts when writing long series of conjunctions.Fourth, we may similarly leave out parentheses between each pair of disjunctswhen writing long series of disjunctions.ch. 8 quantified logic 1558.15 Common student errorsA sentence that says everything with one property also has another propertyshould be translated as a universal governing a conditional. Using the obviousinterpretation key:. ‘Every student is working’: ∀x(Sx ⊃Wx). ‘Every student has a friend’: ∀x(Sx ⊃ ∃yFxy). ‘Only students with friends are working’: ∀x[(Sx&Wx) ⊃ ∃yFxy]One common error is to translate sentences of this form with a different kind ofshape — for example, as a universal governing a conjunction, or as an existentialgoverning a conditional. These are very inaccurate translations of these Englishsentences:. ∀x(Sx&Wx). ∀x(Sx&∃yFxy). ∀x[(Sx&Wx) &∃yFxy]These say that every object in the UD is a student with the properties inquestion. Everyone is a student that is working; everyone is a student with afriend; everyone is a working student who has a friend. Any time you have auniversal governing a conjunction, you are making a very strong claim — you’renot just talking about objects with a particular property, you’re saying thatmultiple things are true about every single object in the domain. Be very carefulif you find yourself offering a universal over a conjunction, and make sure youdon’t mean to use a conditional instead.It is also a serious mistake to use an existential instead of a universal for sentenceslike these:. ∃x(Sx ⊃Wx). ∃x(Sx ⊃ ∃yFxy). ∃x[(Sx&Wx) ⊃ ∃yFxy]These are very weak claims. They say that there is some object in the domainthat satisfies a certain conditional. For example, ∃x(Sx ⊃ Wx) says there is156 forallx (UBC edition)something in the domain such that, if it is a student, it is working. Given thetruth conditions for the material conditional, this will be true if there is evenone object in the domain that is not a student, regardless of who is and isn’tworking; it will also be true if there is even one object in the domain that isworking, regardless of who is and isn’t a student.If you find yourself offering, as a translation of some English sentence, anexistential governing a conditional, you are almost certainly making a mistake.This is not a reasonable translation of any ordinary English sentence. Youprobably want either a universal over a conditional (everything with one propertyhas another property) or an existential over a conjunction (there is somethingwith the following properties).Practice ExercisesPart A Using the symbolization key given, translate each English-languagesentence into QL. Hint: all of these sentences are well-translated as universalsgoverning conditionals.UD: all humans and animals in the world of Bojack HorsemanDx: x is a dog.Cx: x is a cat.Hx: x is a horse.Bx: x is a human being.Mx: x is a movie star.Lxy: x lives with y.Rxy: x represents y (as y’s agent).Wxy: x worked on a movie with y.b: Bojackc: Princess Carolined: Diannep: Mr. Peanutbutter1. Every movie star is a dog.2. Every movie star is a dog or a cat.3. All dog movie stars live with a human being.4. Everyone who lives with Dianne is a movie star.5. Princess Caroline represents every dog movie star.6. Anyone who worked on a movie with Bojack lives with a movie star.7. Only humans live with Mr. Peanutbutter.8. Everyone who has ever worked on a movie with Bojack is either a dog, acat, or someone who lives with a movie star.ch. 8 quantified logic 157? Part B Using the same symbolization key, translate each English-languagesentence into QL. Hint: all of these sentences are well-translated as existentialsgoverning conjunctions.1. Mr. Peanutbutter lives with a human.2. Dianne lives with a dog who worked on a movie with Bojack.3. Princess Caroline represents a horse who lives with a human being.4. Some human being who worked on a movie with Mr. Peanutbutter liveswith a dog or a cat.5. Bojack worked on a movie with a human movie star.6. Bojack worked on a movie with a nonhuman movie star who lives withDianne.Part C Using the same symbolization key, translate each English-languagesentence into QL. Hint: these should have different forms than the cases above.1. If Mr. Peanutbutter is a movie star, then all dogs are movie stars.2. A dog lives with Dianne and Princess Caroline represents a horse.3. Princess Caroline represents a horse and a dog.4. Princess Caroline represents everyone.5. Dianne doesn’t live with anyone.6. No movie star has ever worked with both Bojack and a cat on a movie.7. Princess Caroline is a cat, but she doesn’t represent any cats.? Part D Using the symbolization key given, translate each English-languagesentence into QL.UD: all animalsAx: x is an alligator.Mx: x is a monkey.Rx: x is a reptile.Zx: x lives at the zoo.Lxy: x loves y.a: Amosb: Bouncerc: Cleo1. Amos, Bouncer, and Cleo all live at the zoo.2. Bouncer is a reptile, but not an alligator.3. If Cleo loves Bouncer, then Bouncer is a monkey.4. If both Bouncer and Cleo are alligators, then Amos loves them both.5. Some reptile lives at the zoo.6. Every alligator is a reptile.7. Any animal that lives at the zoo is either a monkey or an alligator.158 forallx (UBC edition)8. There are reptiles which are not alligators.9. Cleo loves a reptile.10. Bouncer loves all the monkeys that live at the zoo.11. All the monkeys that Amos loves love him back.12. If any animal is a reptile, then Amos is.13. If any animal is an alligator, then it is a reptile.14. Every monkey that Cleo loves is also loved by Amos.15. There is a monkey that loves Bouncer, but Bouncer does not reciprocatethis love.Part E These are syllogistic figures identified by Aristotle and his successors,along with their medieval names. Translate each argument into QL.Barbara All Bs are Cs. All As are Bs. .˙. All As are Cs.Baroco All Cs are Bs. Some A is not B. .˙. Some A is not C.Bocardo Some B is not C. All As are Bs. .˙. Some A is not C.Celantes No Bs are Cs. All As are Bs. .˙. No Cs are As.Celarent No Bs are Cs. All As are Bs. .˙. No As are Cs.Cemestres No Cs are Bs. No As are Bs. .˙. No As are Cs.Cesare No Cs are Bs. All As are Bs. .˙. No As are Cs.Dabitis All Bs are Cs. Some A is B. .˙. Some C is A.Darii All Bs are Cs. Some A is B. .˙. Some A is C.Datisi All Bs are Cs. Some A is B. .˙. Some A is C.Disamis Some B is C. All As are Bs. .˙. Some A is C.Ferison No Bs are Cs. Some A is B. .˙. Some A is not C.Ferio No Bs are Cs. Some A is B. .˙. Some A is not C.Festino No Cs are Bs. Some A is B. .˙. Some A is not C.Baralipton All Bs are Cs. All As are Bs. .˙. Some C is A.Frisesomorum Some B is C. No As are Bs. .˙. Some C is not A.? Part F Using the symbolization key given, translate each English-languagesentence into QL.UD: all animalsch. 8 quantified logic 159Dx: x is a dog.Sx: x likes samurai movies.Lxy: x is larger than y.b: Bertiee: Emersonf: Fergis1. Bertie is a dog who likes samurai movies.2. Bertie, Emerson, and Fergis are all dogs.3. Emerson is larger than Bertie, and Fergis is larger than Emerson.4. All dogs like samurai movies.5. Only dogs like samurai movies.6. There is a dog that is larger than Emerson.7. If there is a dog larger than Fergis, then there is a dog larger than Emerson.8. No animal that likes samurai movies is larger than Emerson.9. No dog is larger than Fergis.10. Any animal that dislikes samurai movies is larger than Bertie.11. There is an animal that is between Bertie and Emerson in size.12. There is no dog that is between Bertie and Emerson in size.13. No dog is larger than itself.14. For every dog, there is some dog larger than it.15. There is an animal that is smaller than every dog.Part G For each argument, write a symbolization key and translate the argumentinto QL.1. Nothing on my desk escapes my attention. There is a computer on mydesk. As such, there is a computer that does not escape my attention.2. All my dreams are black and white. Old TV shows are in black and white.Therefore, some of my dreams are old TV shows.3. Neither Holmes nor Watson has been to Australia. A person could see akangaroo only if they had been to Australia or to a zoo. Although Watsonhas not seen a kangaroo, Holmes has. Therefore, Holmes has been to azoo.4. No one expects the Spanish Inquisition. No one knows the troubles I’veseen. Therefore, anyone who expects the Spanish Inquisition knows thetroubles I’ve seen.5. An antelope is bigger than a bread box. I am thinking of something thatis no bigger than a bread box, and it is either an antelope or a cantaloupe.As such, I am thinking of a cantaloupe.6. All babies are illogical. Nobody who is illogical can manage a crocodile.Berthold is a baby. Therefore, Berthold is unable to manage a crocodile.? Part H Using the symbolization key given, translate each English-languagesentence into QL.160 forallx (UBC edition)UD: candiesCx: x has chocolate in it.Mx: x has marzipan in it.Sx: x has sugar in it.Tx: Boris has tried x.Bxy: x is better than y.1. Boris has never tried any candy.2. Marzipan is always made with sugar.3. Some candy is sugar-free.4. The very best candy is chocolate.5. No candy is better than itself.6. Boris has never tried sugar-free chocolate.7. Boris has tried marzipan and chocolate, but never together.8. Any candy with chocolate is better than any candy without it.9. Any candy with chocolate and marzipan is better than any candy thatlacks both.? Part I Using the symbolization key given, translate each English-languagesentence into QL.UD: people and dishes at a potluckRx: x has run out.Tx: x is on the table.Fx: x is food.Px: x is a person.Lxy: x likes y.e: Elif: Francescag: the guacamole1. All the food is on the table.2. If the guacamole has not run out, then it is on the table.3. Everyone likes the guacamole.4. If anyone likes the guacamole, then Eli does.5. Francesca only likes the dishes that have run out.6. Francesca likes no one, and no one likes Francesca.7. Eli likes anyone who likes the guacamole.8. Eli likes everyone who likes anyone that he likes.9. If there is a person on the table already, then all of the food must haverun out.? Part J Using the symbolization key given, translate each English-languagesentence into QL.ch. 8 quantified logic 161UD: peopleDx: x dances ballet.Fx: x is female.Mx: x is male.Cxy: x is a child of y.Sxy: x is a sibling of y.e: Elmerj: Janep: Patrick1. All of Patrick’s children are ballet dancers.2. Jane is Patrick’s daughter.3. Patrick has a daughter.4. Jane is an only child.5. All of Patrick’s daughters dance ballet.6. Patrick has no sons.7. Jane is Elmer’s niece.8. Patrick is Elmer’s brother.9. Patrick’s brothers have no children.10. Jane is an aunt.11. Everyone who dances ballet has a sister who also dances ballet.12. Every man who dances ballet is the child of someone who dances ballet.Part K Identify which variables are bound and which are free.1. ∃xLxy&∀yLyx2. ∀xAx&Bx3. ∀x(Ax&Bx) &∀y(Cx&Dy)4. ∀x∃y[Rxy ⊃ (Jz&Kx)] ∨Ryx5. ∀x1(Mx2 ≡ Lx2x1) &∃x2Lx3x2? Part L1. Identify which of the following are substitution instances of ∀xRcx: Rac,Rca, Raa, Rcb, Rbc, Rcc, Rcd, Rcx2. Identify which of the following are substitution instances of ∃x∀yLxy:∀yLby, ∀xLbx, Lab, ∃xLxaChapter 9A formal semantics for QLIn this chapter, we describe a formal semantics for QL. This corresponds to thediscussion of interpretations and truth in SL given in Chapter 4. Like truth inSL, truth in QL is defined relative to a particular interpretation; entailment is amatter of truth in all interpretations. In SL, models corresponded to assignmentsof truth values to atomic sentences. For example, here is an SL model:I = P = 0Q = 1R = 0Interpretation I settles the truth value of any SL sentence one can construct fromP , Q, and R. Because QL involves richer notions and a more complex vocabularythan SL, it requires correspondingly richer models and interpretations.9.1 Interpretations in QLWhat is an interpretation in QL? Like a symbolization key for QL, an inter-pretation requires a universe of discourse, a schematic meaning for each of thepredicates, and an object that is picked out by each constant. For example:UD: Marvel charactersHx: x is a hero.Sx: x has spider powers.m: Miles Moralesp: Peter Parker162ch. 9 a formal semantics for ql 163r: The Red Skulls: Susan Stormu: Ultimate Spider-ManThis interpretation is given in terms of English descriptions. To apply it, youneed to know some details about the characters in question. For example, Sm istrue on this interpretation, because Miles Morales does have spider powers. Butthe interpretation itself doesn’t tell us that — to get that information from thisway of setting out the interpretation, you need to know some details about whathappens in the story. You need to know, for example, that Miles Morales, likehis more famous mentor Peter Parker, also has spider powers. If you do knowa bit about Marvel comics, you may know that Miles Morales is actually theUltimate Spider-Man. So u and m in this interpretation are two different namesfor the same member of the UD. There is no rule against having multiple namesfor the same member. (We’ll discuss this issue in much more detail in Chapter12.)We want our QL models to encode this kind of information too. Like a goodSL model, a QL model shouldn’t require prior knowledge of comic books. Oneway we could try to do this would be to just give a truth value assignment, aswe did for SL. The truth value assignment would assign 0 or 1 to each atomicwff: Sm = 1, Sp = 1, Sr = 0, and so on. If we were to do that, however, wemight just as well translate the sentences from QL to SL by replacing Sp andSm with sentence letters. We could then rely on the definition of truth for SL,but at the cost of ignoring all the logical structure of predicates and terms. Inwriting a symbolization key for QL, we do not give separate definitions for Spand Sm. Instead, we give meanings to the components S, p, and m. This isessential because we want to be able to reflect the logical relationships betweene.g. Sp and ∃xSx.Our interpretations should include explanations for predicates and names, notjust for sentences. We cannot use a truth value assignment for this, because apredicate by itself is neither true nor false. In the interpretation given above,H is true of Peter Parker (i.e., Hp is true), but it makes no sense at all toask whether H on its own is true. It would be like asking whether the Englishlanguage fragment ‘. . .is a hero’ is true.What does an interpretation do for a predicate, if it does not make it true orfalse? An interpretation helps to pick out the objects to which the predicateapplies. Interpreting Hx to mean ‘x is a hero’ picks out some characters as thethings that are Hs. Formally, this is a set of members of the UD to which thepredicate applies; this set is called the extension of the predicate.Some predicates have indefinitely large extensions. It would be impractical totry and write down all of the Marvel characters individually, so instead we usean English language expression to interpret the predicate. This is somewhat164 forallx (UBC edition)imprecise, because the interpretation alone does not tell you which members ofthe UD are in the extension of the predicate. In order to figure out whether aparticular member of the UD is in the extension of the predicate (to figure outwhether the Red Skull is a hero, for instance), you need to know about comicbooks. (As you might guess from his name, he’s not.) In general, the extensionof a predicate is the result of an interpretation along with some facts.Sometimes it is possible to list all of the things that are in the extension of apredicate. Instead of writing a schematic English sentence, we can write downthe extension as a set of things. Suppose we wanted to add a one-place predicateF to the key above, meaning ‘x is a founding member of the Fantastic Four’, sowe write the extension as a set of characters:extension(F ) = {Reed Richards, Susan Storm, Johnny Storm, Ben Grimm}You do not need to know anything about comic books to be able to determinethat, on this interpretation, Fs is true: Susan Storm, whose name is given as s,is just specified to be one of the things that is F . Similarly, ∃xFx is obviouslytrue on this interpretation: There is at least one member of the UD that is anF — in fact, there are four of them.What about the sentence ∀xFx? The sentence is false, because it is not truethat all members of the UD are F . It requires the barest minimum of knowledgeabout comic books to know that there are other characters besides just thesefour. Although we specified the extension of F in a formally precise way, we stillspecified the UD with an English language description. Formally speaking, aUD is just a set of members.The formal significance of a predicate is determined by its extension, but whatshould we say about constants like m and s? The meaning of a constantdetermines which member of the UD is picked out by the constant. The individualthat the constant picks out is called the referent of the constant. Both mand u have the same referent, since they both refer to the same comic bookcharacter. You can think of a constant letter as a name and the referent as thething named. In English, we can use the different names ‘Miles’ and ‘UltimateSpider-Man’ to refer to the same comic book character. In this interpretation,we also use the different constants ‘m’ and ‘u’ to refer to the same member ofthe UD.9.2 SetsWe use curly brackets ‘{’ and ‘}’ to denote sets. The members of the set can belisted in any order, separated by commas. This means that {foo, bar} and {bar,foo} are the same set.ch. 9 a formal semantics for ql 165It is possible to have a set with no members in it. This is called the empty set.The empty set is sometimes written as {}, but usually it is written as the singlesymbol ∅.9.3 ModelsAs we have seen, an interpretation in QL is only formally significant insofar asit determines a UD, an extension for each predicate, and a referent for eachconstant. We call this formal structure a model for QL.To see how this works, consider this symbolization key:UD: The first ten natural numbersPx: x is prime.n4: 4Given some basic mathematical knowledge, it is obvious that Pn4 is false. Let’sconsider the model this interpretation suggests, to show why it makes this wfffalse. Instead of just giving a description in the UD, we can list the members asa set. We also define the extension of the predicate P , and the referent of theconstant n4:UD = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}extension(P ) = {2, 3, 5, 7}referent(n4) = 4This is not a full model for this interpretation, but it is a detailed enough partialmodel to show that Pn4 is false. You do not need to know anything aboutmathematics to see that this sentence is false in this model. The UD membernamed by n4 is not in the extension of P . In this way, the model captures all ofthe formal significance of the interpretation.Suppose we enrich this interpretation with more predicates:UD: The first ten natural numbersEx: x is even.Nx: x is negative.Lxy: x is less than y.Txyz: x times y equals z.What do we need to add to the model for our new predicates?The extension of E is the subset {2, 4, 6, 8, 10}. There are no negative numbersin the UD, so N has an empty extension; i.e. extension(N) = ∅.166 forallx (UBC edition)Sometimes it will be convenient to represent extensions graphically, similar tothe way we did with truth tables. We can represent the extensions just describedfor P , E, and N thus:P E N1 0 0 02 1 1 03 1 0 04 0 1 05 1 0 06 0 1 07 1 0 08 0 1 09 0 0 010 0 1 0The members of the UD are listed as rows; the predicates are given as columns.The 0s and 1s indicate whether each member satisfies each predicate. Noticethat the same information is conveyed in this chart as in the three sets of orderedpairs described above. Either is an acceptable way of indicating the extension ofthe predicates.The extension of a two-place predicate like L is more complicated. No individualnumber falls under the extension of this predicate; it is about the relation betweenmembers. Note also that sets of pairs numbers aren’t suitable for the extensionof L either, because 1 is less than 8, but 8 is not less than 1. And the set {1,8}is the very same set as the set {8,1}. The solution is to have the extension ofL consist in a set of ordered pairs of numbers. An ordered pair is like a setwith two members, except that the order does matter. We write ordered pairswith angle brackets ‘<’ and ‘>’. The ordered pair is different thanthe ordered pair . The extension of L is a collection of ordered pairs,all of the pairs of numbers in the UD such that the first number is less than thesecond. Writing this out completely:extension(L) = {<1, 2>, <1, 3>, <1, 4>, <1, 5>, <1, 6>, <1, 7>, <1, 8>,<1, 9>, <1, 10>, <2, 3>, <2, 4>, <2, 5>, <2, 6>, <2, 7>, <2, 8>, <2, 9>,<2, 10>, <3, 4>, <3, 5>, <3, 6>, <3, 7>, <3, 8>, <3, 9>, <3, 10>, <4, 5>,<4, 6>, <4, 7>, <4, 8>, <4, 9>, <4, 10>, <5, 6>, <5, 7>, <5, 8>, <5, 9>,<5, 10>, <6, 7>, <6, 8>, <6, 9>, <6, 10>, <7, 8>, <7, 9>, <7, 10>, <8, 9>,<8, 10> <9, 10>}Formally, the extension of a two-place predicate is a set of ordered pairs. Some-times we will find it easier to represent extensions in a chart, with the twovariable positions represented on the two axes. For example, the extension abovecould be expressed via a table like the one below. The ‘0’ in the first cell of thetable says that <1, 1> is not in the extension of L; the next cell in the first rowch. 9 a formal semantics for ql 167says that <1, 2> is. Etc. Sometimes drawing out a chart like this will be theeasiest way to represent models for the extensions of two-place predicates.Lxy 1 2 3 4 5 6 7 8 9 101 0 1 1 1 1 1 1 1 1 12 0 0 1 1 1 1 1 1 1 13 0 0 0 1 1 1 1 1 1 14 0 0 0 0 1 1 1 1 1 15 0 0 0 0 0 1 1 1 1 16 0 0 0 0 0 0 1 1 1 17 0 0 0 0 0 0 0 1 1 18 0 0 0 0 0 0 0 0 1 19 0 0 0 0 0 0 0 0 0 110 0 0 0 0 0 0 0 0 0 0The extension of a three-place predicate is a set of ordered triples where thepredicate is true of those three things in that order. So the extension of T in thismodel will contain ordered triples like <2, 4, 8>, because 2× 4 = 8. Becausethe surface of a sheet of paper is for all intents and purposes two-dimensional, itis usually not convenient to represent 3-or-more place predicates with tables.Generally, the extension of an n-place predicate is a set of all ordered n-tuples〈a1, a2, . . . , an〉 such that a1–an are members of the UD and the predicate istrue of a1–an in that order.9.4 Working with modelsWe will use the double turnstile symbol for QL much as we did for SL. ‘Φ |= Ψ ’means that ‘Φ entails Ψ ’: When Φ and Ψ are two sentences of QL, Φ |= Ψ meansthat there is no model in which Φ is true and Ψ is false. |= Φ is shorthandfor ∅ |= Φ, which means that Φ is true in every model. This allows us to givedefinitions for various concepts in QL. In fact, we can use the same definitionsoffered in Chapter 4.A tautology in ql is a sentence Φ that is true in every model; i.e.,|= Φ.A contradiction in ql is a sentence Φ that is false in every model;i.e., |= ¬Φ.A sentence is contingent in ql if and only if it is neither a tautologynor a contradiction.An argument “ P1,P2, · · · , .˙. Ω ” is valid in ql if and only if thereis no model in which all of the premises are true and the conclusionis false; i.e., {P1,P2, · · · } |= Ω. It is invalid in ql otherwise.168 forallx (UBC edition)Two sentences Φ and Ψ are logically equivalent in ql if andonly if both Φ |= Ψ and Ψ |= Φ.The set {Φ1, Φ2, Φ3, · · · } is consistent in ql if and only if there isat least one model in which all of the sentences are true. The set isinconsistent in ql if and only if there is no such model.9.5 Constructing modelsSuppose we want to show that ∀xAxx ⊃ Bd is not a tautology. This requiresshowing that the sentence is not true in every model. If we can provide anexample of a model in which the sentence is false, then we will have shown thatthe sentence is not a tautology.What would such a model look like? In order for ∀xAxx ⊃ Bd to be false, theantecedent (∀xAxx) must be true, and the consequent (Bd) must be false.To construct such a model, we start with a UD. It will be easier to specifyextensions for predicates if we have a small UD, so start with a UD that hasjust one member. Formally, this single member might be anything. Let’s say itis Miles Morales.We want ∀xAxx to be true, so we want all members of the UD to be pairedwith themself in the extension of A; this means that the extension of A must be{}.We want Bd to be false, so the referent of d must not be in the extension of B.We give B an empty extension.Since Miles is the only member of the UD, it must be the referent of d. Themodel we have constructed looks like this:UD = {Miles Morales}extension(A) = {}extension(B) = ∅referent(d) = Miles MoralesStrictly speaking, a model specifies an extension for every predicate of QL and areferent for every constant. As such, it is generally impossible to write down acomplete model. That would require writing down infinitely many extensions andinfinitely many referents. However, we do not need to consider every predicatein order to show that there are models in which ∀xAxx ⊃ Bd is false. Predicateslike H and constants like f13 make no difference to the truth or falsity of thissentence. It is enough to specify extensions for A and B and a referent for d, aswe have done. This provides a partial model in which the sentence is false.ch. 9 a formal semantics for ql 169Perhaps you are wondering: What does the predicate A mean in English? Itdoesn’t really matter. For formal purposes, the existence of models like the onesdescribed above are enough to show that ∀xAxx ⊃ Bd is not a tautology. Butwe can offer an interpretation in English if we like. How about this one?UD: Miles MoralesAxy: x knows y’s biggest secret.Bx: x’s powers derive from gamma radiation.d: Miles MoralesThis is one way we can interpret the model above. Add is true, because Milesdoes know Miles’s biggest secret. (It’s that he’s the Ultimate Spider-Man. Nowyou know it too!) Bd is false: Miles’s powers came from a genetically enhancedspider bite, not from gamma radiation. But the partial model constructed aboveincludes none of these interpretative details. All it says is that A is a predicatewhich is true of Miles and Miles, and that B is a predicate which does notapply to Miles. There are indefinitely many predicates in English that havethis extension. Axy might instead translate ‘x is the same size as y’ or ‘x andy live in the same city’; Bx might translate ‘x is a billionaire’ or ‘x’s unclewas killed by a robber’ or ‘Donald Trump has written a tweet about x’. Inconstructing a model and giving extensions for A and B, we do not specify whatEnglish predicates A and B should be used to translate. We are concerned withwhether the ∀xAxx ⊃ Bd comes out true or false, and all that matters for truthand falsity in QL is the information in the model: the UD, the extensions ofpredicates, and the referents of constants.We can just as easily show that ∀xAxx ⊃ Bd is not a contradiction. We needonly specify a model in which ∀xAxx ⊃ Bd is true; i.e., a model in which either∀xAxx is false or Bd is true. Here is one such partial model:UD = {The Red Skull}extension(A) = {}extension(B) = {The Red Skull}referent(d) = The Red SkullI’ve switched our object from Miles Morales to The Red Skull to emphasize thatit doesn’t matter what object you pick. (Changing the examples all back toMiles would make no difference.) On this model, ∀xAxx ⊃ Bd is true, because itis a conditional with a true consequent (as well as a true antecedent). We havenow shown that ∀xAxx ⊃ Bd is neither a tautology nor a contradiction. By thedefinition of ‘contingent in QL,’ this means that ∀xAxx ⊃ Bd is contingent. Ingeneral, showing that a sentence is contingent will require two models: one inwhich the sentence is true and another in which the sentence is false.Suppose we want to show that ∀xSx and ∃xSx are not logically equivalent. Weneed to construct a model in which the two sentences have different truth values;170 forallx (UBC edition)we want one of them to be true and the other to be false. We start by specifyinga UD. Again, we make the UD reasonably small so that we can specify extensionseasily. But this time we will need at least two members. If we only had onemember of the domain, we wouldn’t be able to illustrate the difference betweenall and some. Let’s let our UD be {The Red Skull, Miles Morales}.We can make ∃xSx true by including something in the extension of S, and wecan make ∀xSx false by leaving something out of the extension of S. It doesnot matter which one we include and which one we leave out. Making Miles theonly S, we get a partial model that looks like this:UD = {Miles, The Red Skull}extension(S) = {Miles}This partial model shows that the two sentences are not logically equivalent.∃xSx is assigned 1 on this model, but ∀xSx is assigned 0.Back on p. 149, we said that this argument would be invalid in QL:(Rc&K1c) &Tc.˙. T c&K2cNow we can prove that this is so. To show that this argument is invalid, we needto show that there is some model in which the premise is true and the conclusionis false. We can construct such a model deliberately. Here is one way to do it:UD = {Reed Richards}extension(T ) = {Reed Richards}extension(K1) = {Reed Richards}extension(K2) = ∅extension(R) = {Reed Richards}referent(c) = Reed RichardsSimilarly, we can show that a set of sentences is consistent by constructing amodel in which all of the sentences are true.9.6 Reasoning about all modelsWe can show that a sentence is not a tautology just by providing one carefullyspecified model: a model in which the sentence is false. To show that somethingis a tautology, on the other hand, it would not be enough to construct ten, onehundred, or even a thousand models in which the sentence is true. It is onlya tautology if it is true in every model, and there are infinitely many models.ch. 9 a formal semantics for ql 171Table 9.1: It is relatively easy to answer a question if you can do it by constructinga model or two. It is much harder if you need to reason about all possible models.This table shows when constructing models is enough.YES NOIs Φ a tautology? show that Φ must betrue in any modelconstruct a model inwhich Φ is falseIs Φ a contradiction? show that Φ must befalse in any modelconstruct a model inwhich Φ is trueIs Φ contingent? construct two models,one in which Φ is trueand another in which Φis falseeither show that Φ is atautology or show thatΦ is a contradictionAre Φ and Ψ equiva-lent?show that Φ and Ψmust have the sametruth value in anymodelconstruct a model inwhich Φ and Ψ have dif-ferent truth valuesIs the set A consistent? construct a model inwhich all the sentencesin A are trueshow that the sentencesin A could not all betrue in any modelIs the argument‘P , .˙. Ω’ valid?show that any model inwhich P is true must bea model in which Ω istrueconstruct a model inwhich P is true and Ωis falseThis cannot be avoided just by constructing partial models, because there areinfinitely many partial models.Consider, for example, the sentence Raa ≡ Raa. There are two logically distinctpartial models of this sentence that have a 1-member UD. There are 32 distinctpartial models that have a 2-member UD. There are 1526 distinct partial modelsthat have a 3-member UD. There are 262,144 distinct partial models that havea 4-member UD. And so on to infinity. In order to show that this sentence is atautology, we need to show something about all of these models. There is nohope of doing so by dealing with them one at a time.Nevertheless, Raa ≡ Raa is obviously a tautology. We can prove it with a simpleargument:There are two kinds of models: those with 〈referent(a), referent(a)〉is in the extension of R and those in which it is not. In the firstkind of model, Raa is true; by the truth table for the biconditional,Raa ≡ Raa is also true. In the second kind of model, Raa is false;172 forallx (UBC edition)this makes Raa ≡ Raa true. Since the sentence is true in both kindsof model, and since every model is one of the two kinds, Raa ≡ Raais true in every model. Therefore, it is a tautology.This is a sound argument; it should convince us of its conclusion. But note thatit is not an argument in QL. Rather, it is an argument in English about QL; it isan argument in the metalanguage. There is no formal procedure for evaluatingor constructing natural language arguments like this one. The imprecision ofnatural language is the very reason we began thinking about formal languages.There are further difficulties with this approach.Consider the sentence ∀x(Rxx ⊃ Rxx), another obvious tautology. It mightbe tempting to reason in this way: ‘Rxx ⊃ Rxx is true in every model, so∀x(Rxx ⊃ Rxx) must be true.’ The problem is that Rxx ⊃ Rxx is not true inevery model. It is not a sentence, and so it is neither true nor false. We do notyet have the vocabulary to say what we want to say about Rxx ⊃ Rxx. In thenext section, we introduce the concept of satisfaction; after doing so, we will bebetter able to provide an argument that ∀x(Rxx ⊃ Rxx) is a tautology.It is necessary to reason about an infinity of models to show that a sentenceis a tautology. Similarly, it is necessary to reason about an infinity of modelsto show that a sentence is a contradiction, that two sentences are equivalent,that a set of sentences is inconsistent, or that an argument is valid. There areother things we can show by carefully constructing a model or two. Table 9.1summarizes which things are which.9.7 Truth in QLFor SL, we split the definition of truth into two parts: a truth value assignment(a) for sentence letters and a truth function (v) for all sentences. The truthfunction covered the way that complex sentences could be built out of sentenceletters and connectives.In the same way that truth for SL is always truth given a truth value assignment,truth for QL is truth in a model. The atomic sentences, again, are n-placepredicates followed by n constants, like Pj. It is true in a model M if and onlyif the referent of j is in the extension of P in M.We could go on in this way to define truth for all atomic sentences that containonly predicates and constants: Consider any sentence of the form Ra1 . . . anwhere R is an n-place predicate and the as are constants. It is true in M if andonly if 〈referent(a1), . . . , referent(an)〉 is in extension(R ) in M.ch. 9 a formal semantics for ql 173We could then define truth for sentences built up with sentential connectives inthe same way we did for SL. For example, the sentence (Pj ⊃Mda) is true inM if either Pj is false in M or Mda is true in M.Unfortunately, this approach will fail when we consider sentences containingquantifiers. Consider ∀xPx. When is it true in a model M? The answer cannotdepend on whether Px is true or false in M, because the x in Px is a free variable.Px is not a sentence. It is neither true nor false.We were able to give a recursive definition of truth for SL because every well-formed formula of SL has a truth value. This is not true in QL, so we cannotdefine truth by starting with the truth of atomic sentences and building up. Wealso need to consider the atomic formulae which are not sentences. In orderto do this we will define satisfaction; every well-formed formula of QL will besatisfied or not satisfied, even if it does not have a truth value. We will then beable to define truth for sentences of QL in terms of satisfaction.9.8 SatisfactionThe formula Px says, roughly, that x is one of the P s. This cannot be quiteright, however, because x is a variable and not a constant. It does not name anyparticular member of the UD. Instead, its meaning in a sentence is determinedby the quantifier that binds it. The variable x must stand-in for every memberof the UD in the sentence ∀xPx, but it only needs to stand-in for one memberin ∃xPx. Since we want the definition of satisfaction to cover Px without anyquantifier whatsoever, we will start by saying how to interpret a free variablelike the x in Px.We do this by introducing a variable assignment. Formally, this is a function thatmatches up each variable with a member of the UD. Call this function ‘a’. (The‘a’ is for ‘assignment’, but this is not the same as the truth value assignmentthat we used in defining truth for SL.)The formula Px is satisfied in a model M by a variable assignment a if and onlyif a(x), the object that a assigns to x, is in the extension of P in M.When is ∀xPx satisfied? It is not enough if Px is satisfied in M by a, becausethat just means that a(x) is in extension(P ). ∀xPx requires that every othermember of the UD be in extension(P ) as well.So we need another bit of technical notation: For any member pi of the UDand any variable x , let a[pi|x ] be the variable assignment that assigns pi to xbut agrees with a in all other respects. We have used pi, the Greek letter pi,to underscore the fact that it is some member of the UD and not some symbol174 forallx (UBC edition)of QL. Suppose, for example, that the UD is presidents of the United States.The function a[Grover Cleveland|x] assigns Grover Cleveland to the variable x,regardless of what a assigns to x; for any other variable, a[Grover Cleveland|x]agrees with a.We can now say concisely that ∀xPx is satisfied in a model M by a variableassignment a if and only if, for every object pi in the UD of M, Px is satisfied inM by a[pi|x].The intuitive thought here is that wff satisfaction is relative to a variableassignment. A variable assignment is a way of treating each variable as if it werea name for some object or other; a wff is satisfied by a in a given model iff, inthat model, treating the variables the way a suggests would yield a true wff.You may worry that our statement of satisfaction by a variable assignment in amodel is circular, because it gives the satisfaction conditions for the sentence∀xPx using the phrase ‘for every object.’ However, it is important to rememberthe difference between a logical symbol like ‘∀’ and an English language word like‘every.’ The word is part of the metalanguage that we use in defining satisfactionconditions for object language sentences that contain the symbol.We can now give a general definition of satisfaction, extending from the cases wehave already discussed. We define a function s (for ‘satisfaction’) in a model Msuch that for any wff Φ and variable assignment a, s(Φ, a) = 1 if Φ is satisfied inM by a; otherwise s(Φ, a) = 0.1. If Φ is an atomic wff of the form Pt1 . . . tn and pii is the object picked outby ti, thens(Φ, a) ={1 if 〈pi1 . . . pin〉 is in extension(P ) in M,0 otherwise.For each term ti: If ti is a constant, then pii = referent(ti). If ti is avariable, then pii = a(ti).2. If Φ is ¬Ψ for some wff Ψ , thens(Φ, a) ={1 if s(Ψ, a) = 0,0 otherwise.3. If Φ is (Ψ &Ω) for some wffs Φ, Ψ , thens(Φ, a) ={1 if s(Ψ, a) = 1 and s(Ω, a) = 1,0 otherwise.4. If Φ is (Ψ ∨Ω) for some wffs Φ, Ψ , thens(Φ, a) ={0 if s(Ψ, a) = 0 and s(Ω, a) = 0,1 otherwise.ch. 9 a formal semantics for ql 1755. If Φ is (Ψ ⊃ Ω) for some wffs Φ, Ψ , thens(Φ, a) ={0 if s(Ψ, a) = 1 and s(Ω, a) = 0,1 otherwise.6. If Φ is (Ψ ≡ Ω) for some wffs Φ, Ψ , thens(Φ, a) ={1 if s(Ψ, a) = s(Ω, a),0 otherwise.7. If Φ is ∀x Ψ for some wff Ψ and some variable x , thens(Φ, a) ={1 if s(Ψ, a[pi|x ]) = 1 for every member pi of the UD,0 otherwise.8. If Φ is ∃x Ψ for some wff Ψ and some variable x , thens(Φ, a) ={1 if s(Ψ, a[pi|x ]) = 1 for at least one member pi of the UD,0 otherwise.This definition follows the same structure as the definition of a wff for QL, sowe know that every wff of QL will be covered by this definition. For a model Mand a variable assignment a, any wff will either be satisfied or not. No wffs areleft out or assigned conflicting values.9.9 Truth in QLConsider a simple sentence like ∀xPx. By part 7 in the definition of satisfaction,this sentence is satisfied if a[pi|x] satisfies Px in M for every pi in the UD. Inother words, assign that x to any object in the UD you like, and the resultantwff will come out true. By part 1 of the definition, this will be the case if everypi is in the extension of P . Whether ∀xPx is satisfied does not depend on theparticular variable assignment a. If this sentence is satisfied, then it is true. Thisis a formalization of what we have said all along: ∀xPx is true if everything inthe UD is in the extension of P .The same thing holds for any sentence of QL. Because all of the variables arebound, a sentence is satisfied or not regardless of the details of the variableassignment. So we can define truth in this way: A sentence Φ is true in M ifand only if some variable assignment satisfies Φ in M ; Φ is false in M otherwise.Truth in QL is truth in a model. Sentences of QL are not flat-footedly trueor false as mere symbols, but only relative to a model. A model provides themeaning of the symbols, insofar as it makes any difference to truth and falsity.176 forallx (UBC edition)9.10 Reasoning about all models (reprise)At the end of section 9.6, we were stymied when we tried to show that ∀x(Rxx ⊃Rxx) is a tautology. Having defined satisfaction, we can now reason in this way:Consider some arbitrary modelM. Now consider an arbitrary memberof the UD; for the sake of convenience, call it pi. It must be the caseeither that 〈pi, pi〉 is in the extension of R or that it is not. If 〈pi, pi〉 isin the extension of R, then Rxx is satisfied by a variable assignmentthat assigns pi to x (by part 1 of the definition of satisfaction); sincethe consequent of Rxx ⊃ Rxx is satisfied, the conditional is satisfied(by part 5). If 〈pi, pi〉 is not in the extension of R, then Rxx is notsatisfied by a variable assignment that assigns pi to x (by part 1); sinceantecedent of Rxx ⊃ Rxx is not satisfied, the conditional is satisfied(by part 5). In either case, Rxx ⊃ Rxx is satisfied. This is true forany member of the UD, so ∀x(Rxx ⊃ Rxx) is satisfied by any truthvalue assignment (by part 7). So ∀x(Rxx ⊃ Rxx) is true in M (bythe definition of truth). This argument holds regardless of the exactUD and regardless of the exact extension of R, so ∀x(Rxx ⊃ Rxx)is true in any model. Therefore, it is a tautology.Giving arguments about all possible models typically requires clever combinationof two strategies:1. Divide cases between two possible kinds, such that every case must be onekind or the other. In the argument on p. 171, for example, we distinguishedtwo kinds of models based on whether or not a specific ordered pair was inextension(R). In the argument above, we distinguished cases in which an orderedpair was in extension(R) and cases in which it was not.2. Consider an arbitrary object as a way of showing something more general. Inthe argument above, it was crucial that pi was just some arbitrary member of theUD. We did not assume anything special about it. As such, whatever we couldshow to hold of pi must hold of every member of the UD — if we could showit for pi, we could show it for anything. In the same way, we did not assumeanything special about M, and so whatever we could show about M must holdfor all models.Consider one more example. The argument ∀x(Hx& Jx) .˙.∀xHx is obviouslyvalid. We can only show that the argument is valid by considering what mustbe true in every model in which the premise is true.Consider an arbitrary model M in which the premise ∀x(Hx& Jx)is true. The conjunction Hx& Jx is satisfied regardless of what isch. 9 a formal semantics for ql 177assigned to x, so Hx must be also (by part 3 of the definition ofsatisfaction). As such, ∀xHx is satisfied by any variable assignment(by part 7 of the definition of satisfaction) and true in M (by thedefinition of truth). Since we did not assume anything about Mbesides ∀x(Hx& Jx) being true, ∀xHx must be true in any modelin which ∀x(Hx& Jx) is true. So ∀x(Hx& Jx) |= ∀xHx.Even for a simple argument like this one, the reasoning is somewhat complicated.For longer arguments, the reasoning can be insufferable. The problem arisesbecause talking about an infinity of models requires reasoning things out inEnglish. What are we to do? The answer won’t surprise readers of the first halfof the book: we’ll make use of some formal proof systems. We have seen twokinds of proof systems for SL: the tree method, and natural deduction proofs.In the coming chapters, we’ll extend both kinds of systems to QL as well.Practice Exercises? Part A Determine whether each sentence is true or false in the model given.UD = {Corwin, Benedict}extension(A) = {Corwin, Benedict}extension(B) = {Benedict}extension(N) = ∅referent(c) = Corwin1. Bc2. Ac ≡ ¬Nc3. Nc ⊃ (Ac ∨Bc)4. ∀xAx5. ∀x¬Bx6. ∃x(Ax&Bx)7. ∃x(Ax ⊃ Nx)8. ∀x(Nx ∨ ¬Nx)9. ∃xBx ⊃ ∀xAx? Part B Determine whether each sentence is true or false in the model given.UD = {Waylan, Willy, Johnny}extension(H) = {Waylan, Willy, Johnny}extension(W ) = {Waylan, Willy}extension(R) = {,,}referent(m) = Johnny178 forallx (UBC edition)1. ∃x(Rxm&Rmx)2. ∀x(Rxm ∨Rmx)3. ∀x(Hx ≡Wx)4. ∀x(Rxm ⊃Wx)5. ∀x[Wx ⊃ (Hx&Wx)]6. ∃xRxx7. ∃x∃yRxy8. ∀x∀yRxy9. ∀x∀y(Rxy ∨Ryx)10. ∀x∀y∀z[(Rxy&Ryz) ⊃ Rxz]? Part C Determine whether each sentence is true or false in the model given.UD = {Lemmy, Courtney, Eddy}extension(G) = {Lemmy, Courtney, Eddy}extension(H) = {Courtney}extension(M) = {Lemmy, Eddy}referent(c) = Courtneyreferent(e) = Eddy1. Hc2. He3. Mc ∨Me4. Gc ∨ ¬Gc5. Mc ⊃ Gc6. ∃xHx7. ∀xHx8. ∃x¬Mx9. ∃x(Hx&Gx)10. ∃x(Mx&Gx)11. ∀x(Hx ∨Mx)12. ∃xHx&∃xMx13. ∀x(Hx ≡ ¬Mx)14. ∃xGx&∃x¬Gx15. ∀x∃y(Gx&Hy)? Part D Write out the model that corresponds to the interpretation given.UD: natural numbers from 10 to 13Ox: x is odd.Sx: x is less than 7.Tx: x is a two-digit number.Ux: x is thought to be unlucky.Nxy: x is the next number after y.ch. 9 a formal semantics for ql 179Part E Show that each of the following is contingent.1.? Da&Db2.? ∃xTxh3.? Pm&¬∀xPx4. ∀zJz ≡ ∃yJy5. ∀x(Wxmn ∨ ∃yLxy)6. ∃x(Gx ⊃ ∀yMy)? Part F Show that the following pairs of sentences are not logically equivalent.1. Ja, Ka2. ∃xJx, Jm3. ∀xRxx, ∃xRxx4. ∃xPx ⊃ Qc, ∃x(Px ⊃ Qc)5. ∀x(Px ⊃ ¬Qx), ∃x(Px&¬Qx)6. ∃x(Px&Qx), ∃x(Px ⊃ Qx)7. ∀x(Px ⊃ Qx), ∀x(Px&Qx)8. ∀x∃yRxy, ∃x∀yRxy9. ∀x∃yRxy, ∀x∃yRyxPart G Show that the following sets of sentences are consistent.1. {Ma, ¬Na, Pa, ¬Qa}2. {Lee, Lef , ¬Lfe, ¬Lff}3. {¬(Ma& ∃xAx), Ma ∨ Fa, ∀x(Fx ⊃ Ax)}4. {Ma ∨Mb, Ma ⊃ ∀x¬Mx}5. {∀yGy, ∀x(Gx ⊃ Hx), ∃y¬Iy}6. {∃x(Bx ∨Ax), ∀x¬Cx, ∀x[(Ax&Bx) ⊃ Cx]}7. {∃xXx, ∃xY x, ∀x(Xx ≡ ¬Y x)}8. {∀x(Px ∨Qx), ∃x¬(Qx&Px)}9. {∃z(Nz&Ozz), ∀x∀y(Oxy ⊃ Oyx)}10. {¬∃x∀yRxy, ∀x∃yRxy}Part H Construct models to show that the following arguments are invalid.1. ∀x(Ax ⊃ Bx), .˙. ∃xBx2. ∀x(Rx ⊃ Dx), ∀x(Rx ⊃ Fx), .˙. ∃x(Dx&Fx)3. ∃x(Px ⊃ Qx), .˙.∃xPx4. Na&Nb&Nc, .˙. ∀xNx5. Rde, ∃xRxd, .˙. Red6. ∃x(Ex&Fx), ∃xFx ⊃ ∃xGx, .˙. ∃x(Ex&Gx)7. ∀xOxc, ∀xOcx, .˙. ∀xOxx180 forallx (UBC edition)8. ∃x(Jx&Kx), ∃x¬Kx, ∃x¬Jx, .˙. ∃x(¬Jx&¬Kx)9. Lab ⊃ ∀xLxb, ∃xLxb, .˙. LbbPart I1. Many logic books define consistency and inconsistency in this way: “ A set{Φ1, Φ2, Φ3, · · · } is inconsistent if and only if {Φ1, Φ2, Φ3, · · · } |= (Ψ &¬Ψ)for some sentence Ψ . A set is consistent if it is not inconsistent.”Does this definition lead to any different sets being consistent than thedefinition on p. 70? Explain your answer.2.? Our definition of truth says that a sentence Φ is true in M if and only ifsome variable assignment satisfies Φ in M . Would it make any difference ifwe said instead that Φ is true inM if and only if every variable assignmentsatisfies Φ in M? Explain your answer.Chapter 10QL TreesIn Chapter 9 we saw that the more structured models of QL make reasoningabout all possible models rather complex. In SL, where a model was just anassignment of truth values to atomic sentences, reasoning about all models wasrelatively straightforward: we could simply survey them via truth tables. In QLthere isn’t really as simple a method for considering claims about validity orsatisfiability as the truth table method. So we must rely more heavily on proofsystems.This chapter considers a tree method for QL. Like the tree method for SLdiscussed in Chapter 5, the QL tree method is a method that attempts togenerate an interpretation satisfying a given set of sentences. When it succeeds,it shows a way to satisfy them; when it fails, it shows that they are unsatisfiable.As we did in the initial presentation of trees in Chapter 5, we’ll begin by workingthrough a couple of examples at an intuitive level, before laying out the formaltree rules.10.1 Trees with fixed domainsLet’s begin with a simplifying assumption. Let’s suppose that we’re only inter-ested in models with particular fixed domain sizes. (By the time we get to theformal tree rules later in this chapter, we’ll dispense with this assumption, but it’shelpful for introducing the general idea.) Let’s confine our attention to modelsthat have two members in the UD. Consider whether there are any such modelsthat satisfy the following QL sentences: {∀x(Fx ⊃ Gx),¬∃xGx,∃x(Fx ∨Gx)}.To evaluate this question, we put those sentences in the root of a tree.181182 forallx (UBC edition)1.2.3.∀x(Fx ⊃ Gx)¬∃xGx∃x(Fx ∨Gx)Let’s start by considering the universal claim on line 1. A universal says thateach of its instances are true. Since at the moment we are assuming that thereare two objects in our UD, let’s also assume we have just two names, a and b,corresponding to our two objects. So we can just write down the two instancesas a linear development. We also put a check at line 1 to indicate that we’veresolved that sentence.1.2.3.4.5.∀x(Fx ⊃ Gx) X¬∃xGx∃x(Fx ∨Gx)Fa ⊃ GaFb ⊃ GbNext let’s consider the negated existential at line 2. It says that each instanceis false. So it develops into the negation of both instances, and we add a checkmark at line 2.1.2.3.4.5.6.7.∀x(Fx ⊃ Gx) X¬∃xGx X∃x(Fx ∨Gx)Fa ⊃ GaFb ⊃ Gb¬Ga¬GbThere is one more quantified sentence to resolve, on line 3, but before we do that,let’s handle the conditionals on lines 4 and 5. We do these exactly the same waywe handled conditionals in SL trees. They each branch into the negation of theantecedent and the consequent; the latter branch closes each time.ch. 10 ql trees 1831.2.3.4.5.6.7.8.9.∀x(Fx ⊃ Gx) X¬∃xGx X∃x(Fx ∨Gx)Fa ⊃ Ga XFb ⊃ Gb X¬Ga¬Gb¬Fa¬Fb Gb×7, 9Ga×6, 8Now let’s look at line 3. It is an existential claim, so it says that at least one ofits instances are true. Again, we’re assuming for now that there are only twoinstances. So we branch into them those two disjunctions at line 10. After that,we apply the approach to disjunctions from SL trees, and the tree closes.1.2.3.4.5.6.7.8.9.10.11.∀x(Fx ⊃ Gx) X¬∃xGx X∃x(Fx ∨Gx)Fa ⊃ Ga XFb ⊃ Gb X¬Ga¬Gb¬Fa¬FbFa ∨GaFa×8, 11Ga×6, 11Fb ∨GbFb×9, 11Gb×7, 11Gb×7, 9Ga×6, 8The closed tree indicates that our attempt to find a model satisfying the roothas failed. We’re not quite in a position to say that we’ve proven the root184 forallx (UBC edition)entirely unsatisfiable, because we’ve been assuming for simplicity that we’re onlyinterested in models that have two members of the UD. But we’ve shown at leastthat it’s impossible to satisfy the root with any model with a two-object domain.We’ll consider a more generalized procedure soon.Before moving on to the official tree rules, let’s look at one more example thatassumes a two-object domain.Is there any model with a two-object UD that satisfies {∀xFxa,∀y(Gy ⊃ ¬Fya)}?To find out, we put both sentences in the root of a tree, and continue as before.We start by taking both instances of each universal claim, then process the twoconditionals. This tree remains open.1.2.3.4.5.6.7.8.∀xFxa X∀y(Gy ⊃ ¬Fya) XFaaFbaGa ⊃ ¬Faa XGb ⊃ ¬Fba X¬Ga¬Gb↑¬Fba×4, 8¬Faa×3, 7As in SL trees, a completed QL tree with an open branch describes a way ofsatisfying the root. We’re assuming a two-object UD; for simplicity, let’s let ourobjects be the letters ‘a’ and ‘b’, and have ‘a’ be the referent of the QL name a,and ‘b’ be the referent of the QL name b. We look to the two QL atoms andtwo negated QL atoms in the branch; they indicate what needs to be includedin the extension of each predicate in our model. Since Faa is in the branch, weneed to include in the extension of F . Since ¬Ga is in the branch, weneed for a not to be in the extension of G.Here is a partial model that suffices to satisfy the root of the above tree:ch. 10 ql trees 185UD = {a, b}extension(G) =Ga 0b 0extension(F ) =Fxy a ba 1 -b 1 -Notice that we’ve put blanks for two of the cells in the extension of Fxy. That isto indicate that it doesn’t matter whether or not or are includedin the extension of Fxy. (This is also why neither Fab, Fbb, ¬Fab, nor ¬Fbbappeared in the open branch.) So the tree method for QL is similar to that forSL in this respect as well. An open branch in a completed tree describes a wayto satisfy the root.10.2 Generalizing the tree methodTrees are supposed to show something about all possible models, not only thosewith two-object UDs. So we don’t want our official tree rules to build in the ideathat there are just two objects. For a three-object UD, we’d want our universalstatements to develop into all three instances, and we’d want our existentials tobranch into the three instances. For a ten-object UD like the one described onpage 165, we’d need to give ten instances. For an infinite UD like the set of allnatural numbers, we’d need an infinite number of developments. Obviously suchrules would be impractical, to say the least. Worse than that, we often won’tknow at the start of the process what size domain we should be thinking of. Sowe need a version of the tree rules that is flexible as to domain size.In order to state our rule formally, we need one new bit of formalism. For anyQL wff Φ, a constant c, and variable x , define Φ x ⇒ c to mean the wff thatwe get by replacing every occurrence of x in Φ with c. So for example,Fx x⇒ a = FaFor any quantified QL sentence ∀x Φ or ∃x Φ, we call Φ x ⇒ c a substitutioninstance of ∀xΦ and ∃xΦ, and c is called the instantiating constant.This, more formally, is the same notion we were working with in our discussionof trees for models with two-object UDs. For example:. Aa ⊃ Ba, Af ⊃ Bf , and Ak ⊃ Bk are all substitution instances of∀x(Ax ⊃ Bx); the instantiating constants are a, f , and k, respectively.186 forallx (UBC edition). Raj, Rdj, and Rjj are substitution instances of ∃zRzj; the instantiatingconstants are a, d, and j, respectively.With this terminology in hand, we can begin with our official tree rules.10.3 ExistentialsThis will be our official tree rule for existentials:∃x Φ XaΦ x ⇒ a where a is newJust as Φ stands in for any sentence of QL, the x is meant to stand in for anyvariable, and a stands in for any name. The requirement that the name be newmeans that the name chosen for the substitution instance must be a name thathas not appeared anywhere in the tree so far.What is the rationale for this restriction? The existential claim says that somesubstitution instance is true, but it doesn’t say which one it is. If we knew howmany objects we were looking at in the UD, we could make a new branch foreach object — that’s what we did in the examples above — but here we see adifferent way to model this kind of flexibility. Remember, there is no prohibitionin QL of two different names referring to the very same object. So by using aninstance involving a new name, we are remaining open-minded about whether itis also a name for some object we’ve already been discussing.For example, consider this tree:1.2.3.Fa∃x¬Fx Xb¬Fb 2 ∃The appropriate substitution instance of the existential on line 2 is Fb, not Fa,because a is a name that is already in use. We want to say that somethingis F , not necessarily that a is. Had we taken the a instance, the tree wouldhave closed, even though the root is satisfiable. Note also that in marking theexistential as having been processed, we also make a note of the instantiatingconstant. That’s why there is a b next to the check mark in line 2.ch. 10 ql trees 18710.4 UniversalsHere is the tree rule for universals.∀x Φ \aΦ x ⇒ a for any aThe universal rule, like the existential rule, invites you to take an instance. Butunlike the existential rule, we don’t limit ourselves to instances with new names.The other major difference is that one can use this rule multiple times. Thatmakes this rule different from all the other tree rules we’ve seen so far. You’renot necessarily finished processing a universal sentence after you’ve performedthe rule once. We mark this by indicating a \ instead of a Xnext to the universalsentence.Here is an example illustrating the importance of taking multiple instances. Hereis a partially completed tree.1.2.3.4.5.6.∀x(Fx&Gx) \a¬Fa ∨ ¬Gb XFa&Ga XFaGa¬Fa×¬Gb1 ∀3 &2 ∨The root of this tree is pretty obviously unsatisfiable, but so far, the tree hasn’tclosed. On line 3 we take the a instance of the universal at line 1, but the treewon’t close until we take the b instance as well. We can do so at any time. Inthis way of developing the tree, we take that instance at line 7, and then thetree will close:188 forallx (UBC edition)1.2.3.4.5.6.7.8.9.∀x(Fx&Gx) \a, b¬Fa ∨ ¬Gb XFa&Ga XFaGa¬Fa×¬GbFb&Gb XFbGb×1 ∀3 &2 ∨1 ∀7 &Of course, we could also have taken the b instance at the same time as we tookthe a instance on line 3, and the tree would have worked just as well. But takingthings in this order illustrates that sometimes we can go back to universal claimsthat have already been considered, to take an additional instance. (Later on, wewill see some examples of trees that require this kind of backtracking.)10.5 Negated existentialsNegated existential claims are quite similar to universals. They say that noinstance is true, which is another way of saying that every instance is false. Solike the universal rule, the negated existential rule will allow us to take as manyinstances as we like; this time, we take the instances of the negation of the wffin the scope of the quantifier.¬∃x Φ \a¬Φ x ⇒ a for any aFor example, if ¬∃xFx is in the tree, one can use the negated existential rule todevelop a tree branch with ¬Fa, ¬Fb, etc.ch. 10 ql trees 18910.6 Negated universalsConversely, a negated universal is similar to an existential. It says that not allinstances are true, so at least one instance is false. So like the existential rule,we only perform it once (marking it with a check), and ensuring that we use anew name, so as not to assume anything in particular about the instance chosen.¬∀x Φ Xa¬Φ x ⇒ a where a is new10.7 Whither branching?When we learned the tree rules for sentential connectives, there was an importantdistinction between linear rules and branching rules. We retained this whenwe assumed two-object domains in §10.1, where universal quantifiers had linearrules, and existentials had branching rules. Our new quantifier rules, however,all have the same linear shape. Nothing branches. Why not?In SL, and for QL models where finite domains were specified in advance, weused branching to represent the particular possible ways one could satisfy theformula above. For some interpretation I to satisfy P ∨Q, either I (P ) = 1 orI (Q) = 1. If we assume there are only two objects, a and b, then in order forI to satisfy ∃xFx, I must either have a in the extension of F or have b in theextension of F . So we can branch the tree to represent those two possibilities.But if we do not assume anything in particular about the UD, we can’t justlist all the possible ways to satisfy an existential (or a negated universal). We’dneed as many branches as there are objects in the UD. So instead we just takeone instance, but give it a new name, which may or may not be co-referentialwith another name we’ve already considered. The novelty of the name is playingthe functional role of branching.10.8 The other development rulesSo far this chapter we’ve introduced four tree development rules for QL: we haverules for universals, existentials, negated universals, and negated existentials.190 forallx (UBC edition)There are nine more resolution rules, but they are all familiar: they are the samenine resolution rules given in Chapter 5. We’ll use all the same resolution rules forconjunction, negated conjunction, disjunction, negated disjunction, conditional,negated conditional, biconditional, negated biconditional, and double negationas we did before. They were introduced on pp. 77–81.10.9 Branch closure rulesOur branch closure rules also remain unchanged. A branch closes if and only if itcontains some sentence Φ along with its negation, ¬Φ. We mark closed brancheswith an ‘×’.10.10 Tree completion rulesThe completion rules, however, require a modification. The SL tree completionrule, given in §5.6, said that a branch was complete if every sentence, other thanatoms and negated atoms, has been resolved. In the QL tree system we need forour tree completion rules to take account of the fact that some sentences needto be acted on multiple times. The universal rule, for example, lets one takewhatever instance one likes; we need our completion rules to ensure that we’vetaken enough instances.For example, this should not count as a completed tree:1.2.3.4.5.∀x(Fx ⊃ Ga) \a¬GaFbFa ⊃ Ga X¬Fa Ga×Our universal rule can be processed multiple times. This tree has only taken thea instance of (Fx ⊃ Ga); if it also took the b instance, the tree would close:ch. 10 ql trees 1911.2.3.4.5.6.7.∀x(Fx ⊃ Ga) \a, b¬GaFbFa ⊃ Ga X¬FaFb ⊃ Ga X¬Fb×Ga×Ga×We don’t want this tree to count as complete until it’s taken both the a and theb instance. The universal says that every instance must be true; we don’t ensurethat this is satisfied if only some instances have been taken.If we knew how many names our model used, we could simply require thatevery instance be taken. But we don’t know in advance how many names ourmodels are going to use. QL trees construct their interpretations as they go.(Remember that resolving existentials and negated universals involves introducingnew names.) Here is the requirement we want: we require that we’ve taken theinstance corresponding to every name in the branch. If, after taking an instanceor two, we go on to introduce a new name within the same branch, that branchisn’t complete until we’ve taken that instance as well. Once again, this is whywe use the \ instead of a X. We can’t be sure that we’re finished with that lineuntil we’ve done the rest of the tree.A wff is resolvable if it is the kind of wff that gets a check mark upon processingit. (That’s everything except for atoms, negated atoms, universals, and negatedexistentials.) A wff is general if it is a universal or a negated existential. Hereis our new tree completion rule:A branch is complete if and only if either (i) it is closed, or (ii)every resolvable sentence in the branch has been resolved, andfor every general sentence and every name a in the branch, the ainstance of that general sentence has been taken.A tree is complete if and only if every branch in that tree iscomplete.192 forallx (UBC edition)10.11 Resolution orderAs in the SL tree system, the rules do not indicate any particular resolutionorder. But — also as in the SL system — some strategies are going to be moreefficient than others. In fact, given the increased complexity of our rules and thefact that some resolutions introduce new names, it is even more important, froman efficiency perspective, than it used to be to employ sensible strategies. It isstill a good idea to do linear rules before branching rules, and when possible, toperform a branching rule when you can see that one branch will close right away.Now, we can use similar strategies to take instances from general sentences in astrategic way. If you’re not sure which instance to take first, look a step aheadand see if one of the instances will close a branch.Here is a simple tree illustrating the point. Let’s consider whether Gb ⊃ ∀x¬Fxand ∃xFx entail ¬Gb. We begin by putting the first two sentences and thenegation of the third in the root of a tree:1.2.3.Gb ⊃ ∀x¬Fx∃xFx¬¬GbOrdinarily, it is efficient not to branch too early. But in this case, if we lookahead a step, we can see that resolving the conditional on line 1 will close onebranch immediately. So that’s a good place to start.1.2.3.4.Gb ⊃ ∀x¬Fx X∃xFx¬¬Gb¬Gb×3, 4∀x¬Fx 1 ⊃From here we have three choices to consider: lines 2, 3, and 4. Line 3 prettyobviously won’t get us anywhere interesting, so let’s consider either the existentialat line 2 or the universal at line 4. It’s usually a good rule of thumb to dothe new instances first — that way we’ll have more information when we makedecisions about which instances to take for the general rules. So let’s take the ainstance of line 2.ch. 10 ql trees 1931.2.3.4.5.Gb ⊃ ∀x¬Fx X∃xFx Xa¬¬Gb¬Gb×3, 4∀x¬FxFa1 ⊃2 ∃Now we turn our attention to the universal at line 4. We can take whateverinstance we want. Let’s be intelligent about it. The b instance wouldn’t doanything interesting; Fb won’t interact with that ¬¬Gb in any helpful way. Butthe a instance will close the tree. So we take that one.1.2.3.4.5.6.Gb ⊃ ∀x¬Fx X∃xFx Xa¬¬Gb¬Gb×3, 4∀x¬Fx \aFa¬Fa×5, 61 ⊃2 ∃4 ∀Planning ahead can help make your trees more efficient. The general principleyou should keep in mind is, if you have a choice for what resolvable sentence toresolve, or what instance of a general sentence to take, make a choice that willallow you to close branches more quickly.10.12 Infinite treesIn the SL tree system, every development of the tree took us closer to a completedtree, because we checked sentences above as resolved, and wrote simpler sentencesin one or two branches below. SL trees are guaranteed to be completed in afinite number of steps — they will either close, or they will reach a point whereeverything resolvable has been resolved.Because of the way our QL system works, it is possible for trees to continueindefinitely without closing. General sentences require that instances be takenfor all names in the branch; but the rules for existentials and negated universalsrequire that new names be introduced. Often, we can introduce the new namesbefore dealing with the general instances, as in this completed tree:194 forallx (UBC edition)1.2.3.4.∃xFx Xa∀xGx \aFaGa↑1 ∃2 ∀In this tree, we introduce the name a for the existential, then take the instance forthe one and only name in the branch for the universal, and the tree is complete.But things won’t always be so simple. Consider a tree with this single sentencein its root: ∀x∃yRxy. In this case, we must begin by performing the universalrule, as our one and only sentence is a universal. This rule allows us to take aninstance with any name. Ordinarily, we use the names that already exist in thebranch, but since there are no names, we’ll just take the a instance from thestart, putting that instance — an existential — on line 2.To resolve that existential, we must use a new name. So on line 3, we take the binstance from line 2.1.2.3.∀x∃yRxy \a∃yRay XbRab1 ∀2 ∃This tree has not closed, but note that it is not complete. We haven’t takenthe b instance of the universal on line 1. When we do so, we will find ourselveswith another existential on line 4; it in turn requires an instance involving a newname, and the process will continue indefinitely.ch. 10 ql trees 1951.2.3.4.5.6.7.8.∀x∃yRxy \a, b, c∃yRay XbRab∃yRby XcRbc∃yRcy XdRcd...1 ∀2 ∃1 ∀4 ∃1 ∀5 ∃As the pattern makes clear, this tree will never close, but it will also never becomplete. It will just keep taking instances of an existential with new names, thentaking those instances of the universal, which in turn require a new existential,and so on. (Eventually we will run out of letters, but we have subscripts availabletoo, so we’ll never run out of names.) Because the tree will never close, its openbranch is describing a model that satisfies the root, but the model suggested isan infinite one. Line 3 tells us that should be in the extension of R; line5 says that should too. There is a clear pattern that tells us what R’sextension must be like:extension(R) = {,,,, . . .}Or we can represent it graphically, thus:extension(R) =Rxy a b c d . . .a - 1 - - -b - - 1 - -c - - - 1 -d - - - - 1... - - - - -The blanks indicate that it doesn’t matter whether those pairs fall under theextension of R or not.What this suggests is that if one had a UD comprising an infinite list of objects,with each object related via R to the next item in the list, this model wouldsatisfy the root of the tree. For example, suppose that our UD were the set of all196 forallx (UBC edition)natural numbers, and that we interpreted Rxy as x+ 1 = y. For every numberx, there is a number y that is equal to x+ 1.Note, however, that this is only one of many possible ways to construct a modelmeeting the constraint described. In fact, we don’t necessarily require that ourUD be infinite. Our tree ended up using an infinite number of names, but there’sno rule requiring that each name be assigned to a unique object.The extension of R requires that certain pairs be included. Note that it doesn’trequire that any pairs be excluded. There are no 0s in the chart above. Theblanks mean it doesn’t matter whether you include them. A model that includedevery ordered pair in the extension of R would be an example of the kind ofmodel indicated:Rxy a b c d . . .a 1 1 1 1 1b 1 1 1 1 1c 1 1 1 1 1d 1 1 1 1 1... 1 1 1 1 1Once we put it this way, we can see that the R predicate needn’t draw anydistinctions between the various objects in the domain. Each object could bethe same, as far as R is concerned. So we could even offer a simpler model, witha UD comprising a simple object that is R-related to itself.UD = {a}extension(R) = {}When a tree continues indefinitely, the root is satisfiable; you can describe amodel for it using an infinite domain, taking advantage of the pattern indicatedby the infinite branch; or you can often find a way to simplify it into a modelwith a finite UD.Let’s consider another example of an infinite tree. This time, we’ll use nameswith subscripts. Everything is straightforward up until line 6; at line 7, we needto go back and take a second instance of line 2, and the infinite cycle begins.ch. 10 ql trees 1971.2.3.4.5.6.7.8.9.10.11.12.13.14.15.∃x∀y∃z(¬Rxy&Rzx) Xa1∀y∃z(¬Ra1y&Rza1) \a1, a2, a3∃z(¬Ra1a1 &Rza1) Xa2¬Ra1a1 &Ra2a1 X¬Ra1a1Ra2a1∃z(¬Ra1a2 &Rza1) Xa3¬Ra1a2 &Ra3a1 X¬Ra1a2Ra3a1∃z(¬Ra1a3 &Rza1) Xa4¬Ra1a3 &Ra4a1 X¬Ra1a3Ra4a1...1 ∃2 ∀3 ∃4 &2 ∀8 ∃9 &2 ∀11 ∃9 &If we continued drawing the tree, we’d take the a4 instance of the universal atline 2 on line 15, then introduce a new name by resolving that universal. Thetree will continue forever, again with a clear pattern. Lines 5, 9, 13, etc. tellus that for every integer i, Ra1ai=0. Lines 6, 10, 14, etc. tell us that for everyi > 1, Raia1=1. Or in chart form:Rxy a1 a2 a3 a4 . . .a1 0 0 0 0 0a2 1 - - - -a3 1 - - - -a4 1 - - - -... 1 - - - -This chart describes the extension of R that, as part of a model with the infinite198 forallx (UBC edition)UD {a1, a2, a3, . . . }, would satisfy ∃x∀y∃z(¬Rxy&Rzx). As in the previousexample, we do not need an infinite model to satisfy this sentence; the extensionof R here can be collapsed into a finite one. In the previous example, ourextension required no distinctions between any of the objects; here, we do havesome required differences. Ra1a1 = 0, but Ra2a1 = 1. So our model is treatinga1 and a2 differently. But it requires no further distinctions than that. Everyobject after a2 can be the same as a2, as far as this model is concerned. So atwo-object model would also suffice to satisfy this sentence. Here is a graphicalrepresentation of the denotation for R:Rxy a1 a2a1 0 0a2 1 -Or, describing the model in question with sets and n-tuples:UD = {a1, a2}extension(R) = {}The reasoning involved in collapsing models with infinite domains into finiteones centrally involves reasoning about what distinctions are required: do weneed to suppose that the objects named are different objects, or can we treat itas a case of multiple names for the same thing. In Chapter 12, we’ll foregroundthis kind of question in more detail, and introduce a way to talk about it in QL.Note that, for purposes of answering questions about entailment, you needn’tgo through the reasoning that allows you to move from infinite models to finiteones that would serve just as well. The tree method, most fundamentally, isa question about whether there is any possible model of the root. If a treecontinues infinitely, then it is answering that question: yes, there is an infinitemodel for it. It’s not necessary, for the question of whether there is any modelsatisfying the root, to determine whether there is a finite model that does so.10.13 Common student errors10.13.1 Using old namesThe most frequent errors for trees in QL involve ignoring the restrictions onnames involved in taking instances. Particular sentences (existentials and negateduniversals) require a new name — if the name appears anywhere in the branchat the time the existential is taken (including below it), it is not suitable forsubstitution via these rules.For example, this tree is a mistake:ch. 10 ql trees 1991.2.3.4.∀x∃y¬Rxy \aRab∃y¬Ray Xb¬Rab×2, 41 ∀3 ∃Line 3 here is fine; but at line 4, this tree takes the b instance of the existentialon line 3, even though b appeared in line 2. The existential rule requires anew name. The correct version of this tree will remain open. (It will extendinfinitely.)10.13.2 Main ConnectivesAlways work on the main connectives. If you have a sentence like∀x∃y¬Ryx& (Fa ⊃ Fb)don’t use the universal rule. This sentence is a conjunction, so you must performthe conjunction rule on it. That will give you a universal, at which point youcan take an instance.Practice Exercises? Part A Use a tree to test whether the following sentences are tautologies. Ifthey are not tautologies, describe a model on which they are false.1. ∀x∀y(Gxy ⊃ ∃zGxz)2. ∀xFx ∨ ∀x(Fx ⊃ Gx)3. ∀x(Fx ⊃ (¬Fx ⊃ ∀yGy))4. ∃x(Fx ∨ ¬Fx)5. ∃xJx ≡ ¬∀x¬Jx6. ∀x(Fx ∨Gx) ⊃ (∀yFy ∨ ∃xGx)? Part B Use a tree to test whether the following arguments are valid. If theyare not, give a model as a counterexample.1. Fa, Ga, .˙. ∀x(Fx ⊃ Gx)2. Fa, Ga, .˙. ∃x(Fx&Gx)200 forallx (UBC edition)3. ∀x∃yLxy, .˙. ∃x∀yLxy4. ∃x(Fx&Gx), Fb ≡ Fa, Fc ⊃ Fa, .˙. Fa5. ∀x∃yGyx, .˙. ∀x∃y(Gxy ∨Gyx)Part C Translate each argument into QL, specifying a UD, then use a tree toevaluate for validity. If they are invalid, give a model as a counterexample.1. Every logic student is studying. Deborah is not studying. Therefore,Deborah is not a logic student.2. Kirk is a white male Captain. Therefore, some Captains are white.3. The Red Sox are going to win the game. Every team who wins the gamewill be celebrated. Therefore, the Red Sox will be celebrated.4. The Red Sox are going to win the game. Therefore, the Yankees are notgoing to win the game.5. All cats make Frank sneeze, unless they are hairless. Some hairless catsare cuddly. Therefore, some cuddly things make Frank sneeze.Chapter 11Soundness andCompleteness for QL TreesIn Chapter 6 we proved that our SL tree method works the way it is supposedto: any tree that closed was guaranteed to have a root that was unsatisfiablein SL, and any completed tree that remained open was guaranteed to describean SL model that satisfies the root. These two theorems together are calledsoundness and completeness.soundness: If a tree closes, that guarantees that its root is unsat-isfiable. In other words:X ` ⊥ ⇒ X |= ⊥completeness: If a root is unsatisfiable, that guarantees that thetree will close. In other words:X |= ⊥ ⇒ X ` ⊥These definitions were first given on pages 89 and 97. They are equally applicableto our extended tree system for QL. The system introduced in Chapter 10 is alsosound and complete. The rules are designed in a way so as to guarantee that ifthe tree closes, there is no model for the root (sound), and to guarantee that anopen branch describes a model for the root (complete). The proofs for these twotheorems are structurally very similar to the proofs given in Chapter 6.201202 forallx (UBC edition)11.1 SoundnessIf our tree method is sound, then there is no possible set of QL sentences X thatare satisfiable, where a tree with root X closes. Not every possible tree methodis sound. For example, suppose we dropped the requirement, for a negateduniversal, that one take an instance with a new name.In other words, suppose we replaced this rule (introduced on page 189)¬∀x Φ Xa¬Φ x ⇒ a where a is newwith this alternate one:¬∀x Φ Xa¬Φ x ⇒ a for any aIf we had this rule, there would be counterexamples to soundness — trees withsatisfiable roots, which nevertheless closed. Here is an example:1.2.3.Fa¬∀xFx¬Fa×1, 32 alt˙¬∀This tree is a counterexample to the soundness of the alternate tree system justdescribed. To prove that our system is sound is to prove that our actual rulesdo not allow for this kind of tree. To prove this, we’ll begin by assuming that Xis satisfiable, and demonstrate from that assumption that at least one branch ofany tree that develops according to our rules will remain open. As in the case ofthe parallel discussion in Chapter 6, our proof will be a recursive one. We willdemonstrate that, if the tree starts with a set of satisfiable sentences, then, foreach legal way the tree may develop, at least one branch will continue to have aset of satisfiable sentences. This is effectively to show that such a branch willnever close, since branches only close when they contain some sentences Φ and¬Φ, which are of course never jointly satisfiable.ch. 11 soundness and completeness for ql trees 203Suppose, then, we have some satisfiable set of sentences X in the root of a tree.If the root is satisfiable, then there is some model that satisfies it. Call thisinterpretation I. We prove that, if our tree follows the rules given in Chapter 10,then I doesn’t just satisfy the root — it satisfies every sentence in any completedbranch. We will prove this recursively.11.1.1 RootWe assume that the tree begins with a satisfiable root. Given this assumption,I is just our name for one of the interpretations we are assuming must exist. SoI trivially satisfies everything in the root.Now we must prove, for each possible way of developing the tree, that if thesentences in the branch we begin with are satisfiable, then the sentences we haveafter applying the rule are satisfiable too. There are thirteen possible ways atree can develop, corresponding to the thirteen kinds of non-atomic sentencesin QL, each of which has a particular processing rule. The thirteen kinds ofsentences are:. double negation. conjunction. negated conjunction. disjunction. negated disjunction. conditional. negated conditional. biconditional. negated biconditional. existential. negated existential. universal. negated universalFortunately, we’ve already proven what we need to prove for the first nine itemson the list. Our QL tree method uses all the same rules as the SL method did204 forallx (UBC edition)for the sentential connectives; we proved in §6.4.2, for each of those rules, that ithas the key property: if some interpretation I satisfies what comes above therule, then the development below it is also satisfiable. (Indeed, we proved therethat the very same interpretation, I, satisfied it.)So to extend our soundness proof to our QL system, we need only prove thesame thing for the four rules for quantifiers. This is the project of the next foursubsections.11.1.2 ExistentialsSuppose some satisfiable set of QL sentences X is developed according to theexistential rule:∃x Φ XaΦ x ⇒ a where a is newWe assume that I models X , which includes some existential ∃x Φ. We want toprove that there is a model for the expanded branch which comprises both X andΦ x ⇒ a , i.e., X ∪ Φ x ⇒ a (the set containing X and the new developmentsentence too). Unlike in the parallel proof for the sentential rules, we cannotassume that I itself satisfies our new development, because our new developmentintroduces a new name; we cannot assume that I included any assignment forthe new name a. But we can be assured that I can be expanded into a new,similar interpretation, I*, which does include a. Moreover, since we know that Isatisfied the existential ∃x Φ, we know that there was some object in I’s domainthat satisfied Φ. So it will be possible to construct our new interpretation I* sothat it includes the new name, and assigns it to that object. This will ensurethat I*(Φ x ⇒ a ) = 1. And since I* is just like I with respect to everythingother than a — and since we are assured that a was not in X (the rule requiresthat it be new) — I* will satisfy X in just the same way that I did.In other words, assuming that I satisfies every sentence in a branch before theexistential rule is performed, there is guaranteed to be a new interpretation, I*,which is an extension of I that includes a new name attached to an object inthe UD satisfying the existential, which satisfies everything in the branch upto the point after the existential rule is performed. That is to say, like the ninesentential rules considered in Chapter 6, the existential rule can never take youfrom a satisfiable branch to an unsatisfiable one.ch. 11 soundness and completeness for ql trees 20511.1.3 UniversalsSuppose a branch of a tree uses the rule for universals:∀x Φ \aΦ x ⇒ a for any aAssume that the set of QL sentences X above this development is satisfiable.Then some model I makes it true. The universal rule allows an instance to bedeveloped using any name, so, as before, we cannot guarantee that I makesthe development true, because I may or may not interpret the name a; butas before, we can be assured if it doesn’t, we can extend the interpretation toinclude it. So consider a new model I*, which includes the name a. If a wasn’tinterpreted by I, then it can be assigned to any element of the UD. Since therest of I is unchanged, and since I(∀x Φ) = 1, we know that our new extendedinterpretation will satisfy Φ x ⇒ a too.That is to say, once again, if we assume that I satisfies every sentence in abranch before the universal rule is performed, there is guaranteed to be a model— either the very same one, I, or a modification of it, I*, which assigns a newname to any object in the UD, which satisfies everything in the branch up to thepoint after the universal rule is performed. In other words, the universal rulecan never take you from a satisfiable branch to an unsatisfiable one.Note that the fact that we can perform this rule multiple times does not interferewith the soundness proof. We have proven that each time you perform it, youare guaranteed not to switch from a satisfiable branch to an unsatisfiable one.So no matter how many times you take an instance of a universal, you won’t beable to go from a satisfiable set of sentences to an unsatisfiable one.11.1.4 Negated ExistentialThe reasoning behind soundness for the negated existential rule is exactly parallelto that for the universal rule. We begin by assuming that some negated universal¬∃x Φ, is satisfied by I. Here is the rule for negated existentials:¬∃x Φ \a¬Φ x ⇒ a for any a206 forallx (UBC edition)We want to prove that the result of this rule is also satisfied, either by I itself (if awas interpreted in I), or by an extension of it I*, that preserves the satisfactionof everything above (if a was a new name). Since I satisfies ¬∃x Φ, it makesevery substitution instance for x of Φ false. If a was interpreted by I already,then I(Φ x ⇒ a ) = 0. If it wasn’t, the new model I* will assign the new nameto some object in the UD of the original model; since no object in that modelsatisfied Φ, I*(Φ x ⇒ a ) = 0. Either way, our interpretation falsifies Φ x ⇒ a ,and so satisfies that sentence’s negation, which is the continuation of the branch.So this rule too can never take us from a satisfiable set of QL sentences to anunsatisfiable one.11.1.5 Negated UniversalNegated universals are similar to existentials. Assume that a negated universalis part of a set of sentences satisfied by I, and that this rule is then applied:¬∀x Φ Xa¬Φ x ⇒ a where a is newConstruct a new interpretation I*, which differs from I only in that it includesan interpretation of the new name a, and assigns that name to some object thatfalsifies Φ. We know there is at least one such object because we are assumingthat I satisfies the negated universal. Then our new interpretation I* satisfiesthe new development of the branch. It also satisfies everything above the branch,just like I did, because nothing above the branch included the name a.That last bit of reasoning relied centrally on the requirement that we’re taking anew name. We saw in the introduction to this chapter that if we do not includethat requirement, soundness would be violated.11.1.6 Summarizing the soundness proofWe have now shown, for our four quantifier rules, that each of them has thefollowing property: you can never start with a branch that is satisfiable, and usethat rule to extend that branch into one that is unsatisfiable. Since we’ve alsoshown that the nine sentential rules also have this property, we’ve effectivelyshown that there is no possible way to start with a satisfiable set of sentencesand develop the branch into one that is not satisfiable. This in turn means thatif the branch starts with a satisfiable set of sentences, the branch will neverch. 11 soundness and completeness for ql trees 207close. But that’s just what soundness says: if the root is satisfiable, the tree isguaranteed to remain open. Soundness is proven.11.2 CompletenessCompleteness says that if a branch of a completed tree remains open, then theroot is satisfiable. We prove this by assuming that we have an open completedbranch, and use it to construct an interpretation that satisfies every sentence inthat branch, which includes the root. The proof for completeness of our QL treesystem is structurally just like the one given in Chapter 6.Given a completed open branch, we construct a model I, based on that branch,as follows: for any predicate F , if some QL atom Fa1, . . . , an is in the branch— i.e., if P or Fa or Rab is in the branch — then I makes that atom true,by putting in the extension of F . And if ¬Fa1, . . . , an is in thebranch, I excludes from the extension of F , thus making thenegation of the atom true. This is of course just the way that we constructinterpretations from open branches of completed trees.Now we will prove, for every sentence in QL, that if it is in the open branch, itis made true by I. The QL atoms trivially meet this criterion — I was designedprecisely to satisfy them. We will prove by induction that every possible QLsentence also meets this criterion. In §6.6 we showed, for each propositionalconnective, if you construct a more complex SL sentence out of simpler SLsentences that have this criterion, the more complex one will too. That proofcarries on unchanged here. So it remains only to show that the same is true ofour four quantifier rules.11.2.1 ExistentialConsider an existential — a QL sentence of the form ∃x Φ. We need to provethat if it is in the open, completed branch, I satisfies it. Since the branch iscomplete, we know that the existential rule has been performed to resolve thissentence. So the branch includes a substitution instance of Φ that used a newname. For our present purposes, it doesn’t actually matter whether the namewas new — the fact that there is some instance of Φ in the branch already isenough to prove what we need to prove. Since there is an instance of Φ in thebranch, if it is satisfied by I, the existential ∃x Φ must be satisfied by I too.So, just as we showed for the nine sentential rules, the existential rule hasthis important property: in a completed tree, any interpretation that satisfiesthe simpler sentences below the existential development, must also satisfy the208 forallx (UBC edition)existential above it.11.2.2 UniversalSuppose a universal sentence ∀x Φ appears in a completed open branch. Since thebranch is complete, that means that, for every name a in the branch, Φ x ⇒ ais also in the branch. We therefore assume that I satisfies each Φ x ⇒ a ; so Imust also satisfy ∀x Φ. Because the UD for I includes only those names thatoccur in the branch, every instance of Φ is included, so the universal is true.Once again, any interpretation that satisfies everything below the universaldevelopment, must also satisfy the universal above it.11.2.3 Negated ExistentialNegated existentials work just like universals. If ¬∃x Φ is in a completed openbranch, then for every name a in I, ¬Φ x ⇒ a is below it in the branch. Andif I satisfies each of these negations, it will also satisfy the negated existential.11.2.4 Negated UniversalNegated universals work just like existentials. If ¬∀x Φ is in the branch, thensome instance of the negation ¬Φ is in the branch below. If I satisfies someinstance of ¬Φ, then, given the definition of truth for negation and universals inQL, it will also satisfy ¬∀x Φ.11.2.5 Summarizing the completeness proofThe sentence shapes just considered, combined with the nine shapes consideredin §6.6, correspond to all the possible QL sentences. So we have proven that, forany possible QL sentence Φ, if an interpretation satisfies the simpler sentencesbelow it in the branch, that interpretation also satisfies Φ itself. Since we alsohave a recipe for constructing an interpretation I that is guaranteed to satisfythe atoms, we can prove by induction that it can satisfy everything in thebranch, including the root. A completed open branch guarantees a satisfiableroot. Completeness is proven.ch. 11 soundness and completeness for ql trees 209Practice Exercises? Part A Following are possible modifications to our QL tree system. For each,imagine a system that is like the system laid out in this chapter, except for theindicated change. Would the modified tree system be sound? If so, explain howthe proof given in this chapter would extend to a system with this rule; if not,give a tree that is a counterexample to the soundness of the modified system.1. Change the rule for existentials to this rule:∃x Φ XaΦ x ⇒ a for any a2. Change the rule for existentials to this rule:∃x Φ XdΦ x ⇒ d (whether or not d is new)3. Change the rule for existentials to this rule:∃x Φ XΦ x ⇒ aΦ x ⇒ bΦ x ⇒ cfor 3 different names, old or new4. Change the rule for universals to this rule:∀x Φ XΦ x ⇒ aΦ x ⇒ bΦ x ⇒ cfor 3 different names, old or new210 forallx (UBC edition)5. Change the rule for existentials to this rule:∃x Φ XΦ x ⇒ aΦ x ⇒ bΦ x ⇒ cfor 3 new names6. Change the rule for universals to this rule:∀x Φ XaΦ x ⇒ a where a is new7. Change the rule for conjunction to this rule:Φ&Ψ X∃x ΦΨwhere x does not occur in Φ8. Change this requirement (given on page 191)...A branch is complete if and only if either (i) it is closed, or(ii) every resolvable sentence in every branch has been resolved,and for every general sentence and every name a in the branch,the a instance of that general sentence has been taken....to this one:A branch is complete if and only if either (i) it is closed, or(ii) every resolvable sentence in every branch has been resolved,and for every general sentence, at least one instance of thatgeneral sentence has been taken.9. Change the branch completion requirement to:. . . and for every general sentence and every name a that isabove that general sentence in the branch, the a instance ofthat general sentence has been taken.10. Change the branch completion requirement to:ch. 11 soundness and completeness for ql trees 211. . . and for every general sentence and every name a in thebranch, the a instance of that general sentence has been taken,and at least one additional new instance of that general sentencehas also been taken.? Part B For each of the rule modifications given in Part A, would the modifiedtree system be complete? If so, explain how the proof given in this chapter wouldextend to a system with this rule; if not, give a tree that is a counterexample tothe completeness of the modified system.Chapter 12IdentityThis chapter extends QL by adding logical resources for discussion of identity. Inthe logical sense, identity is about which object something is. Two things aren’t‘identical’ in our sense just because they look exactly alike. (Identical twinsaren’t logically identical.) Neither does identity, in our logical sense, have to dowith the way an individual conceives of oneself. (We’re not getting into whetherone identifies as a woman, etc.) Instead, we will develop logical vocabulary totalk about whether, e.g., the person who robbed the bank is the same person asthe person who shot Uncle Ben.It is already possible to talk about identity in QL, by assigning identity andrelated concepts to predicates in interpretation keys. (In this sense, it’s possibleto talk about pretty much anything in QL.) However, there are a few reasonswhy we may wish to have a more robust logical notion of identity than that.We begin by examining some of these motivations. Identity will strengthen QL,increasing its expressive power in many ways.12.1 Motivating identity as a logical categoryConsider this sentence:1. Only you can prevent forest fires.How should we translate it into QL? Here is one simple option. We could usethis symbolization key:212ch. 12 identity 213UD: peoplePx: only x can prevent forest firesu: youAnd then we could translate sentence 1 straightforwardly as Pu. This might beadequate for some purposes, but this translation misses out on a key fact aboutsentence 1. Notice that sentence 1 is inconsistent with sentence 2:2. Only your mother can prevent forest fires.If we add a name for your mother to the symbolization key above,m: your motherwe can translate sentence 2 as Pm. The problem is that Pu and Pm areobviously consistent in QL, as this simple model demonstrates:UD = {u, m}extension(P ) = {u, m}The ‘only’ in each sentence has just been made a part of the predicate, in away that ignores its logical force. (Compare the analogous discussion in theIntroduction to Chapter 8 that motivated extending SL with predicates andquantifiers.) If only you can prevent forest fires, that means that no one else —i.e., no one who is not you — can prevent forest fires. We don’t yet have thelogical vocabulary to express these thoughts in a strong enough way.These sentences give rise to a related issue:3. Peter Parker lives in Queens.4. Peter Parker is Spider-Man.5. Spider-Man lives in Queens.Here is a natural interpretation key for these sentences, given the resources we’veseen so far.UD: people and placesLxy: x lives in yIxy: x is yp: Peter Parkers: Spider-Manq: Queens214 forallx (UBC edition)Then sentences 3–5 can be translated thus:3. Lpq4. Ips5. LsqBut the argument from the English 3 and 4 to 5 seems valid, whereas thecorresponding argument form is obviously invalid in QL. This model satisfiesthe premises and falsifies the conclusion:UD = {p, s, q}extension(L) =Lxy p s qp - - 1s - - 0q - - -extension(I) =Ixy p s qp - 1 -s - - -q - - -Given the interpretation key above, this is a model where Peter is Spider-Man,but where Peter and Spider-Man do not live in the same place. This doesnot correspond to a genuine possibility; if Peter and Spider-Man are the sameperson, then they must live in the same place. This is an important logical factabout identity. But in the formalism given, I is just another predicate like L,susceptible to many different interpretations and extensions.Here is a third example.6. Mozart composed Le Nozze di Figaro.7. Mozart composed Don Giovanni.8. Mozart composed at least two things.We introduce the obvious interpretation key:UD: people and operasCxy: x composed ym: Mozartn: Le Nozze di Figarog: Don GiovanniThen we might translate the argument thus:ch. 12 identity 215CmnCmg.˙. ∃xCmx& ∃yCmyThis is a valid argument form. But the problem is that ∃xCmx& ∃yCmy is avery bad translation of ‘Mozart composed at least two things’. Notice that it isthe conjunction of two existentials, each of which says that Mozart composed atleast one thing. Suppose he composed only one thing; then each conjunct is true,so the conjunction is true. To say that he composed at least two things, it’s notenough to say that he composed at least one thing, and then to say again thathe composed at least one thing — we need some way to ensure that the x and yare not the same thing.What all three of these cases — the forest fire case, the Spider-Man case, andthe Mozart case — show us is that if we want to reflect the logical relationshipshaving to do with identity, we need special logical vocabulary to do so. Just aswe introduced the ‘∀’ and ‘∃’ formalisms to reflect logical facts about quantifiers,we also need a special symbol for identity. That symbol will be ‘=’. Like ourother logical symbols — ‘⊃’, ‘¬’, ‘∀’, etc. — its meaning will be fixed as part ofour language. Interpretation keys tell us what predicates like Fx mean; theydon’t get to specify the meaning of the logical terms.When we introduced quantifiers and predicates, we described the shift as switch-ing to a new language, from SL to QL. In fact, however, QL isn’t so much achange away from SL as it is an addition to it. (All SL sentences are QL sentences,and all SL entailments are also QL entailments.) Here we will make one moreaddition, corresponding to identity. We won’t introduce a whole new name forour extended version of QL; it can be specified descriptively as ‘quantified logicwith identity’. (The choice about whether to introduce a new language name isa purely conventional one; we’re here just following ordinary practice.)12.2 =We revise the definitions of grammaticality and truth in QL to add the followingstipulation:For any two names a and b , ‘a=b ’ is an atomic sentence in QL. It is true in amodel just in case, in that model, the referent of a is the same object as thereferent of b ; i.e., the model assigns a and b to the same object. For example,here is a model:216 forallx (UBC edition)UD = {Peter Parker, Aunt May}referent(p) = Peter Parkerreferent(s) = Peter Parkerreferent(m) = Aunt MayIn this model, p=s is true, and p=m and s=m are false. We’ll introduce a familiarshorthand for negating identity claims; s6=m is our preferred abbreviation for¬s=m. (Remember that ‘s=m’ is an atomic sentence, so ‘¬s=m’ is the negationof that sentence. It is tempting, when one looks at that sentence, to attach thenegation only to s, so that the sentence says that the ‘negation of s’ is identicalto m. But this way of reading it ultimately makes no sense; names do not havenegations in QL. Only sentences do. This is part of the reason we will usuallyprefer to write s6=m, rather than the potentially confusing ¬s=m. But if you dosee or write the latter, remember that the negation applies to the whole identityclaim.)A logical notion of identity is just what is needed to overcome the shortcomingsof the translations in the introduction to this chapter.12.3 Identity and ‘no one else’Consider again sentence 1:1. Only you can prevent forest fires.In the introduction we tried translating this into QL with a predicate ‘onlycan prevent forest fires’; now we can reflect its logical structure more fully.So we’ll let our predicate just indicate the ability to prevent forest fires:UD: peoplePx: x can prevent forest firesu: youIf only you can prevent forest fires, that means that you can prevent forest fires,and, moreover, no one else — no one who is not you — can prevent forest fires.The first part is simple. You can prevent forest fires: Pu. Now we must say thatno one else can prevent forest fires. Another way to say that is this: it is notthe case that there is someone who is not you who can prevent forest fires. Soyou can think of that as a negated existential: ¬∃x(x 6=u&Px). Putting thesetwo pieces together as a conjunction, we can translate sentence 1 as:1. Pu&¬∃x(x 6=u&Px)ch. 12 identity 217A second way of characterizing the ‘no one else’ clause would translate it as auniversal claim, saying, of everything, that if it is not you, then it cannot preventforest fires: ∀x(x 6=u ⊃ ¬Px). This is equivalent to ¬∃x(x 6=u&Px); either is anacceptable translation. (By the end of this chapter we’ll be able to prove thisequivalence.)Using similar reasoning, one can translate other kinds of sentences that talkabout objects other than ones just mentioned. For example, suppose we want tosay that Rebecca only thinks about herself. She thinks about herself, and thereis no one else that she thinks about. Using r for Rebecca and Txy for ‘x thinksabout y’, we could say:Trr&¬∃x(x 6=r&Trx)There are more examples of similar kinds of translation exercises in the practiceproblems at the end of this chapter.12.4 Identical objects satisfy identical predicatesIn §12.1, we saw that treating identity as a simple predicate prevented thevalidity of intuitively valid arguments like this one:3 Peter Parker lives in Queens.4 Peter Parker is Spider-Man.5 Spider-Man lives in Queens.The problem was that if the extension of the identity predicate can simply bepresented in a model, without regard for the other features of the model, therewas no guarantee that p and s — the names for Peter Parker and Spider-Man —were related in any particular way, just because happened to be assigned in the extension of the I predicate.But if we translate the identity predicate with the QL identity claim p=s, thisrequires that the names s and p be assigned to the very same object in themodel. Consequently, for any sentence Φ involving the name p, Φ p⇒ s — thesentence that results from replacing each p in Φ with an s — will have the sametruth value as Φ. If Fp is true, then the object named by p is in the extension ofF . And if p=s, then the object named by p is the very same object as the objectnamed by s. So of course the object named by s (that same object) must alsobe in the extension of F . In general, identical objects have identical properties.(The converse is not true in general. There is no rule in QL that says that if allthe same wffs are true of two different names, they must be two names for thesame object.)218 forallx (UBC edition)If Peter is Spider-Man, then everything true of Peter is also true of Spider-Man.So this argument form, unlike the first translation we attempted, is valid in QL:Lpqp=s.˙. LsqIn §12.7 below we’ll extend our tree method to claims with identity, which willgive us a formal means for proving the validity of arguments like this one. (Thenatural deduction system for QL given in Chapter 13 will give us another.)12.5 QuantityIdentity also lets us talk about quantity in a way that wasn’t available before.In §12.1 we observed some challenges with translating claims about the numberof objects that satisfy a particular description. For example, we struggled withsentences like ‘Mozart composed more than one opera’. But identity makes suchclaims straightforward.UD: people and operasCxy: x composed yOx: x is an operam: Mozartn: Le Nozze di Figarog: Don GiovanniWe want to say that there are two things, that have the following features: theyare different from each other, they are both operas, and they are both composedby Mozart. A double-existential sentence, listing those conjuncts, will suffice:∃x∃y(x6=y&Ox&Oy&Cmx&Cmy)Strictly speaking, QL conjunctions only have two conjuncts, so to write thissentence scrupulously, we’d need to include brackets to signify that we’re talkingabout conjunctions of conjunctions:∃x∃y(x 6=y& [(Ox&Oy) & (Cmx&Cmy)])In sentences like these — and the longer ones listing more conditions that willfeature soon — it is usually preferable for readability to leave these internalbrackets out. We will write as if our system allows arbitrary numbers of conjuncts,with the understanding that, should the need arise, we could always put themback in explicitly. (We’ll rarely have occasion to do so.)ch. 12 identity 219This sentence says that Mozart wrote at least two operas. To say that he wroteat least three operas would require another quantifier with a new variable z, andthe specification that z be an opera, written by Mozart, and non-identical toboth of the ones previously mentioned:∃x∃y∃z(x 6=y&x 6=z& y 6=z&Ox&Oy&Oz&Cmx&Cmy&Cmz)As you can see, these sentences get long and complicated rather quickly.Suppose we wanted to say that Mozart composed exactly two operas — no more,and no fewer. There are two reasonably natural choices for how to say that. Oneis to conjoin the first QL sentence of this section — Mozart composed at leasttwo operas — with the negation of the second — Mozart did not compose atleast three operas. The result is:∃x∃y(x 6=y&Ox&Oy&Cmx&Cmy) &¬∃x∃y∃z(x 6=y&x 6=z& y 6=z&Ox&Oy&Oz&Cmx&Cmy&Cmz)A slightly more efficient way of saying the same thing, rather than conjoining anew negated triple existential, would be to add, within the scope of the first twoquantifiers, a conjunct to the effect that every opera composed by Mozart is oneof the two mentioned in that first conjunction. To say that any opera composedby Mozart is either x or y is to say that, for any z, if z is an opera composed byMozart, then it is either x or y. In other words: ∀z[(Oz&Cmz) ⊃ (z=x∨ z=y)].So another way to say that Mozart composed exactly two operas is:∃x∃y((x 6=y&Ox&Oy&Cmx&Cmy) &∀z[(Oz&Cmz) ⊃ (z=x ∨ z=y)])So identity gives us a way to talk about quantities, albeit not a particularlyefficient one.12.6 Definite descriptionsIn 1905, Bertrand Russell famously characterized definite descriptions in termsof identity. A definite description is a description that implies that only oneobject satisfies it. Paradigmatically, definite descriptions are ones that use thedefinite article ‘the’. If I say ‘the baby is hungry’, I’m saying that the one andonly baby we might be talking about is hungry. Russell was motivated in part bythe apparent fact that one can use this sort of language in a meaningful wayeven if one is wrong about whether there’s any baby around. If there is no baby— the crying I’m hearing is a recording — my statement is false, but it’s still220 forallx (UBC edition)meaningful. For this reason, Russell was reluctant to suppose that we shouldunderstand ‘the baby’ as a name. Remember, in QL, all names have to refer toobjects in the UD. Instead, my sentence can be understood as an existentiallyquantified claim about a unique baby. If I say ‘the baby is hungry’, according toRussell, I’m in effect saying three things: there is a baby, there’s no other babythan that one, and that baby is hungry. That second element, the uniquenessclaim, can be expressed in QL with identity.UD: People in this houseBx: x is a baby.Hx: x is hungry.j: JonathanA sentence like ‘Jonathan is hungry’ is straightforwardly translated as Hj.According to Russell’s theory of definite descriptions, ‘the baby is hungry’ has amuch more complex logical form: there is a baby x, there is no baby other thanx, and x is hungry.∃x[Bx&¬∃y(By& y 6=x) &Hx]One of the interesting features of Russell’s theory is that ‘the baby is not hungry’is not the negation of ‘the baby is hungry’. Instead, the negation applies only tothe hunger predication in the last conjunct:∃x[Bx&¬∃y(By& y 6=x) &¬Hx]The reason Russell designed his theory this way was that he thought that bothof these sentences equally implied that there is a baby. If there is no baby,then you’d be mistaken in saying either ‘the baby is hungry’ or ‘the baby is nothungry’. Consequently, one can’t be the negation of the other. As a treatmentof the truth conditions of English sentences, Russell’s theory is controversial.Some philosophers of language think that sentences that seem to presuppose theexistence of something that isn’t there aren’t straightforwardly false, but arerather defective in some other way — perhaps they fail to be meaningful at all,or perhaps they take on some truth value other than true or false. These mattersare beyond our present scope. We will remain neutral on whether Russell’s theoryis an accurate treatment of English; it is relevant for us because it provides aninteresting and useful case for translation of sentences into QL with identity.The general statement of Russell’s theory of definite descriptions is, if you havea sentence applying a predicate P to the referent of a definite description D,translate it thus: there is something x that is D, there is nothing other than xthat is D, and x is P .ch. 12 identity 22112.7 Identity and treesBecause we have extended QL to include claims of identity, we also need tomodify our tree system. Notice that if we do not do so, the system will not becomplete. For example, the rules offered in Chapter 11 do not provide a way todemonstrate that Fa,¬Fb, a=b |= ⊥, even though those sentences are obviouslyinconsistent. Since each of those sentences are atoms or negated atoms in QL,there is nothing to resolve, and there is no contradiction to close the tree. Weneed to add new rules to deal with identity.Here is an obvious one:If a branch contains an identity claim a=b and some sentence Φ,then you may add Φ a ⇒ b or Φ b ⇒ a to the branch.Recall that ‘Φ a ⇒ b ’ just means the sentence you get by replacing every ‘a’ inΦ with ‘b ’. With this rule, we can easily demonstrate that the set of sentencesmentioned above closes:{Fa,¬Fb, a=b} ` ⊥1.2.3.4.Fa¬Fba=bFb×2, 41, 3 =We will add this rule to our tree system. Notice that it is not a resolution rule —we do not put a check mark next to either sentence referenced. They may beused again.Since our tree system is supposed to be complete, we need to ensure that thetree will close; it’s not enough that we have a rule that allows that it might. So,as in the case of the general rules discussed in Chapter 10, we will also add anadditional requirement governing branch completion. What we need is to makesure that we’ve made enough identity substitutions to demonstrate everythingthat needs demonstrating. The way we’ll do this is by requiring, for everyidentity claim in the branch, that, for at least one of its directions (left-to-rightor right-to-left) the substitution has been performed in that direction on everyatomic sentence or negated atomic sentence in the branch.More precisely stated, an open branch is complete only if, for any identity claim222 forallx (UBC edition)a=b in the branch, either, for any atomic sentence or negated atomic sentenceΦ containing a in the branch, Φ a ⇒ b is in the branch, or, for any atomicsentence or negated atomic sentence Φ containing b in the branch, Φ b ⇒ a isin the branch.So, adding this requirement to the definition of branch completion given on page191, we get:A branch is complete if and only if either (i) it is closed, or (ii)all of the following conditions are met:1. Every resolvable sentence in the branch has been resolved;2. For every general quantified sentence Φ and every name a inthe branch, the a instance of Φ is in the branch; and3. For every identity claim a=b in the branch, either:(a) for every atomic sentence or negated atomic sentence Φcontaining a in the branch, Φ a ⇒ b is in the branch, or(b) for every atomic sentence or negated atomic sentence Φcontaining b in the branch, Φ b ⇒ a is in the branch.We must also add one more rule to our tree system. Previously we said thatbranches only close when they contain some sentence along with its negation.We add one more condition that suffices for branch closure. If any branchcontains a sentence a 6=a, then that branch closes. In no model can somethingbe non-identical with itself.Adding this to the rule given on page 190:A branch closes if and only if either (i) it contains some sentenceΦ along with its negation, ¬Φ, or (ii) it contains, for some name a,the sentence a 6=a.We need this rule, for example, to prove with a tree that ∀x a=x is inconsistentwith ∃x x 6=a:ch. 12 identity 223{∀x a=x, ∃x x 6=a} ` ⊥1.2.3.4.5.∀x a=x \b∃x x 6=a Xbb 6=aa=bb6=b×52 ∃1 ∀3, 4 =We must also take care, when constructing models from completed open branches,to respect identity claims. If for example the sentence b=c appears in an openbranch, we will not construct a model with two different objects, b and c —instead, we’ll just pick one of them as the object indicated, and assign bothnames to it. For example, consider this tree:1.2.3.4.5.6.7.8.9.10.11.12.∀x(Rax ⊃ x=b) \a, b, c∃xRax XcRacRaa ⊃ a=b XRab ⊃ b=b XRac ⊃ c=b X¬Rac×3, 7c=bRab¬Rab×8, 9b=b¬Raa↑a=bRbbRbc↑2 ∃1 ∀1 ∀1 ∀6 ⊃3, 7 =5 ⊃4 ⊃8, 10 =3, 10 =We end up with two open branches. We only need one to demonstrate that theroot is satisfiable, but it is useful practice to consider both. Let’s begin with224 forallx (UBC edition)the shorter open branch to the left, that completes at line 10. Examining thatbranch, we see three names, a, b, and c, and three atomic sentences concerningthe extension of R: Rac, Rab, and ¬Raa. Using our previous method forconstructing models, we would have posited a three-object UD and related thoseobjects via R in the appropriate way. Now, however, we also have identity claimsto consider. There are two identity claims in the branch. b=b doesn’t tell usanything interesting — of course that object is identical with itself — but thepresence of c=b indicates that our model will not include separate objects band c. We’ll only posit one object in the UD for those two names, and anotherobject for a. We can pick either letter for the object; let’s call it b. (Remember,it doesn’t matter what kinds of objects we pick for the UD; it could be ReedRichards for all it matters. We typically pick letters because they are easy tokeep track of.)So the model suggested by the first open branch here is:UD = {a, b}referent(a) = areferent(b) = breferent(c) = bextension(R) =Rxy a ba 0 1b - -The longer open branch includes additional identity claims. It includes a=b andc=b, which of course implies that all three of these names are names for thesame object. So the UD for the model corresponding to the right branch willhave only one object, which all three names denote.UD = {a}referent(a) = areferent(b) = areferent(c) = aextension(R) = Rxy aa 1It it sometimes helpful, for an intuitive grip on what is going on in these trees,to think about an English translation alongside the formalisms. For example,suppose that Rxy is interpreted as ‘x loves y’. Line 1 of the tree says that persona loves at most one person; line 2 says that a loves at least one person. The treeworks out two ways that might be: the one and only person a loves could be a,or it could be someone else.ch. 12 identity 225Practice Exercises? Part A Using the symbolization key given, translate each English-languagesentence into QL with identity. For sentences containing definite descriptions,assume Russell’s theory.UD: peopleKx: x knows the combination to the safe.Sx: x is a spy.Vx: x is a vegetarian.Txy: x trusts y.h: Hofthori: Ingmar1. Hofthor is a spy, but no vegetarian is a spy.2. No one knows the combination to the safe unless Ingmar does.3. No spy knows the combination to the safe.4. Neither Hofthor nor Ingmar is a vegetarian.5. Hofthor trusts a vegetarian.6. Everyone who trusts Ingmar trusts a vegetarian.7. Everyone who trusts Ingmar trusts someone who trusts a vegetarian.8. Only Ingmar knows the combination to the safe.9. Ingmar trusts Hofthor, but no one else.10. The person who knows the combination to the safe is a vegetarian.11. The person who knows the combination to the safe is not a spy.? Part B Using the symbolization key given, translate each English-languagesentence into QL with identity. For sentences containing definite descriptions,assume Russell’s theory.UD: cards in a standard deckBx: x is black.Cx: x is a club.Dx: x is a deuce.Jx: x is a jack.Mx: x is a man with an axe.Ox: x is one-eyed.Wx: x is wild.1. All clubs are black cards.2. There are no wild cards.3. There are at least two clubs.4. There is more than one one-eyed jack.226 forallx (UBC edition)5. There are at most two one-eyed jacks.6. There are exactly two black jacks.7. There are exactly four deuces.8. The deuce of clubs is a black card.9. One-eyed jacks and the man with the axe are wild.10. If the deuce of clubs is wild, then there is exactly one wild card.11. The man with the axe is not a jack.12. The deuce of clubs is not the man with the axe.? Part C Using the symbolization key given, translate each English-languagesentence into QL with identity.UD: people, generations, and monstersGx: x is a generation.Hx: x is human.Sx: x is a slayer.Vx: x is a vampire.Dx: x is a demon.Wx: x is a werewolf.Fx: x is a force of darkness.Axy: x will stand against y.Bxy: x is born in generation y.Kxy: x will kick y.b: Buffyf: Faithw: Willow1. Buffy and Willow were born unto the same generation.2. There is no more than one slayer born in each generation.3. A slayer other than Buffy is one of the forces of darkness.4. Willow will stand against any force of darkness other than a werewolf.5. Faith will kick everyone except herself.6. Buffy will kick anyone who stands against a slayer, unless they are alsokicking vampires or demons.7. In every generation a slayer is born.8. In every generation a slayer is born. She will stand against the vampires,demons, and forces of darkness.9. In every generation a slayer is born. She alone will stand against thevampires, demons, and forces of darkness.Part D Using the symbolization key given, translate each English-languagesentence into QL with identity. For sentences containing definite descriptions,assume Russell’s theory.UD: animals in the worldch. 12 identity 227Bx: x is in Farmer Brown’s field.Hx: x is a horse.Px: x is a Pegasus.Wx: x has wings.1. There are at least three horses in the world.2. There are at least three animals in the world.3. There is more than one horse in Farmer Brown’s field.4. There are three horses in Farmer Brown’s field.5. There is a single winged creature in Farmer Brown’s field; any othercreatures in the field must be wingless.6. The Pegasus is a winged horse.7. The animal in Farmer Brown’s field is not a horse.8. The horse in Farmer Brown’s field does not have wings.? Part E Construct a model to demonstrate each of the following. If you wish,you can draw a tree to help you answer these questions; however, it is goodconceptual practice to tackle some of these questions directly by thinking aboutjust what you’d need to put in your model.1. Show that {¬Raa,∀x(x=a ∨Rxa)} is consistent.2. Show that {∀x∀y∀z(x=y ∨ y=z ∨ x=z),∃x∃y x 6=y} is consistent.3. Show that {∀x∀y x=y,∃x x 6=a} is inconsistent.4. Show that ∃x(x=h&x=i) is contingent.5. Show that {∃x∃y(Zx&Zy&x=y), ¬Zd, d=s} is consistent.6. Show that ‘∀x(Dx ⊃ ∃yTyx) .˙. ∃y∃z y 6=z’ is invalid.? Part F Construct a tree to test the following entailment claims. If they arefalse, provide a model that demonstrates this.1. |= ∀x∀y(x=y ⊃ y=x)2. |= ∀x∃y x=y3. |= ∃x∀y x=y4. ∃x∀y x=y |= ∀x∀y(Rxy ≡ Ryx)5. |= ¬∀x∀y∀z[(Axy&Azx& y=z) ⊃ Axx]6. ∀x∀y x=y |= ∃xFx ≡ ∀xFx7. ∀x(x=a ∨ x=b), Ga ≡ ¬Gb |= ¬∃x∃y∃z(Gx&Gy&¬Gz)8. ∀x(Fx ⊃ x=f),∃x(Fx ∨ ∀y y=x) |= Ff9. ∃x∃yDxy |= ∀x1∀x2∀x3∀x4[(Dx1x2 &Dx3x4) ⊃ (x2 6=x3 ∨Dx1x4)]Part G In §12.5 we looked at two different translations of ‘Mozart wrote exactlytwo operas’. Use trees to prove that they are equivalent.Part H Translate these arguments into QL with identity, and evaluate them forvalidity with a tree. (Don’t be surprised or discouraged if some of these treesend up very complex.)228 forallx (UBC edition)1. Dudley will threaten anyone who threatens anyone. Therefore, Dudley willthreaten himself.2. The exam is easy. Therefore every exam Sheila took was easy. (UseRussell’s theory of definite descriptions.)3. Three wise men visited Jesus. Every wise man who visited Jesus gaveJesus a gift. Therefore, Jesus received more than one gift.4. Worf is the only Klingon in Starfleet. Everyone in Starfleet is brave. Allbrave Klingons are warriors. Therefore, there is at least one brave warriorin Starfleet.5. Worf is the only Klingon in Starfleet. Everyone in Starfleet is brave. Allbrave Klingons are warriors. Therefore, there is exactly one brave warriorin Starfleet.6. Every person likes every kind of sandwich that is tasty. Jack is a person.Jack likes exactly one kind of sandwich. Therefore, no more than one kindof sandwich is tasty.Chapter 13Natural Deduction Proofsin QLTrees employ a kind of ‘brute force’ strategy for proving entailment claims. Whenthe logical structure of the relevant sentences are rather simple, as they are inSL and in some QL cases, it can be an effective strategy, but in some cases theycan become very tedious and complex. It is useful to have a proof system thatallows one to reason in a more targeted way — especially if you already have anintuitive understanding of why a given argument should be expected to turn outvalid. As we did in Chapter 7 for SL, in this chapter we will consider a naturaldeduction system for quantified logic. As in the case of trees, our QL naturaldeduction system is an extension of the one we learned previously for SL.Like our SL natural deduction system, our system can only be used to demon-strate that an argument is valid. We do not have a formal method, as part ofour natural deduction system, for demonstrating an argument invalid. In thisrespect natural deduction differs from trees. We won’t go through the proof inthis book, but the natural deduction is both sound and complete. That meansthat there are natural deductions proofs corresponding to all and only the validarguments in QL.13.1 Natural Deduction: the basicsIn a natural deduction system, one begins by writing down the assumptions onebegins with — these correspond to the premises of an argument — then addsa series of additional sentences, justified via a series of particular rules by thesentences above. Additional assumptions may be made and discharged along the229230 forallx (UBC edition)way. If one succeeds in writing down the conclusion of the argument on a newline, with no additional undischarged assumptions, consistent with the naturaldeduction rules, one has proven that the conclusion follows from those premises.If you need a refresher on how that works, review Chapter 7.Our QL system will use all the same sentential rules that our SL system did. Allthe introduction and elimination rules for the sentential connectives, as well asthe derived and replacement rules, will be part of this system too. (These aresummarized, along with our new rules, in the Quick Reference guide at the endof this book.) We’ll add introduction and elimination rules for the existentialand universal quantifiers, and some other derived rules as well.13.2 Basic quantifier rulesRecall the relationship between quantified sentences and their instances. Thesentence Pa is a particular instance of the general claim ∀xPx. For any wff Φ,constant c, and variable x , we define Φ x ⇒ c to mean the wff that we get byreplacing every occurrence of x in Φ with c. Φ x ⇒ c is called a substitutioninstance of ∀xΦ and ∃xΦ, and c is called the instantiating constant. Thisshould be familiar from our discussion of our tree rules for quantifiers in Chapter10. We will also use this notation to describe our quantifier rules.13.2.1 Universal eliminationIf you have ∀xAx, it is legitimate to infer, of anything, that it is A. You caninfer Aa, or Ab, or Az, or Ad3, etc.. This is, you can infer any substitutioninstance – in short, you can infer Ac for any constant c. This is the general formof the universal elimination rule (∀E):m ∀x ΦΦ x⇒ c ∀E mRemember that the box mark for a substitution instance is not a symbol of QL,so you cannot write it directly in a proof. Instead, you write the substitutedsentence with the constant c replacing all occurrences of the variable x in Φ, asin this example:ch. 13 natural deduction proofs in ql 2311 ∀x(Mx ⊃ Rxd)2 Ma ⊃ Rad ∀E 13 Md ⊃ Rdd ∀E 1This rule is very similar to the tree rule for universals, which, in our tree system,allowed one to develop a branch containing a universal with any instance of itone likes. Here you are permitted to write down any instance you like on a newline.13.2.2 Existential introductionWhen is it legitimate to infer ∃xAx? If you know that something is an A — forinstance, if you have Aa available in the proof.This is the existential introduction rule (∃I):m Φ∃x Φ c⇒ x ∃I mIt is important to notice that Φ c⇒ x is not necessarily a substitution instance.We write it with double boxes to show that the variable x does not need toreplace all occurrences of the constant c. You can decide which occurrencesto replace and which to leave in place. For example, each of lines 2–6 can bejustified by ∃I:1 Ma ⊃ Rad2 ∃x(Ma ⊃ Rax) ∃I 13 ∃x(Mx ⊃ Rxd) ∃I 14 ∃x(Mx ⊃ Rad) ∃I 15 ∃y∃x(Mx ⊃ Ryd) ∃I 46 ∃z∃y∃x(Mx ⊃ Ryz) ∃I 513.2.3 Universal introductionA universal claim like ∀xPx would be proven if every substitution instance of ithad been proven, if every sentence Pa, Pb, . . . were available in a proof. Alas,232 forallx (UBC edition)there is no hope of proving every substitution instance. That would requireproving Pa, Pb, . . ., Pj2, . . ., Ps7, . . ., and so on to infinity. There are infinitelymany constants in QL, and so this process would never come to an end.Consider a simple argument: ∀xMx, .˙. ∀yMyIt makes no difference to the meaning of the sentence whether we use the variablex or the variable y, so this argument is obviously valid. Suppose we begin inthis way:1 ∀xMx want ∀yMy2 Ma ∀E 1We have derived Ma. Nothing stops us from using the same justification toderive Mb, . . ., Mj2, . . ., Ms7, . . ., and so on until we run out of space orpatience. We have effectively shown the way to prove Mc for any constant c.From this, ∀yMy follows.1 ∀xMx2 Ma ∀E 13 ∀yMy ∀I 2It is important here that a was just some arbitrary constant. We had not madeany special assumptions about it. If a had already been mentioned, say as apremise in the argument, then this would not show anything about all y. Forexample:1 ∀xRxa2 Raa ∀E 13 ∀yRyy not allowed!This is the schematic form of the universal introduction rule (∀I):m Φ∀x Φ c∗ ⇒ x ∀I m∗ The constant c must not occur in any undischarged assumption.Note that we can do this for any constant that does not occur in an undischargedassumption and for any variable.Note also that the constant may not occur in any undischarged assumption, butch. 13 natural deduction proofs in ql 233it may occur as the assumption of a subproof that we have already closed. Forexample, here is a valid proof of ∀z(Dz ⊃ Dz) that does not use any premises.1 Df want Df2 Df R 13 Df ⊃ Df ⊃I 1–24 ∀z(Dz ⊃ Dz) ∀I 313.2.4 Existential eliminationA sentence with an existential quantifier tells us that there is some memberof the UD that satisfies a formula. For example, ∃xSx tells us (roughly) thatthere is at least one S. It does not tell us which member of the UD satisfies S,however. We cannot immediately conclude Sa, Sf23, or any other substitutioninstance of the sentence. What can we do?Suppose that we knew both ∃xSx and ∀x(Sx ⊃ Tx). We could reason in thisway:Since ∃xSx, there is something that is an S. We do not know whichconstants refer to this thing, if any do, so call this thing ‘Ishmael’.From ∀x(Sx ⊃ Tx), it follows that if Ishmael is an S, then it is aT . Therefore, Ishmael is a T . Because Ishmael is a T , we know that∃xTx.In this paragraph, we introduced a name for the thing that is an S. We gaveit an arbitrary name (‘Ishmael’) so that we could reason about it and derivesome consequences from there being an S. Since ‘Ishmael’ is just a bogus nameintroduced for the purpose of the proof and not a genuine constant, we couldnot mention it in the conclusion. Yet we could derive a sentence that doesnot mention Ishmael; namely, ∃xTx. This sentence does follow from the twopremises.We want the existential elimination rule to work in a similar way. Yet sinceEnglish language worlds like ‘Ishmael’ are not symbols of QL, we cannot usethem in formal proofs. Instead, just as we did in the analogous rule within ourtree system, we will use names that are new — names which do not appearanywhere else in the proof. (This includes the conclusion you are aiming for.)A constant that is used to stand in for whatever it is that satisfies an existentialclaim is called a proxy. Reasoning with the proxy must all occur inside a234 forallx (UBC edition)subproof, and the proxy cannot be a constant that is doing work elsewhere inthe proof.This is the schematic form of the existential elimination rule (∃E):m ∃x Φn Φ c∗ ⇒ x for ∃Ep ΨΨ ∃E m, n–p∗ The constant c must not appear outside the subproof.Remember that the proxy constant cannot appear in Ψ , the sentence you proveusing ∃E. (It would actually be enough just to require that the proxy constantnot appear in ∃x Φ, in Ψ , or in any undischarged assumption. In recognition ofthe fact that it is just a place holder that we use inside the subproof, though,we require an entirely new constant which does not appear anywhere else in theproof.)The existential elimination rule, like the rules for conditional introduction andnegation introduction and elimination, is a rule that involves discharging anassumption. Assume a proxy instance, and see what would follow from thatinstance; if you have the existential, then, you can stop making the assumptionabout the proxy, and help yourself to what would have followed from it. As withthose other assumption-involving rules, instead of a justification, one includes anote — in this case, ‘for ∃E’ — about the role of the assumption in the proof.Remember that assumptions must be discharged before your proof is complete,so you should only make an assumption that goes beyond your premises whenyou have a plan for discharging it.With this rule, we can give a formal proof that ∃xSx and ∀x(Sx ⊃ Tx) togetherentail ∃xTx.1 ∃xSx2 ∀x(Sx ⊃ Tx) want ∃xTx3 Sa for ∃E4 Sa ⊃ Ta ∀E 25 Ta ⊃E 3, 46 ∃xTx ∃I 57 ∃xTx ∃E 1, 3–6ch. 13 natural deduction proofs in ql 23513.2.5 Quantifier negation¬∃x ¬Φ is logically equivalent to ∀x Φ. The first says that nothing falsifies Φ; thesecond says everything satisfies Φ. In QL, they are provably equivalent. Here isa proof schema for half of that equivalence via a natural deduction reductio. Forany wff Φ, variable x , and new name a:1 ∀x Φ want ¬∃x ¬Φ2 ∃x ¬Φ for reductio3 ¬Φ x ⇒ a∗ for ∃E4 ∀x Φ for reductio5 Φ x ⇒ a∗ ∀E 16 ¬Φ x ⇒ a∗ R 37 ¬∀x Φ ¬I 4–68 ¬∀x Φ ∃E 2, 3–79 ∀x Φ R 110 ¬∃x ¬Φ ¬I 2–9, 2–8∗ Where name a does not appear outside the subproof.This is a proof schema — it is not itself a proof in QL, as its lines are not QLsentences. But it describes how a proof of this form can be given. For example,here is one instance of the above schema:1 ∀yAy want ¬∃y¬Ay2 ∃y¬Ay for reductio3 ¬Ac for ∃E4 ∀yAy for reductio5 Ac ∀E 16 ¬Ac R 37 ¬∀yAy ¬I 4–68 ¬∀yAy ∃E 2, 3–79 ∀yAy R 110 ¬∃y¬Ay ¬I 2–9, 2–8236 forallx (UBC edition)(Note that this proof encodes the same form of reasoning one would employ todemonstrate via a tree that ∀yAy |= ¬∃y¬Ay. As an exercise: draw out the treeto compare it.)In order to fully demonstrate that ¬∃x ¬Φ is logically equivalent to ∀x Φ, wewould also need a second proof that assumes ¬∃x ¬Φ and derives ∀x Φ. We leavethat proof as an exercise for the reader.It will often be useful to translate between quantifiers by adding or subtractingnegations in this way, so we add two derived rules for this purpose. These rulesare called quantifier negation (QN):¬∀x Φ⇐⇒ ∃x ¬Φ¬∃x Φ⇐⇒ ∀x ¬Φ QNQN is a replacement rule. Like our SL replacement rules (DeMorgan, DoubleNegation, etc.), it can be used on whole sentences or on subformulae.13.3 Identity IntroductionThe introduction rule for identity is very simple. Everything is identical to itself;so, for any name a, one may write — regardless of what one has on the previouslines of the proof — that a=a:a = a =IThe =I rule is unlike our other rules in that it does not require referring to anyprior lines of the proof. We need only cite the rule itself; it does not referenceany previous line numbers.13.4 Identity EliminationIf you have shown that a=b, then anything that is true of a must also be true ofb. For any sentence with a in it, you can replace some or all of the occurrences ofa with b and produce an equivalent sentence. For example, if you already knowRaa, then you are justified in concluding Rab, Rba, Rbb. Recall that Φ a⇒ bis the sentence produced by replacing a in Φ with b. This is not the same as asubstitution instance, because b may replace some or all occurrences of a. Theidentity elimination rule (=E) justifies replacing terms with other terms thatare identical to it:ch. 13 natural deduction proofs in ql 237m Φ=Ψn ΦΦ a⇒ b =E m, nΦ b⇒ a =E m, nHere is a simple proof of an instance of the transitivity of identity. Let’s provethat if a=b and b=c, then a=c:1 a=b& b=c want a=c2 a=b & E 13 b=c & E 14 a=c =E 2, 35 (a=b& b=c) ⊃ a=c ⊃I 1–4At line 4, we took advantage of the identity claim b=c on line 3, and replacedthe b in line 2 with a c. Then we used the familiar ⊃I rule to discharge theassumption of line 1, proving the conditional we were aiming for.13.5 Translation and evaluationConsider this argument: There is only one button in my pocket. There is a bluebutton in my pocket. So there is no non-blue button in my pocket.We begin by defining a symbolization key:UD: buttons in my pocketBx: x is blue.Because we have no need to discuss anything other than buttons in my pocket,we’ve restricted the UD accordingly. If we included other things (buttonselsewhere and/or things other than buttons), we’d need predicates correspondingto being a button and things’ locations. The simple version here is adequate forour present needs. The argument is translated as:1. ∀x∀y x=y2. ∃xBx3. .˙. ¬∃x¬Bx238 forallx (UBC edition)So the set-up for a natural deduction proof will be:1 ∀x∀y x=y2 ∃xBx want ¬∃x¬Bx3There are various strategies one might employ. Here are two clues that pointtoward one promising strategy. Note again that we have an existential on line 2— this suggests existential elimination as a possible strategy. Note also that weare aiming for ¬∃x¬Bx, which equivalent to ¬¬∀xBx by QN. This in turn isequivalent, by DN, to ∀xBx, which suggests that universal introduction is goingto be an important step. If we introduce an assumption with a proxy instance of∃xBx, we’ll be able to work toward a generic instance of Bx. In this example,we’ll take e as our proxy, and show that Bf follows from ∃xBx:1 ∀x∀y x=y2 ∃xBx want ¬∃x¬Bx3 Be for ∃E4 ∀y e=y ∀E 15 e=f ∀E 46 Bf =E 5, 37 Bf ∃E 2, 3–6By line 7, we have discharged the assumption about the proxy — we won’tuse the name e any more — we have established that Bf follows from thetwo premises. Since f is an arbitrary name — one that does not appear inany undischarged assumption — we can perform universal introduction on thatinstance. This in turn lets us complete the proof via the two substitution rulesmentioned above:ch. 13 natural deduction proofs in ql 2391 ∀x∀y x=y2 ∃xBx want ¬∃x¬Bx3 Be for ∃E4 ∀y e=y ∀E 15 e=f ∀E 46 Bf =E 5, 37 Bf ∃E 2, 3–68 ∀xBx ∀I 79 ¬¬∀xBx DN 810 ¬∃x¬Bx QN 913.6 Natural deduction strategyAll the strategy advice given in §7.4 is equally applicable to natural deductionproofs in QL. Review the suggestions there for general advice for natural de-duction proofs. Applied to our new QL rules, if you have a universal, you canthink about taking any instances that look useful. (Taking random instances isunlikely to be useful.) If you have an existential, consider using the existentialelimination rule, which begins by assuming an instance with a proxy, thenderiving a conclusion that does not contain that proxy name.Showing ¬∃xΦ can also be hard, and it is often easier to show ∀x¬Φ and usethe QN rule.13.7 Soundness and completenessThe proofs for soundness and completeness of our natural deduction system arebeyond the scope of this textbook. But if you are interested in thinking throughhow those proofs would go, here are a few hints to get you started. Soundnessin a natural deduction system amounts to the claim that if any sentence Φ isderivable from a set of sentences X , then X |= Φ. To prove this, you would needto demonstrate that any possible natural deduction proof meets this constraint.This is trivial for ‘proofs’ that only contain premises; you’d next have to show,for every possible way of extending the proof (i.e., everything permitted by anyone of our rules), that any newly added lines with no undischarged assumptions240 forallx (UBC edition)are entailed by the premises.Undischarged assumptions would require special treatment. You can think of anassumption as being similar to a ‘temporary premise’ — what you really wantto prove is that, for every possible line in a proof, that line is entailed by thepremises in addition to any undischarged assumptions.The completeness proof is more complex. We need some way to guarantee thatthere is a proof corresponding to every QL entailment. The way to do this is tofind an algorithmic procedure that is guaranteed to find a proof if one exists,and to prove that this is so. One good way to do this is to take advantage of theproven completeness of our tree system for QL, presented in Chapter 11, andfind a way to demonstrate that any tree proof can be converted to a naturaldeduction proof. Here is a hint if you’d like to undertake that project: the treemethod encodes the same kind of reasoning that reductio proofs do.Practice Exercises? Part A Provide a justification (rule and line numbers) for each line of proofthat requires one.1 ∀x∃y(Rxy ∨Ryx)2 ∀x¬Rmx3 ∃y(Rmy ∨Rym)4 Rma ∨Ram5 ¬Rma6 Ram7 ∃xRxm8 ∃xRxm1 ∀x(∃yLxy ⊃ ∀zLzx)2 Lab3 ∃yLay ⊃ ∀zLza4 ∃yLay5 ∀zLza6 Lca7 ∃yLcy ⊃ ∀zLzc8 ∃yLcy9 ∀zLzc10 Lcc11 ∀xLxxch. 13 natural deduction proofs in ql 2411 ∀x(Jx ⊃ Kx)2 ∃x∀yLxy3 ∀xJx4 ∀yLay5 Ja6 Ja ⊃ Ka7 Ka8 Laa9 Ka&Laa10 ∃x(Kx&Lxx)11 ∃x(Kx&Lxx)1 ¬(∃xMx ∨ ∀x¬Mx)2 ¬∃xMx&¬∀x¬Mx3 ¬∃xMx4 ∀x¬Mx5 ¬∀x¬Mx6 ∃xMx ∨ ∀x¬Mx? Part B Provide a natural deduction proof of each claim.1. ` ∀xFx ∨ ¬∀xFx2. {∀x(Mx ≡ Nx),Ma&∃xRxa} ` ∃xNx3. {∀x(¬Mx ∨ Ljx),∀x(Bx ⊃ Ljx),∀x(Mx ∨Bx)} ` ∀xLjx4. ∀x(Cx&Dt) ` ∀xCx&Dt5. ∃x(Cx ∨Dt) ` ∃xCx ∨DtPart C Provide a proof of the argument about Billy on p. 149.Part D Look back at Part E on p. 158. Provide proofs to show that each of theargument forms is valid in QL.? Part E Provide a natural deduction proof of each claim.1. ∀x∀yGxy ` ∃xGxx2. ∀x∀y(Gxy ⊃ Gyx) ` ∀x∀y(Gxy ≡ Gyx)3. {∀x(Ax ⊃ Bx),∃xAx} ` ∃xBx4. {Na ⊃ ∀x(Mx ≡Ma),Ma,¬Mb} ` ¬Na5. ` ∀z(Pz ∨ ¬Pz)6. ` ∀xRxx ⊃ ∃x∃yRxy7. ` ∀y∃x(Qy ⊃ Qx)Part F Show that each pair of sentences is provably equivalent.242 forallx (UBC edition)1. ∀x(Ax ⊃ ¬Bx), ¬∃x(Ax&Bx)2. ∀x(¬Ax ⊃ Bd), ∀xAx ∨Bd3. ∃xPx ⊃ Qc, ∀x(Px ⊃ Qc)Part G Show that each of the following is provably inconsistent.1. {Sa ⊃ Tm, Tm ⊃ Sa, Tm&¬Sa}2. {¬∃xRxa, ∀x∀yRyx}3. {¬∃x∃yLxy, Laa}4. {∀x(Px ⊃ Qx), ∀z(Pz ⊃ Rz), ∀yPy, ¬Qa&¬Rb}? Part H Write a symbolization key for the following argument, translate it,and prove it:There is someone who likes everyone who likes everyone that firstperson likes. Therefore, there is someone who likes themself.Part I Provide a proof of each claim.1. {Pa ∨Qb,Qb ⊃ b=c,¬Pa} ` Qc2. {m=n ∨ n=o,An} ` Am ∨Ao3. {∀x x=m,Rma} ` ∃xRxx4. ¬∃x x 6=m ` ∀x∀y(Px ⊃ Py)5. ∀x∀y(Rxy ⊃ x=y) ` Rab ⊃ Rba6. {∃xJx, ∃x¬Jx} ` ∃x∃y x 6=y7. {∀x(x=n ≡Mx),∀x(Ox ∨ ¬Mx)} ` On8. {∃xDx,∀x(x=p ≡ Dx)} ` Dp9. {∃x[Kx&∀y(Ky ⊃ x=y) &Bx],Kd} ` Bd10. ` Pa ⊃ ∀x(Px ∨ x 6=a)? Part J For each of the following pairs of sentences: If they are logicallyequivalent in QL, give proofs to show this. If they are not, construct a model toshow this.1. ∀xPx ⊃ Qc, ∀x(Px ⊃ Qc)2. ∀xPx&Qc, ∀x(Px&Qc)3. Qc ∨ ∃xQx, ∃x(Qc ∨Qx)4. ∀x∀y∀zBxyz, ∀xBxxx5. ∀x∀yDxy, ∀y∀xDxy6. ∃x∀yDxy, ∀y∃xDxy? Part K For each of the following arguments: If it is valid in QL, give a proof.If it is invalid, construct a model to show that it is invalid.ch. 13 natural deduction proofs in ql 2431. ∀x∃yRxy, .˙. ∃y∀xRxy2. ∃y∀xRxy, .˙. ∀x∃yRxy3. ∃x(Px&¬Qx), .˙. ∀x(Px ⊃ ¬Qx)4. ∀x(Sx ⊃ Ta), Sd, .˙. Ta5. ∀x(Ax ⊃ Bx), ∀x(Bx ⊃ Cx), .˙. ∀x(Ax ⊃ Cx)6. ∃x(Dx ∨ Ex), ∀x(Dx ⊃ Fx), .˙. ∃x(Dx&Fx)7. ∀x∀y(Rxy ∨Ryx), .˙. Rjj8. ∃x∃y(Rxy ∨Ryx), .˙. Rjj9. ∀xPx ⊃ ∀xQx, ∃x¬Px, .˙. ∃x¬Qx10. ∃xMx ⊃ ∃xNx, ¬∃xNx, .˙. ∀x¬MxPart L Look at the arguments given in Chapter 10, Problem Part C (page 200).For those arguments whose QL translations are valid, prove their validity vianatural deduction.Part M Look at the entailment claims given in Chapter 12, Problem Part F(page 227). For those entailment claims that are true, prove them via naturaldeduction.Part N Look at the arguments given in Chapter 12, Problem Part H (page 227).For those arguments whose QL translations are valid, prove their validity vianatural deduction.Appendix ASymbolic notationIn the history of formal logic, different symbols have been used at different timesand by different authors. Often, authors were forced to use notation that theirprinters could typeset.In one sense, the choice of symbols used for various logical constants is arbitrary.There is nothing written in heaven that says that ‘¬’ must be the symbol fortruth-functional negation. We might have specified a different symbol to playthat part. Once we have given definitions for well-formed formulae (wff) and fortruth in our logic languages, however, using ‘¬’ is no longer arbitrary. That isthe symbol for negation in this textbook, and so it is the symbol for negationwhen writing sentences in our languages SL or QL.This appendix presents some common symbols, so that you can recognize themif you encounter them in an article or in another book.summary of symbolsnegation ¬, ∼conjunction &, ∧, •disjunction ∨conditional →, ⊃biconditional ↔, ≡ Negation Two commonly used symbols are the hoe, ‘¬’, and the swung dash,‘∼.’ In some more advanced formal systems it is necessary to distinguish betweentwo kinds of negation; the distinction is sometimes represented by using both ‘¬’and ‘∼.’Disjunction The symbol ‘∨’ is typically used to symbolize inclusive disjunc-tion.Conjunction Conjunction is often symbolized with the ampersand, ‘&.’ Theampersand is actually a decorative form of the Latin word ‘et’ which means ‘and’;it is commonly used in English writing. As a symbol in a formal system, theampersand is not the word ‘and’; its meaning is given by the formal semanticsfor the language. Perhaps to avoid this confusion, some systems use a differentsymbol for conjunction. For example, ‘∧’ is a counterpart to the symbol usedfor disjunction. Sometimes a single dot, ‘•’, is used. In some older texts, there isno symbol for conjunction at all; ‘A and B’ is simply written ‘AB.’Material Conditional There are two common symbols for the material con-ditional: the arrow, ‘→’, and the hook, ‘⊃.’244appendix: symbolic notation 245Material Biconditional The double-headed arrow, ‘↔’, is used in systemsthat use the arrow to represent the material conditional. Systems that use thehook for the conditional typically use the triple bar, ‘≡’, for the biconditional.Quantifiers The universal quantifier is typically symbolized as an upside-downA, ‘∀’, and the existential quantifier as a backwards E, ‘∃.’ Quantifiers are given insome systems in parentheses, as in ‘(∀x)(∃x)Fxy’, and in some systems without,as in ‘∀x∃xFxy’.In some texts, there is no separate symbol for the universal quantifier. Instead,the variable is just written in parentheses in front of the formula that it binds.For example, ‘all x are P ’ is written (x)Px.In some systems, the quantifiers are symbolized with larger versions of thesymbols used for conjunction and disjunction. Although quantified expressionscannot be translated into expressions without quantifiers, there is a conceptualconnection between the universal quantifier and conjunction and between theexistential quantifier and disjunction. Consider the sentence ∃xPx, for example.It means that either the first member of the UD is a P , or the second one is, orthe third one is, . . . . Such a system uses the symbol ‘∨’ instead of ‘∃.’Polish notationThis section briefly discusses sentential logic in Polish notation, a system ofnotation introduced in the late 1920s by the Polish logician Jan Lukasiewicz.Lower case letters are used as sentence letters. The capital letter N is used fornegation. A is used for disjunction, K for conjunction, C for the conditional, Efor the biconditional. (‘A’ is for alternation, another name for logical disjunction.‘E’ is for equivalence.)notation Polishof SL notation¬ N& K∨ A⊃ C≡ EIn Polish notation, a binary connective is written before the two sentences thatit connects. For example, the sentence A&B of SL would be written Kab inPolish notation.The sentences ¬A ⊃ B and ¬(A ⊃ B) are very different; the main logicaloperator of the first is the conditional, but the main connective of the second isnegation. In SL, we show this by putting parentheses around the conditional inthe second sentence. In Polish notation, parentheses are never required. Theleft-most connective is always the main connective. The first sentence wouldsimply be written CNab and the second NCab.This feature of Polish notation means that it is possible to evaluate sentencessimply by working through the symbols from right to left. If you were constructing246 forallx (UBC edition)a truth table for NKab, for example, you would first consider the truth-valuesassigned to b and a, then consider their conjunction, and then negate the result.The general rule for what to evaluate next in SL is not nearly so simple. InSL, the truth table for ¬(A&B) requires looking at A and B, then looking inthe middle of the sentence at the conjunction, and then at the beginning ofthe sentence at the negation. Because the order of operations can be specifiedmore mechanically in Polish notation, variants of Polish notation are used asthe internal structure for many computer programming languages.Appendix BSolutions to selected exercisesMany of the exercises may be answered correctly in different ways. Where thatis the case, the solution here represents one possible correct answer.Chapter 1Chapter 1 Part A1, 2, 4, 5, 6, 8, and 10 are sentences.Chapter 1 Part C1. consistent2. inconsistent3. consistent4. consistentChapter 1 Part D1. Possible. For example:Vancouver is in Canada.Seattle is in Canada..˙. Seattle and Vancouver are both in Canada.2. Possible. The previous argument is an example.3. Possible. For example:Vancouver is in Canada.Vancouver is not in Canada.˙. Vancouver is in Canada and it is not in Canada.4. Impossible. If the conclusion is a tautology, the argument is guaranteed tobe valid.5. Impossible. By definition, a tautology is not contingent.6. Possible. Any two tautologies will do.7. Impossible. Anything logically equivalent to a tautology is a tautology.8. Possible. For example:247248 forallx (UBC edition)Vancouver is in Canada and Vancouver is not in Canada.Seattle is in Canada and Seattle is not in Canada.9. Impossible. If it contains a contradiction, it is inconsistent.10. Possible. For example:Vancouver is in Canada or Vancouver is not in Canada.Seattle is in Canada and Seattle is not in Canada.Chapter 2Chapter 2 Part A1. ¬M2. M ∨ ¬M3. G ∨ C4. ¬C &¬G5. C ⊃ (¬G&¬M)6. M ∨ (C ∨G)Chapter 2 Part C1. E1 &E22. F1 ⊃ S13. F1 ∨ E14. E2 &¬S25. ¬E1 &¬E26. E1 &E2 &¬(S1 ∨ S2)7. S2 ⊃ F28. (¬E1 ⊃ ¬E2) & (E1 ⊃ E2)9. S1 ≡ ¬S210. (E2 &F2) ⊃ S211. ¬(E2 &F2)12. (F1 &F2) ≡ (¬E1 &¬E2)Chapter 2 Part DA: Alice is a spy.B: Bob is a spy.C: The code has been broken.solutions for ch. 3 249G: The German embassy will be in an uproar.1. A&B2. (A ∨B) ⊃ C3. ¬(A ∨B) ⊃ ¬C4. G ∨ C5. (C ∨ ¬C) &G6. (A ∨B) &¬(A&B)Chapter 2 Part EG: Gregor plays first base.L: The team will lose.M: There is a miracle.C: Gregor’s mom will bake cookies.1. G ⊃ L2. ¬M ⊃ L3. (L ∨ ¬L) &G4. C ≡ G5. M ⊃ ¬CChapter 2 Part G1. (a) no (b) no2. (a) no (b) yes3. (a) yes (b) yes4. (a) no (b) no5. (a) yes (b) yes6. (a) no (b) no7. (a) no (b) yes8. (a) no (b) yes9. (a) no (b) noChapter 3Chapter 3 Part A1. tautology2. contradiction250 forallx (UBC edition)3. contingent4. tautology5. tautology6. contingent7. tautology8. contradiction9. tautology10. contradiction11. tautology12. contingent13. contradiction14. contingent15. tautology16. tautology17. contingent18. contingentChapter 3 Part B2, 3, 5, 6, 8, and 9 are logically equivalent.Chapter 3 Part C11, 13, 16, 17, and 18 are consistent.Chapter 3 Part D21, 23, 26, and 28 are valid.Chapter 3 Part E1. Φ and Ψ have the same truth value on every line of a complete truth table,so Φ ≡ Ψ is true on every line. It is a tautology.2. The sentence is false on some line of a complete truth table. On that line,Φ and Ψ are true and Ω is false. So the argument is invalid.3. Since there is no line of a complete truth table on which all three sentencesare true, the conjunction is false on every line. So it is a contradiction.4. Since Φ is false on every line of a complete truth table, there is no line onwhich Φ and Ψ are true and Ω is false. So the argument is valid.5. Since Ω is true on every line of a complete truth table, there is no line onwhich Φ and Ψ are true and Ω is false. So the argument is valid.solutions for ch. 5 2516. Not much. (Φ ∨ Ψ) is a tautology if Φ and Ψ are tautologies; it is a contra-diction if they are contradictions; it is contingent if they are contingent.7. Φ and Ψ have different truth values on at least one line of a complete truthtable, and (Φ ∨ Ψ) will be true on that line. On other lines, it might betrue or false. So (Φ∨ Ψ) is either a tautology or it is contingent; it is not acontradiction.Chapter 3 Part F1. ¬A ⊃ B2. ¬(A ⊃ ¬B)3. ¬[(A ⊃ B) ⊃ ¬(B ⊃ A)]Chapter 5Chapter 5 Part A1. (a) {P, P ⊃ Q,Q ⊃ ¬P}, (b) true2. (a) ¬((P ⊃ Q) ≡ (Q ⊃ P )), (b) true3. (a) {P &Q,¬R ⊃ ¬Q,¬(P &R)}, (b) true4. (a) {A ∨B,B ⊃ C,A ≡ C,¬C}, (b) true5. (a) A ≡ ¬A, (b) true6. (a) {P, P ⊃ Q,¬Q,¬A}, (b) true7. (a) {P ⊃ Q,¬P ∨ ¬Q,Q ⊃ P}, (b) false.252 forallx (UBC edition)Chapter 5 Part BThe argument is valid.{A ≡ B,¬B ⊃ (C ∨D), E ⊃ ¬C, (¬D&F ) ∨G,¬A&E} ` H ∨G1.2.3.4.5.6.7.8.9.10.11.12.13.14.15.16.17.18.A ≡ B X¬B ⊃ (C ∨D) XE ⊃ ¬C X(¬D&F ) ∨G X¬A&E X¬(H ∨G) X¬AE¬H¬G¬D&F X¬DF¬E×¬CAB×¬A¬B¬¬B×C ∨D XC×D×G×5 &6 ¬∨4 ∨11 &3 ⊃1 ≡2 ⊃17 ∨solutions for ch. 5 253Chapter 5 Part C1. True.{P, P ⊃ Q,Q ⊃ ¬P} ` ⊥1.2.3.4.5.PP ⊃ Q XQ ⊃ ¬P X¬P×Q¬Q×¬P×2. False.1.2.3.4.5.6.7.8.¬((P ⊃ Q) ≡ (Q ⊃ P )) XP ⊃ Q X¬(Q ⊃ P ) XQ¬P¬P↑Q↑¬(P ⊃ Q) XQ ⊃ P XP¬Q¬Q↑P↑1 ¬≡3 ¬⊃2 ¬⊃2 ⊃3 ⊃{P = 0, Q = 1} and {P = 1, Q = 0} each falsify ((P ⊃ Q) ≡ (Q ⊃ P )), soit is not a tautology.3. True.1.2.3.4.5.6.7.P &Q X¬R ⊃ ¬Q X¬(P &R) XPQ¬P×¬R¬¬R×¬Q×1 &3 ¬&2 ⊃254 forallx (UBC edition)4. True.1.2.3.4.5.6.7.8.A ∨B XB ⊃ C XA ≡ C X¬C¬BAAC×¬A¬C×B×C×2 ⊃1 ∨3 ≡5. True.1.2.3.A ≡ ¬A XA¬A×¬AA×1 ≡6. True.1.2.3.4.5.P XP ⊃ Q X¬Q X¬A¬P×Q×2 ⊃7. True. {P = 0, Q = 0} satisfies these sentences.1.2.3.4.5.6.P ⊃ Q X¬P ∨ ¬Q XQ ⊃ P X¬P¬P¬Q↑P×Q¬Q×P×¬Q¬P¬Q↑P×Q×2 ∨1 ⊃3 ⊃solutions for ch. 6 255Chapter 6Chapter 6 Part A1. Yes. Assuming that the root Φ&Ψ is satisfiable, for some interpretationI, I(Φ&Ψ) = 1, so I(Φ) = 1 and I(Ψ) = 1. So if what comes above issatisfiable, at least one branch is guaranteed to be satisfiable after this rulehas been performed. (In fact, both are.)2. Yes. Assuming that the root Φ&Ψ is satisfiable, for some interpretationI, I(Φ&Ψ) = 1, so I(Φ) = 1. So if what comes above is satisfiable,the continuation is guaranteed to be satisfiable after this rule has beenperformed.3. No. It is possible to satisfy the root Φ&Ψ without satisfying the extensionof the tree. Here is a counterexample to the soundness of the revisedsystem. This tree has a satisfiable root, but closes if we use the rule inquestion.BA&B XA¬B×4. No. It is possible to satisfy the root Φ∨Ψ without satisfying the extensionof the tree. Here is a counterexample to the soundness of the revisedsystem. This tree has a satisfiable root, but closes if we use the rule inquestion.¬AA ∨B XAB×5. Yes. Assume I(Φ∨Ψ) = 1. Then either I(Φ) = 1 or I(Ψ) = 1. If I(Φ) = 1,then I satisfies the left branch. If I(Ψ) = 1, then I satisfies the middlebranch. So, assuming that I satisfies the sentences above this resolutionrule, it must satisfy at least one branch below it. So this rule will nevertake us from a satisfiable branch to a tree with no satisfiable branches.256 forallx (UBC edition)6. Yes. Assume I(Φ ⊃ Ψ) = 1. Then either I(¬Φ) = 1 or I(Ψ) = 1. IfI(¬Φ) = 1, then I(Φ) = 0, so I satisfies the left branch. If I(Ψ) = 1,then I satisfies the right branch, as the rule for disjunctions is if at leastone disjunct is satisfied, the disjunction is satisfied. So, assuming that Isatisfies the sentences above this resolution rule, it must satisfy at least onebranch below it. So this rule will never take us from a satisfiable branchto a tree with no satisfiable branches.7. Yes. Assume I(Φ ⊃ Ψ) = 1. Then either I(¬Φ) = 1 or I(Ψ) = 1. Ineither case, the single branch is satisfied, as the rule for disjunctions is ifat least one disjunct is satisfied, the disjunction is satisfied. So, assumingthat I satisfies the sentences above this resolution rule, it must satisfy atleast one branch below it. So this rule will never take us from a satisfiablebranch to a tree with no satisfiable branches.8. No. Here is a counterexample to the soundness of the revised system. Thistree has a satisfiable root, but closes if we use the rule in question.P ≡ Q X¬PPQ×9. Yes. Assume I(Φ∨Ψ) = 1. Then either I(Φ) = 1 or I(Ψ) = 1. If I(Φ) = 1,then I satisfies the left branch. If I(Ψ) = 1, then I satisfies the middlebranch. So, assuming that I satisfies the sentences above this resolutionrule, it must satisfy at least one branch below it. So this rule will nevertake us from a satisfiable branch to a tree with no satisfiable branches.solutions for ch. 7 257Chapter 7Chapter 7 Part A1 W ⊃ ¬B2 A&W3 B ∨ (J &K)4 W & E 25 ¬B ⊃E 1, 46 J &K ∨E 3, 57 K & E 61 L ≡ ¬O2 L ∨ ¬O3 ¬L for reductio4 ¬O ∨E 2, 35 L ≡E 1, 46 ¬L R 37 L ¬E 3–5, 3–6258 forallx (UBC edition)1 Z ⊃ (C &¬N)2 ¬Z ⊃ (N &¬C)3 ¬(N ∨ C) for reductio4 ¬N &¬C DeM 35 Z for reductio6 C &¬N ⊃E 1, 57 C & E 68 ¬C & E 49 ¬Z ¬I 5–7, 5–810 N &¬C ⊃E 2, 911 N & E 1012 ¬N & E 413 N ∨ C ¬E 3–11, 3–12Chapter 7 Part B1. 1 K &L want K ≡ L2 K want L3 L & E 14 K ⊃ L ⊃I 2–35 L want K6 K & E 17 L ⊃ K ⊃I 5–68 K ≡ L ≡I 4, 7solutions for ch. 7 2592. 1 A ⊃ (B ⊃ C) want (A&B) ⊃ C2 A&B want C3 A & E 24 B ⊃ C ⊃E 1, 35 B & E 26 C ⊃E 4, 57 (A&B) ⊃ C ⊃I 2–63. 1 P & (Q ∨R)2 P ⊃ ¬R want Q ∨ E3 P & E 14 ¬R ⊃E 2, 35 Q ∨R & E 16 Q ∨E 4, 57 Q ∨ E ∨I 64. 1 (C &D) ∨ E want E ∨D2 ¬E want D3 C &D ∨E 1, 24 D & E 35 ¬E ⊃ D ⊃I 2–46 E ∨D MC 5260 forallx (UBC edition)5. 1 ¬F ⊃ G2 F ⊃ H want G ∨H3 ¬G want H4 ¬¬F MT 1, 35 F DN 46 H ⊃E 2, 57 ¬G ⊃ H ⊃I 3–68 G ∨H MC 76. 1 (X &Y ) ∨ (X &Z)2 ¬(X &D)3 D ∨M want M4 ¬X for reductio5 ¬X ∨ ¬Y ∨I 46 ¬(X &Y ) DeM 57 X &Z ∨E 1, 68 X & E 79 ¬X R 410 X ¬E 4–8, 4–911 ¬M for reductio12 D ∨E 3, 1113 X &D & I 10, 1214 ¬(X &D) R 215 M ¬E 11–13, 11–14solutions for ch. 7 261Chapter 7 Part D1.O ⊃ O1 O want O2 O R 13 O ⊃ O ⊃I 1–22.N ∨ ¬N1 ¬(N ∨ ¬N) for reductio2 ¬N &¬¬N DeM 13 ¬N & E 24 ¬¬N & E 25 N ∨ ¬N ¬E 1–3, 1–43.¬(P &¬P )1 P &¬P for reductio2 P & E 13 ¬P & E 14 ¬(P &¬P ) ¬I 1–2, 1–34. 1 ¬(A ⊃ ¬C) want A ⊃ C2 A want C3 ¬(¬A ∨ ¬C) MC 14 ¬¬A&¬¬C DeM 35 ¬¬A&C DN 46 C & E 57 A ⊃ C ⊃I 2–68 ¬(A ⊃ ¬C) ⊃ (A ⊃ C) ⊃I 1–7262 forallx (UBC edition)5. 1 J want J ∨ (L&¬L)2 J ∨ (L&¬L) ∨I 13 J ⊃ [J ∨ (L&¬L)] ⊃I 1–24 J ∨ (L&¬L) want J5 ¬J for reductio6 L&¬L ∨E 4, 57 L & E 68 ¬L & E 69 J ¬E 5–7, 5–810 (J ∨ (L&¬L)) ⊃ J ⊃I 4–911 J ≡ [J ∨ (L&¬L)] ≡I 3, 10Chapter 7 Part F1. 1 M & (¬N ⊃ ¬M)2 M & E 13 ¬N ⊃ ¬M & E 14 ¬¬M DN 25 ¬¬N MT 3, 46 N DN 57 N &M & I 6, 28 (N &M) ∨ ¬M ∨I 7solutions for ch. 7 2632. 1 C ⊃ (E&G)2 ¬C ⊃ G3 ¬G for reductio4 ¬¬C MT 2, 35 C DN 46 E&G ⊃E 1, 57 G & E 68 ¬G R 39 G ¬E 3–7, 3–83. 1 (Z &K) ≡ (Y &M)2 D& (D ⊃M)3 D ⊃M & E 24 D & E 25 M ⊃E 3, 46 Y want Z7 Y &M & I 6, 58 Z &K ≡E 1, 79 Z & E 810 Y ⊃ Z ⊃I 6–9264 forallx (UBC edition)4. 1 (W ∨X) ∨ (Y ∨ Z)2 X ⊃ Y3 ¬Z4 ¬(W ∨ Y ) for reductio5 ¬W &¬Y DeM 46 ¬Y & E 57 ¬W & E 58 ¬X MT 2, 69 ¬W &¬X & I 7, 810 ¬(W ∨X) DeM 911 Y ∨ Z ∨E 1, 1012 Z ∨E 11, 613 ¬Z R 314 W ∨ Y ¬E 4–12, 4–13Chapter 8Chapter 8 Part B1. ∃x(Lpx&Bx)2. ∃x(Ldx&Dx&Wxb)3. ∃x[Rcx&Hx&∃y(By&Lxy)]4. ∃x(Bx&Wxp& [∃y(Dy&Lxy) ∨ ∃y(Cy&Lxy)])5. ∃x(Bx&Mx&Wbx)6. ∃x(¬Bx&Mx&Wbx&Lxd)Chapter 8 Part D1. Za&Zb&Zc2. Rb&¬Ab3. Lcb ⊃Mb4. (Ab&Ac) ⊃ (Lab&Lac)5. ∃x(Rx&Zx)solutions for ch. 8 2656. ∀x(Ax ⊃ Rx)7. ∀x[Zx ⊃ (Mx ∨Ax)]8. ∃x(Rx&¬Ax)9. ∃x(Rx&Lcx)10. ∀x[(Mx&Zx) ⊃ Lbx]11. ∀x[(Mx&Lax) ⊃ Lxa]12. ∃xRx ⊃ Ra13. ∀x(Ax ⊃ Rx)14. ∀x[(Mx&Lcx) ⊃ Lax]15. ∃x(Mx&Lxb&¬Lbx)Chapter 8 Part F1. Db&Sb2. Db&De&Df3. Leb&Lfe4. ∀x(Dx ⊃ Sx)5. ∀x(Sx ⊃ Dx)6. ∃x(Dx&Lxe)7. ∃x(Dx&Lxf) ⊃ ∃x(Dx&Lxe)8. ∀x(Lxe ⊃ ¬Sx)9. ¬∃x(Lxf &Dx)10. ∀x(¬Sx ⊃ Lxb)11. ∃x((Lxb&Lex) ∨ (Lxe&Lbx))12. ¬∃x(Dx& [(Lxb&Lex) ∨ (Lxe&Lbx)])13. ¬∃x(Dx&Lxx)14. ∀x(Dx ⊃ ∃y(Dy&Lyx))15. ∃x(∀y(Dy ⊃ Lyx))Chapter 8 Part H1. ¬∃xTx2. ∀x(Mx ⊃ Sx)3. ∃x¬Sx4. ∃x[Cx&¬∃yByx]5. ¬∃xBxx6. ¬∃x(Cx&¬Sx&Tx)7. ∃x(Cx&Tx) &∃x(Mx&Tx) &¬∃x(Cx&Mx&Tx)8. ∀x[Cx ⊃ ∀y(¬Cy ⊃ Bxy)]9. ∀x((Cx&Mx) ⊃ ∀y[(¬Cy&¬My) ⊃ Bxy])266 forallx (UBC edition)Chapter 8 Part I1. ∀x(Fx ⊃ Tx)2. ¬Rg ⊃ Tg3. ∀x(Px ⊃ Lxg)4. ∃x(Px&Lxg) ⊃ Leg5. ∀x((Fx&Lfx) ⊃ Rx)6. ¬∃x(Px&Lfx) &¬∃y(Py&Lyf)7. ∀x((Px&Lxg) ⊃ Lex)8. ∀x∀y((Px&Py&Lxy&Ley) ⊃ Lex)9. ∃x(Px&Tx) ⊃ ∀y(Fy ⊃ Ry)Chapter 8 Part J1. ∀x(Cxp ⊃ Dx)2. Cjp&Fj3. ∃x(Cxp&Fx)4. ¬∃xSxj5. ∀x[(Cxp&Fx) ⊃ Dx]6. ¬∃x(Cxp&Mx)7. ∃x(Cjx&Sxe&Fj)8. Spe&Mp9. ∀x[(Sxp&Mx) ⊃ ¬∃yCyx]10. ∃x(Sxj&∃yCyx&Fj)11. ∀x[Dx ⊃ ∃y(Sxy&Fy&Dy)]12. ∀x[(Mx&Dx) ⊃ ∃y(Cxy&Dy)]Chapter 8 Part L1. Rca, Rcb, Rcc, and Rcd are substitution instances of ∀xRcx.2. Of the expressions listed, only ∀yLby is a substitution instance of ∃x∀yLxy.Chapter 9Chapter 9 Part A2, 3, 4, 6, 8, and 9 are true in the model.solutions for ch. 9 267Chapter 9 Part B13, 14, and 16 are true in the model.Chapter 9 Part C20, 22, 23, 24, 25, 27, 28, 29, 30, 31, 32, and 34 are true in the model.Chapter 9 Part DUD = {10,11,12,13}extension(O) = {11,13}extension(S) = ∅extension(T ) = {10,11,12,13}extension(U) = {13}extension(N) = {<11,10>,<12,11>,<13,12>}Chapter 9 Part E1. The sentence is true in this model:UD = {Stan}extension(D) = {Stan}referent(a) = Stanreferent(b) = StanAnd it is false in this model:UD = {Stan}extension(D) = ∅referent(a) = Stanreferent(b) = Stan2. The sentence is true in this model:UD = {Stan}extension(T ) = {}referent(h) = StanAnd it is false in this model:UD = {Stan}extension(T ) = ∅referent(h) = Stan3. The sentence is true in this model:UD = {Stan, Ollie}extension(P ) = {Stan}referent(m) = Stan268 forallx (UBC edition)And it is false in this model:UD = {Stan}extension(P ) = ∅referent(m) = StanChapter 9 Part FThere are many possible correct answers. Here are some:1. Making the first sentence true and the second false:UD = {alpha}extension(J) = {alpha}extension(K) = ∅referent(a) = alpha2. Making the first sentence true and the second false:UD = {alpha, omega}extension(J) = {alpha}referent(m) = omega3. Making the first sentence false and the second true:UD = {alpha, omega}extension(R) = {}4. Making the first sentence false and the second true:UD = {alpha, omega}extension(P ) = {alpha}extension(Q) = ∅referent(c) = alpha5. Making the first sentence true and the second false:UD = {iota}extension(P ) = ∅extension(Q) = ∅6. Making the first sentence false and the second true:UD = {iota}extension(P ) = ∅extension(Q) = {iota}7. Making the first sentence true and the second false:UD = {iota}extension(P ) = ∅extension(Q) = {iota}8. Making the first sentence true and the second false:UD = {alpha, omega}extension(R) = {, }solutions for ch. 10 2699. Making the first sentence false and the second true:UD = {alpha, omega}extension(R) = {, }Chapter 9 Part I2. No, it would not make any difference. The satisfaction of a formula withone or more free variables depends on what the variable assignment doesfor those variables. Because a sentence has no free variables, however, itssatisfaction does not depend on the variable assignment. So a sentencethat is satisfied by some variable assignment is satisfied by every othervariable assignment as well.Chapter 10Chapter 10 Part A1. ∀x∀y(Gxy ⊃ ∃zGxz) is a tautology.1.2.3.4.5.6.¬∀x∀y(Gxy ⊃ ∃zGxz) Xa¬∀y(Gay ⊃ ∃zGaz) Xb¬(Gab ⊃ ∃zGaz) XGab¬∃zGaz \b¬Gab×4, 61 ¬∀2 ¬∀3 ¬⊃270 forallx (UBC edition)2. ∀xFx ∨ ∀x(Fx ⊃ Gx) is not a tautology.1.2.3.4.5.6.7.¬(∀xFx ∨ ∀x(Fx ⊃ Gx)) X¬∀xFx Xa¬∀x(Fx ⊃ Gx) Xb¬Fa¬(Fb ⊃ Gb) XFb¬Gb↑1 ¬∨2 ¬∀3 ¬∀5 ¬⊃UD={a, b}, extension(F )={b}, extension(G)=∅3. ∀x(Fx ⊃ (¬Fx ⊃ ∀yGy)) is a tautology.1.2.3.4.5.6.¬∀x(Fx ⊃ (¬Fx ⊃ ∀yGy)) Xa¬(Fa ⊃ (¬Fa ⊃ ∀yGy)) XFa¬(¬Fa ⊃ ∀yGy) X¬Fa¬∀yGy×3, 51 ¬∀2 ¬⊃4 ¬⊃4. ∃x(Fx ∨ ¬Fx) is a tautology.1.2.3.4.¬∃x(Fx ∨ ¬Fx) \a¬(Fa ∨ ¬Fa) X¬Fa¬¬Fa×3, 41 ¬∃2 ¬∨solutions for ch. 10 2715. ∃xJx ≡ ¬∀x¬Jx is a tautology.1.2.3.4.5.6.7.8.¬(∃xJx ≡ ¬∀x¬Jx) X∃xJx Xa¬¬∀x¬Jx XJa∀x¬Jx \a¬Ja×5, 8¬∃xJx \a¬∀x¬Jx Xa¬¬Ja¬Ja×4, 61 ≡3 ¬∀2 ∃2 ¬∃3 ¬¬7 ∀6. ∀x(Fx ∨Gx) ⊃ (∀yFy ∨ ∃xGx) is a tautology.1.2.3.4.5.6.7.8.9.¬(∀x(Fx ∨Gx) ⊃ (∀yFy ∨ ∃xGx)) X∀x(Fx ∨Gx) \a¬(∀yFy ∨ ∃xGx) X¬∀yFy Xa¬∃xGx \a¬Fa¬GaFa ∨Ga XFa×Ga×1 ¬⊃3 ¬∨4 ¬∀5 ¬∃2 ∀8 ∨272 forallx (UBC edition)Chapter 10 Part B1. Fa, Ga, .˙. ∀x(Fx ⊃ Gx) is invalid.1.2.3.4.5.6.FaGa¬∀x(Fx ⊃ Gx) Xb¬(Fb ⊃ Gb) XFb¬Gb↑3 ¬∀4 ¬⊃UD={a, b}, extension(F )={a, b}, extension(G)={a}2. Fa, Ga, .˙. ∃x(Fx&Gx) is valid.1.2.3.4.5.FaGa¬∃x(Fx&Gx) \a¬(Fa&Ga) X¬Fa×¬Ga×3 ¬∃4 ¬&solutions for ch. 10 2733. ∀x∃yLxy, .˙. ∃x∀yLxy is invalid.1.2.3.4.5.6.7.8.9.10.11.12.13.14.15.∀x∃yLxy \a, b, c, ...¬∃x∀yLxy \a, b, c, ...∃yLay XbLab¬∀yLay Xc¬Lac∃yLby XdLbd¬∀yLby Xe¬Lbe∃yLcy XfLcf¬∀yLcy Xg¬Lcg...1 ∀3 ∃2 ¬∃5 ¬∀1 ∀7 ∃2 ¬∃9 ¬∀1 ∀11 ∃2 ¬∃13 ¬∀This tree continues in an infinite way; each new name requires additionalinstances at line 1 and line 2, which in turn require two more new names.We can represent a model with an infinite UD, giving the extension ofL as either a chart or an infinite set of ordered pairs. It is also possibleto collapse this interpretation into a finite one; examining the chart, wecan see that there is no reason the names couldn’t be repeating namesfor the same two objects; we could let d and e be additional names for band c, respectively, and so on for f and g, etc. So a two-object model willalso suffice to satisfy the premises and falsify the conclusion, with UD={b,c} and extension(L)={, }. This is also shown in a secondchart.274 forallx (UBC edition)Lxy a b c d e f g . . .a - 1 0b 1 0c 1 0d 1 0...Lxy b cb 1 0c 1 04. ∃x(Fx&Gx), Fb ≡ Fa, Fc ⊃ Fa, .˙. Fa is invalid.1.2.3.4.5.6.7.8.9.10.∃x(Fx&Gx) XdFb ≡ Fa XFc ⊃ Fa X¬FaFd&Gd XFdGdFbFa×4, 9¬Fb¬Fa¬Fc↑Fa×9, 101 ∃5 &2 ≡3 ⊃UD={a, b, c, d}, extension(F )={d}, extension(G)={d}solutions for ch. 11 2755. ∀x∃yGyx, .˙. ∀x∃y(Gxy ∨Gyx) is valid.1.2.3.4.5.6.7.8.∀x∃yGyx \a¬∀x∃y(Gxy ∨Gyx) Xa¬∃y(Gay ∨Gya) \b∃yGya XbGba¬(Gab ∨Gba) X¬Gab¬Gba×5, 82 ¬∀1 ∀4 ∃3 ¬∃6 ¬∨Chapter 11Chapter 11 Part A1. This system would not be sound. The satisfiability of an existential does notimply the satisfiability of any arbitrary instance. Here is a counterexample.This tree has a satisfiable root, but can close, given the proposed rule.∃xFx¬FaFa×2. This system would not be sound. Satisfying the existential doesn’t guaran-tee satisfying its d instance. Here is a counterexample:∃xFx¬FdFd×3. This system would not be sound. The satisfiability of an existential doesnot imply the satisfiability of any three arbitrary instances. Here is a276 forallx (UBC edition)counterexample. This tree has a satisfiable root, but can close, given theproposed rule.∃xFx¬FaFaFbFc×4. This system would still be sound; the proof extends. If we assume ∀x Φis satisfiable, then some interpretation I satisfies it. This interpretationeither satisfies all three instances (if the names were interpreted by I), or itcan easily be extended into a new interpretation I* that satisfies them, byadding any new names and assigning them to any objects in I’s UD, all ofwhich we know satisfy Φ. So this rule will never take us from a satisfiablebranch to an unsatisfiable one.5. This system would still be sound; the proof extends. If we assume ∃x Φ issatisfiable, then some interpretation I satisfies it. Create a new interpreta-tion I*, which includes the three names, and assign them each to an objectin I’s UD that satisfies Φ. We know there is at least one such object, sinceI(∃x Φ) = 1. (Remember there is no prohibition on assigning multiplenames to the same object.) I* is guaranteed to satisfy the extension ofthe branch along with that which came above. So this rule will never takeus from a satisfiable branch to an unsatisfiable one.6. This system would still be sound; the proof extends. If we assume ∀x Φ issatisfiable, then some interpretation I satisfies it. Extend the interpretationto include the new name, and assign it to any object in the UD, whichwe know will satisfy Φ; the new interpretation satisfies the extension andwhat came before. So this rule will never take us from a satisfiable branchto an unsatisfiable one.7. This system would still be sound; the proof extends. If x does not occurin Φ, then ∃x Φ is logically equivalent to Φ. (Every substitution instancefor x of Φ will trivially just be Φ— since there is no x in that sentence,replacing ‘every’ instance of that variable with any name results in nochange at all.) So this rule is equivalent to the original conjunction rule,which is sound.8. This system would still be sound. The soundness proof does not rely onthe branch completion rules, so changing those rules will never interferewith the soundness proof.9. This system would still be sound. The soundness proof does not rely onthe branch completion rules, so changing those rules will never interferewith the soundness proof.10. This system would remain sound. Adding an additional requirement forcompletion will never make it easier to close branches.solutions for ch. 11 277Chapter 11 Part B1. This system would remain complete. If an existential is in a completedopen branch, and this rule has been performed, then some instance of thatexistential is also in that branch. Any interpretation that satisfies thatinstance will also satisfy the existential.2. This system would still be complete. If an existential is in a completedopen branch, and this rule has been performed, then the d instance of thatexistential is also in that branch. Any interpretation that satisfies thatinstance will also satisfy the existential.3. This system would still be complete. If an existential is in a completedopen branch, and this rule has been performed, then three instances ofthat existential are also in that branch. Any interpretation that satisfiesthem will also satisfy the existential.4. This system would not be complete. Satisfying three substitution instancesfor x of Φ doesn’t guarantee satisfying ∀x Φ. Here is a counterexample tocompleteness— a tree with an unsatisfiable root that remains open.∀xFx X¬FaFbFcFd5. This system would still be complete; the proof extends. Satisfying sub-stitution instances for x of Φ— whether new or not, and no matter howmany times— guarantees satisfying ∃x Φ.6. This system would not be complete. Satisfying an instance of a universal(whether or not the name is new) is no guarantee that the universal willbe satisfied. Here is a counterexample:∀xFx X¬FaFb7. This system would still be complete; the proof extends. If x does not occurin Φ, then ∃x Φ is logically equivalent to Φ. (Every substitution instancefor x of Φ will trivially just be Φ— since there is no x in that sentence,replacing ‘every’ instance of that variable with any name results in nochange at all.) So this rule is equivalent to the original conjunction rule,which is complete.8. This system would not be complete. Taking one instance isn’t enough toensure that the universal is satisfied. Here is a counterexample:278 forallx (UBC edition)∀xFx X¬FaFb9. This system would not be complete. If a name is introduced later on inthe tree and that instance of the universal hasn’t been taken, satisfyingthe instances corresponding to the old names is not enough to guaranteesatisfying the universal. Here is a counterexample to completeness:1.2.3.4.5.6.7.Fa∀x∀y¬Rxy \a∀x∃yRxy \a∀y¬Ray \a∃yRay Xb¬RaaRab1 ∀2 ∀3 ¬∀4 ∃10. This system would remain complete. The addition is simply the newrequirement that at least one new name be introduced via this rule; thereasoning that applied for completeness of the original system is unchanged.Satisfying every instance corresponding to a name in the branch will guar-antee satisfying the general claims (the universals or negated existentials),since the UDs are constructed based on the names in the branch.Chapter 12Chapter 12 Part A1. Sh&¬∃x(V x&Sx)2. ¬Ki ⊃ ¬∃xKx3. ¬∃x(Sx&Kx)4. ¬V h&¬V i5. ∃x(V x&Thx)6. ∀x(Txi ⊃ ∃y(Txy&V y))7. ∀x(Txi ⊃ ∃y(Txy&∃z(Tyz&V z)))8. Ki&¬∃x(Kx&x6=i)9. Tih&¬∃x(Tix&x 6=h)10. ∃x(Kx&¬∃y(Ky&x 6=y) &V x)11. ∃x(Kx&¬∃y(Ky&x 6=y) &¬Sx)solutions for ch. 12 279Chapter 12 Part B1. ∀x(Cx ⊃ Bx)2. ¬∃xWx3. ∃x∃y(Cx&Cy&x 6=y)4. ∃x∃y(Jx&Ox& Jy&Oy&x 6=y)5. ∀x∀y∀z[(Jx&Ox& Jy&Oy& Jz&Oz) ⊃ (x=y ∨ x=z ∨ y=z)]6. ∃x∃y(Jx&Bx& Jy&By&x 6=y&∀z[(Jz&Bz) ⊃ (x=z ∨ y=z)])7. ∃x1∃x2∃x3∃x4(Dx1 &Dx2 &Dx3 &Dx4 &x1 6=x2 &x1 6=x3 &x1 6=x4 &x2 6=x3 &x2 6=x4 &x3 6=x4&¬∃y(Dy& y 6=x1 & y 6=x2 & y 6=x3 & y 6=x4))8. ∃x(Dx&Cx&∀y[(Dy&Cy) ⊃ x=y] &Bx)9. ∀x[(Ox& Jx) ⊃Wx]&∃x[Mx&∀y(My ⊃ x=y) &Wx]10. ∃x(Dx&Cx&∀y[(Dy&Cy) ⊃ x=y] &Wx) ⊃ ∃x∀y(Wx ≡ x=y)11. ∃x[Mx&∀y(My ⊃ x=y) &¬Jx]12. ∃x∃z(Dx&Cx&Mz&∀y[(Dy&Cy) ⊃ x=y] &∀y[(My ⊃ z=y) &x 6=z])Chapter 12 Part C1. ∃x(Bbx&Bwx)2. ∀x∀y[∃z(Gz&Sx&Sy&Bxz&Byz) ⊃ x=y]3. ∃x(Sx&x 6=b&Fx)4. ∀x[(Fx&¬Wx) ⊃ Awx]5. ∀x(x 6=f ⊃ Kfx)6. ∀x(∃y(Sy&Axy&¬∃z((V z ∨Dz) &Kxz)) ⊃ Kbx)7. ∀x(Gx ⊃ ∃y(Sy&Byx))8. ∀x(Gx ⊃ ∃y(Sy&Byx&∀z[(V z ∨Dz ∨ Fz) ⊃ Ayz]))9. ∀x(Gx ⊃ ∃y(Sy&Byx&∀z[(V z ∨Dz ∨ Fz) ⊃ Ayz] &∀y2(∀z[(V z ∨Dz ∨ Fz) ⊃ Ay2z] ⊃ y2=y)))Chapter 12 Part E1. There are many possible answers. Here is one:UD = {Harry, Sally}extension(R) = {}referent(a) = Harry2. There are no predicates or constants, so we only need to give a UD. AnyUD with 2 members will do.3. We need to show that it is impossible to construct a model in which theseare both true. Suppose ∃x x 6=a is true in a model. There is something inthe universe of discourse that is not the referent of a. So there are at leasttwo things in the universe of discourse: referent(a) and this other thing.280 forallx (UBC edition)Call this other thing β— we know a 6=β. But if a6=β, then ∀x∀y x=y isfalse. So the first sentence must be false if the second sentence is true. Assuch, there is no model in which they are both true. Therefore, they areinconsistent.4. A model where this is true:UD = {Harry}referent(h) = Harryreferent(i) = HarryA model where this is false:UD = {Harry, Ivan}referent(h) = Harryreferent(i) = Ivan5. This is one of many possible answers:UD = {Dorothy, Harry}extension(Z) = {}referent(d) = Dorothyreferent(h) = Harryreferent(s) = Dorothy6. This is one of a number of possible answers. Any model with a 1 objectUD that satisfies the premise will work.UD = {Dorothy}extension(D) = {}extension(T ) = {}referent(d) = DorothyChapter 12 Part F1. |= ∀x∀y(x=y ⊃ y=x) is true.1.2.3.4.5.6.¬∀x∀y(x=y ⊃ y=x) Xa¬∀y(a=y ⊃ y=a) Xb¬(a=b ⊃ b=a) Xa=bb 6=ab 6=b×61 ¬∀2 ¬∀3 ¬⊃4, 5 =solutions for ch. 12 2812. |= ∀x∃y x=y is true.1.2.3.¬∀x∃y x=y Xa¬∃y y=a \aa6=a×31 ¬∀2 ¬∃3. |= ∃x∀y x=y is false.1.2.3.4.5.6.¬∃x∀y x=y \a, b¬∀y a=y Xba6=b¬∀y b=y Xcb6=c...1 ¬∃2 ¬∀1 ¬∃4 ¬∀UD = {a, b, c, . . . }This infinite tree describes a model with an infinite UD. Since there areno predicates we needn’t provide any extensions. Note that a two-objectdomain would also have sufficed to satisfy the root, even though the treewas not complete after line 3, since we hadn’t taken the b instance of thenegated existential. The tree will be infinite since at line 6 we would takethe c instance from line 1, etc.282 forallx (UBC edition)4. ∃x∀y x=y |= ∀x∀y(Rxy ≡ Ryx) is true.1.2.3.4.5.6.7.8.9.10.11.∃x∀y x=y Xa¬∀x∀y(Rxy ≡ Ryx) Xb∀y a=y \b, c¬∀y(Rby ≡ Ryb) Xc¬(Rbc ≡ Rcb) Xa=ba=c¬(Rac ≡ Rca)¬(Raa ≡ Raa) XRaa¬Raa×10, 11Raa¬Raa×10, 111 ∃2 ¬∀4 ¬∀3 ∀3 ∀5, 6 =5, 7 =9 ¬≡5. |= ¬∀x∀y∀z[(Axy&Azx& y=z) ⊃ Axx] is false.1.2.3.4.5.6.7.¬¬∀x∀y∀z((Axy&Azx& y=z) ⊃ Axx) X∀x∀y∀z((Axy&Azx& y=z) ⊃ Axx) \a∀y∀z((Aay&Aza& y=z) ⊃ Aaa) \a∀z((Aaa&Aza& a=z) ⊃ Aaa) \a(Aaa&Aaa& a=a) ⊃ Aaa X¬(Aaa&Aaa& a=a) X¬Aaa↑¬Aaa↑a 6=a×7Aaa↑1 ¬¬2 ∀3 ∀4 ∀5 ⊃6 ¬&This tree is complete. Three open branches describe these two models:UD = {a}extension(A) = ∅UD = {a}extension(A) = {}solutions for ch. 12 2836. ∀x∀y x=y |= ∃xFx ≡ ∀xFx is true.1.2.3.4.5.6.7.8.9.10.11.∀x∀y x=y \a¬(∃xFx ≡ ∀xFx) X∃xFx Xa¬∀xFx XbFa¬Fb∀y a=y \ba=b¬Fa×7, 11¬∃xFx \a∀xFx \a¬FaFa×2 ≡3 ¬∃4 ∀3 ∃4 ¬∀1 ∀9 ∀10, 8 =284 forallx (UBC edition)7. ∀x(x=a ∨ x=b), Ga ≡ ¬Gb |= ¬∃x∃y∃z(Gx&Gy&¬Gz) is false.1.2.3.4.5.6.7.8.9.10.11.12.13.14.15.16.17.18.19.20.21.22.23.24.∀x(x=a ∨ x=b) \c, d, e, a, bGa ≡ ¬Gb X¬¬∃x∃y∃z(Gx&Gy&¬Gz) X∃x∃y∃z(Gx&Gy&¬Gz) Xc∃y∃z(Gc&Gy&¬Gz) Xd∃z(Gx&Gd&¬Gz) XeGc&Gd&¬Ge XGcGd¬Gec=a ∨ c=b Xc=aGaGa¬Gbd=a ∨ d=b Xd=ae=a ∨ e=b Xe=a¬Ga×13, 20e=ba=a ∨ a=b Xa=ab=a ∨ b=b Xb=aGb×15, 24b=b↑a=bd=b¬GaGb×13, 14c=b3 ¬¬4 ∃5 ∃7 &1 ∀11 ∨8, 12 =2 ≡1 ∀16 ∨1 ∀18 ∨1 ∀20 ∨1 ∀22 ∨This tree will end up with multiple open branches; since one open branch isenough to falsify the entailment claim, this version focuses on the left-mostsolutions for ch. 12 285open branch at every point, leaving the other open branches incomplete.It describes this model:UD = {a, b}referent(a) = areferent(b) = breferent(c) = areferent(d) = areferent(e) = bextension(G) = {a}8. ∀x(Fx ⊃ x=f),∃x(Fx ∨ ∀y y=x) |= Ff is false.1.2.3.4.5.6.7.8.9.10.11.12.∀x(Fx ⊃ x=f) \a, f∃x(Fx ∨ ∀y y=a) Xa¬FfFa ∨ ∀y y=a XFa ⊃ a=f X¬FaFa×∀y y=a \f, af=aa=aFf ⊃ f=f X¬Ff¬Faf=f↑a=f2 ∃1 ∀5 ⊃4 ∨7 ∀7 ∀1 ∀8, 11 =We have one completed open branch. (Another open branch has not beencompleted.) It gives this model:UD = {a}referent(a) = areferent(f) = aextension(F ) = ∅286 forallx (UBC edition)9. ∃x∃yDxy |= ∀x1∀x2∀x3∀x4[(Dx1x2 &Dx3x4) ⊃ (x2 6=x3∨Dx1x4)] is false.1.2.3.4.5.6.7.8.9.10.11.12.13.14.15.16.17.∃x∃yDxy Xa¬∀x1∀x2∀x3∀x4((Dx1x2 &Dx3x4) ⊃ (x2 6=x3 ∨Dx1x4)) Xc∃yDay XbDab¬∀x2∀x3∀x4((Dcx2 &Dx3x4) ⊃ (x2 6=x3 ∨Dcx4)) Xd¬∀x3∀x4((Dcd&Dx3x4) ⊃ (d6=x3 ∨Dcx4)) Xe¬∀x4((Dcd&Dex4) ⊃ (d 6=e ∨Dcx4)) Xf¬((Dcd&Def) ⊃ (d 6=e ∨Dcf)) XDcd&Def X¬(d 6=e ∨Dcf) XDcdDef¬d6=e¬Dcfd=eDdfDce↑1 ∃3 ∃2 ¬∀5 ¬∀6 ¬∀7 ¬∀8 ¬⊃9 &10 ¬⊃13 ¬¬12, 15 =11, 15 =UD = {a, b, c, d, f}referent(a) = areferent(b) = breferent(c) = creferent(d) = dreferent(e) = dreferent(f) = fextension(D) =Dxy a b c d fa - 1 - - -b - - - - -c - - - 1 0d - - - - 1f - - - - -solutions for ch. 13 287Chapter 13Chapter 13 Part A1 ∀x∃y(Rxy ∨Ryx)2 ∀x¬Rmx3 ∃y(Rmy ∨Rym) ∀E 14 Rma ∨Ram for ∃E5 ¬Rma ∀E 26 Ram ∨E 4, 57 ∃xRxm ∃I 68 ∃xRxm ∃E 3, 4–71 ∀x(∃yLxy ⊃ ∀zLzx)2 Lab3 ∃yLay ⊃ ∀zLza ∀E 14 ∃yLay ∃I 25 ∀zLza ⊃E 3, 46 Lca ∀E 57 ∃yLcy ⊃ ∀zLzc ∀E 18 ∃yLcy ∃I 69 ∀zLzc ⊃E 7, 810 Lcc ∀E 911 ∀xLxx ∀I 10288 forallx (UBC edition)1 ∀x(Jx ⊃ Kx)2 ∃x∀yLxy3 ∀xJx4 ∀yLay for ∃E5 Ja ∀E 36 Ja ⊃ Ka ∀E 17 Ka ⊃E 6, 58 Laa ∀E 49 Ka&Laa & I 7, 810 ∃x(Kx&Lxx) ∃I 911 ∃x(Kx&Lxx) ∃E 2, 4–101 ¬(∃xMx ∨ ∀x¬Mx) for reductio2 ¬∃xMx&¬∀x¬Mx DeM 13 ¬∃xMx & E 24 ∀x¬Mx QN 35 ¬∀x¬Mx & E 26 ∃xMx ∨ ∀x¬Mx ¬E 1–4, 1–5Chapter 13 Part B1. 1 ¬(∀xFx ∨ ¬∀xFx) for reductio2 ¬∀xFx&¬¬∀xFx DeM 13 ¬∀xFx & E 24 ¬¬∀xFx & E 25 ∀xFx ∨ ¬∀xFx ¬E 1–3, 1–4solutions for ch. 13 2892. 1 ∀x(Mx ≡ Nx)2 Ma& ∃xRxa want ∃xNx3 Ma ≡ Na ∀E 14 Ma & E 25 Na ≡E 3, 46 ∃xNx ∃I 53. 1 ∀x(¬Mx ∨ Ljx)2 ∀x(Bx ⊃ Ljx)3 ∀x(Mx ∨Bx) want ∀xLjx4 ¬Ma ∨ Lja ∀E 15 Ma ⊃ Lja MC 46 Ba ⊃ Lja ∀E 27 Ma ∨Ba ∀E 38 Lja DIL 7, 5, 69 ∀xLjx ∀I 84. 1 ∀x(Cx&Dt) want ∀xCx&Dt2 Ca&Dt ∀E 13 Ca & E 24 ∀xCx ∀I 35 Dt & E 26 ∀xCx&Dt & I 4, 5290 forallx (UBC edition)5. 1 ∃x(Cx ∨Dt) want ∃xCx ∨Dt2 Ca ∨Dt for ∃E3 ¬(∃xCx ∨Dt) for reductio4 ¬∃xCx&¬Dt DeM 35 ¬Dt & E 46 Ca ∨E 2, 57 ∃xCx ∃I 68 ¬∃xCx & E 49 ∃xCx ∨Dt ¬E 3–7, 3–810 ∃xCx ∨Dt ∃E 1, 2–9Chapter 13 Part E1. 1 ∀x∀yGxy2 ∀yGay ∀E 13 Gaa ∀E 24 ∃xGxx ∃I 32. 1 ∀x∀y(Gxy ⊃ Gyx)2 ∀y(Gay ⊃ Gya) ∀E 13 Gab ⊃ Gba ∀E 24 ∀y(Gby ⊃ Gyb) ∀E 15 Gba ⊃ Gab ∀E 46 Gab ≡ Gba ≡I 3, 57 ∀y(Gay ≡ Gya) ∀I 68 ∀x∀y(Gxy ≡ Gyx) ∀I 7solutions for ch. 13 2913. 1 ∀x(Ax ⊃ Bx)2 ∃xAx want ∃xBx3 Aa for ∃E4 Aa ⊃ Ba ∀E 15 Ba ⊃E 3, 46 ∃xBx ∃I 57 ∃xBx ∃E 2, 3–64. 1 Na ⊃ ∀x(Mx ≡Ma)2 Ma3 ¬Mb want ¬Na4 Na for reductio5 ∀x(Mx ≡Ma) ⊃E 1, 46 Mb ≡Ma ∀E 57 Mb ≡E 2, 68 ¬Mb R 39 ¬Na ¬I 4–7, 4–85. 1 Pa want Pa2 Pa R 13 Pa ⊃ Pa ⊃I 1–24 ¬Pa ∨ Pa MC 35 Pa ∨ ¬Pa Comm 46 ∀z(Pz ∨ ¬Pz) ∀I 56. 1 ∀xRxx want ⊃ ∃x∃yRxy2 Raa ∀E 13 ∃yRay ∃I 24 ∃x∃yRxy ∃I 35 ∀xRxx ⊃ ∃x∃yRxy ⊃I 1–4292 forallx (UBC edition)7. 1 Qa want Qa2 Qa R 13 Qa ⊃ Qa ⊃I 1–24 ∃x(Qa ⊃ Qx) ∃I 35 ∀y∃x(Qy ⊃ Qx) ∀I 4Chapter 13 Part HRegarding the translation of this argument, see p. 151.1 ∃x∀y[∀z(Lxz ⊃ Lyz) ⊃ Lxy]2 ∀y[∀z(Laz ⊃ Lyz) ⊃ Lay] for ∃E3 ∀z(Laz ⊃ Laz) ⊃ Laa ∀E 24 ¬∃xLxx for reductio5 ∀x¬Lxx QN 46 ¬Laa ∀E 57 ¬∀z(Laz ⊃ Laz) MT 3, 68 Lab want Lab9 Lab R 810 Lab ⊃ Lab ⊃I 8–911 ∀z(Laz ⊃ Laz) ∀I 1012 ¬∀z(Laz ⊃ Laz) R 713 ∃xLxx ¬E 4–1214 ∃xLxx ∃E 1, 2–13Chapter 13 Part J2, 3, and 5 are logically equivalent.solutions for ch. 13 293Chapter 13 Part K8, 10, 11, 13, and 16 are valid. Here are complete answers for some of them:1. UD = {mocha, freddo}extension(R) = {, }2. 1 ∃y∀xRxy want ∀x∃yRxy2 ∀xRxa for ∃E3 Rba ∀E 24 ∃yRby ∃I 35 ∀x∃yRxy ∀I 46 ∀x∃yRxy ∃E 1, 2–5Appendix CQuick ReferenceCharacteristic Truth TablesΦ ¬Φ1 00 1Φ Ψ Φ&Ψ Φ∨Ψ Φ⊃Ψ Φ≡Ψ1 1 1 1 1 11 0 0 1 0 00 1 0 1 1 00 0 0 0 1 1SymbolizationSentential Connectives (chapter 2)It is not the case that P . ¬PEither P , or Q. (P ∨Q)Neither P , nor Q. ¬(P ∨Q) or (¬P &¬Q)Both P , and Q. (P &Q)If P , then Q. (P ⊃ Q)P only if Q. (P ⊃ Q)P if and only if Q. (P ≡ Q)Unless P , Q. P unless Q. (P ∨Q)Predicates (chapter 8)All F s are Gs. ∀x(Fx ⊃ Gx)Some F s are Gs. ∃x(Fx&Gx)Not all F s are Gs. ¬∀x(Fx ⊃ Gx) or ∃x(Fx&¬Gx)No F s are Gs. ∀x(Fx ⊃ ¬Gx) or ¬∃x(Fx&Gx)Identity (chapter 12)Only j is G. ∀x(Gx ≡ x=j)Everything besides j is G. ∀x(x6=j ⊃ Gx)The F is G. ∃x(Fx&∀y(Fy ⊃ x=y) &Gx)The F is not G. ∃x(Fx&∀y(Fy ⊃ x=y) &¬Gx)Tree RulesSentential Rules (chapter 5)Φ&ΨΦΨ¬(Φ ∨ Ψ)¬Φ¬ΨΦ ≡ ΨΦΨ¬Φ¬Ψ¬(Φ&Ψ)¬Φ ¬ΨΦ ⊃ Ψ¬Φ Ψ¬(Φ ≡ Ψ)Φ¬Ψ¬ΦΨΦ ∨ ΨΦ Ψ¬(Φ ⊃ Ψ)Φ¬Ψ¬¬ΦΦQuantifier Rules (chapter 8)∃x Φ XaΦ x ⇒ a*a must be new∀x Φ \aΦ x ⇒ afor any a¬∃x Φ \a¬Φ x ⇒ afor any a¬∀x Φ Xa¬Φ x ⇒ a*a must be newIdentity Rules (chapter 12)If a branch contains an identity claim a=b and some sentence Φ, then you mayadd Φ a ⇒ b or Φ b ⇒ a to the branch.A branch is complete if and only if either (i) it is closed, or (ii) all of thefollowing conditions are met:1. Every resolvable sentence in the branch has been resolved;2. For every general quantified sentence Φ and every name a in the branch,the a instance of Φ is in the branch; and3. For every identity claim a=b in the branch, either:(a) for every atomic sentence or negated atomic sentence Φ containing ain the branch, Φ a ⇒ b is in the branch, or(b) for every atomic sentence or negated atomic sentence Φ containing bin the branch, Φ b ⇒ a is in the branch.A branch closes if and only if either (i) it contains some sentence Φ along withits negation, ¬Φ, or (ii) it contains, for some name a, the sentence a 6=a.Natural Deduction RulesBasic RulesReiterationm ΦΦ R mConjunction Introductionm Φn ΨΦ&Ψ & I m, nConjunction Eliminationm Φ&ΨΦ & E mΨ & E mDisjunction Introductionm ΦΦ ∨ Ψ ∨I mΨ ∨ Φ ∨I mDisjunction Eliminationm Φ ∨ Ψn ¬ΨΦ ∨E m, nm Φ ∨ Ψn ¬ΦΨ ∨E m, nConditional Introductionm Φ want Ψn ΨΦ ⊃ Ψ ⊃I m–nConditional Eliminationm Φ ⊃ Ψn ΦΨ ⊃E m, nBiconditional Introductionm Φ ⊃ Ψn Ψ ⊃ ΦΦ ≡ Ψ ≡I m, nBiconditional Eliminationm Φ ≡ Ψn ΨΦ ≡E m, nm Φ ≡ Ψn ΦΨ ≡E m, nNegation Introductionm Φ for reduction Ψo ¬Ψp ¬Φ ¬I m–n, m–oNegation Eliminationm ¬Φ for reduction Ψo ¬Ψp Φ ¬E m–n, m–oQuantifier RulesExistential Introductionm Φ∃x Φ c∗ ⇒ x ∃I m∗ x may replace some or all occurrences of c.Existential Eliminationm ∃x Φn Φ x ⇒ c∗p ΨΨ ∃E m, n–p∗ c must not appear outside the subproof.Universal Introductionm Φ∀x Φ c ⇒ x ∀I mc must not occur in any undischarged assump-tions.Universal Eliminationm ∀x ΦΦ x ⇒ c ∀E mIdentity Rulesc=c =Im c=dn ΦΦ c⇒ d =E m, nΦ d⇒ c =E m, nOne constant may replace some or all occurrencesof the other.Derived RulesDilemmam Φ ∨ Ψn Φ ⊃ Ωp Ψ ⊃ ΩΩ DIL m, n, pModus Tollensm Φ ⊃ Ψn ¬Ψ¬Φ MT m, nHypothetical Syllogismm Φ ⊃ Ψn Ψ ⊃ ΩΦ ⊃ Ω HS m, nReplacement RulesCommutativity (Comm)(Φ&Ψ)⇐⇒ (Ψ &Φ)(Φ ∨ Ψ)⇐⇒ (Ψ ∨ Φ)(Φ ≡ Ψ)⇐⇒ (Ψ ≡ Φ)DeMorgan (DeM)¬(Φ ∨ Ψ)⇐⇒ (¬Φ&¬Ψ)¬(Φ&Ψ)⇐⇒ (¬Φ ∨ ¬Ψ)Double Negation (DN)¬¬Φ⇐⇒ ΦMaterial Conditional (MC)(Φ ⊃ Ψ)⇐⇒ (¬Φ ∨ Ψ)(Φ ∨ Ψ)⇐⇒ (¬Φ ⊃ Ψ)Biconditional Exchange (≡ex)[(Φ ⊃ Ψ) & (Ψ ⊃ Φ)]⇐⇒ (Φ ≡ Ψ)Quantifier Negation (QN)¬∀x Φ⇐⇒ ∃x ¬Φ¬∃x Φ⇐⇒ ∀x ¬ΦIn the Introduction to his volume Symbolic Logic,Charles Lutwidge Dodson advised: “When youcome to any passage you don’t understand, read itagain: if you still don’t understand it, read it again:if you fail, even after three readings, very likely yourbrain is getting a little tired. In that case, put thebook away, and take to other occupations, andnext day, when you come to it fresh, you will verylikely find that it is quite easy.”The same might be said for this volume, althoughreaders are forgiven if they take a break for snacksafter two readings.about the authors:P.D. Magnus is an associate professor of philosophyin Albany, New York. His primary research is inthe philosophy of science.Jonathan Ichikawa is an associate professorof philosophy in Vancouver, British Columbia. Hisprimary research is in epistemology.

forall x (UBC Edition)

Abstract

Similar works

Full text

Available Versions

PhilPapers