A slight modification to one of Tarski's axioms of plane Euclidean geometry
is proposed. This modification allows another of the axioms to be omitted from
the set of axioms and proven as a theorem. This change to the system of axioms
simplifies the system as a whole, without sacrificing the useful modularity of
some of its axioms. The new system is shown to possess all of the known
independence properties of the system on which it was based; in addition,
another of the axioms is shown to be independent in the new system.Comment: 10 page

Makarios, Timothy

English

arXiv

A slight modification to one of Tarski’s axioms of plane absolute geometry is proposed. This modification allows another of the axioms to be omitted from the set of axioms and proven as a theorem. This change to the system of axioms simplifies the system as a whole, without sacrificing the useful modularity of some of its axioms. The new system is shown to possess all of the known independence properties of the system on which it was based; in addition, another of the axioms is shown to be independent in the new system

Makarios, T.J.M.

Università del Salento: ESE - Salento University Publishing

Note di Matematica ISSN 1123-2536, e-ISSN 1590-0932Note Mat. 33 (2013) no. 2, 123–132. doi:10.1285/i15900932v33n2p123A further simplification of Tarski’s axiomsof geometryT. J. M. Makariostjm1983@gmail.comReceived: 28.8.2013; accepted: 22.11.2013.Abstract. A slight modification to one of Tarski’s axioms of plane absolute geometry isproposed. This modification allows another of the axioms to be omitted from the set of axiomsand proven as a theorem. This change to the system of axioms simplifies the system as a whole,without sacrificing the useful modularity of some of its axioms. The new system is shown topossess all of the known independence properties of the system on which it was based; inaddition, another of the axioms is shown to be independent in the new system.Keywords: foundations of geometry, absolute geometry, Tarski’s axioms, independenceMSC 2010 classification: primary 03B30, secondary 51F05, 51F101 BackgroundAlfred Tarski’s axioms of geometry were first described in a course he gaveat the University of Warsaw in 1926–1927. Since then, they have undergonenumerous improvements, with some axioms modified, and other superfluousaxioms removed; for a history of the changes, see [11] (especially Section 2), orfor a summary, see Figure 2 in [4].The axioms are expressed in a one-sorted language, with individual vari-ables to be interpreted as points, and two primitive relations: betweenness andcongruence. Congruence is denoted ab ≡ cd, and can be interpreted as assertingthat the line segment from a to b is congruent to the line segment from c to d.Betweenness is denoted B abc, and can be interpreted as asserting that b lies onthe segment from a to c (and may be equal to a or c).The version of the axioms used in [7] (see pages 10–14) consists of ten first-order axioms, together with either a first-order axiom schema, or a single higher-order axiom. This version has been adopted in later publications, such as [3] (seeSections 2.3 and 2.4) and [4] (see Figure 3), and Victor Pambuccian has calledit the “most polished form” of Tarski’s axioms (see [6], page 122).This semi-canonical version of the system is the result of many simplifica-tions to the original system of twenty axioms plus one axiom schema. At leastone of these simplifications appears to have taken the form of a slight alterationhttp://siba-ese.unisalento.it/ c© 2013 Universita` del Salento124 T. J. M. Makariosto one axiom in order to allow another axiom to be dropped and subsequentlyproven as a theorem.Specifically, in [9] (see note 18), axiom (ix) is a version of what is called theaxiom of Pasch, which states (modulo notational differences):1B apc ∧ B qcb −→ ∃x. B axq ∧ Bxpb (OP)In [10], this has been replaced by axiom A9, which states (again, modulo nota-tional changes):∃x. B apc ∧ B qcb −→ B axq ∧ B bpx (OP′)The significant change between the two is the reversal of the order of points inthe final betweenness relation. They are easily shown to be equivalent using thesymmetry of betweenness, which states:B abc −→ B cba (SB)The interesting thing is that in [9], (SB) is an axiom (axiom (iii)), but in[10], it has been removed from the list of axioms. Haragauri Gupta’s proof of(SB) (see [2], Theorem 2.18) relies on the precise ordering of points in the finalbetweenness relation in (OP′).Also, Wolfram Schwabha¨user, Wanda Szmielew, and Alfred Tarski, on page12 of [7], draw attention to the ordering of points in their version of the axiomof Pasch (labelled (IP) in this paper), noting that it is important until afterthe proof of (SB) (which they call Satz 3.2). Again, their proof of (SB) (whichis essentially the same as this paper’s proof of Lemma 4) relies on the preciseordering of points in (IP).Therefore, although it does not appear to be explicitly acknowledged in thepublished literature, it seems likely that the change from (OP) to (OP′) wasnecessary to allow the removal of (SB) from the set of axioms. It may be thecase that (SB) is a theorem even with (OP), and that this was not known whenit was replaced by (OP′), or that it was known, but (OP′) allows a simpler proof.In any case, it appears that Tarski was willing to reorder points in his axiomsto allow the simplification of the axiom system as a whole, either by removingaxioms or merely by simplifying proofs of theorems.In the tradition of such simplifications, this paper presents one further sim-plification of the axiom system; one of the axioms is slightly modified, allowinganother of the traditional axioms to be proven as a theorem, rather than as-sumed as an axiom.1Throughout this article, universal quantifiers scoped over a whole formula are omitted inorder to improve readability.Simplifying Tarski’s axioms of geometry 1252 The axiomsTarski’s axioms, as stated in [7], pages 10–14, are as follows. The namesare adopted from [3] (Section 2.4), which provides some diagrams and intuitiveexplanations for the axioms.• Reflexivity axiom for equidistanceab ≡ ba (RE)• Transitivity axiom for equidistanceab ≡ pq ∧ ab ≡ rs −→ pq ≡ rs (TE)• Identity axiom for equidistanceab ≡ cc −→ a = b (IE)• Axiom of segment construction∃x. B qax ∧ ax ≡ bc (SC)• Five-segments axioma 6= b ∧ B abc ∧ B a′b′c′ ∧ ab ≡ a′b′ ∧ bc ≡ b′c′ ∧ ad ≡ a′d′ ∧ bd ≡ b′d′−→ cd ≡ c′d′ (FS)• Identity axiom for betweennessB aba −→ a = b (IB)• Axiom of PaschB apc ∧ B bqc −→ ∃x. B pxb ∧ B qxa (IP)• Lower 2-dimensional axiom∃a, b, c. ¬B abc ∧ ¬B bca ∧ ¬B cab (Lo2)• Upper 2-dimensional axiomp 6= q ∧ ap ≡ aq ∧ bp ≡ bq ∧ cp ≡ cq −→ (B abc ∨ B bca ∨ B cab) (Up2)126 T. J. M. Makarios• Euclidean axiomB adt ∧ B bdc ∧ a 6= d −→ ∃x, y. B abx ∧ B acy ∧ Bxty (Eu)• Axiom of continuity(∃a. ∀x, y. x ∈ X ∧ y ∈ Y −→ B axy)−→ (∃b. ∀x, y. x ∈ X ∧ y ∈ Y −→ Bxby) (Co)This collection of eleven axioms, Tarski’s axioms of the continuous Euclideanplane, will be denoted CE2. A similar collection of axioms, denoted CE′2, isobtained by removing (RE) from CE2 and replacing (FS) witha 6= b ∧ B abc ∧ B a′b′c′ ∧ ab ≡ a′b′ ∧ bc ≡ b′c′ ∧ ad ≡ a′d′ ∧ bd ≡ b′d′−→ dc ≡ c′d′ (FS′)The only difference between (FS) and (FS′) is the reversal of the first two pointsin the last congruence relation.This paper will show that CE′2is equivalent to CE2, and will, in fact, showa stronger result — Theorem 1 — about a smaller set of axioms.One of the features of Tarski’s axiom system is its modularity: some textsomit or delay the introduction of (Co) (see, for example, [1], page 61); (Eu)can be replaced by another axiom in order to investigate hyperbolic geometry,or omitted entirely, for absolute geometry (see [5], pages 331–333); (Lo2) and(Up2) can be replaced by other axioms that characterize other dimensions (see[10], footnote 5). For this reason, let A denote the collection of axioms (RE),(TE), (IE), (SC), (FS), (IB), and (IP). These are Tarski’s axioms of absolutedimension-free geometry without the axiom of continuity. Let A′ denote thecollection of axioms (TE), (IE), (SC), (FS′), (IB), and (IP).The stronger result that this paper shows is that A′ is equivalent to A.Thus, the modularity of axioms (Lo2), (Up2), (Eu), and (Co) is unaffected bythe proposed change to Tarski’s axiom system.3 Proof of equivalenceLemma 1. If (TE) and (SC) hold, then given any points a and b, we haveab ≡ ab.Proof. Given a and b, (SC) lets us obtain a point x such that ax ≡ ab. Usingthis twice in (TE) gives us ab ≡ ab. QEDSimplifying Tarski’s axioms of geometry 127Lemma 2. If (TE) and (SC) hold and a, b, c, and d are points such thatab ≡ cd, then cd ≡ ab.Proof. By Lemma 1, we have ab ≡ ab. Using ab ≡ cd and ab ≡ ab, (TE) tells usthat cd ≡ ab. QEDLemma 3. If (IE) and (SC) hold, then given any points a and b, we haveB abb.Proof. Given a and b, (SC) lets us obtain a point x such that B abx and bx ≡ bb.Then (IE) tells us that b = x, so B abb. QEDLemma 4. (IE), (SC), (IB), and (IP) together imply (SB).Proof. Suppose we are given points a, b, and c such that B abc. We also haveB bcc, by Lemma 3. Then (IP) lets us obtain a point x such that B bxb andB cxa. According to (IB), the former implies b = x, so the latter tells us thatB cba. QEDLemma 5. A′ implies (RE).Proof. Given arbitrary points a and b, (SC) lets us obtain a point x such thatB bax and ax ≡ ba. We consider two cases: x = a and x 6= a.If x = a, then aa ≡ ba. By Lemma 2, we have ba ≡ aa, so by (IE), we haveb = a. Substituting this back into the congruence as necessary gives us ab ≡ ba,as desired.Suppose, on the other hand, that x 6= a. Lemma 4 and B bax tell us thatBxab. Lemma 1 tells us that xa ≡ xa, ab ≡ ab, and aa ≡ aa. We makethe following substitutions in (FS′): a, a′ 7→ x; b, b′, d, d′ 7→ a; and c, c′ 7→ b.Then all of the hypotheses of (FS′) are satisfied, and its conclusion is thatab ≡ ba. QEDLemma 6. If (RE) and (TE) hold, then (FS′) is equivalent to (FS).Proof. Because the hypotheses of (FS) and (FS′) are identical, we need onlyshow that their conclusions are equivalent.(RE) tells us that cd ≡ dc and dc ≡ cd.If cd ≡ c′d′, then cd ≡ dc together with this fact and (TE) let us concludethat dc ≡ c′d′.Similarly, if dc ≡ c′d′, then dc ≡ cd together with this fact and (TE) let usconclude that cd ≡ c′d′.Therefore (FS′) is equivalent to (FS). QEDTheorem 1. A′ is equivalent to A.128 T. J. M. MakariosProof. By Lemmas 5 and 6, A′ implies (RE) and (FS). A′ contains all of theother axioms of A, so A′ implies A.By Lemma 6, A implies (FS′), and it contains all of the other axioms of A′,so A implies A′.Therefore A′ is equivalent to A. QEDAs an immediate corollary, we have the following:Corollary 1. CE′2is equivalent to CE2. QED4 Independence resultsThe first part of Section 5 of [11] (see pages 199 and 200) concerns theindependence of Tarski’s axioms. One problem seen there is that the varioushistorical changes to Tarski’s axioms often force a reconsideration of previ-ously established independence results. This paper’s suggested simplificationof Tarski’s axioms is no exception, so this section aims to establish which of theknown independence results apply to the specific set of axioms CE′2.Because CE2 and CE′2differ only in their subsets A and A′, Theorem 1 tellsus that the axioms in CE′2but not in A′ are independent if and only if they areindependent in CE2. In fact, we can go further than this.Theorem 2. Suppose that (Ax) is an axiom of CE′2other than (TE) or(FS′). If (Ax) is independent in CE2 then (Ax) is also independent in CE′2.Proof. Note that (Ax) is not (TE), because this was explicitly excluded; noris it (RE) or (FS), because these are not axioms of CE′2, from which (Ax) waschosen. Therefore CE2 \ { (Ax) } contains (RE), (TE), and (FS), so by Lemma6, CE2 \ { (Ax) } ⊢ (FS′).All of the axioms of CE′2other than (FS′) are also axioms of CE2, soCE2 \ { (Ax) } ⊢ CE′2 \ { (Ax) }. Therefore, if CE′2 \ { (Ax) } ⊢ (Ax), thenCE2 \ { (Ax) } ⊢ (Ax). Taking the contrapositive of this statement, we see thatif (Ax) is independent in CE2 then it is independent in CE′2. QEDThis allows us to adopt almost all of the independence results known forCE2.Corollary 2. (SC), (IB), (Lo2), (Up2), (Eu), and (Co) are each individu-ally independent in CE′2.Proof. The independence of each of these axioms in CE2 is noted in [7], page26. QEDThe remaining independence result noted in [7] can also be adapted to CE′2:Simplifying Tarski’s axioms of geometry 129Theorem 3. (FS′) is independent in CE′2.Proof. The existence of a modelM demonstrating the independence of (FS) inCE2 is noted in [7], page 26. Because M satisfies (RE) and (TE), but violates(FS), we can conclude, using Lemma 6, that M also violates (FS′). M satisfiesevery other axiom of CE′2, because all such axioms are also axioms of CE2 (andare not equal to (FS)); therefore, M demonstrates the independence of (FS′)in CE′2. QEDBecause of the absence of (RE) from CE′2, we can easily show the indepen-dence of (TE) in that axiom system:Theorem 4. (TE) is independent in CE′2.Proof. We proceed as is usual in independence proofs, by defining a model Mof every axiom of CE′2except (TE). As in the real Cartesian plane (the standardmodel of CE2, and hence of CE′2), we take R2 to be the set of points, and definebetweenness so that B abc if and only if b = ta + (1− t) c, for some t with0 ≤ t ≤ 1. Departing from the standard model, we define congruence so thatab ≡ cd if and only if a = b.Because the real Cartesian plane is a model of CE′2, and M differs from thereal Cartesian plane only in its definition of congruence, we can conclude thatM is a model of all of the axioms of CE′2that make no mention of congruence.Those axioms are (IB), (IP), (Lo2), (Eu), and (Co).The definition of congruence ensures that M trivially satisfies (IE).Choosing x = a ensures that (SC) is satisfied.The hypotheses of (FS′) include a 6= b and ab ≡ a′b′, which implies a = b;this contradiction in the hypotheses means that (FS′) is vacuously true.The hypotheses of (Up2) imply that a = b = c = p; it is always the casethat B ppp, so (Up2) is satisfied.Finally, in M, it is the case that (0, 0) (0, 0) ≡ (0, 0) (0, 1), but not the casethat (0, 0) (0, 1) ≡ (0, 0) (0, 1), so M does not satisfy (TE).Because M satisfies every axiom of CE′2except (TE), we can conclude that(TE) is independent in CE′2. QEDNotice thatM in the proof of Theorem 4 also violates (RE) (because it is notthe case that (0, 0) (0, 1) ≡ (0, 1) (0, 0)), so this model would not demonstratethe independence of (TE) in CE2, which is, as far as the author is aware, stillan open question.We now have independence results for all of the axioms of CE′2except (IE)and (IP). As far as the author knows, the independence of these in CE2 isstill an open question (see also [11], pages 199 and 200), although there are130 T. J. M. Makariosindependence results relating to these axioms in other versions of Tarski’s axiomsystem.Gupta shows the independence of (IE) (his axiom A5; see [2], pages 41and 41a), but only in the context of a particular variant of Tarski’s axiomsystem. This variant uses a weaker but more complicated form of the upper 2-dimensional axiom, Gupta’s A′12;2 his independence model for (IE) also violates(Up2), which is trivially equivalent to his original axiom A12.Les law Szczerba established the independence of a version of the axiom ofPasch within a certain variant of Tarski’s axiom system (see [8]). This variantused, instead of (Eu), an axiom essentially asserting that any three non-collinearpoints have a circumcentre.35 ConclusionsAlthough this paper has not answered the question of whether (RE) is inde-pendent in the axiom system of [7], it has demonstrated that (RE) is superfluousto Tarski’s axioms of geometry. Even in a proper subset of the axioms, a slightmodification to (FS) allows (RE) to be proven as a theorem, and therefore re-moved from the set of axioms. This simplification of the axiom system doesnot diminish its deductive power or the important ways in which it exhibitsmodularity.Besides removing one of the axioms not known to be independent in theunmodified axiom system, the modified system allows an easy proof of the inde-pendence of (TE), which was also not known to be independent in the unmodi-fied system. The two remaining independence questions for the new system are,as far as the author knows, still open questions for the old one; furthermore,if either axiom is shown to be independent in the old system, then Theorem 2would immediately establish its independence in the new one.As well as trying to resolve these remaining independence questions, futurework might seek other slight modifications to the axioms that may allow evenknown independent axioms to be dropped. That this may be possible can be2It seems that A′12 in Gupta’s thesis ought to include u 6= v among the hypotheses, as A12does. Taken exactly as it is printed, A′12 is violated by the real Cartesian plane whenever x,y, and z are any non-collinear points and u = v.3There appears to be a typographical error in the statement of this axiom in [8] (page 492).His axiom A8′ states (in our notation):∃p. ¬ (B abc ∨ B bca ∨ B cab) −→ ap ≡ bp ∧ bp ≡ cbIt seems that the second congruence relation in the consequent should state bp ≡ cp; see also[11], pages 199 and 184.Simplifying Tarski’s axioms of geometry 131seen by considering Gupta’s fully independent version of Tarski’s axioms forplane Euclidean geometry (see [2], pages 41–41c).His version had eleven axioms,4 some of which were deliberately made morecomplicated in order to allow easy proofs of the independence of other axioms(see, for example, his note on the independence of his axiom A7 on page 41b).The system adopted by [7] already has simpler axioms than Gupta’s system(which he used only for the demonstration that a fully independent system ispossible), but this article’s further simplification now shows that a reductionin the number of axioms is also possible, without making any of them morecomplex, despite Gupta’s system being fully independent.Acknowledgements. Thanks to Rob Goldblatt and Victor Pambuccianfor direct and indirect encouragement and advice; thanks also the referee forsome specific suggestions.References[1] P. Ehrlich: From completeness to Archimedean completeness: An essay in the founda-tions of Euclidean geometry, Synthese, 110 (1997), 57–76, http://www.ohio.edu/people/ehrlich/FromCompletenessToArch.pdf.[2] H. N. Gupta: Contributions to the axiomatic foundations of geometry, Ph.D. Thesis,University of California, Berkeley, 1965.[3] T. J. M. Makarios: A mechanical verification of the independence of Tarski’s Euclideanaxiom, Master’s Thesis, Victoria University of Wellington, New Zealand, 2012, http://researcharchive.vuw.ac.nz/handle/10063/2315.[4] J. Narboux: Mechanical theorem proving in Tarski’s geometry, in F. Botana, T. Recio,(editors): “Automated deduction in geometry”, Lecture Notes in Computer Science, 4869,Springer Berlin–Heidelberg, 2007, pp. 139–156.[5] V. Pambuccian: Axiomatizations of hyperbolic geometry: A comparison based on languageand quantifier type complexity, Synthese, 133 (2002), 331–341.[6] V. Pambuccian: Axiomatizations of hyperbolic and absolute geometries, in A. Pre´kopa,E. Molna´r (editors): “Non-euclidean geometries”, Mathematics and Its Applications, 581,Springer US, 2006, pp. 119–153.[7] W. Schwabha¨user, W. Szmielew, A. Tarski: Metamathematische methoden in dergeometrie, Springer-Verlag, 1983.[8] L. W. Szczerba: Independence of Pasch’s axiom, Bulletin de l’Academie Polonaise desSciences, Se´rie des Sciences Mathe´matiques, Astronomiques et Physiques, 18, n. 8, (1970),491–498.4Strictly speaking, as he presented it, it had infinitely many axioms, but if his axiom schemaA′13 is replaced by a comparable second-order axiom, then there are eleven axioms in total,and the second-order axiom is shown to be independent by his example on page 41c.132 T. J. M. Makarios[9] A. Tarski: A decision method for elementary algebra and geometry, The RAND Corpo-ration, 1957. First published August 1948.[10] A. Tarski: What is elementary geometry? in L. Henkin, P. Suppes, A. Tarski, (editors):“The axiomatic method: with special reference to geometry and physics”, North-HollandPublishing Company, 1959, pp. 16–29.[11] A. Tarski, S. Givant: Tarski’s system of geometry, Bull. Sympolic Logic, 5, n. 2, (1999),175–214.

A further simplification of Tarski’s axioms of geometry

ESE - Salento University Publishing

Note di Matematica ISSN 1123-2536, e-ISSN 1590-0932Note Mat. 33 (2013) no. 2, 123–132. doi:10.1285/i15900932v33n2p123A further simplification of Tarski’s axiomsof geometryT. J. M. Makariostjm1983@gmail.comReceived: 28.8.2013; accepted: 22.11.2013.Abstract. A slight modification to one of Tarski’s axioms of plane absolute geometry isproposed. This modification allows another of the axioms to be omitted from the set of axiomsand proven as a theorem. This change to the system of axioms simplifies the system as a whole,without sacrificing the useful modularity of some of its axioms. The new system is shown topossess all of the known independence properties of the system on which it was based; inaddition, another of the axioms is shown to be independent in the new system.Keywords: foundations of geometry, absolute geometry, Tarski’s axioms, independenceMSC 2010 classification: primary 03B30, secondary 51F05, 51F101 BackgroundAlfred Tarski’s axioms of geometry were ﬁrst described in a course he gaveat the University of Warsaw in 1926–1927. Since then, they have undergonenumerous improvements, with some axioms modiﬁed, and other superﬂuousaxioms removed; for a history of the changes, see [11] (especially Section 2), orfor a summary, see Figure 2 in [4].The axioms are expressed in a one-sorted language, with individual vari-ables to be interpreted as points, and two primitive relations: betweenness andcongruence. Congruence is denoted ab ≡ cd, and can be interpreted as assertingthat the line segment from a to b is congruent to the line segment from c to d.Betweenness is denoted B abc, and can be interpreted as asserting that b lies onthe segment from a to c (and may be equal to a or c).The version of the axioms used in [7] (see pages 10–14) consists of ten ﬁrst-order axioms, together with either a ﬁrst-order axiom schema, or a single higher-order axiom. This version has been adopted in later publications, such as [3] (seeSections 2.3 and 2.4) and [4] (see Figure 3), and Victor Pambuccian has calledit the “most polished form” of Tarski’s axioms (see [6], page 122).This semi-canonical version of the system is the result of many simpliﬁca-tions to the original system of twenty axioms plus one axiom schema. At leastone of these simpliﬁcations appears to have taken the form of a slight alterationhttp://siba-ese.unisalento.it/ c© 2013 Universita` del Salento124 T. J. M. Makariosto one axiom in order to allow another axiom to be dropped and subsequentlyproven as a theorem.Speciﬁcally, in [9] (see note 18), axiom (ix) is a version of what is called theaxiom of Pasch, which states (modulo notational diﬀerences):1B apc ∧ B qcb −→ ∃x. B axq ∧ Bxpb (OP)In [10], this has been replaced by axiom A9, which states (again, modulo nota-tional changes):∃x. B apc ∧ B qcb −→ B axq ∧ B bpx (OP′)The signiﬁcant change between the two is the reversal of the order of points inthe ﬁnal betweenness relation. They are easily shown to be equivalent using thesymmetry of betweenness, which states:B abc −→ B cba (SB)The interesting thing is that in [9], (SB) is an axiom (axiom (iii)), but in[10], it has been removed from the list of axioms. Haragauri Gupta’s proof of(SB) (see [2], Theorem 2.18) relies on the precise ordering of points in the ﬁnalbetweenness relation in (OP′).Also, Wolfram Schwabha¨user, Wanda Szmielew, and Alfred Tarski, on page12 of [7], draw attention to the ordering of points in their version of the axiomof Pasch (labelled (IP) in this paper), noting that it is important until afterthe proof of (SB) (which they call Satz 3.2). Again, their proof of (SB) (whichis essentially the same as this paper’s proof of Lemma 4) relies on the preciseordering of points in (IP).Therefore, although it does not appear to be explicitly acknowledged in thepublished literature, it seems likely that the change from (OP) to (OP′) wasnecessary to allow the removal of (SB) from the set of axioms. It may be thecase that (SB) is a theorem even with (OP), and that this was not known whenit was replaced by (OP′), or that it was known, but (OP′) allows a simpler proof.In any case, it appears that Tarski was willing to reorder points in his axiomsto allow the simpliﬁcation of the axiom system as a whole, either by removingaxioms or merely by simplifying proofs of theorems.In the tradition of such simpliﬁcations, this paper presents one further sim-pliﬁcation of the axiom system; one of the axioms is slightly modiﬁed, allowinganother of the traditional axioms to be proven as a theorem, rather than as-sumed as an axiom.1Throughout this article, universal quantifiers scoped over a whole formula are omitted inorder to improve readability.Simplifying Tarski’s axioms of geometry 1252 The axiomsTarski’s axioms, as stated in [7], pages 10–14, are as follows. The namesare adopted from [3] (Section 2.4), which provides some diagrams and intuitiveexplanations for the axioms.• Reﬂexivity axiom for equidistanceab ≡ ba (RE)• Transitivity axiom for equidistanceab ≡ pq ∧ ab ≡ rs −→ pq ≡ rs (TE)• Identity axiom for equidistanceab ≡ cc −→ a = b (IE)• Axiom of segment construction∃x. B qax ∧ ax ≡ bc (SC)• Five-segments axioma 6= b ∧ B abc ∧ B a′b′c′ ∧ ab ≡ a′b′ ∧ bc ≡ b′c′ ∧ ad ≡ a′d′ ∧ bd ≡ b′d′−→ cd ≡ c′d′ (FS)• Identity axiom for betweennessB aba −→ a = b (IB)• Axiom of PaschB apc ∧ B bqc −→ ∃x. B pxb ∧ B qxa (IP)• Lower 2-dimensional axiom∃a, b, c. ¬B abc ∧ ¬B bca ∧ ¬B cab (Lo2)• Upper 2-dimensional axiomp 6= q ∧ ap ≡ aq ∧ bp ≡ bq ∧ cp ≡ cq −→ (B abc ∨ B bca ∨ B cab) (Up2)126 T. J. M. Makarios• Euclidean axiomB adt ∧ B bdc ∧ a 6= d −→ ∃x, y. B abx ∧ B acy ∧ Bxty (Eu)• Axiom of continuity(∃a. ∀x, y. x ∈ X ∧ y ∈ Y −→ B axy)−→ (∃b. ∀x, y. x ∈ X ∧ y ∈ Y −→ Bxby) (Co)This collection of eleven axioms, Tarski’s axioms of the continuous Euclideanplane, will be denoted CE2. A similar collection of axioms, denoted CE′2, isobtained by removing (RE) from CE2 and replacing (FS) witha 6= b ∧ B abc ∧ B a′b′c′ ∧ ab ≡ a′b′ ∧ bc ≡ b′c′ ∧ ad ≡ a′d′ ∧ bd ≡ b′d′−→ dc ≡ c′d′ (FS′)The only diﬀerence between (FS) and (FS′) is the reversal of the ﬁrst two pointsin the last congruence relation.This paper will show that CE′2is equivalent to CE2, and will, in fact, showa stronger result — Theorem 1 — about a smaller set of axioms.One of the features of Tarski’s axiom system is its modularity: some textsomit or delay the introduction of (Co) (see, for example, [1], page 61); (Eu)can be replaced by another axiom in order to investigate hyperbolic geometry,or omitted entirely, for absolute geometry (see [5], pages 331–333); (Lo2) and(Up2) can be replaced by other axioms that characterize other dimensions (see[10], footnote 5). For this reason, let A denote the collection of axioms (RE),(TE), (IE), (SC), (FS), (IB), and (IP). These are Tarski’s axioms of absolutedimension-free geometry without the axiom of continuity. Let A′ denote thecollection of axioms (TE), (IE), (SC), (FS′), (IB), and (IP).The stronger result that this paper shows is that A′ is equivalent to A.Thus, the modularity of axioms (Lo2), (Up2), (Eu), and (Co) is unaﬀected bythe proposed change to Tarski’s axiom system.3 Proof of equivalenceLemma 1. If (TE) and (SC) hold, then given any points a and b, we haveab ≡ ab.Proof. Given a and b, (SC) lets us obtain a point x such that ax ≡ ab. Usingthis twice in (TE) gives us ab ≡ ab. QEDSimplifying Tarski’s axioms of geometry 127Lemma 2. If (TE) and (SC) hold and a, b, c, and d are points such thatab ≡ cd, then cd ≡ ab.Proof. By Lemma 1, we have ab ≡ ab. Using ab ≡ cd and ab ≡ ab, (TE) tells usthat cd ≡ ab. QEDLemma 3. If (IE) and (SC) hold, then given any points a and b, we haveB abb.Proof. Given a and b, (SC) lets us obtain a point x such that B abx and bx ≡ bb.Then (IE) tells us that b = x, so B abb. QEDLemma 4. (IE), (SC), (IB), and (IP) together imply (SB).Proof. Suppose we are given points a, b, and c such that B abc. We also haveB bcc, by Lemma 3. Then (IP) lets us obtain a point x such that B bxb andB cxa. According to (IB), the former implies b = x, so the latter tells us thatB cba. QEDLemma 5. A′ implies (RE).Proof. Given arbitrary points a and b, (SC) lets us obtain a point x such thatB bax and ax ≡ ba. We consider two cases: x = a and x 6= a.If x = a, then aa ≡ ba. By Lemma 2, we have ba ≡ aa, so by (IE), we haveb = a. Substituting this back into the congruence as necessary gives us ab ≡ ba,as desired.Suppose, on the other hand, that x 6= a. Lemma 4 and B bax tell us thatBxab. Lemma 1 tells us that xa ≡ xa, ab ≡ ab, and aa ≡ aa. We makethe following substitutions in (FS′): a, a′ 7→ x; b, b′, d, d′ 7→ a; and c, c′ 7→ b.Then all of the hypotheses of (FS′) are satisﬁed, and its conclusion is thatab ≡ ba. QEDLemma 6. If (RE) and (TE) hold, then (FS′) is equivalent to (FS).Proof. Because the hypotheses of (FS) and (FS′) are identical, we need onlyshow that their conclusions are equivalent.(RE) tells us that cd ≡ dc and dc ≡ cd.If cd ≡ c′d′, then cd ≡ dc together with this fact and (TE) let us concludethat dc ≡ c′d′.Similarly, if dc ≡ c′d′, then dc ≡ cd together with this fact and (TE) let usconclude that cd ≡ c′d′.Therefore (FS′) is equivalent to (FS). QEDTheorem 1. A′ is equivalent to A.128 T. J. M. MakariosProof. By Lemmas 5 and 6, A′ implies (RE) and (FS). A′ contains all of theother axioms of A, so A′ implies A.By Lemma 6, A implies (FS′), and it contains all of the other axioms of A′,so A implies A′.Therefore A′ is equivalent to A. QEDAs an immediate corollary, we have the following:Corollary 1. CE′2is equivalent to CE2. QED4 Independence resultsThe ﬁrst part of Section 5 of [11] (see pages 199 and 200) concerns theindependence of Tarski’s axioms. One problem seen there is that the varioushistorical changes to Tarski’s axioms often force a reconsideration of previ-ously established independence results. This paper’s suggested simpliﬁcationof Tarski’s axioms is no exception, so this section aims to establish which of theknown independence results apply to the speciﬁc set of axioms CE′2.Because CE2 and CE′2diﬀer only in their subsets A and A′, Theorem 1 tellsus that the axioms in CE′2but not in A′ are independent if and only if they areindependent in CE2. In fact, we can go further than this.Theorem 2. Suppose that (Ax) is an axiom of CE′2other than (TE) or(FS′). If (Ax) is independent in CE2 then (Ax) is also independent in CE′2.Proof. Note that (Ax) is not (TE), because this was explicitly excluded; noris it (RE) or (FS), because these are not axioms of CE′2, from which (Ax) waschosen. Therefore CE2 \ { (Ax) } contains (RE), (TE), and (FS), so by Lemma6, CE2 \ { (Ax) } ⊢ (FS′).All of the axioms of CE′2other than (FS′) are also axioms of CE2, soCE2 \ { (Ax) } ⊢ CE′2 \ { (Ax) }. Therefore, if CE′2 \ { (Ax) } ⊢ (Ax), thenCE2 \ { (Ax) } ⊢ (Ax). Taking the contrapositive of this statement, we see thatif (Ax) is independent in CE2 then it is independent in CE′2. QEDThis allows us to adopt almost all of the independence results known forCE2.Corollary 2. (SC), (IB), (Lo2), (Up2), (Eu), and (Co) are each individu-ally independent in CE′2.Proof. The independence of each of these axioms in CE2 is noted in [7], page26. QEDThe remaining independence result noted in [7] can also be adapted to CE′2:Simplifying Tarski’s axioms of geometry 129Theorem 3. (FS′) is independent in CE′2.Proof. The existence of a modelM demonstrating the independence of (FS) inCE2 is noted in [7], page 26. Because M satisﬁes (RE) and (TE), but violates(FS), we can conclude, using Lemma 6, that M also violates (FS′). M satisﬁesevery other axiom of CE′2, because all such axioms are also axioms of CE2 (andare not equal to (FS)); therefore, M demonstrates the independence of (FS′)in CE′2. QEDBecause of the absence of (RE) from CE′2, we can easily show the indepen-dence of (TE) in that axiom system:Theorem 4. (TE) is independent in CE′2.Proof. We proceed as is usual in independence proofs, by deﬁning a model Mof every axiom of CE′2except (TE). As in the real Cartesian plane (the standardmodel of CE2, and hence of CE′2), we take R2 to be the set of points, and deﬁnebetweenness so that B abc if and only if b = ta + (1− t) c, for some t with0 ≤ t ≤ 1. Departing from the standard model, we deﬁne congruence so thatab ≡ cd if and only if a = b.Because the real Cartesian plane is a model of CE′2, and M diﬀers from thereal Cartesian plane only in its deﬁnition of congruence, we can conclude thatM is a model of all of the axioms of CE′2that make no mention of congruence.Those axioms are (IB), (IP), (Lo2), (Eu), and (Co).The deﬁnition of congruence ensures that M trivially satisﬁes (IE).Choosing x = a ensures that (SC) is satisﬁed.The hypotheses of (FS′) include a 6= b and ab ≡ a′b′, which implies a = b;this contradiction in the hypotheses means that (FS′) is vacuously true.The hypotheses of (Up2) imply that a = b = c = p; it is always the casethat B ppp, so (Up2) is satisﬁed.Finally, in M, it is the case that (0, 0) (0, 0) ≡ (0, 0) (0, 1), but not the casethat (0, 0) (0, 1) ≡ (0, 0) (0, 1), so M does not satisfy (TE).Because M satisﬁes every axiom of CE′2except (TE), we can conclude that(TE) is independent in CE′2. QEDNotice thatM in the proof of Theorem 4 also violates (RE) (because it is notthe case that (0, 0) (0, 1) ≡ (0, 1) (0, 0)), so this model would not demonstratethe independence of (TE) in CE2, which is, as far as the author is aware, stillan open question.We now have independence results for all of the axioms of CE′2except (IE)and (IP). As far as the author knows, the independence of these in CE2 isstill an open question (see also [11], pages 199 and 200), although there are130 T. J. M. Makariosindependence results relating to these axioms in other versions of Tarski’s axiomsystem.Gupta shows the independence of (IE) (his axiom A5; see [2], pages 41and 41a), but only in the context of a particular variant of Tarski’s axiomsystem. This variant uses a weaker but more complicated form of the upper 2-dimensional axiom, Gupta’s A′12;2 his independence model for (IE) also violates(Up2), which is trivially equivalent to his original axiom A12.Les law Szczerba established the independence of a version of the axiom ofPasch within a certain variant of Tarski’s axiom system (see [8]). This variantused, instead of (Eu), an axiom essentially asserting that any three non-collinearpoints have a circumcentre.35 ConclusionsAlthough this paper has not answered the question of whether (RE) is inde-pendent in the axiom system of [7], it has demonstrated that (RE) is superﬂuousto Tarski’s axioms of geometry. Even in a proper subset of the axioms, a slightmodiﬁcation to (FS) allows (RE) to be proven as a theorem, and therefore re-moved from the set of axioms. This simpliﬁcation of the axiom system doesnot diminish its deductive power or the important ways in which it exhibitsmodularity.Besides removing one of the axioms not known to be independent in theunmodiﬁed axiom system, the modiﬁed system allows an easy proof of the inde-pendence of (TE), which was also not known to be independent in the unmodi-ﬁed system. The two remaining independence questions for the new system are,as far as the author knows, still open questions for the old one; furthermore,if either axiom is shown to be independent in the old system, then Theorem 2would immediately establish its independence in the new one.As well as trying to resolve these remaining independence questions, futurework might seek other slight modiﬁcations to the axioms that may allow evenknown independent axioms to be dropped. That this may be possible can be2It seems that A′12 in Gupta’s thesis ought to include u 6= v among the hypotheses, as A12does. Taken exactly as it is printed, A′12 is violated by the real Cartesian plane whenever x,y, and z are any non-collinear points and u = v.3There appears to be a typographical error in the statement of this axiom in [8] (page 492).His axiom A8′ states (in our notation):∃p. ¬ (B abc ∨ B bca ∨ B cab) −→ ap ≡ bp ∧ bp ≡ cbIt seems that the second congruence relation in the consequent should state bp ≡ cp; see also[11], pages 199 and 184.Simplifying Tarski’s axioms of geometry 131seen by considering Gupta’s fully independent version of Tarski’s axioms forplane Euclidean geometry (see [2], pages 41–41c).His version had eleven axioms,4 some of which were deliberately made morecomplicated in order to allow easy proofs of the independence of other axioms(see, for example, his note on the independence of his axiom A7 on page 41b).The system adopted by [7] already has simpler axioms than Gupta’s system(which he used only for the demonstration that a fully independent system ispossible), but this article’s further simpliﬁcation now shows that a reductionin the number of axioms is also possible, without making any of them morecomplex, despite Gupta’s system being fully independent.Acknowledgements. Thanks to Rob Goldblatt and Victor Pambuccianfor direct and indirect encouragement and advice; thanks also the referee forsome speciﬁc suggestions.References[1] P. Ehrlich: From completeness to Archimedean completeness: An essay in the founda-tions of Euclidean geometry, Synthese, 110 (1997), 57–76, http://www.ohio.edu/people/ehrlich/FromCompletenessToArch.pdf.[2] H. N. Gupta: Contributions to the axiomatic foundations of geometry, Ph.D. Thesis,University of California, Berkeley, 1965.[3] T. J. M. Makarios: A mechanical verification of the independence of Tarski’s Euclideanaxiom, Master’s Thesis, Victoria University of Wellington, New Zealand, 2012, http://researcharchive.vuw.ac.nz/handle/10063/2315.[4] J. Narboux: Mechanical theorem proving in Tarski’s geometry, in F. Botana, T. Recio,(editors): “Automated deduction in geometry”, Lecture Notes in Computer Science, 4869,Springer Berlin–Heidelberg, 2007, pp. 139–156.[5] V. Pambuccian: Axiomatizations of hyperbolic geometry: A comparison based on languageand quantifier type complexity, Synthese, 133 (2002), 331–341.[6] V. Pambuccian: Axiomatizations of hyperbolic and absolute geometries, in A. Pre´kopa,E. Molna´r (editors): “Non-euclidean geometries”, Mathematics and Its Applications, 581,Springer US, 2006, pp. 119–153.[7] W. Schwabha¨user, W. Szmielew, A. Tarski: Metamathematische methoden in dergeometrie, Springer-Verlag, 1983.[8] L. W. Szczerba: Independence of Pasch’s axiom, Bulletin de l’Academie Polonaise desSciences, Se´rie des Sciences Mathe´matiques, Astronomiques et Physiques, 18, n. 8, (1970),491–498.4Strictly speaking, as he presented it, it had infinitely many axioms, but if his axiom schemaA′13 is replaced by a comparable second-order axiom, then there are eleven axioms in total,and the second-order axiom is shown to be independent by his example on page 41c.132 T. J. M. Makarios[9] A. Tarski: A decision method for elementary algebra and geometry, The RAND Corpo-ration, 1957. First published August 1948.[10] A. Tarski: What is elementary geometry? in L. Henkin, P. Suppes, A. Tarski, (editors):“The axiomatic method: with special reference to geometry and physics”, North-HollandPublishing Company, 1959, pp. 16–29.[11] A. Tarski, S. Givant: Tarski’s system of geometry, Bull. Sympolic Logic, 5, n. 2, (1999),175–214.

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A further simplification of Tarski's axioms of geometry

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