This article develops a logic with two fundamental components: objects and labels. We compare the properties of the two universes that can be constructed from these building blocks and show how they naturally resolve a class of linguistic paradoxes. We conclude with a application to modal logics involving context fields

Harvey, Tom

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REFERENTIAL AND SUBSTANTIAL LOGICSTOM HARVEYAbstract. This article develops a logic with two fundamental components: ob-jects and labels. We compare the properties of the two universes that can beconstructed from these building blocks and show how they naturally resolve a classof linguistic paradoxes. We conclude with a application to modal logics involvingcontext fields.1. DefinitionsConsider an interpretation (O,L, []1, 〈〉1). O is a set of objects and L is a set oflabels. We also have two operators:For l ∈ L and o ∈ O,[]1 : L→ P (O), [l]1 = {the set of all o ∈ O referred to by l}〈〉1 : O → P (L), 〈o〉1 = {the set of all l ∈ L that refers to o}.From these we define the following:For m ∈M ⊆ L, p ∈ P ⊆ O[] : P (L)→ P (O), [M ] =⋃m∈M[m]1,〈〉 : P (O)→ P (L), 〈P 〉 =⋃p∈P〈p〉1.where P (X) is the powerset (set of all subsets) of X, and extend the definition so that[l] = [{l}] and〈o〉 = 〈{o}〉.2. ConstructionsThe universe in which we wish to reason can be constructed in two ways:A referential construction is obtained by starting with the set L and then defining Oas follows:Date: October 21, 2022.2010 Mathematics Subject Classification. Primary: 03B65. Secondary: 91F20.Key words and phrases. referential logic, substantial logic, objects, labels, paradox, context.12 TOM HARVEY(1) O = [L]i.e. O consists of anything to which a label refers. We will refer to such a construc-tion as R. A substantial construction is obtained by starting with the set O and thendefining L as follows:(2) L = 〈O〉i.e. L consists of anything referring to an object. We will refer to such a con-struction as S. Where necessary we will delineate specific labels with "" and specificobjects with ''.3. Properties common to R and SFor what follows, l ∈ L, o ∈ O. The following three properties hold for both R andS:(3) o ∈ [l] ⇐⇒ l ∈ 〈o〉Proof. =⇒ : Suppose l /∈ 〈o〉. Then l is not in the set of objects referring to o. So ois not in the set of objects referred to by l, i.e. o /∈ [l], a contradiction.⇐= : Suppose o /∈ [l]. Then o is not in the set of objects referred to by l. So l is notin the set of objects referring to o, i.e. l /∈ 〈o〉, a contradiction. (4) l ∈ 〈[l]〉Proof. l /∈ 〈[l]〉 ⇐⇒ [l] /∈ [l] by (3), a contradiction. (5) o ∈ [〈o〉]Proof. o /∈ [〈o〉] ⇐⇒ 〈o〉 /∈ 〈o〉 by (3), a contradiction. For example consider o = ′  ′. Now 〈o〉 contains labels like "diamond" and"lozenge", which are specific selectors, but also contains the general selector "quadri-lateral", which refers to many other objects as well, such as ′′, etc. So [〈o〉] is muchlarger than the single element o.As another example take l = "a rock". This can be a label for any individual rock,each of which may have many other specific labels, making 〈[l]〉 much larger than thesingle element l. The label l is acting as a general selector.REFERENTIAL AND SUBSTANTIAL LOGICS 34. Additional properties of R(6) L ⊆ OIn R, every label can be referred to by another label e.g. if l ∈ L, x ="the label l",then l ∈ [x], so by (1), l ∈ O.Let us investigate [x], wherex ="An object to which no label refers".Assume o is an object to which no label refers. Then 〈o〉 = ∅. But o ∈ [x], sox ∈ 〈o〉 by (3). This is a contradiction, so o /∈ O by (1) and [x] = ∅.Now considery ="A label that does not refer to any object"We have just shown that x ∈ [y]. So there are redundant labels in R, but no in-accessible objects. Call the set of redundant labels L∅ = L \ 〈O〉.5. Additional properties of S(7) L 6⊆ OWhilst O may be declared to contain L, this is not a fundamental property forgeneral O and we cannot a priori assume this. We will instead assume the opposite,(7). This means labels cannot be referred to in a sentence in S.Considery ="A label that does not refer to any object"Assume y ∈ L. Then [y] 6= ∅ so there exists some z ∈ [y] such that [z] = ∅.But z ∈ 〈O〉 by (2) =⇒ [z] 6= ∅, contradiction. So y /∈ L and there are no redundantlabels in S.Now consider [x], wherex ="an object to which no label refers"Assume x ∈ L. By (2), [x] 6= ∅ so there exists some o ∈ [x] and x ∈ 〈o〉 by (3).But o ∈ [x] =⇒ 〈o〉 = ∅, a contradiction. So x /∈ L.4 TOM HARVEYCall the set of inaccessible objects O∅ = O \ [L].Open question: Is it possible to say anything about O∅ at all? Does O∅ = ∅?One thing we can try is referring to the set O∅:X = "The set of objects to which no label refers"An attempted investigation might go as follows:Assume o1, o2, o3, · · · is an enumeration of all objects to which no label refers., i.e.∀oi, 〈oi〉 = ∅.So [X] = ∪oi.But X ∈ 〈[X]〉 and (4) givesX ∈ 〈[X]〉 = 〈∪oi〉 = ∪〈oi〉1 (by the definition of 〈〉)Now ∀oi, 〈oi〉 = ∅ =⇒ 〈{oi}〉 = ∪〈oi〉1 = 〈oi〉1 = ∅,So ∪〈oi〉1 = ∪∅ = ∅, a contradiction.The problem is that we cannot enumerate the objects oi in this way. Their definingproperty is that they cannot be directly referred to, so we gain no real insight fromthis approach.6. Limit points of the [] and 〈〉 operatorsUse the following notation to indicate repeated application of the [] and 〈〉 opera-tors:[· · · [[l]]· · · ] n times = [l]n〈· · · 〈〈o〉〉 · · · 〉ntimes = 〈o〉nDefine the limit of the operators [], 〈〉 as[l]∞ = limn→∞[l]n〈o〉∞ = limn→∞〈o〉nREFERENTIAL AND SUBSTANTIAL LOGICS 5If an application of the [] or 〈〉 operators is undefined, for example [o] whereo ∈ P (O \ L), then the limit has the value of its input (o in this case).For a set X, we will call [X]∞ \X the object boundary of X, and denote it ∂[]X.For a set X, we will call 〈X〉∞ \X the label boundary of X, and denote it ∂〈〉X.7. General properties of limit pointsSince [[l]∞]∞ = [l]∞, ∂[](∂[]X) = [∂[]X]∞ \ ∂[]X = ∂[]X \ ∂[]X = ∅.Since 〈〈l〉∞〉∞ = 〈l〉∞, ∂〈〉(∂〈〉X) = 〈∂〈〉X〉∞ \ ∂〈〉X = ∂〈〉X \ ∂〈〉X = ∅.By (3), [l]∞ = 〈[l]∞〉 and 〈l〉∞ = [〈l〉∞]8. Limit points in RThere are three possibilities for [l]∞:(1) [l]m ⊆ [l]k where k ≤ m, which would result in recursion. For example ["thislabel"]m ="this label" for all m. [l]∞ can be defined as the set⋃i∈[k,m][l]i.(2) [l]m ∈ P (O \L). In this case the sequence terminates since [l]m+1 is undefinedand the limit is [l]m.(3) [l]∞ /∈ P (O). For example l ="1 the label whose first word is a number onegreater than this label’s first word", yields the infinite sequence [l]m ="m thelabel whose first word is a number one greater than this label’s first word",with "m" in the label replaced by the index m as a number.The set of these type 3 limits is ∂[]O and [L]∞ = [O]∞. ∂[]L = ∂[]O∪ (O \L). Notethat the elements of L∅ map to ∅ under [], which is absorbed by the union operationwhen calculating [O]∞ and [L]∞.In contrast, there is only one possibility for 〈o〉∞:All elements of L have type 3 limits, 〈o〉∞ /∈ P (O), by the example given for (6), andthe elements of O \L map to L under 〈〉. So 〈o〉∞ are all of type 3 and 〈O〉∞ ⊂ 〈L〉∞.〈O〉∞ 6= 〈L〉∞ since we have previously shown in §4 that redundant labels can existin L. ∂〈〉L = ∂〈〉O ∪ 〈L∅〉∞.6 TOM HARVEYFigure 1. The boundary of ∂〈〉O in R is represented by the dashedline.It is tempting to think that all type 3 elements of ∂[]L and ∂〈〉L are in O since theyhave a label referring to them, however what is that label? It is not one that can beconstructed a priori and therefore cannot be declared to be in L in general.9. Limit points in SBy (7), all elements of L have type 2 [l]∞ limits, [l]m ∈ P (O \ L). In fact[L]∞ = [L] ⊆ O = [O]∞. This gives ∂[]O = ∅, ∂[]L = O \O∅.All elements o of O \ O∅ map to L under 〈〉, so by (7), all 〈o〉∞ are of type 3,〈o〉∞ /∈ P (O). All elements of O∅ map to ∅ under 〈〉, which is absorbed by the unionoperation when calculating 〈O〉∞. This gives 〈O〉∞ = 〈L〉∞ = L = ∂〈〉O and ∂〈〉L = ∅.10. Resolution of ParadoxesConsider linguistic "paradoxes" of the form:a = 3thereforea dog = 3 dog,which rely on slavish substitution of identicals. Reformulated in terms of labelsand objects:[a] = 3REFERENTIAL AND SUBSTANTIAL LOGICS 7therefore〈′a′〉 dog = [a] dogis clearly false.11. ApplicationsDefine the context field to be the set of all possible worlds and times, and a context(w, t), to be a subset of this space, where we will use ws to indicate the worlds and tsto indicate the times. We will use subscripts to indicate the context in which a clauseis being considered. If either w and/or t is omitted it is assumed that the clause istrue for all worlds and/or times.A worldly example:[2]w1 + [3]w1 = [7]w1 is false,[2]w2 + [3]w2 = [7]w2 is true,〈[2]w1〉w2 + 〈[3]w1〉w2 = 〈[7]w1〉w2 is false.In w2 different meanings have been assigned to the labels 2, 3, and/or 7 than inw1 ("our" world). Operators (+,=) are labels we are assuming here are equivalentin w1 and w2.A temporal example:c = "lump of clay", something that can be moulded into multiple forms.〈o〉t0 ='lump of clay' [c]t0 = o〈o〉t1 ='pottery' [c]t1 = ∅〈o〉t2 ='mug' [c]t2 = ∅〈o〉t3 ='lump of clay' [c]t3 = oAs the clay changes shape it is no longer referred to by its original label.12. SummaryWe have developed referential and substantial logics and explored the topologicaland other properties of these constructions, highlighting their similarities and dif-ferences. We have also shown interesting applications in resolving paradoxes and tomodal logic.

Referential and Substantial Logics

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Referential and Substantial Logics

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