MUST SYNONYMOUS PREDICATES BE COEXTENSIVE? Charles SAYWARD In answering this question it is best to distinguish two cases. In one case two predicates belong to two distinct languages. Here I think a persuasive and straight-forward argument shows they might be synonymous but not coextensive. In the second case the predicates belong to the same language. Here the issue is more involved; but a , reasonable case can be made for the same conclusion. I Consider a first-order language S, in Polish notation, whose voc- abulary is made up of constants N, A, II, E and variables x, y, z, ... and whose domain ofdiscourse is some set of sets of which the axioms of Zermelo-Fraenkel hold. Let MS be a fragment of English suitable for a Tarski-style definition of truth for S. Following Tarski we state in a metametalanguage what each constant of S means in MS: (1) 'N' means 'it is not the case that' (2) 'A' means 'or' (3) 'II' means 'for all' (4) 'E' means 'is a member of' On the basis of (1)-(4) plus the syntax of Sand MS, the MS translation of each S sentence is determined. For example, the S sentence (5) IIxNExx is translated in MS by 432 CHARLES SA YWARD (6) Nothing is a member of itself. I am supposing M5 is suitable for a Tarski-style definition of truth for 5. This means that M5's domain of discourse must include all infinite sequences of members of 5's domain of discourse (among other things). This, in turn, implies that the extension of 'e:' in 5 is not the same as the extension 'is a member of in M5. But, by (4), the meaning of '10' in 5 is the same as the meaning of 'is a member of in M5. So, unless there is something wrong with (4), it is clearly possible for two predicates belonging to different languages to have both the same meaning and different extensions. I have followed the same procedure Tarski followed in determining metalinguistic translations: the metalinguistic translation of each object language sentence is determined by matching up each constant in the object language with a constant in the metalanguage, and stipulating that the one constant means the same as the other. The sentences (1)-(4) are stipulations. They confer meaning on the con- stants of 5. So (4) cannot be said to be false. It might be rejected on other grounds, such as leading to a contradiction or some other incoherency. But I can see no such grounds for rejecting (1)-(4). So, I conclude, '10' and 'is a member of are synonymous but are not coextensive. Other examples reinforce the conclusion. Let K and H be two languages whose distinct (possibly overlapping) domains each consist of persons. Might there not be predicates FI and F2 belonging to language K and H, respectively, each of which mean, respectively in K and H, the same as 'is the father of means English? Clearly F 1 and F 2 need not be coextensive. This raises the question: Can there be two synonymous, non- coextensive predicates belonging to one language with a single domain of discourse? I now consider this. II Saul Kripke suggests that in general a homophonic truth theory may be produced from a non-homophonic one by expanding the vocabul- ary of the original metalanguage (so as to include the vocabulary of the object language) and adding certain biconditionals as axioms: ~AYWARD :If. or a Tarski-style definition of truth . ain of discourse must include all S's domain of discourse (among that the extension of 'E' in S is not !mber of in MS. But, by (4), the :he meaning of 'is a member of in vrong with (4), it is clearly possible ferent languages to have both the ions. ure Tarski followed in determining letaIinguistic translation of each ined by matching up each constant mstant in the metalanguage, and neans the same as the other. The They confer meaning on the cono be false. It might be rejected on o a contradiction or some other : grounds for rejecting (1)-(4). So, I of are synonymous but are not :onclusion. Let K and H be two overlapping) domains each consist *redicates F 1 and F2 belonging to lch of which mean, respectively in of means English? Clearly F 1 and there be two synonymous, non- ) one language with a single domain :ral a homophonic truth theory may l1ic one by expanding the vocabul- o as to include the vocabulary of the n biconditionals as axioms: MUST SYNONYMOUS PREDICATIVES BE COEXISTENSIVE? 433 ... let me mention a more or less mechanical way in which a non-homophonic truth theory can be made homophonic. First extend the metalanguage so that it contains the object language . Next, add to the old truth theory as axioms all statements of the form ~ ==~', where ~ is in the object l~nguage and ~' is its translation into the metalanguage. Then, smce T(~) followed from the old axioms, T (~) ==~ follows from the new ones. ([1], p. 358) By reflecting upon Kripke's ideas, I hope to show that it is possible for synonymous predicates to have different extensions even if they belong to one language with a single domai~of discourse. I shall follow Kripke in his use of the bar: ~ is the designation in the metalanguage of the object language sentence~. I also shall under- stand Kripke to be presupposing that the object language and metalanguage do not share any expressions. (Otherwise, chaos is possible. Suppose, for example, the sign of disjunction in the object language is the same as the sign for conjunction in the metalanguage.) Let L be the same as S , described above, except the domain is made up of sets such that no set in the domain is a member of any set in t~e domain. Now consider a semantical metalanguage ML whose domam includes S 's domain as a subset plus all infinite sequences of elements of S's domain. I shall suppose that the stipulations (1)-(4), which match the constants ofS with synonyms in MS, apply to Land ML as well. I shall also suppose ML contains Greek letters 'a', '13', etc. which range over the elements of L's domain. Finally, suppose an adequate, Tarski-style definition of truth for L is formulated in ML. Then among ML's theorems will be this biconditional: (7) T llx lly N EXY == for all a, and for all ~, it is not the case that a is a member of ~. Following Kripke's suggestion we can extend ML to contain L, and add to the axioms of M L such biconditionals as: (8) llx lly N €xy == for all a, and for all ~, it is not the case that a is a member of ~. From (7) and (8) it follows that a theorem of the extended metalan- guage ('ML +L', for short) is this: 434 CHARLES SAYWARD (9) T ilx ily NEXY ilx I1y NEXy Kripke does not say that in shifting from L to ML + L the variables ofL have to be restricted in M L +L to range over the same entities as they do in L. So suppose that in Land M L + L the variables 'x', 'y', 'z', etc. are unrestricted in range. I shall also assume that the axioms of ML are true in ML. From these suppositions this follows: Theorem; If Kripke's suggestion is correct, then 'c' and 'is a member of are not coextensive in ML + L. Proof: Kripke's construction requires adding each sentence ~ of L to ML to form ML + L. The construction is purely syntactic. So if the axioms of ML are true in ML, they are also true in ML + L. Suppose now 'E' and 'is a member of are coextensive. Then, since 'x' and 'y' are unrestricted in ML + L, (10) ilx ily Ncxy == nothing is a member of anything is true in ML +L. But the right side of (10) is false in ML + L. So 'ilx ily NEXY' is false in M L + L while being true in L. Since (7) is true in ML +L (it is a theorem of ML), the right side of (8) is true in ML +L. So the axiom (8) is false in ML +L. Hence, Kripke's suggestion is not correct. For some of the axioms of the homophonic truth theory obtained by his method will be false in that theory. From stipulations (1)-(4) we have: Corollary: If Kripke's suggestion is correct, 'E' and 'is a member of are synonymous and not coextensive in ML + L. Is Kripke's suggestion incorrect? The following is intended as a defense of his suggestion. By the axioms of ML, carried over to ML + L, for each sentence ~ of L there is a sentence~! of ML such that Tl f is a theorem of ML. Further~! is an ML translation of~. We are assuming the axioms of ML are true in ML. Then they shaH also be true in ML + L. Hence the theorems of ML are true in both ML and ML + L. Thus each theorem T~ == ~! is true in ML +L. Now Kripke says that, for each such ~ and ~f, ~ == ~f is to be an axiom ofML +L. If this axiom is true in ML + L, then, since Tl ~' IS tf 'ilx H, trutC trutr c( Of'E 'is a ML relat of L men; 'is a ML The [I] SB ; SAYWARD NEXY fting from L to ML + L the variables L to range over the same entities as Land ML + L the variables 'x', 'y', . I shall also assume that the axioms ollows: lon is correct, then 'E' and 'is a :oextensive in ML + L. luires adding each sentence ~ of L to 'uction is purely syntactic. So if the ey are also true in ML + L. Suppose coextensive. Then, since 'x' and 'y' > a member of anything side of (10) is false in ML + L. So ~ while being true in L. Since (7) is If ML), the right side of (8) is true in false in ML +L. Hence, Kripke's ne of the axioms of the homophonic hod will be false in that theory. lYe: on is correct, 'E' and 'is a member of' d not coextensive in ML + L. ect? The following is intended as a over to ML + L, for each sentence ~ [ such that T (I) == ~' is a theorem of ion of~. We are assuming the axioms r shall also be true in ML + L. Hence 1 both ML and ML + L. Thus each I-L. lch such ~ and ~', ~ ~' is to be an >true in ML +L, then, since T(I) == ~' MUST SYNONYMOUS PREDICATIVES BE COEXISTENSIVE? 435 is true in ML + L, ~ is true in L iff ~ is true in ML + L. Thus 'IIx IIy N EXY' cannot be false in M L + L if it is true in L. How could ~ ~f be false in M[ + L ? The axiom determines the truth conditions of ~ in ML -i"L. By this axiom ~ gains in ML + L the truth conditions of f in M L + L. Collectively the axioms ~ == ~f of M L +L determine the extension of 'E' in ML + L. Its extension is the same as it is in L. By (4), 'E' and 'is a member of are synonymous in ML +L. By the axioms of M L +L the extension of 'E' is the restriction of the membership relation to a proper subset of the domain of M L + L, viz., the domain of L. The extension of 'is a member of in ML + L, however, is the membership relation defined on the domain of ML +L. Thus, 'E' and 'is a member of mean the same in ML + L but are not coextensive in ML +L. The University of Nebraska-Lincoln Charles SAYWARD REFERENCE [J 1 Saul KRIPKE, 'Substitutional Quantification', Truth alld Meaning, edited by Gareth Evans and john McDowell, Oxford, 1976.