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单词 TarskisResultOnTheUndefinabilityOfTruth
释义

Tarski’s result on the undefinability of truth


Assume 𝐋 is a logic which is under contradictory negationMathworldPlanetmath and has the usual truth-functional connectivesMathworldPlanetmath. Assume also that 𝐋 has a notion of formulaMathworldPlanetmathPlanetmath with one variable and of substitution. Assume that T is a theory of 𝐋 in which we can define surrogates for formulae of 𝐋, and in which all true instances of the substitution relationMathworldPlanetmath and the truth-functional connective relations are provable. We show that either T is inconsistent or T can’t be augmented with a truth predicateMathworldPlanetmathPlanetmath 𝐓𝐫𝐮𝐞 for which the following T-schema holds

𝐓𝐫𝐮𝐞(ϕ)ϕ

Assume that the formulae with one variable of 𝐋 have been indexed by some suitable set that is representable in T (otherwise the predicate 𝐓𝐫𝐮𝐞 would be next to useless, since if there’s no way to speak of sentencesMathworldPlanetmath of a logic, there’s little hope to define a truth-predicate for it). Denote the i:th element in this indexing by Bi. Consider now the following open formula with one variable

𝐋𝐢𝐚𝐫(x)=¬𝐓𝐫𝐮𝐞(Bx(x))

Now, since 𝐋𝐢𝐚𝐫 is an open formula with one free variableMathworldPlanetmathPlanetmath it’s indexed by some i. Now consider the sentence 𝐋𝐢𝐚𝐫(i). From the T-schema we know that

𝐓𝐫𝐮𝐞(𝐋𝐢𝐚𝐫(i))𝐋𝐢𝐚𝐫(𝐢)

and by the definition of 𝐋𝐢𝐚𝐫 and the fact that i is the of 𝐋𝐢𝐚𝐫(x) we have

𝐓𝐫𝐮𝐞(𝐋𝐢𝐚𝐫(i))¬𝐓𝐫𝐮𝐞(𝐋𝐢𝐚𝐫(𝐢))

which clearly is absurd. Thus there can’t be an of T with a predicate 𝐓𝐫𝐮𝐭𝐡 for which the T-schema holds.

We have made several assumptionsPlanetmathPlanetmath on the logic 𝐋 which are crucial in order for this proof to go through. The most important is that 𝐋 is closed under contradictory negation. There are logics which allow truth-predicates, but these are not usually closed under contradictory negation (so that it’s possible that 𝐓𝐫𝐮𝐞(𝐋𝐢𝐚𝐫(i)) is neither true nor false). These logics usually have stronger notions of negation, so that a sentence ¬P says more than just that P is not true, and the propositionPlanetmathPlanetmathPlanetmath that P is simply not true is not expressible.

An example of a logic for which Tarski’s undefinability result does not hold is the so-called Independence Friendly logic, the semantics of which is based on game theory and which allows various generalised quantifiersMathworldPlanetmath (the Henkin branching quantifier, etc.) to be used.

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更新时间:2025/5/4 18:44:04