Wikipedia:Reference desk/Archives/Mathematics/2021 April 14

= April 14 =

Translation in classical logic?
Is there something like the Gödel–Gentzen translation for a translation from Many-valued logic in classical logic?--82.82.233.47 (talk) 20:05, 14 April 2021 (UTC)


 * It depends a bit on what you demand of the translation and which version of many-valued logic is taken as the source logic of the translation. I assume we want the translation to be compositional. To explain what I mean by this, let $$\odot$$ stand for any logical operator of the source logic – for simplicity I only consider the case of a dyadic (binary) operator, taking two logical formulas $$\varphi$$ and $$\psi$$ as operands to form a new formula $$\varphi\odot\psi.$$ Denoting translation by a superscript "${}^\mathsf{T}$", for this translation to be compositional it should be the case that $$(\varphi\odot\psi)^\mathsf{T}=\varphi^\mathsf{T}\odot^\mathsf{T}\psi^\mathsf{T},$$ where $$\odot^\mathsf{T}$$ denotes an operation that is definable using the target logic, here classical logic. It is a must that the translation is representation insensitive, meaning that it maps equivalent formulas in the source formalism to equivalent results in the target formalism. I also assume that we exclude trivial translations, such as the one that defines $$\varphi^\mathsf{T}=\mathrm{T}$$ for all $$\varphi.$$ I'll concentrate on Kleene's $$K_3.$$ The translation has to be a bit more complicated, since a propositional formula in the source logic – excluding trivial translations that map every formula to the same classical formula – cannot simply be mapped compositionally to a formula in the target language. Proof of this impossibility is by showing that a compositional translation is trivial. Let $${}^\mathsf{T}$$ be any compositional translation. Put $$\mathrm{T}^\mathsf{T}=\mathrm{X},$$ $$\mathrm{I}^\mathsf{T}=\mathrm{Y},$$ and $$\mathrm{F}^\mathsf{T}=\mathrm{Z}.$$ Since $$\neg^\mathsf{T} \mathrm{Y}=\neg^\mathsf{T}\mathrm{I}^\mathsf{T}=(\neg\mathrm{I})^\mathsf{T}=\mathrm{I}^\mathsf{T}=\mathrm{Y},$$ the operation $$\neg^\mathsf{T}$$ has a fixpoint. There are three classical logical operation that have a fixpoint, constant $$\mathrm{T},$$ constant $$\mathrm{F},$$ and the identity. If $$\neg^\mathsf{T}$$ is the identity, $$(\neg\varphi)^\mathsf{T}=\neg^\mathsf{T}\varphi^\mathsf{T}=\varphi^\mathsf{T}$$, so $$\mathrm{X}=\mathrm{T}^\mathsf{T}=(\neg\mathrm{T})^\mathsf{T}=\mathrm{F}^\mathsf{T}=\mathrm{Z}.$$ Then also $$\mathrm{Y}=\mathrm{I}^\mathsf{T}=$$ $$(\mathrm{I}\land\mathrm{T})^\mathsf{T}=$$ $$\mathrm{Y}\land^\mathsf{T}\mathrm{X}=$$ $$\mathrm{Y}\land^\mathsf{T}\mathrm{Z}=$$ $$(\mathrm{I}\land\mathrm{F})^\mathsf{T}=$$ $$\mathrm{F}^\mathsf{T}=\mathrm{Z}.$$ The constant operations can likewise be excluded. However, a translation is possible if we translate a source formula $$\varphi$$ to a pair of formulas as follows:
 * $$\mathrm{T}^\mathsf{T}=(\mathrm{T},\mathrm{F});$$
 * $$\mathrm{I}^\mathsf{T}=(\mathrm{F},\mathrm{F});$$
 * $$\mathrm{F}^\mathsf{T}=(\mathrm{F},\mathrm{T});$$
 * $$\neg^\mathsf{T}(\varphi_0,\varphi_1)=(\varphi_1,\varphi_0);$$
 * $$(\varphi_0,\varphi_1)\land^\mathsf{T}(\psi_0,\psi_1)=(\varphi_0\land\psi_0,\varphi_1\lor\psi_1).$$
 * The other operations follow from the usual identities, such as $$\varphi\to\psi\equiv(\neg\varphi)\lor\psi.$$ The first component of the pair can be interpreted as "definitely true", and the second as "definitely false". Then $$(\mathrm{F},\mathrm{F})$$ means: "neither definitely true nor definitely false". --Lambiam 00:38, 15 April 2021 (UTC)