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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 $\phi$ and $\psi$ as operands to form a new formula $\phi \odot \psi .$ Denoting translation by a superscript Template:Nowrap, for this translation to be compositional it should be the case that $(\phi \odot \psi {)}^{𝖳}={\phi }^{𝖳}{\odot }^{𝖳}{\psi }^{𝖳},$ where ${\odot }^{𝖳}$ 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 ${\phi }^{𝖳}=\mathrm{T}$ for all $\phi .$ 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 ${}^{𝖳}$ be any compositional translation. Put ${\mathrm{T}}^{𝖳}=\mathrm{X},$ ${\mathrm{I}}^{𝖳}=\mathrm{Y},$ and ${\mathrm{F}}^{𝖳}=\mathrm{Z}.$ Since ${\mathrm{\neg }}^{𝖳}\mathrm{Y}={\mathrm{\neg }}^{𝖳}{\mathrm{I}}^{𝖳}=(\mathrm{\neg }\mathrm{I}{)}^{𝖳}={\mathrm{I}}^{𝖳}=\mathrm{Y},$ the operation ${\mathrm{\neg }}^{𝖳}$ has a fixpoint. There are three classical logical operation that have a fixpoint, constant $\mathrm{T},$ constant $\mathrm{F},$ and the identity. If ${\mathrm{\neg }}^{𝖳}$ is the identity, $(\mathrm{\neg }\phi {)}^{𝖳}={\mathrm{\neg }}^{𝖳}{\phi }^{𝖳}={\phi }^{𝖳}$, so $\mathrm{X}={\mathrm{T}}^{𝖳}=(\mathrm{\neg }\mathrm{T}{)}^{𝖳}={\mathrm{F}}^{𝖳}=\mathrm{Z}.$ Then also $\mathrm{Y}={\mathrm{I}}^{𝖳}=$ $(\mathrm{I}\wedge \mathrm{T}{)}^{𝖳}=$ $\mathrm{Y}{\wedge }^{𝖳}\mathrm{X}=$ $\mathrm{Y}{\wedge }^{𝖳}\mathrm{Z}=$ $(\mathrm{I}\wedge \mathrm{F}{)}^{𝖳}=$ ${\mathrm{F}}^{𝖳}=\mathrm{Z}.$ The constant operations can likewise be excluded. However, a translation is possible if we translate a source formula $\phi$ to a pair of formulas as follows:

${\mathrm{T}}^{𝖳}=(\mathrm{T},\mathrm{F});$

${\mathrm{I}}^{𝖳}=(\mathrm{F},\mathrm{F});$

${\mathrm{F}}^{𝖳}=(\mathrm{F},\mathrm{T});$

${\mathrm{\neg }}^{𝖳}({\phi }_0,{\phi }_1)=({\phi }_1,{\phi }_0);$

$({\phi }_0,{\phi }_1){\wedge }^{𝖳}({\psi }_0,{\psi }_1)=({\phi }_0\wedge {\psi }_0,{\phi }_1\vee {\psi }_1).$

The other operations follow from the usual identities, such as $\phi \to \psi \equiv (\mathrm{\neg }\phi )\vee \psi .$ 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)