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Can there exist a consistent, recursively axiomatizable theory $T$, such that $\forall \phi, TA\vdash \phi \Rightarrow$ $T\vdash \tau(\phi)$, where $\tau$ is some suitable translation from the language of $TA$ to that of $T$?

Edit: By "suitable translation" I was thinking about a translation that preserves the intended meaning of $\phi$ in $TA$. For example, for any theorem $\phi$ of $PA$, you can also prove it ($\tau(\phi)$) in the language of $ZFC$, and $\phi$ and $\tau(\phi)$ share the same intended meaning in the standard model $\mathcal{N}$ of arithmetic. Only now instead of $PA$ and $ZFC$, we're thinking about $TA$ and any recursively axiomatizable theory.

(I apologize if this sounds too sloppy. I had some trouble getting this part straight in my head, but I hope the point is understood)

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  • $\begingroup$ What’s “suitable”? $\endgroup$
    – Monroe Eskew
    Commented Jul 5, 2020 at 8:48
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    $\begingroup$ As long as the translation $\tau$ is computable (or even arithmetical) this isn't possible. For any arithmetical $\tau$ and recursively axiomatizable $T$ the set $\{\varphi\mid T\vdash \tau(\varphi)\}$ will be arithmetical and hence couldn't coincide with the set of theorems of $\mathsf{TA}$. $\endgroup$
    – Fedor Pakhomov
    Commented Jul 5, 2020 at 9:39
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    $\begingroup$ @FedorPakhomov Does not the question ask about inclusion rather than coincidence? Maybe it is trivial to encompass this in your comment, but... $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Jul 5, 2020 at 9:56
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    $\begingroup$ @მამუკაჯიბლაძე Right, I missed it. So we would need to require a bit more from $\tau$. For example, we could only consider $\tau$ that preserve negation, i.e. $T\vdash \lnot \tau(\varphi)\leftrightarrow\tau(\lnot\varphi)$. Then we should have $\{\varphi\mid \mathsf{TA}\vdash\varphi\}=\{\varphi\mid T\vdash \tau(\varphi)\}$. Otherwise due to completeness of $\mathsf{TA}$ there would be $\psi$ such that $T\vdash\tau(\psi)$ and $T\vdash\tau(\lnot\psi)$, which would imply inconsistency of $T$. $\endgroup$
    – Fedor Pakhomov
    Commented Jul 5, 2020 at 10:09
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    $\begingroup$ @Eric Assume for a contradiction that $\{\varphi\mid \mathsf{TA}\vdash\varphi\}\subsetneq \{\varphi\mid T\vdash \tau(\varphi)\}$. Then there is $\psi$ s.t. $\mathsf{TA}\nvdash \psi$ and $T\vdash \tau(\psi)$. However, since $\mathsf{TA}$ is complete, we would have $\mathsf{TA}\vdash \lnot\psi$. Thus we would have $T\vdash \tau(\lnot \psi)$ and further $T\vdash \lnot \tau(\psi)$ (by preservation of negation). Contradiction, since $T$ was consistent and $T\vdash \tau(\psi)$. $\endgroup$
    – Fedor Pakhomov
    Commented Jul 5, 2020 at 15:32

1 Answer 1

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Without further requirements on $\tau$, this is trivial. Let $T$ be any first-order theory and let $\top$ be any tautology. Define $\tau(\phi)= \top$.

If you want $\tau$ to be injective, then let $T$ be any first-order theory in the language of $TA$ and define $\tau(\phi)=\top\vee\phi$.

If you also want equivalence, then take $T$ to be any first-order theory in the language of $TA$ (such that $T\subseteq TA$) and define $\tau$ as follows:

-$\tau(\phi)=\phi$ if $\phi$ is not a theorem of $TA$

-$\tau(\phi)=\top\vee\phi$ if $\phi$ is a theorem of $TA$

This $\tau$ is injective and gives you the equivalence: $\forall \phi(TA\vdash\phi\leftrightarrow T\vdash\tau(\phi))$.

If you want $\tau$ to be an interpretation in the usual sense, as the title of the question suggests, then this is not possible as Fedor Pakhomov has explained in two comments (intepretations preserve negation).

So, you must specify what do you mean by a suitable $\tau$, as Monroe Eskew has already asked in a comment.

EDIT:

The edited version of the question suggests that the OP wants $\tau$ to be an interpretation in the usual sense. There is no such $\tau$ (this was already clear from the comments and the previous version of the answer, but I will unify the argument here for convenience).

Indeed, assume without loss of generality that the signature of $TA$ is finite, and let $\tau$ be an interpretation of $TA$ in $T$. In this case, $\tau$ is computable. Therefore, the set of sentences $\phi$ such that $T\vdash \tau(\phi)$ is RE. However, the set of those sentences is just the set of true sentences of arithmetic, for if $TA\vdash \phi$ then $T\vdash\tau(\phi)$, and if $TA\not\vdash\phi$, then $TA\vdash\neg\phi$, $T\vdash\neg\tau(\phi)$, and $T\not\vdash\tau(\phi)$. This is a contradiction with Tarski's theorem.

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  • $\begingroup$ Entirely true, and answers the question in its present form perfectly. Still, it is formulated in such way that it is a comment rather than answer. $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Jul 5, 2020 at 11:50
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    $\begingroup$ I agree, I have expanded the answer with more information. $\endgroup$
    – Rodrigo Freire
    Commented Jul 5, 2020 at 12:10
  • $\begingroup$ I edited the question to address the indeterminacy. But suppose $\tau$ preserves negation, why does this make $T$ impossible? $\endgroup$
    – Eric
    Commented Jul 5, 2020 at 14:09
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    $\begingroup$ Since the language of arithmetic is finite, interpretations in the usual sense are automatically computable (hence arithmetical). $\endgroup$
    – Emil Jeřábek
    Commented Jul 5, 2020 at 14:29
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    $\begingroup$ This does not matter. You can’t interpret any extension of TA to any language you want, because you cannot even interpret its finite-language fragment in the usual language of arithmetic. $\endgroup$
    – Emil Jeřábek
    Commented Jul 5, 2020 at 18:37

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