Decidability of Interpreted Theory

Question

If a theory $T'$ is interpretable in a theory $T$ and $T$ is decidable, is it true that $T'$ must be decidable as well?

Answer 1

No. For example, the theory $T'$ of groups is known to be undecidable, but it is clearly interpretable in the theory $T$ of torsion-free divisible abelian groups $(\mathbb{Q}$-vector spaces), which is decidable.

On the other hand, if $T'$ a complete theory which is interpretable in $T$, via a computable interpretation, and $T$ is decidable, then $T'$ is decidable.

And if $T'$ and $T$ are bi-interpretable, via computable interpretations, and $T$ is decidable, then $T'$ is decidable.

Computable here means that the function assigning to each symbol of the language $L'$ of $T'$ its corresponding formula in the language $L$ of $T$ must be computable. It follows that the map $\varphi\mapsto \hat{\varphi}$ from $L'$-formulas to corresponding $L$-formulas is computable.

So given an $L'$-sentence $\varphi$, we know that if $T'\vdash \varphi$, then $T\vdash \hat{\varphi}$. If we knew the converse, that if $T\vdash \hat{\varphi}$, then $T'\vdash \varphi$, then we could computably reduce the question of whether $T\vdash \varphi$ to the question of whether $T\vdash \hat{\varphi}$.

If $T'$ is complete, then if $T'\not\vdash \varphi$, then $T'\vdash \lnot \varphi$, so $T\vdash \lnot \varphi$, and $T\not \vdash \varphi$. So we have the desired converse. If $T'$ is not complete, but $T'$ and $T$ are bi-interpretable, we can also get the converse: If $T\vdash \hat{\varphi}$, then $T'\vdash \hat{\hat{\varphi}}$, which implies $T\vdash \varphi$ (using the definition of bi-interpretation).