Decidability of Finite Extensions In his Undecidable Theories, Alfred Tarski has shown the following:
Let $T_1$ and $T_2$ are theories that have the same constants and $T_2$ is a finite extension of $T_1$. If $T_1$ is decidable, then so is $T_2$. 
My question is that what if $T_2$ has more constants than $T_1$ but $T_2$ is still a finite extension of $T_1$. Is it still possible to show the same result, i.e., if $T_1$ is decidable, then so is $T_2$?
 A: No, it is not. Keep in mind that there are finitely axiomatizable undecidable theories; let $T$ be such a theory, let $S$ be any decidable theory in a disjoint language, and consider $T\cup S$ as a finite extension (in a larger langauge) of the decidable theory $S$.
Looking at the proof of Tarski's result clarifies the issue. Suppose $A$ is a finite extension of $B$ in the same language and $B$ is decidable. Let $\varphi$ be such that $B\cup\{\varphi\}$ is equivalent to $A$ (such a $\varphi$ exists since $A$ is a finite extension of $B$). Then for all $\theta$ we have $$A\vdash\theta\quad\iff\quad B\vdash(\varphi\rightarrow\theta).$$ Since $B$ is decidable and the map sending $\theta$ to $\varphi\rightarrow\theta$ is computable, $A$ is also decidable.
The crucial point is that it makes sense to ask whether $B\vdash(\varphi\rightarrow\theta)$ in the first place. If we allow $A$ to have a bigger language, then $\varphi$ might not be a sentence in the language of $B$, and so the whole argument breaks down.

Now let me be a bit more precise (especially since the last couple sentences suggested that it sometimes doesn't make sense to ask whether $X\vdash\alpha$, which is bonkers!). When we talk about the decidability of a theory $T$, we are implicitly considering it within a given language. This matters a lot - for instance, as a theory in the empty language the empty theory is decidable, but as a theory in the language with a single binary relation symbol it isn't. So here's what we can say:


*

*If $B$ is a decidable $L$-theory and $A$ is a finite extension of $B$ also in the language $L$, then $A$ is also decidable $L$-theory.

*However, if $B$ is a decidable $L$ theory and $A$ is a finite extension of $B$ in some larger language $L'$ then $A$ might not be a decidable $L'$-theory since $B$ might not be a decidable $L'$-theory.
