Finitely axiomatizable theories Let $T_1$ and $T_2$ be two theories having the same set of symbols. 
Assume that any interpretation of $T_1$ is a model of $T_1$ if and only if it is not a model of $T_2$. Then:
$T_1$ and $T_2$ are finitely axiomatizable.
(i.e. there are finite sets of sentences $A_1$ and $A_2$ such that, for any sentence $S$:
$T_1$ proves $S$ if and only if $A_1$ proves $S$, and $T_2$ proves $S$ if and only if $A_2$ proves $S$).
/The proof will be by contradiction; assume $T_1$ or $T_2$ are not finitely axiomatizable, then .....?/
Any one have any idea of how to prove this argument?
 A: There's a result using compactness that says that if two sets of sentences, $T_1,T_2$ have no common model then there exists a sentence $\phi$ such that $T_1\models\phi$ and $T_2\models\lnot\phi$. To see this use the contrapositive: If for every sentence $\phi$ we have $T_1\models\phi$ has as a consequence that $T_2\cup\{\phi\}$ is satisfiable, then for every sentence $\phi$ that is a conjunction of sentences of $T_1$ we have $T_2\cup\{\phi\}$ is satisfiable and through compactness $T_1$ and $T_2$ share a model.
Now let $\phi$ be such that $T_1\models\phi$ and $T_2\models\lnot\phi$. Then $\phi$ is an axiomatization of $T_1$ and $\lnot\phi$ is an axiomatization of $T_2$. This is because, if a model satisfies $\phi$ then it can't satisfy $T_2$ and therefore satisfies $T_1$ and vice versa.
That's the answer I came up with, but maybe there's a more direct way to it.
A: The union $T_1\cup T_2$ has no models, and so by the Compactness theorem there is a finite subtheory with no models. This amounts to finite $A_1\subset T_1$ and $A_2\subset T_2$ such that $A_1\cup A_2$ has no models. Any model $M$ of $A_1$ is therefore not a model of $A_2$ and so $M$ is not a model of $T_2$ and hence by your assumption it is a model of $T_1$. So $A_1\vdash T_1$ and similarly $A_2\vdash T_2$, so both theories are finitely axiomatizable. 
