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?
Best Answer
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.