As suggested, $\Sigma_{1}\cup\Sigma_{2}$ is unsatisfiable. According to the compactness theorem, there is a finite unsatisfiable subset$\Delta\subseteq\Sigma_{1}\cup\Sigma_{2}$ . Let $\Delta_{i}=\Sigma_{i}\cap\Delta$ , and $\tau_{i}$ be the conjunction of sentences in $\Delta_{i}$ (which is finite), or any valid sentence (an axiom of the language) if$\Delta_{i}$ is empty. Then, $\tau_{i}$ is logically equivalent to $\Delta_{i}$ , and $\Sigma_{i}\vDash\Delta_{i}$ , hence, $\Sigma_{i}\vDash\tau_{i}$ and $\mbox{Mod}\Sigma_{i}\subseteq\mbox{Mod}\tau_{i}$ (well, if $\Delta_{i}$ is empty we can either assume that every structure is a model of $\Delta_{i}$ , so all relations hold, or simply argue that the latter relation holds in this case as $\tau_{i}$ is valid). Now, $\tau_{1};\tau_{2}$ is not satisfiable (again, even if one of them is valid), and, hence, it is not consistent, $\tau_{2}\vdash\tau_{1}$ , and $\tau_{2}\vDash\neg\tau_{1}$ . In particular, $\mbox{Mod}\Sigma_{2}\subseteq\mbox{Mod}\tau_{2}\subseteq\mbox{Mod}\neg\tau_{1}$ .

Note. In the above proof we could have considered the case when one of $\Sigma_{i}$ is inconsistent separately, and set $\tau$ to be either a valid sentence or the negation of a valid sentence, and then repeat the proof for the case when both $\Sigma_{1}$ and $\Sigma_{2}$ are satisfiable to avoid the necessity to mention the case when one of the $\Delta_{i}$ ’s is empty (if both $\Sigma_{1}$ and $\Sigma_{2}$ are satisfiable, then the unsatisfiable $\Delta$ cannot be a subset of either one).