[Math] Show that if the set of formulas $\Sigma$ is finitely satisfiable then the same is true for $\Sigma ; \alpha$ or $\Sigma ; \neg \alpha$

Question

Does it suffice to state the law of excluded middle here?

Assume $\Sigma; \alpha$ is not finitely satisfiable. Then there must be some [finite] $\Sigma_0 \subseteq \Sigma$ such that $\Sigma_0 ; \alpha$ is not satisfiable. Let $f$ be a truth assignment which satisfies every finite subset of $\Sigma$ – since $\Sigma$ is finitely satisfiable such an $f$ must exist. If $f$ does not satisfy $\Sigma_0 ; \alpha$, then it must satisfy $\Sigma_0 ; \neg \alpha$, and similarly for the remaining case.

Is that correct? If not, why?

Answer 1

You want to show that $\Sigma,\alpha$ or $\Sigma,\neg\alpha$ is finitely satisfiable. Hint: Assume what would happen if neither of them was finitely satisfiable (you then use some ideas from your post). The full proof is in the end of this post.

I have some problems understanding your proof. It seems to me that you use that $\Sigma$ is satisfiable, instead of finitely satisfiable (secret use of compactness?). Of course, if you assume that $\Sigma$ satisfiable then adding one of $\alpha,\neg\alpha$ is trivial.

Suppose both $\Sigma,\alpha$ and $\Sigma,\neg\alpha$ are not finitely satisfiable. Then there are two finite subsets $\Delta,\Delta'\subseteq \Sigma$ such that $\Delta,\alpha$ and $\Delta',\neg\alpha$ are not satisfiable. However now $\Delta\cup\Delta'$ are not satisfiable, since otherwise one of $\Delta,\alpha$ or $\Delta',\neg\alpha$ would be. Therefore $\Sigma$ is not finitely satisfiable. Q.E.D.