Prove that if Σ1∪Σ2 is a not satisfiable set of propositional formulas then there is ϕ so as Σ1⊨ϕ and Σ2⊨¬ϕ

Question

Let $\Sigma_1$, $\Sigma_2$ be (possibly infinite) sets of proposational formulas so as $\Sigma_1\cup\Sigma_2$ is not satisfiable.

Prove that there is a propositional formula $\phi$ so as $\Sigma_1\vDash\phi$ and $\Sigma_2\vDash\lnot\phi$

I think that the compactness theorem has to be used due to the possibly infinite sets and also that the general format of the proof will be proof by contradiction.

Do you have any hints to offer?

Answer 1

You are right that you have to use compactness here. I will provide a hint, and complete the argument step by step in spoiler tags so that you (and other users) can view the complete answer if you get stuck again.

Since $\Sigma_1 \cup \Sigma_2$ is inconsistent, then by compactness there must be a finite $\Delta \subseteq \Sigma_1 \cup \Sigma_2$ that is inconsistent. Consider now $\Delta \cap \Sigma_1$...

Step 1.

Let $\phi$ be the conjunction of the formulas in $\Delta \cap \Sigma_1$. Clearly $\Sigma_1 \models \phi$.

Step 2.

We must have $\Sigma_2 \models \neg \phi$, because otherwise there is a valuation $v$ that satisfies $\Sigma_2$ such that $v(\phi) = 1$. But then $v(\delta) = 1$ for all $\delta \in \Delta$. This is a contradiction, because $\Delta$ was inconsistent.