Let $\Sigma$ be any set of propositional sentences. If $\Sigma$ is finitely satisfiable, then for every propositional sentence $\theta$, either $\Sigma\cup\{\theta\}$ is finitely satisfiable or $\Sigma\cup\{\neg\theta\}$ is finitely satisfiable.

Can someone please tell me the error in my proof?

Suppose $\Sigma$ is finitely satisfiable, and $\theta$ is an arbitrary sentence. Let $\Pi\subset\Sigma$ be any finite subset, then there exists a truth evaluation $e$ such that $e(p)=1$ for every $p\in\Pi$. Now either $e(\theta)=1$, or $e(\neg\theta)=1$, so either $e$ satisfies $\Pi\cup\{\theta\}$ or $e$ satisfies $\Pi\cup\{\neg\theta\}$.

But $\Pi$ was chosen arbitrarily, hence the result follows.


The problem is this, suppose that for $\Pi_1$ and $\Pi_2$ you have two assignments, $e_1$ and $e_2$ which give "True" and "False" to $\theta$ respectively.

You have no control over this, because these were taken arbitrarily. So you haven't shown that $\Pi_2\cup\{\theta\}$ is satisfiable. What you have shown is that every finite set of propositions can be satisfied with either $\theta$ or its negation, which is trivially true.

What you need here is to argue that there is a way to satisfy all the formulas in $\Sigma$ at once. And luckily, the condition you have is exactly enough to argue that. Compactness is your only friend.

  • $\begingroup$ Thanks a lot! I never thought it through until now. $\endgroup$ – Sid Caroline Aug 6 '17 at 11:50

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