compactness theorem for PL(3)

we'll prove compactness theorem and its corollary is equivalent.

 

it sufficies to show corollary -> compactness thereom(we'll prove if $\Sigma$ is finitely satisfiable, then $\Sigma$ is satisfiable).

 

 

Suppose not. That is, suppose corollary holds and $\Sigma$ is finitely satisfiable, but $\Sigma$ is unsatisfiable.

 

let $\varphi \in \Sigma$. define $\bar{\Sigma}=\Sigma ;(\neg(\neg \varphi))$. clearly $\bar{\Sigma} \models (\neg(\neg\varphi))$. then $\bar{\Sigma} \models (\neg(\neg\varphi))$. then by corollary, there exist some finite $\Sigma_0 \subseteq \bar{\Sigma}$ such that $\Sigma_0 \models (\neg(\neg\varphi))$.

 

check $\bar{\Sigma}$ is finitely satisfiable but unsatisfiable. I leave out this for readers. since $\Sigma ;\neg\neg\varphi$ is unsatisfiable, we get $\Sigma \models \neg\varphi$(refer to compactness theorem(2)). by corollary, we get some finite set $\Sigma_0 \subseteq \Sigma$ such that $\Sigma_0 \models\neg\varphi$(*).

 

consider $\Sigma_0 ;\neg\neg\varphi \subseteq \bar{\Sigma}$. this set is finite thus satisfiable(**).

 

by (*), for every truth assignment $v$ for $\Sigma_0$ and $\varphi$, $\bar{v}(\Sigma_0)=\{T\} \rightarrow \bar{v}(\neg\varphi)=T$. but by (**), there exists some truth assignment $v$ such that $\bar{v}(\Sigma_0)=T$ and $\bar{v}(\neg\neg\varphi)=T$. this yields contradiction.

 

this proves equivalence of compactness theorem and its corollary.