 Proof: If \(M\) satisfies every member of \(\Gamma\), then \(M\) satisfies every member of each finite subset of \(\Gamma\). For the converse, suppose that \(\Gamma\) is not satisfiable. Then we show that some finite subset of \(\Gamma\) is not satisfiable. By Completeness (Theorem 20), \(\Gamma\) is inconsistent. By Theorem 9 (and Weakening), there is a finite subset \(\Gamma'\subseteq \Gamma\) such that \(\Gamma'\) is inconsistent. By Corollary \(19, \Gamma'\) is not satisfiable.