 Proof: Suppose that \(\Gamma \vDash \theta\). Then there is no interpretation \(M\) such that M satisfies every member of \(\Gamma\) but does not satisfy \(\theta\). So the set \(\Gamma,\neg \theta\) is not satisfiable. By Completeness (Theorem 20), \(\Gamma,\neg \theta\) is inconsistent. So there is a sentence \(\phi\) such that \(\Gamma,\neg \theta \vdash \phi\) and \(\Gamma,\neg \theta \vdash \neg \phi\). By \((\neg\)I), \(\Gamma \vdash \neg \neg \theta\), and by (DNE) \(\Gamma \vdash \theta\).