 Proof: Suppose that \(\Gamma\) is satisfiable. So let \(M\) be an interpretation such that \(M\) satisfies every member of \(\Gamma\). Assume that \(\Gamma\) is inconsistent. Then there is a sentence \(\theta\) such that \(\Gamma \vdash \theta\) and \(\Gamma \vdash \neg \theta\). By soundness (Theorem 18), \(\Gamma \vDash \theta\) and \(\Gamma \vDash \neg \theta\). So we have that \(M\vDash \theta\) and \(M\vDash \neg \theta\). But this is impossible, given the clause for negation in the definition of satisfaction.