 Proof: (a)\(\Rightarrow\)(b): Suppose that \(\Gamma\) is satisfiable and let \(\theta\) be any sentence. There is an interpretation \(M\) such that \(M\vDash \psi\) for every member \(\psi\) of \(\Gamma\). By the clause for negations, we cannot have both \(M\vDash \theta\) and \(M\vDash \neg \theta\). So either \(\langle \Gamma,\theta \rangle\) is not valid or else \(\langle \Gamma,\neg \theta \rangle\) is not valid. (b)\(\Rightarrow\)(c): This is immediate. (c)\(\Rightarrow\)(a): Suppose that it is not the case that \(\Gamma \vDash \psi\). Then there is an interpretation \(M\) such that \(M\vDash \theta\), for every sentence \(\theta\) in \(\Gamma\) and it is not the case that \(M\vDash \psi\). A fortiori, \(M\) satisfies every member of \(\Gamma\), and so \(\Gamma\) is satisfiable.