 To see that Henkin’s theorem implies Strong completeness, let us assume the antecedent, \(\Gamma \models \varphi\). Therefore, \(\Gamma \cup \{ \lnot \varphi \}\) is not satisfiable, it has no model. Using Henkin’s theorem we conclude that \(\Gamma \cup \{ \lnot \varphi \}\) is contradictory and so \(\Gamma \cup \{ \lnot \varphi \} \vdash \varphi\). The calculus rules allow us to get rid of \(\lnot \varphi\) and infer \(\Gamma \vdash \varphi\).