I know that, a set $\Gamma $ of formulas is satisfiable if there exists an interpretation that satisfies all formulas in $\Gamma$, and unsatisfiable otherwise. A formula $F$ is satisfiable if $\{F\}$ is satisfiable
How can i prove that for any set $\Gamma$ of formulas and any formula $F, \Gamma \models F$ iff the set $\Gamma \cup \{\lnot F\}$ is unsatisfiable.
First, suppose that $\Gamma \models F$. This means that for every interpretation $I$ such that $\phi[I]$ is true for all $\phi \in \Gamma$ we have that $F[I]$ is true. In other words, there is no interpretation $I$ such that $\phi[I]$ is true for all $\phi \in \Gamma$ but $\neg F[I]$ is true (as the latter would mean that $F[I]$ is false). Hence $\Gamma \cup \{ \neg F \}$ is unsatisfiable.
On the other hand, if $\Gamma \cup \{ \neg F \}$ is unsatisfiable, then for every interpretation $I$ such that $\phi[I]$ is true for every $\phi \in \Gamma$ we must have that $\neg F[I]$ is false, i.e. $F[I]$ is true. By the definition of $\models$ this means that $\Gamma \models F$.