David Marker's Model Theory:
If not $Th_{A}(M) \models\phi( \overline{v} ) \rightarrow \psi( \overline{v} )$, then there is an elementary extension $N$ of $M$ and $\overline{a} \in N$ such that $ N \models \phi( \overline{a} ) \wedge \neg \psi( \overline{a} )$.
Could you explain why there exists such an elementary extension?
Would be grateful for your help.
I believe what you meant is if $Th_A(\mathcal{M}) \not \models \phi(\bar{v}) \rightarrow \psi(\bar{v})$, then there is an elementary extension $\mathcal{N}$ of $\mathcal{M}$ with $\bar{a} \in N$ such that $\mathcal{N} \models \phi(\bar{a}) \wedge \lnot \psi(\bar{a})$.
The point is that if $Th_A(\mathcal{M}) \not \models \phi(\bar{v}) \rightarrow \psi(\bar{v})$, then, by the completeness theorem, there is a model $\mathcal{N}_0 \models Th_A(\mathcal{M})$ such that $\mathcal{N}_0 \models \exists \bar{v} \phi(\bar{v}) \wedge \lnot \psi(\bar{v})$, and hence $Th_A(\mathcal{M}) \cup \{ \phi(\bar{v}), \lnot \psi(\bar{v}) \}$ is satisfiable. By proposition 4.1.3 in that same section, there is an elementary extension of $\mathcal{M}$ where the type $\{ \phi(\bar{v}), \lnot \psi(\bar{v}) \}$ is realized.