Assuming $\varphi$ is a formula, $\Gamma$ is a set of formulas, $\varphi^{\forall} = \forall x_1 \forall x_2 \dots \forall x_n \varphi$ where $x_1,x_2, \dots, x_n$ are the free variables in $\varphi$ and $\Gamma^{\forall} = \{ \varphi^{\forall} | \varphi \in \Gamma \}$
Using the facts I know
$M \models \varphi $ if and only if $M \models \varphi^{\forall}$
$M \models \Gamma $ if and only if $M \models \Gamma^{\forall}$
I've tried the following:
If $M \models \Gamma$ then $M \models \varphi$ hence $M \models \varphi^{\forall}$ hence for all valuation $v$ $M,v \models \varphi^{\forall}$ then by $TT_{\neg}$ $M,v \not\models \neg(\varphi^{\forall})$ hence indeed for all valuation $v$ we have $M,v \not\models \Gamma^{\forall} \cup \{ \neg(\varphi^{\forall}) \}$
The second case doesn't seem to work out, if $M \not\models \Gamma$ then i know $M \not\models \Gamma^{\forall}$ but that still doesn't mean there is no valuation $v$ such that $M,v \models \Gamma^{\forall}$.
I will be happy for any help with the second case, or somehow proving it differently.
$M \not\models \Gamma^{\forall}$ implies there exist $v_0$ such that $M,v_0 \not\models \Gamma^{\forall}$ but $\Gamma^{\forall}$ contains formulas without any free variables hence a different valuation doesn't change anything (no assignment is made). i.e, for all valuations $v(\psi) = v_0(\psi)$ for all $\psi \in \Gamma^{\forall}$ hence for all $v$ we have $M,v \not\models \Gamma^{\forall}$ finishing the proof.