In the proof of Proposition $7.3$ of 'A Course in Mathematical Logic' Manin says that the quantifier $\forall$ reduces to $\exists$ in the natural way.'
I am trying to prove the case for the quantifier $\forall$ but facing a bit difficulty, here is my attempt -
Say $P$ is $(M,N)$-absolute.
i) Assume that $|\forall xP|_M(\mathcal{E})=0$, $\mathcal{E}\in\overline{M}$, then there exists a variation $\mathcal{E}'$ of $\mathcal{E}$ along $x$ such that $|P|_M(\mathcal{E}')=0$. But then from $(M,N)$-absoluteness of $P(x)$, $|P|_N(\mathcal{E}')=0$. Hence $|\forall xP|_N(\mathcal{E})=0$.
ii) Assume $|\forall xP|_M(\mathcal{E})=1$, thus if $\mathcal{E}'$ is a variation of $\mathcal{E}$ along $x$ in $M$ then $|P|_M(\mathcal{E'})=1$. If $|\forall xP|_N(\mathcal{E})=0$ then as the logical axioms are true in any interpretation, $|\neg\exists x\neg P|_N(\mathcal{E})=0$. Hence there is a variation $\eta'$ of $\mathcal{E}$ along $x$ in $N$ such that $|P|_N(\eta')=0$. As $P$ can have at most finitely many free variables thus there exists a $k$ such that $y^{\mathcal{E}}\in M_k$ if $y$ is free in $P$. Thus we can choose one $\eta\in \overline{M_k}$ such that $y^{\eta}=y^{\mathcal{E}}$ whenever $x$ is free in $P$. Hence $|\forall xP|_M(\mathcal{E})=|\forall xP|_M(\eta)=1$.( I am unable to procees further)