I'm trying to prove (or disprove, although I think this is correct) that $$\mathcal{M}\models\forall x\phi\wedge\psi\implies\mathcal{M}\models\forall x(\phi\wedge\psi)$$
I think I was able to do so, but there are 2 parts of my proof where I'm a little skeptical it is correct. My proof goes as follows:
Assume that $\mathcal{M}\models\forall x\phi\wedge\psi$ this means that $\mathcal{M}\models\phi(x)$ for all interpretations of $x$ and $\mathcal{M}\models \psi$. If $x$ is not in the formula $\psi$, then $\mathcal{M}\models\forall x(\phi\wedge\psi)$.
(1) if $x$ is in the formula $\psi$ change all it's ocuurences for a new variable $x'$ that is not in either $\psi$ or $\phi$ and get a new formula $\psi'$.
then $\mathcal M \models \psi\iff \mathcal M \models \psi'$ and, as such we have that $$\mathcal M\models\forall x(\phi\wedge\psi')$$ witch implies that $$\mathcal M\models\forall x(\phi\wedge\psi)$$ as the formulas $\psi$ and $\psi'$ are equivalent.
My question is, is this correct? I'm I allowed to just assume that $x$ is not in $\psi$ by doing (1)?