How to show that $\phi \vdash \psi$ implies $\phi^{\mathbf{M}} \vdash \psi^{\mathbf{M}}$

Question

Given a class $\mathbf{M} \neq \emptyset$ and a formula $\phi$ from the language of set theory, let $\phi^{\mathbf{M}}$ denote the relativizatization of $\phi$ to $\mathbf{M}$. I want to show the if $\phi \vdash \psi$, then $\phi^{\mathbf{M}} \vdash \psi^{\mathbf{M}}$. I tried proving that if $\vdash \phi$, then $\vdash \phi^{\mathbf{M}}$, but I'm getting stucked in the induction steps - and one of the induction steps seems to be exactly what I want to prove. Anyone can shed some light?

Answer 1

For any structure $A$ in the language of set theory, the class $\mathbf{M}$ can be interpreted as a subset $\mathbf{M}^A$ of $A$, which we view as a substructure of $A$ by restricting $\in$ to $\mathbf{M}^A$.

Then given a formula $\varphi$, the defining property of the formula $\varphi^\mathbf{M}$ is that for any structure $A$, we have $A\models \varphi^\mathbf{M}$ if and only if $\mathbf{M}^A\models \varphi$.

Now suppose $\varphi\vdash \psi$. By soundness and completeness, we have $\varphi\models \psi$, and it suffices to show $\varphi^\mathbf{M}\models \psi^\mathbf{M}$. So let $A$ be any model of $\varphi^\mathbf{M}$. Then $\mathbf{M}^A\models \varphi$. Since $\varphi\models \psi$, we have $\mathbf{M}^A\models \psi$, so $A\models \psi^\mathbf{M}$.