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

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?

• Show that your inference rules relativize, and then mechanically relativize the proof. – Asaf Karagila Dec 26 '18 at 8:26

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 $$A\models \psi^\mathbf{M}$$.
• nice. I was trying to do it without using $\vDash$ - only the formal deductions. I would prefer to do it in this way, but it is just personal preference. – Nuntractatuses Amável Dec 26 '18 at 6:47
• I tried proving the defining property you mentioned, and I encountered some dificulties. Mainly, the point that if $\phi$ is the sentence $\exists x \psi$, then $\psi$ is not necessarily a sentence - in which case I am not sure what the defining property means. I then tried proving for arbitrary formulas, with an assignment of values for the variables: then it struck me that, for this to make any sense, the variables can only assume values in $\mathbf{M}^A$. Is this the case? I've seen relativization only in the context of set theory - in Kunen's book. I'm slightly confused. – Nuntractatuses Amável Dec 26 '18 at 8:07
• Yes, to prove the "defining property", you need to do an induction over formulas. Here the statement you want to prove by induction is: For any formula $\varphi(x)$, any structure $A$, and any tuple $a\in (\mathbf{M}^A)^x$, we have $A\models \varphi^{\mathbf{M}}(a)$ if and only if $\mathbf{M}^A\models \varphi(a)$. This restricts to the property I wrote above when $\varphi$ is a sentence. – Alex Kruckman Dec 26 '18 at 15:50