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?

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    $\begingroup$ Show that your inference rules relativize, and then mechanically relativize the proof. $\endgroup$ – 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 $A\models \psi^\mathbf{M}$.

  • $\begingroup$ 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. $\endgroup$ – Nuntractatuses Amável Dec 26 '18 at 6:47
  • $\begingroup$ 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. $\endgroup$ – Nuntractatuses Amável Dec 26 '18 at 8:07
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    $\begingroup$ 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. $\endgroup$ – Alex Kruckman Dec 26 '18 at 15:50
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    $\begingroup$ If you prefer a purely syntactic proof, you can follow Asaf's hint above. $\endgroup$ – Alex Kruckman Dec 26 '18 at 15:52
  • $\begingroup$ I managed to prove it (the "defining property"), thank you :). I'll try to do it in a purely syntactical way, now. $\endgroup$ – Nuntractatuses Amável Dec 28 '18 at 0:33

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