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I want to show that the definition of satisfiability is consistent with the definition given by relativization, i.e. Let $L=\{\in\}$. Let $M$ be a definable set and let $E\subset{M\times{M}}$. Let $\sigma(\bar{x})\in FM_{L}$. Let $\bar{a}\in{M^{lg{\bar{x}}}}$. I want to show that $\sigma^{M,E}(\bar{a}) \iff (M,E)\models \sigma(\bar{a})$. I can see that this should be done by induction on formulas but I'm having trouble seeing how. For example if $\sigma(\bar{a}):=a_{1}=a_{2}$ and suppose that $ZFC\vdash{a_{1}=a_{2}}$ but now $M$ may not interpret equality as set equality (or it might I'm not really sure). Also what about the converse, i.e if $M\models a_{1}=a_{2}$, then why does $ZFC\vdash{a_{1}=a_{2}}$?

$\sigma^{M,E}$ is defined using provability as found in standard texts like Jech or Kunen. This result appears in the third edition (pg 162) of Jech: "Let $\sigma$ be a formula of set theory. If M is a set and E is a binary relation on M and if $a_{1},...,a_{n}$ are elements of $M$, then $\sigma^{M,E}(\bar{a}) \iff (M,E)\models \sigma(\bar{a})$"

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  • $\begingroup$ How do you define $\sigma^{M,E}$? (Are you using $\vdash$ for provability or something else?) $\endgroup$ Sep 21, 2013 at 21:50
  • $\begingroup$ $\sigma^{M,E}$ is defined using provability as found in standard texts like Jech or Kunen. (In fact Jech states this result without proving it). $\endgroup$
    – UserB1234
    Sep 21, 2013 at 22:02
  • $\begingroup$ You should include the reference or the definition. (The latter should help you answer your own question.) It's still unclear what you mean by $ZFC \vdash$: a theory proves sentences but $a_1 = a_2$ isn't one. $\endgroup$ Sep 21, 2013 at 22:05
  • $\begingroup$ This appears in the third edition (pg 162) of Jech: "If $M$ is a set and $E$ is a binary relation on $M$ and if $a_{1},...,a_{n}$ are elements of $M$, then $\sigma^{M,E}(\bar{a}) \iff (M,E)\models \sigma(\bar{a})$. Now I'm considering the formula $x=y$ with $a_{1},a_{2}\in{M}$ $\endgroup$
    – UserB1234
    Sep 21, 2013 at 22:11
  • $\begingroup$ OK. If you write down the definition of $\sigma^{M,E}$ for that case, you should get $a_1 = a_2$. If you write down the definition of $(M,E) \vDash a_1 = a_2$ you should get the same thing. $\endgroup$ Sep 21, 2013 at 22:29

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