In the first order logic system, on p121 of Ebbinghaus' Mathematical Logic

We call two sets $\Phi$ and $\Psi$ of $S$-sentences equivalent if $Mod^S\Phi = Mod^S\Psi$. Then, in particular, $\Phi \models \phi$ iff $\Phi \models \phi$ for all $\phi \in L^S$.

  1. Are $\Phi$ and $\Psi$ equivalent , if and only if $\Phi \models \phi$ iff $\Psi \models \phi$ for all $\phi \in L^S$? (The last sentence in the quote says only "only if".)

  2. Are two $S$-sentences $\phi$ and $\psi$ equivalent, if and only if $\models (\phi \leftrightarrow \psi)$? (This is used as the definition of logical equivalence between two formulas in propositional logic, on p202.)


  • $\begingroup$ Small typo in your question: both in the quote and in point 1 the "iff $\Phi \models \phi$" should be "iff $\Psi \models \phi$". $\endgroup$ – Mark Kamsma Sep 2 '20 at 8:09
  • $\begingroup$ Did you try to prove either of these yourself? If so, where did you get stuck? $\endgroup$ – Noah Schweber Sep 2 '20 at 16:53
  1. Yes. Let $\mathscr M \in Mod^S\Phi$. Since $\Psi \models \psi$ for all $\psi \in \Psi$, we have that $\Phi \models \psi$ for all $\psi\in \Psi$, and hence $\Phi \models \Psi$. Therefore, since $\mathscr M \models \Phi$, it follows that $\mathscr M \models \Psi$, so $\mathscr M \in Mod^S\Psi$ and thus $Mod^S \Phi \subseteq Mod^S\Psi$. The other direction is similar.
  2. Yes again. Note that $\models \phi \leftrightarrow \psi$ means that for any $S$-structure $\mathscr M$ we have that $\mathscr M \models \phi \leftrightarrow \psi$, i.e. for any $S$-structure $\mathscr M$, $\mathscr M \models \phi$ if and only if $\mathscr M \models \psi$. Therefore $\models \phi \leftrightarrow \psi$ is equivalent to the statement that the models of $\phi$ are exactly the same as the models of $\psi$, which is precisely the definition of $Mod^S \phi = Mod^S\psi$.

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