# Are these two characterizations of two (sets of) formulas being equivalent?

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.)

Thanks.

• Small typo in your question: both in the quote and in point 1 the "iff $\Phi \models \phi$" should be "iff $\Psi \models \phi$". – Mark Kamsma Sep 2 '20 at 8:09
• Did you try to prove either of these yourself? If so, where did you get stuck? – 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 $$Mod^S \phi = Mod^S\psi$$.