confusion on a statement about truth I have to prove the following: $$(\exists x) \phi \to (\exists x)\psi \vDash (\exists x)(\phi\to\psi)$$ and prove the the opposite direction is wrong. However i think i ve misunderstood something cause i think it s the other way around.
What i have tried is: (I ll use "a" instead of gothic a to save some time here, meaning an L-structure).
Let $a\vDash (\exists x) \phi \to (\exists x)\psi $, so we have $a\vDash (\exists x) \phi \Rightarrow a\vDash(\exists x)(\psi) $, thus [$a\vDash \phi(c_a)$ for some $ a \in A \Rightarrow a\vDash \psi(c_b)$ for some  $b\in A$] but from here we cannot infer that $a \vDash \psi (c_a)$. Instead if we have the opposite direction we have that if
$a\vDash (\exists x)(\phi\to\psi)$ then..... $a\vDash \phi(c_a) \rightarrow \psi(c_a)$ for some $a \in A$ and thus $a \vDash(\exists x) \phi \to (\exists x)\psi $
What am i doing wrong here? Could you also provide a counterexample to get a better understanding?
 A: Hints for one possible approach to the forward argument: assuming the universe of discourse $\mathfrak{a}$ is nonempty, let $x_0$ be an object in $\mathfrak{a}$.  The proof will then have two cases, based on whether or not $\mathfrak{a} \models \phi[x:=x_0]$:


*

*Suppose $\mathfrak{a} \models \phi[x := x_0]$.  Then $\mathfrak{a} \models (\exists x) \phi$, so $\mathfrak{a} \models (\exists x) \psi$.  Let $y_0$ be a witness, so that $\mathfrak{a} \models \psi[x := y_0]$.  Now, why does it follow that $\mathfrak{a} \models \exists x (\phi \rightarrow \psi)$?

*Suppose $\mathfrak{a} \models \lnot \phi[x := x_0]$.  Then you can immediately conclude $\mathfrak{a} \models \exists x (\phi \rightarrow \psi)$ (why?).
For a counterexample for the reverse direction: as indicated earlier in the comments, use the model where the universe of discourse is $\{ a, b \}$, with:
$$ \mathfrak{a} \models \phi[x := a], \lnot \psi[x := a], \lnot \phi[x := b], \lnot \psi[x := b]. $$
Then the right hand side is satisfied (with witness $b$); however, the left hand side is not satisfied, since $(\exists x) \phi$ is satisfied with witness $a$, while $(\exists x) \psi$ is not satisfied.
