Can we imply $\exists x\phi(x)\rightarrow \exists x\psi(x)$ from $\exists x(\phi(x)→\psi(x))$ let $\Sigma$  be a consistent set of formulas in first-order logic and implies $\exists x(\phi (x)→\psi(x))$. which one of these statements is logical implication from $\Sigma$?

a) $\forall x\phi(x)→\forall y\psi(y) $
b) $\exists x\phi(x)→\forall y\psi(y)$
c) $\exists x\phi(x)→\exists x\psi(x) $
d) $\forall x\phi (x)→\exists y\psi(y)$

I think This sentence implies $\exists x\phi(x)→\exists x\psi(x) $ but the answer is $\forall x\phi (x)→\exists y\psi(y)$ how we can imply that for all x exist a y which $y\psi(y)$ ?! I can't understand it.
 A: The inference $\exists x (\phi(x) \to \psi(x)) \nvDash \exists x\phi(x) \to \exists x \psi(x)$ is invalid because there exists a counter model:
Let the domain be $\{a, b\}$ and $\phi$ true of $a$ but not of $b$, and $\psi$ true of none of the elements.
Then for $x \mapsto b$, $\phi(x)$ is false so $\phi(x) \to \psi(x)$ is true, so $\exists x (\phi(x) \to \psi(x))$ is true.
But $\exists x \phi(x)$ is true with $x \mapsto a$ whereas $\exists x \psi(x)$ is false since $\psi$ is true of neither $a$ nor $b$, so $\exists x \phi(x) \to \exists x \psi(x)$ is false.
Since in this structure the premise is true but the conclusion is false, the inference is invalid.
The inference $\exists x (\phi(x) \to \psi(x)) \vDash \forall x\phi(x) \to \exists x \psi(x)$ is valid by the following reasoning:
By assumption, $\phi(x) \to \psi(x)$ holds of some object; let it be $y$. Assume $\forall x \phi(x)$, then in particular $\phi$ holds of $y$. By modus ponens, $\psi$ must also be true of $y$. Hence there exists an object of which $\psi$ holds, so $\exists x \psi(x)$ is true. Thus $\forall x \phi(x) \to \exists x \psi(x)$. Since an object such as $y$ is assumed to exist, $\forall x \phi(x) \to \exists x \psi(x)$ holds regardless of which particular object $\phi(x) \to \psi(x)$ is true of.
As an exercise, you can now try to translate this informal proof into a natural deduction proof.
A: For a counter-example to (c), you just need a single $x$ for which $\phi(x)$ is false. Then for this $x$, we have $\phi(x)\to$ anything at all. So $\exists x(\phi(x)\to\psi(x))$ is true. But we have no reason to infer that $\exists x(\psi(x))$.
As for your final paragraph: it looks like you are interpreting
$$\forall x\phi (x)\to\exists y\psi(y)$$
as
$$\forall x(\phi (x)\to\exists y\psi(y))$$
whereas the correct interpretation is
$$(\forall x\phi (x))\to(\exists y\psi(y))$$
