In predicate logic, is it possible to distribute quantifiers Is possible to establish that $\forall x \,\exists y\,(Fx \rightarrow Gy)$ is logically equal to $\forall x\,Fx \rightarrow \exists y\,Gy$?
If it does not work, why not?
 A: It depends on whether you mean $\;(\forall x\,Fx) \rightarrow \exists y\,Gy\;$ or $\;\forall x\,(Fx \rightarrow \exists y\,Gy)\;$.
The original expression is equivalent to the latter, but not to the former.
Here is a proof, using a sightly different notation, in tiny baby steps:
\begin{align}
& \langle \forall x :: \langle \exists y :: F(x) \Rightarrow G(y) \rangle \rangle \\
\equiv & \;\;\;\;\;\text{"rewrite $\;p \Rightarrow q\;$ to $\;\lnot p \lor q\;$ -- the latter is usually easier to manipulate"} \\
& \langle \forall x :: \langle \exists y :: \lnot F(x) \lor G(y) \rangle \rangle \\
\equiv & \;\;\;\;\;\text{"$\;\lor\;$ distributes over $\;\exists\;$"} \\
& \langle \forall x :: \langle \exists y :: \lnot F(x) \rangle \lor \langle \exists y :: G(y) \rangle \rangle \\
\equiv & \;\;\;\;\;\text{"leave out quantification over variable $\;y\;$ which does not occur in $\;\lnot F(x)\;$"} \\
& \langle \forall x :: \lnot F(x) \lor \langle \exists y :: G(y) \rangle \rangle \\
\equiv & \;\;\;\;\;\text{"reintroduce $\;\Rightarrow\;$ -- to achieve our goal"} \\
& \langle \forall x :: F(x) \Rightarrow \langle \exists y :: G(y) \rangle \rangle \\
\end{align}
A: Yes, it is possible to distribute the quantifiers, but the result with the formula $∀x∃y(Fx→Gy)$ is not what you have suggested.
I'll use Enderton's system (see Herbert Enderton, A Mathematical Introduction to Logic (2nd - 2001)).
We need some provable equivalences [see Enderton, page 121 and page 130] :

(Q2A) -- $\vdash (\alpha \rightarrow \forall x \beta) \leftrightarrow \forall x (\alpha \rightarrow \beta)$, if $x$ does not occur free in $\alpha$
(Q2B) -- $\vdash (\alpha \rightarrow \exists x \beta) \leftrightarrow \exists x (\alpha \rightarrow \beta)$, if $x$ does not occur free in $\alpha$
(Q3A) -- $\vdash (\forall x \beta \rightarrow \alpha) \leftrightarrow \exists x (\beta \rightarrow \alpha)$, if $x$ does not occur free in $\alpha$
(Q3B) -- $\vdash (\exists x \beta \rightarrow \alpha) \leftrightarrow \forall x (\beta \rightarrow \alpha)$, if $x$ does not occur free in $\alpha$.

Starting with :

$∀x∃y(Fx \rightarrow Gy)$

we apply first (Q2B), because $y$ is not free in $Fx$, to get :

$∀x(Fx \rightarrow ∃yGy)$.

Then we apply (Q2A), because $x$ is not free in $∃yGy$, to get :


$∃xFx \rightarrow ∃yGy$.


This formula is clearly not equivalent to : $∀xFx \rightarrow ∃yGy$.
A: It is not equivalent. If you just read what the statements are saying it should be clear. The statement
$\displaystyle \forall x\ \exists y\ (f(x) \rightarrow g(y))$
says that for every $x$ there exists some $y$ such that the statement $f(x) \rightarrow g(y)$ is true. The statement
$(\forall x\ f(x)) \rightarrow (\exists y\ g(y))$
says that if $f(x)$ is true for every $x$, then there exists a $y$ for which $g(y)$ is true. Hence the proprositional functions that you need to satisfy are different
A: You could get the following
$\forall x (Fx\rightarrow \exists y Gy)$.
A: The easiest approach to show they aren't equivalent is to exhibit a countermodel:
The latter statement -- assuming you mean $(\forall x Fx)\to \exists y Gy$ -- is true in a model where there's an F and a non-F, but nothing is G. But the former statement -- $\forall x\exists y(Fx\to Gy)$ -- is false in such a model.
