What is the meaning of ∀x(∃y(A(x))) At first English is not my native language if something is not perfectly formulated or described I'm sorry.
Could somebody please tell me what the generally valid statement of this is?
$$
\forall x(\exists y(A(x)))
$$
I personally believe that it could mean something like
Forall x there is one y that I get when i put x into A(x)
 A: For all $x$, there exist an $y$ such that the property $A$ is true on $x$. But this is strange since $y$ is not used... Maybe the statement should be:
$$\forall x, \exists y, A(x,y).$$
which means: for all $x$, there exist an $y$ such that the property $A$ is true on the pair $(x,y)$.
For example, if your property is $A(x,y) \Leftrightarrow x+y = 0$, then the statement $\forall x, \exists y, A(x,y)$ is true. Indeed, for all $x$, choosing $y = -x$ makes $A(x,y)$ true.
A: The way your statement is written now, it simply means:

For all values of $x$, there exists such a value of $y$ that $A(x)$ is true.

This is a strange, but technically completely correct statement, which is always equivalent to the statement
$$\forall x: A(x)$$
A: The details may vary according to the syntactical specifications of the language but, in general, nothing forbid to have $\exists y A(x)$ when $y$ is not free in $A(x)$; the usual recursive definition of formula in FOL is :

(i) if $t_1,\ldots,t_n$ are terms and $P^n$ is an $n$-ary predicate symbol, then $t_1=t_2$ and $P^n(t_1,\ldots,t_n)$ are atomic formulae;
(ii) if $\varphi$ is a formula, then $\lnot \varphi$ is a formula;

[...]

(iv) if $\varphi$ is a formula, then $\forall x \varphi$ and $\exists x \varphi$ are formulae.


Intuitively, if $y$ is not free in $\alpha$, then $\forall y \alpha$ and $\exists y \alpha$ have the "same meaning" of $\alpha$.
We can prove it "formally" showing that :

if $y$ is not free in $\alpha$, then $\vDash \alpha \leftrightarrow \forall y \alpha$,

i.e. $\alpha$ and $\forall y \alpha$ are logically equivalent, and the same for $\exists y \alpha$.

We can apply the above result to : $∀x(∃y(A(x)))$ and we get that it is equivalent to : $∀x(A(x))$.
The semantical specifications for the quantifiers are [according to Herbert Enderton, A Mathematical Introduction to Logic (2nd - 2001)), page 83-84] :
We say that a structure $\mathcal A$ satisfy a formula $\varphi$ with an assignment function $s$, in symbols : $\mathcal A \vDash \varphi [s]$, when :
[...]

$\mathcal A \vDash \forall x \varphi[s]$ iff for every $d \in |\mathcal A|$, we have $\mathcal A \vDash \varphi[s(x|d)]$

and :

$\mathcal A \vDash \exists x \varphi[s]$ iff for some $d \in |\mathcal A|$, we have $\mathcal A \vDash \varphi[s(x|d)]$.

Thus, consider a structure $\mathcal A$ and a function $s$ and apply the above definition; $\mathcal A$ satisfies $\forall x (\exists y A(x))$ with $s$ if

$\mathcal A \vDash \exists y A(x)[s(x|d)]$, for every $d \in |\mathcal A|$.

And this, in turn, means :

$\mathcal A \vDash A(x)[s(x|d)(y|e)]$, for every $d \in |\mathcal A|$ and some $e \in |\mathcal A|$.

But the variable $y$ does not occurr free in $A(x)$ and thus it does not matter what is the "denotation" that $s(x|d)$ assign to it. So we can discard $e$ and we have :

$\mathcal A \vDash A(x)[s(x|d)]$, for every $d \in |\mathcal A|$,

and this amounts to : $\mathcal A \vDash \forall x A(x)[s]$.
