how to prove ∃x(∃yA(y) → A(x)) is valid in classical logic ∃x(∃yA(y) → A(x))
i know how to prove it by natural deduction. but it requires to use model theoretical approach. so i'm not sure how we can prove the validity.
can anyone help me please ? 
 A: Let $\mathcal M$ a structure whatever, with domain $D$. 
Two cases :
(i) $∃yA(y)$ is False.
Thus $∃yA(y) \to A(x)$ is True.
For an assignment $s$ such that $s(x)=a$, for $a \in D$ whatever, we have $\mathcal M \vDash (∃yA(y) \to A(x))[s]$. 
And this means that $\mathcal M \vDash \exists x (∃yA(y) \to A(x))$.
(ii) $∃yA(y)$ is True, i.e. there is an object $a \in D$ such that $A^{\mathcal M}(a)$ holds.
This implies that $\mathcal M \vDash (∃yA(y) \to A(x))[s]$, for an assignment $s$ such that $s(x)=a$, and thus : $\mathcal M \vDash \exists x (∃yA(y) \to A(x))$.
A: Try a proof by contradiction:  
So, suppose there is some structure that makes this sentence false. 
That means that there is not some object $x$ in the domain for which it holds that if there is something which has property $A$, then $x$ has property $A$. 
By pure logic that means that for all objects $x$ in the domain it does not hold that if there is something which has property $A$, then $x$ has property $A$. 
By the semantics of the material implication, this then means that for all objects $x$ in the domain it holds that there is something with property $A$, and that $x$ does not have property $A$. 
But that means that there is something with property $A$ as well as that nothing has property $A$, and we have our contradiction.
So, there is no structure that makes the sentence false, and hence all structures will set this sentence to true.
Hence, the sentence is valid.
