Need help to prove this by using natural deduction. i m concerned to prove these by using Natural Deduction. And i am also concerned to prove it for both sides. $$\exists x(P (x) \implies A) \equiv \forall xP (x) \implies A$$
I have some difficulties to show it by using Natural Deduction. Any kind of help will be appreciated 

$(\Longrightarrow)$ Suppose $\exists x(P (x) \implies A)$. This means we can pick an $x_0$ such that $P (x_0) \implies A$. Now suppose $\forall x P (x)$. In particular, this gives us $P(x_0)$. Then, by modus ponens, $A$. So we have shown that $\forall x P(x) \implies A$.  
$(\Longleftarrow)$ Suppose $\forall x P(x) \implies A$. Either $\forall x P(x)$ or $\neg \forall x P(x)$. We will consider these cases separately:
$\hspace{1cm}$ Case 1. $\forall x P(x)$. By modus ponens, we have $A$. Because $A$ is true, we can pick an $x_0$ at random and declare $\neg P(x_0) \lor A$. This is equivalent to $P (x_0) \implies A$. Therefore, $\exists x(P (x) \implies A)$. 
$\hspace{1cm}$ Case 2. $\neg \forall x P(x)$. This means we can pick an $x_0$ such that $\neg P(x_0)$. Because $\neg P(x_0)$, we must have $\neg P(x_0) \lor A$. This is equivalent to $P (x_0) \implies A$. Therefore, $\exists x(P (x) \implies A)$. 
In either case, we have $\exists x(P (x) \implies A)$. 
