I am struggeling with giving prove for the next statement : $\vdash\exists x (P(x) \rightarrow P(y))$.
This is what I have done but it fails because $\alpha$ isn't a logical sentence.
$\exists x (P(x) \rightarrow P(y)) \equiv \lnot \forall x \lnot (P(x) \rightarrow P(y)) \equiv \alpha$
1) $ \lnot \alpha \equiv \lnot (\forall x \lnot (P(x) \rightarrow P(y))) \equiv \forall x \lnot (P(x) \rightarrow P(y))$
2) $ \forall x \lnot (P(x) \rightarrow P(y)) \rightarrow \lnot (P(x) \rightarrow P(y)) $ (Assignment of x=x + Axiom)
3) $ \lnot (P(x) \rightarrow P(y)) $ (1 + 3 + M.P.)
4) $ P(x) $
5) $ \lnot P(y)$
5 + 6 derived from the following statement: $ \lnot (\alpha \rightarrow \beta) \vdash \alpha , \lnot \beta $
Now $x ,y $ are just random variables so they depends on the specific placement of x or y , and therefore I proved that $ \lnot \alpha \vdash P(x) , \lnot P(x)$ and from lemma we get that: $\vdash \lnot (\lnot\alpha)$ which is exactly as: $\vdash \alpha$
I know how to prove $ \exists x (P(x) \rightarrow \forall P(y))$ but can it help me with this statement?