# how to give a formal prove to $\vdash \exists x (P(x) \rightarrow P(y))$

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?

• Informally, there exists such $x$ because you can pick $x=y$. – Hagen von Eitzen May 24 at 6:13
• @HagenvonEitzen And formally, you just do an existential introduction with $y$ as the witness, then you need to prove $P(y)\to P(y)$ which is pretty straightforward, $\to$ introduction, then by assumption. – Derek Elkins May 24 at 7:43
• You might try assuming the negation, switching the quantifier, etc. and deriving the contradiction $P(y) \land \neg P(y).$ – Dan Christensen May 24 at 14:40