I am trying to prove the expression: $\exists x P(x) \implies Q(y) \Longleftrightarrow \forall x [P(x) \implies P(y)]$
I was hoping I could get some intimation on what to do next. For your time I am beforehand very appreciative. I'll start proving first $[\exists x P(x) \implies Q(y)] \implies \forall x [P(x) \implies P(y)]$
$$\exists x P(x) \rightarrow Q(y)$$ [Existential Instantiation] $$P(a) \to Q(y)$$ [Idempotency] $$P(a) \wedge P(a) \to Q(y)$$ [Extraction] $$P(a) \to (P(a) \to Q(y))$$ [Material Implication] $$P(a) \to ( \neg P(a) ∨ Q(y)) $$ [DeMorgan] $$P(a) \to \neg (P(a) \wedge \neg Q(y)) $$ [Simplification] $$P(a) \to \neg (P(a))$$
This is as far as I've gotten. I seem to be really stuck. If anyone could provide me with suggestion on how to proceed from here I will greatly appreciate it. Thank you kindly.