Show both are equivalent.

Question

$\exists xP(x)\implies P(y)$ is equivalent to $\forall x(P(x)\implies P(y))$

What can I show in this question to prove that they are equivalent. Any help is appreciated.

Thanks.

Answer 1

It's mostly a matter of using DeMorgan rule for quantifiers:

$$\exists x P(x)\rightarrow P(y)$$ $$\neg \exists x P(x)\lor P(y)$$ $$\forall x \neg P(x)\lor P(y)$$ $$\forall x (\neg P(x)\lor P(y))$$ $$\forall x (P(x)\rightarrow P(y))$$

The step between the third and fourth is because $P(y)$ doesn't depend on $x$.

Answer 2

It is intuitionistically provable, although proof by cases or De Morgan laws don't show this. A deductive proof:

($\to$) Suppose that $\exists x P(x)\to Q$. Suppose $P(x)$. Then $\exists x P(x)$, so by MP $Q$ is true. Thus $P(x)\to Q$, and this is true for arbitrary $x$, $\forall x(P(x)\to Q)$.

($\leftarrow$) Suppose that $\forall x(P(x)\to Q)$. Take some $x$, so $P(x)\to Q$. Since $Q$ does not depend on $x$, deduce $\exists x P(x)\to Q$.

Thus $(\exists x P(x)\to Q)\iff \forall x(P(x)\to Q)$.

This theorem is sometimes taken as the "definition" of $\exists x P(x)$ in intuitionistic logic. In classical logic it is sufficient to take $Q=\bot$ so that it simplifies to $\lnot\exists x P(x)\iff \forall x\lnot P(x)$, the De Morgan law for quantifiers, and another contraposition yields an actual definitional theorem, $\exists x P(x)\iff \lnot\forall x\lnot P(x)$. In intuitionistic logic these two are not equivalent, but the version with $Q$ is a useful substitute.