Universal quantifier distributes over implication

Question

Is

$\forall x \forall y: P(x) \to Q(y)$

the same thing as

$(\forall x P(x)) \to (\forall y:Q(y))$

?

If not can someone give an example as to why it isn't?

I'm not getting the whole meaning of $\forall x \forall y:P(x) \to Q(y)$ at all. Does that mean for if $P(x)$ is satisfied, then $Q(y)$ is satisfied for all $y$? Or does it mean that if $P(x)$ is satisfied, then there is some $y$ out there that satisfied $Q(y)$?

Answer 1

Let $P(x)$ be a statement which is sometimes true and sometimes false; likewise for $Q(y)$. Then

• $P(x)\to Q(y)$ is sometimes false, so $\forall x\forall y (P(x)\to Q(y))$ is false;
• but $(\forall x P(x))$ is false, so $(\forall x P(x))\to(\forall y Q(y))$ is true.

So the two formulae do not mean the same thing.

Answer 2

$$ \forall x \forall y : (Px \to Qy) $$ means: If you pick any $x$ you want, and you at the same time independently pick any $y$ you want, it must be true that $P(x) \to Q(y)$. So either $P(x)$ is false for that pair $x,y$, or $P(x)$ is true and $Q(y)$ is true.

This is not equivalent to $(\forall x : Px) \to (\forall y : Qy)$. The latter, as it turns out, is instead equivalent to $$ \exists x \forall y : (Px \to Qy). $$

Answer 3

As per this question

No; they are not.

$\forall x\;\forall y\; (P(x)\to Q(y))$ means "for every two things, if one is $P$ then the other is $Q$". Thus if there's is anything that is $P$, then everything is $Q$.

This is equivalent to $(\exists x\; P(x)) \to (\forall y\; Q(y))\;$.

...

On the other flipper, $( \forall x\; P(x) )\to (\forall y\; Q(y))$ means "if every $x$ is $P$ then every $y$ is $Q$."   That is not the same thing.

---

$$\begin{align} &\forall x\; \forall y\; [P(x) \to Q(y)] \\[1ex] \iff & (\text{implication equivalence}) \\[1ex]&\forall x\;\forall y\; (\neg P(x) \vee Q(y)) \\[1ex] \iff & (y\text{ is free in } P(x)) \\[1ex]&\forall x\; (\neg P(x) \vee (\forall y\; Q(y))) \\[1ex] \iff & (x\text{ is free in }\forall y\;Q(y)) \\[1ex]&(\forall x\; \neg P(x)) \vee (\forall y\; Q(y)) \\[1ex] \iff & (\text{dual negation}) \\[1ex]&(\neg \exists x\; P(x)) \vee (\forall y\; Q(y)) \\[1ex] \iff & (\text{implication equivalence}) \\[1ex]&(\exists x\; P(x)) \to (\forall y\; Q(y)) \end{align}$$