Distributive properties of quanitifiers

Question

What is the difference between

$\forall x~(~P(x) \to Q(x)~)$

and

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

To me they seem to be the same thing, what difference does it make where the quantifiers go?

Answer 1

$\forall x (P(x) \rightarrow Q(x))$ means that for every $x$ that you care to pick, if for that particular $x$ $P(x)$ holds, then $Q(x)$ also holds for that particular $x$. For $x$ that don't satisfy $P(x)$, you know nothing about $Q(x)$.

$\forall x P(x) \rightarrow \forall x Q(x)$ means that if $P(x)$ holds for every $x$ at the same time, then $Q(x)$ also holds for every $x$ at the same time. If there is even a single $x$ for which $P(x)$ does not hold, it doesn't tell you anything about $Q(x)$ for any $x$.

Example: With the domain ${\mathbb N}$, take $P(x) = x \geq 2$ and $Q(x) = x \geq 3$. Then $\forall x (P(x) \rightarrow Q(x))$ is false, since $x = 2$ is a counterexample. However, $\forall x P(x)$ is false, and therefore the statement $\forall x P(x) \rightarrow \forall x Q(x)$ is true.

As an aside, note that the first statement $\forall x (P(x) \rightarrow Q(x))$ implies the second one $\forall x P(x) \rightarrow \forall x Q(x)$.

Answer 2

Suppose some, but not all, $x$ satisfy $P(x)$, and suppose that there is NO $x$ that satisfies $Q(x)$. Then the first statement is false, because there will be some $x$ such that $P(x)$ and not $Q(x)$. The second statement is true, because we have that the premise is false, and the conclusion is false, and any implication with a false premise is by definition, true. Hence the two statements cannot be equivalent.

Answer 3

The first means, 'Everything that is $\;P\;$ is also $\;Q\;$.' The second means, 'If everything is $\;P\;$, then everything is $\;Q\;$ as well.'

So if you have only one thing that is $\;P\;$, and none at all that are $\;Q\;$, then the first statement is false, but the second statement is ("vacuously") true.