why is $\forall x (p(x) \implies q(x)) \not\equiv (\forall x p(x)) \implies (\forall x q(x))$

Question

I'm having a hard time wrapping my head around why $\forall x (p(x) \implies q(x)) \not\equiv (\forall x p(x)) \implies (\forall x q(x))$

Answer 1

The first implication says: for any $x$ such that property $p(x)$ is true, then $q(x)$ is also true. But $p$ could be true only for a single $x$, for example.

The second implication says that "if $p(x)$ is true for all $x$, then $q(x)$ is true for all $x$", or, said differently, "if $p$ always holds, then $q$ always holds".

Concrete example: for $x\in\mathbb N$, let $p(x)$ be "$x$ ends with a $0$" and let $q(x)$ be "$x$ is a multiple of 5". Then, the first implication says that "if $x$ ends with a $0$, then $x$ is a multiple of $5$". The second implication would be "if all $x\in\mathbb N$ end with a zero, then all $x$ are multiples of $5$". Of course all $x\in\mathbb N$ don't end with a 0, so the second implication is not useful, and, more importantly, not equivalent to the first one.

Answer 2

Let $p(x)$ be $x<a$ and $cue(x)$ be $x<b$. LHS is equivalent to $a<b$ . but RHS is false independent of $a, b$.

Answer 3

Consider the domain of real numbers, where $P$ means "equals $\pi$" and $Q$ means "is rational".

Then $\forall x (P(x)\to Q(x))$ means "All real numbers that equal pi are rational", which is blatantly false.

Where as $\forall x~P(x) \to \forall x~Q(x)$ means "If every real number equals pi, then every real number is rational," which is vacuously true.

Thus there is at least one interpretation where the statements are not equivalent.

Answer 4

The scope relations are different here, and hence the meanings.

The first one says: "For all x it holds that: If p(x), then q(x)". The main operator is the universal quantifier. The second one says: "If for all x it holds that p(x), then for all x it holds that p(x)." The main operator is the implication.

Looking at the truth conditions, the two formulas say something different:

The first formula says basically that only if p(x), the consequent of the implication needs to be evaluated, otherwise (i.e. if the antecedent is false), the statement is true anyway; if p(x) is false for any x, then q(x) can either be true or false without this having an effect, as long as for all the x that DO fulfill the condition that p(x) also are true with q(x). So you basically discard those x for which the antecedent does not hold (which means that the statement can be true even if there is no x at all such that p(x), because even then, there is no x for which it does NOT hold that if it where p(x), it still wouldn't be q(x), therefore the formula is not false.
The universal quantifier has scope over the whole rest of the formula, therefore the "main" thing we do is to range over each individual x and check whether the implication is true.

The second formula, in which the implication is the main operator, needs to be evaluated differently: Only if it is true that for all x, it holds that p(x), only then you need to evaluate the second clause. If it is not true anyway that all x are p(x), then the implication is (more or less vacusouly) true. But if the first statement is true, then it needs to be the case that also all those x are q(x) - which then corresponds to the first formula for the x indivually (all x must be q(x) if that x is p(x)), but in the first formula, we look checked for each x whether p(x) and then whether q(x), while in the second, q(x) only needs to hold for all x if ALL x are p(x).
The main operator is the implication, so the univeral quantifiers ranging over the individual x have narrow scope with respect to the implication.

This is why the two statements are different, because in the first formula, we check for each x whether if p(x), then q(x) with the universal quantifier being the main operator, while in the second formula, we first check whether it is true that all x are p(x), and if that is true, we go over to check whether it is true that all x are q(x), without the implication holding between all x individually (because the main operator is not a univeral quantifier ranging over all x, but the implication).