How to show Universal Quantifier distributes over implication?

Question

How to show Universal Quantifier distributes over implication? I've tried to no avail to show $\forall x(P(x) \implies Q(x))$ is equivalent to $\forall x(P(x)) \implies \forall (Q(x))$ but it seems no manipulation of logical connectives and quantifier negations gets me the desired result, also the equivalence between the material conditional and its associated OR form seem to prevent me from showing this equivalence... :/ I have looked online as well and cannot find an understandable answer that doesn't involve some complicated metalogic symbol pushing. Is there some simple process I am not seeing? If not then where should I start reading in order to understand these complicated metalogic derivations?

Answer 1

Let $P(x)$ mean "$x$ is a princess", and let $Q(x)$ mean "x is royalty"

Then $\forall x\Big(P(x)\to Q(x)\Big)$ reads, "anyone, who is a princess, is royalty."

Likewise $\Big(\forall x \; P(x)\Big)\to \Big(\forall x \;Q(x)\Big)$ reads, "if everyone is a princess, then everyone is royalty."

As these don't quite mean the same thing, then it is clear that they are not equivalent.

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Is there a way to show this through syntax? I.e. converting the implication to not p or q and doing something there? Or is that not possible? It seems that way since universal quantified does not distribute over OR statements – skyfire

Exactly so. Both the inability to distribute universal quantification over disjunction, and dual negation, prevent you from equating the first and second statements.

The first becomes, $\forall x\Big(\neg P(x) \vee Q(x)\Big)\;$, which reads "Anyone either is not princess or is royalty." This makes it necessary that any princess must be royalty.

The second becomes $\Big(\neg \forall x \; P(x)\Big)\vee \Big(\forall x \;Q(x)\Big)\;$ which reads "either not everyone is a princess or everybody is royalty."   This allows the possibility of a princess who is not royalty.

Answer 2

An easy counterexample to the equivalence would be to let $P(x)$ be a condition that holds for only some things in the model but not all. (For example, say we're talking about reals and $P(x)$ says "$x$ is an integer". Then $\forall x P(x)$ is false and the formula $(\forall x P(x))\rightarrow (\forall x Q(x))$ is satisfied no matter what $Q$ says. (For example, let $Q(x)$ say "$x=\frac{1}{2}$". Then $\forall x(P(x)\rightarrow Q(x))$ is false, but the other formula is still true.)

However, the distributive direction is true. A proof of it depends on what you're taking your deduction axioms to be though. What text are you studying?

Answer 3

1. ∀x(P(x)→Q(x))
2. (∀xP(x))→(∀xQ(x))

One way to see how the two statements are different, is to see how we can we refute each statement.

To refute the first statement, you only need one item that is a P and not a Q. If there are items in the universe that belong to both P and Q, then this doesn't stop us from refuting the first statement. It is completely legitimate to have items belonging to both P and Q, but still we can refute the statement if we just found one item that is a P and not a Q.

We can refute the second statement if all the items in the universe belonged to P, and only one of the items didn't belong to Q.