Prove or disprove validity: $(\forall x \exists y (P(x) \supset Q(y))) \supset(\exists y \forall x (P (x) \supset Q(y)))$

I have working on this formula $(\forall x \exists y (P(x) \supset Q(y))) \supset (\exists y \forall x (P (x) \supset Q(y)))$ to either prove or disprove it.

First, I tempted to disprove it, but I changed my mind.

I wrote down "for all x that there exists some y satisfies corresponding condition", and "there exists some y that for all x satisfies corresponding condition." I think these statements refer to the same idea.

Any suggestions?

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If you replace the specific clause $P(x)\implies Q(y)$ with an arbitrary formula $\Phi(x,y)$, this formula isn't generally true.  The left-hand side means "for every specific $x$, we can find some $y$ that makes $\Phi(x,y)$ true".  The right-hand side means "there is a particular $y$ that makes $\Phi(x,y)$ true regardless of $x$".  This is a much stronger statement in general.  In other words, the two statements don't refer to the same idea in general. – mjqxxxx Dec 8 '11 at 15:05
– Oleksandr Kozlov Dec 8 '11 at 16:31
I have looked Confused between Nested Quantifiers, but there is only one predicate. I think this is different. – mert Dec 8 '11 at 21:47
There are cases where P(x) and Q(y) have different truth values. – mert Dec 8 '11 at 21:53

2 Answers

In the usual context of nonempty structures, the statement is valid, meaning that it is true in all models. One way to see this is that if there is any $y$ in the structure for which $Q(y)$ holds, then the statement is true, since we may always choose that $y$, and this will make $Q(y)$ true and hence the final implication true, regardless of any $x$, and so the whole implication is true. Otherwise, we are in the case where $Q(y)$ is always false, in which case $P(x)\to Q(y)$ is logically equivalent to $\neg P(x)$, and so the quantification over $y$ becomes irrelevent (provided the structure is nonempty). So again the statement is true. So the statement is true in any nonempty structure.

Edit. Meanwhile, in the context of first-order logic allowing the empty structure (which is a bit unusual, and which is only possible if your language has no constant symbols), then the statement is not valid, since it is false in the empty structure. This is because all universal statements hold vacuously in the empty structure and all existential statements fail in the empty structure, so the antecedent is true and the conclusion is false, and so the implication fails in the empty structure.

Conclusion: your statement is valid if the language has constant symbols; valid if your logic disallows the empty structure; but otherwise it is not valid.

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I need to prove its validity by proof theory. – mert Dec 8 '11 at 12:48
Then you need to consult the course instructor, since he (not us) knows the rules for proofs that you can use. – GEdgar Dec 8 '11 at 13:30

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Ah, I had understood instead that you wanted us merely to prove the statement, which I did. If you want a formal derivation of it in some formal proof system, then we would need to know the details of your system; there are many proof systems. If your proof system is complete, then the existence of a formal derivation follows from the statement's validity. So it seems that you want an actual formal derivation (which I rarely find illuminating, and so I will leave this). But my argument suggests an approach: first derive $(\exists y Q(y))\vee(\forall y \neg Q(y))$ and then break into cases. – JDH Dec 8 '11 at 13:43

$(\forall x \exists y (P(x) \supset Q(y))) \supset (\exists y \forall x (P(x) \supset Q(y)))$ The formula is not valid. Consider the following idea: $P^M$ is the relation that {x : x moves first} $Q^M$ is the relation that {x : x wins the game} For the left part of implication $(\forall x \exists y (P(x) \supset Q(y)))$ it indicates that for all $x$ making the first move, there exists some $y$ that wins the game. However, considering right part of implication $(\exists y \forall x (P(x) \supset Q(y)))$ there exists some $y$ that beats all first moves of $x$. This is obviously different and harder than the left part of implication, which can be false while first part is true so that makes the theory false. The idea is from this page.

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This answer is incorrect. If there is any $y$ that is a winner, then that $y$ will satisfy $Q(y)$, and so the final implication will be true, regardless of anything that is claimed about some $x$ or other, and so the whole implication will be true. Otherwise, there is no $y$ that is a winner, and so $P(x)\to Q(y)$ is logically equivalent to $\neg P(x)$, and the quantification over $y$ becomes irrelevant. This is the argument I give in my answer. I believe that you have in mind a two-place property $P(x,y)$, asserting something like, $y$ is the move that wins the game that began with move $x$. – JDH Dec 9 '11 at 0:02
Thus, you are refuting the implication $(\forall x\exists y P(x,y))\to(\exists y\forall x P(x,y))$, which is indeed not valid. – JDH Dec 9 '11 at 0:03
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