Solution 3.1.9.1.

These statements are NOT logically equivalent. To see this, we should provide an interpretation of the predicate \(P(x,y)\) which makes one of the statements true and the other false.

Let \(P(x,y)\) be the predicate \(x \lt y\text{.}\) It is true, in the natural numbers, that for all \(x\) there is some \(y\) greater than that \(x\) (since there are infinitely many numbers). However, there is not a natural number \(y\) which is greater than every number \(x\text{.}\) Thus it is possible for \(\forall x \exists y P(x,y)\) to be true while \(\exists y \forall x P(x,y)\) is false.

We cannot do the reverse of this though. If there is some \(y\) for which every \(x\) satisfies \(P(x,y)\text{,}\) then certainly for every \(x\) there is some \(y\) which satisfies \(P(x,y)\text{.}\) The first is saying we can find one \(y\) that works for every \(x\text{.}\) The second allows different \(y\)'s to work for different \(x\)'s, but there is nothing preventing us from using the same \(y\) that work for every \(x\text{.}\) In other words, while we don't have logical equivalence between the two statements, we do have a valid deduction rule:

\(\exists y \forall x P(x,y)\)
\(\therefore\) \(\forall x \exists y P(x,y)\)

Put yet another way, this says that the single statement

\begin{equation*} \exists y \forall x P(x,y) \imp \forall x \exists y P(x,y) \end{equation*}

is always true. This is sort of like a tautology, although we reserve that term for necessary truths in propositional logic. A statement in predicate logic that is necessarily true gets the more prestigious designation of a law of logic (or sometimes logically valid, but that is less fun).

in-context