Let x, y ∈ R, show that ∀x∃y(xy = 0) ⇔ ∃y∀x(xy = 0)

To prove that ∀x∃y(xy = 0) ⇔ ∃y∀x(xy = 0), we need to prove both directions of the equivalence:

First, let's prove that if ∀x∃y(xy = 0), then ∃y∀x(xy = 0):

Assume ∀x∃y(xy = 0). We want to show that there exists a y such that for all x, xy = 0.

Let's choose an arbitrary value of y such that for any x, either xy = 0 (if x ≠ 0) or y = 0 (if x = 0). Such a value of y exists since for any nonzero x, there exists a y = 0 such that xy = 0, and for x = 0, any y would satisfy xy = 0.

Now, we need to show that this y satisfies ∀x(xy = 0):

Take an arbitrary x. If x = 0, then xy = 0 (since y can be any value). If x ≠ 0, then y was chosen so that xy = 0. Therefore, in either case, xy = 0, and we have shown that ∃y∀x(xy = 0).

Now let's prove the other direction, that if ∃y∀x(xy = 0), then ∀x∃y(xy = 0):

Assume ∃y∀x(xy = 0). We want to show that for any x, there exists a y such that xy = 0.

Take an arbitrary x. Since ∃y∀x(xy = 0), there exists a y such that for all x, xy = 0. In particular, this y satisfies xy = 0 for our chosen value of x. Therefore, we have shown that for any x, there exists a y such that xy = 0, and thus we have proved ∀x∃y(xy = 0).

Since we have proved both directions of the equivalence, we can conclude that ∀x∃y(xy = 0) ⇔ ∃y∀x(xy = 0).