$P(x, y)$ for which $\exists x\forall yP(x, y)$ and $\exists y\forall xP(x, y)$ are not equivalent

Question

The domain of $x$ and $y$ is the same, real numbers. I was wondering if there was a $P(x)$ for which these two would be non equivalent.

Answer 1

You could let $P(x,y)$ mean $x=0$. (Yes, $y$ deliberately doesn't appear there).

Answer 2

You can also consider:

$$P(x,y)="y\times\lfloor x^2+1\rfloor\text{ is an integer}".$$

Then

$$\exists x,\quad\forall y,\quad P(x,y)$$

is false since the domain you consider is the set of real numbers.

But

$$\exists y,\quad\forall x,\quad P(x,y)$$

is true (you can take $y=0$ or $y=k\in \mathbb Z$ for instance).