Commutivity of unique existence quantifiers

Question

Find an expression P(x,y) to disprove the following equivalence,

$(\exists!x)(\exists!y)P(x,y)\Leftrightarrow(\exists!y)(\exists!x)P(x,y)$

I could only think of a few statements of $P(x,y)$ that made some sense, such as in a commutative ring, $(\exists!x)(\exists!y)(\forall a)(ax=x\wedge ay=a)$ where $x$ is our zero and $y$ is our 1. This doesn't seem to work since it doesn't matter if we change the order of the quantifiers of $x$ and $y$.

Or maybe a version of L'Hopital's rule, $(\exists!x)(\exists!y)\left(\lim_{z\rightarrow a}\frac{f'(z)}{g'(z)}\Rightarrow\lim_{z\rightarrow a}\frac{f(z)}{g(z)}\right)$, which to me seems more plausable since the unique existence of $y$ has no bearing on the unique existence of $x$, as L'Hopital's rule states.

However, determining whether this helps come up with a counter-example is difficult since I'm having trouble interpreting the truth of this statement (maybe I need to specify $a,g$, or $f$).

Feel free to fix either of my examples, or make some counter-examples of your own.

Answer 1

It may be easier to make up a completely abstract example which is designed to do what you want, rather than looking for a "real life" example. On $S=\{1,2,3\}$ define $$P(x,y)\quad\hbox{if and only if}\quad (x,y)\in\{(1,1),(2,1),(2,2),(3,1),(3,2)\}\ .$$ Then for $x=1$ there is a unique $y$ (namely, $y=1$) such that $P(x,y)$; and for any $x\ne1$ there are two values of $y$ for which $P(x,y)$ is true. So $$\exists!\,x\ \exists!\,y\ P(x,y)$$ is true. On the other hand, for any $y\in S$ there are either three, two or no values of $x$ for which $P(x,y)$ is true. Therefore $$\exists\,y\ \exists!\,x\ P(x,y)$$ is false, and consequently $$\exists!\,y\ \exists!\,x\ P(x,y)$$ is false.

Answer 2

Your first example isn't good because $P(x,y)$ can't have any more variables and you're looking at $a$ as a variable. You should look at it as a constant and do not quantify it. Even better would be to replace by a constant, but you need to find an actual example that works.

Your l'Hôpital example doesn't make sense because $\left(\lim_{z\rightarrow a}\frac{f'(z)}{g'(z)}\Rightarrow\lim_{z\rightarrow a}\frac{f(z)}{g(z)}\right)$ is not a well formed formula, it is not a statement.

For an example that works, consider your universe to be $\mathbb R$ and the predicate $P(x,y)\colon x^2=y$.