Commutivity of unique existence quantifiers 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.
 A: 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.
A: 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 $P(x,y)\colon x^2=y$.
