Are $\exists ! (x,y)$ and $\exists ! x \exists ! y$ equivalent? Here I have the following definition of $\exists ! (x,y) P(x,y)$:
$$\exists x \exists y[P(x,y) \land \lnot \exists u \exists v (P (u,v) \land (u \neq x \lor v \neq y))]$$
Are that formula equivalent to $\exists ! x \exists ! yP(x,y)$ ?
I'm using that $\exists ! xP(x)$ is equivalent to:
$$\exists x (P(x) \land \lnot \exists u (P(u) \land u \neq x)$$
I could not prove them equivalent, but don't know how to prove that it's not equivalent.

Let $P(x,y)$ be defined by $y^2 = x$.
Then $\exists!(x,y)P(x,y)$ is false, since there are infinitely many pairs $(x,y)$ that solve the equation.
However, $\exists!x(\exists!yP(x,y))$ is true, since if and only if $x=0$, there is only one $y$ that solves the equation ($y=0$ obviously).
