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.


"There is exactly one pair of people $x$ and $y$ at this party such that $x$ is friends with $y$.

is not the same as

"There is exactly one person $x$ at this party such that there is exactly one person $y$ at this party such that $x$ is friends with $y$.

At the first party, $x$ is the only person with a friend. At the second party, poor $x$ only has one friend, but everybody else is free to have lots of friends (or no friends).
For this "real life" example to  work, the "friends with" relation must not be symmetric. But I think that's not too unrealistic.
