Do the "nothing" and "unique existence" quantifiers respect or at least semi-respect ordered pairs?

Question

Let $Nx$ be the quantifier "for no $x$", and let $\exists!x$ be the quantifier "there is one and only one $x$". Do those quantifiers respect ordered pairs? That is, is $NxNyP(x,y)$ equivalent to $N(x,y)P(x,y)$, and is $\exists!x \exists!y P(x,y)$ equivalent to $\exists!(x,y)P(x,y)$? If either or both of those equivalences do not hold, does at least one direction of the equivalences hold? (That is what I mean by semi-respect).

Answer 1

To give an example where $NxNyP(x,y) \neq N(x,y)P(x,y)$, consider $P(x,y):=x=y$ where the universe from where $x$ and $y$ come is any non-empty set, say $\mathbb N$.

Given an $x$, $NyP(x,y)$ then means that there is no $y$ with $x=y$. This is incrorrect for every $x$, as one can chose simply $y$ equal to $x$. So $NxNyP(x,y)$ is True, there is no $x$ such that $NyP(x,y)$ holds.

OTOH, $N(x,y)P(x,y)$ means there is no ordered pair $(x,y)$ where $x=y$, which is False. Over $\mathbb N$ the pair $(1,1)$ would be a counterexample and such a coounter example exists for any non-empty set.