$\exists!x\exists!yP(x,y)$ is not the same as $\exists!y\exists!xP(x,y)$?

Question

Suppose $P(x,y)$ is a property pertaining to any object $x$ and $y$.
Let $\exists!$ be the unique existential quantifier.

Then I just found that $\exists!x\exists!yP(x,y)$ is not necessarily equivalent to $\exists!y\exists!xP(x,y)$ , generally.

For instance, suppose that $P(1,0),P(2,1),P(2,2)$ are true statements, otherwise $P(x,y)$ is false. Then clearly $\exists!x\exists!yP(x,y)$ is true but $\exists!y\exists!xP(x,y)$ is false.

Then how can I express sentences such as "there exists a unique pair of sets $x$ and $y$ such that $P(x,y)$" , meaning that there exists an object $x$ and an object $y$ such that $P(x,y)$ and that $x=x'\land y=y'$ whenever $P(x,y)\land P(x',y')$ ?

Answer 1

$\exists!x\exists yP(x,y)\land\exists!y\exists xP(x,y)$

Answer 2

It's nearly always a good idea (in my humble opinion) to specify the domain that you are working with—though it sometimes causes extra notational confusion, it can also work wonders for clarity!

For example, assuming that the domain/universe for $x$ is $X$ and the domain/universe for $y$ is $Y$, one way is to write $(\exists ! (x,y) \in X \times Y)P(x,y)$. The set $X \times Y$ is exactly the set of ordered pairs where the first element comes from $X$ and the second from $Y$, and this expression then reads "there is a unique ordered pair of elements $x$ and $y$ from the sets $X$ and $Y$ respectively such that $P(x,y)$ is true".

Answer 3

While bof's:

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

is certainly very efficient, I would suggest:

$\exists x \exists y \forall x' \forall y' (P(x',y') \leftrightarrow (x' = x \land y' = y))$

Note how this expression contains only one $P(x,y)$ formula. So it is definitely more efficient than your:

$\exists x \exists y P(x,y) \land \forall x \forall y \forall x' \forall y' (P(x,y) \rightarrow (x = x' \land y = y'))$

And, if $P(x,y)$ is a complicated formula, my suggested expression could well be more efficient than bof's.

In fact, this method generalizes well to triples, quadruples, etc. For example, to say there is exactly one triple, you get:

$\exists x \exists y \exists z \forall x' \forall y' \forall z' (P(x',y',z') \leftrightarrow (x' = x \land y' = y \land z' = z))$

So, still just one $P(x,y,z)$ formula!

Note how with your method you also still have two formulas:

$\exists x \exists y \exists z P(x,y,z) \land \forall x \forall y \forall z \forall x' \forall y' \forall z' (P(x,y,z) \rightarrow (x = x' \land y = y' \land z = z'))$

But with bof's strategy you now need three:

$\exists ! x \exists y \exists z P(x,y,z) \land \exists ! y \exists x \exists z (P(x,y,z) \land \exists ! z \exists x \exists y (P(x,y,z) $

And finally, if you were to actually use these expressions in, say, some formal proof, having 'plain' existentials is often more practical than having 'unique' existentials.