Existential quantifier question $\forall x \forall y(P(x)\land P(y) \implies x=y)$

Quesiton about existential quantifier The unique quantifier is $!\exists xP(x)$ so only one x is true.

And I am wondering if the following is equal to unique quantifier

$\forall x \forall y(P(x)\land P(y)) \implies x=y)$

So this is saying that for all the x and y which makes P() true then x is identical to y. But this is equivalent to saying

$\forall x \forall y(\neg P(x)\lor \neg P(y) \lor x=y)$

And if for all x P(x) is not true then it cannot be true for one x.

Not sure if I am right.

Solution 1:

$\exists !x P(x)$ means that $P(x)$ is true for exactly one $x$. Your statement $\forall x\forall y((P(x) \wedge P(y)) \to x = y)$ means that $P(x)$ is true for at most one $x$. To say that $P(x)$ is true for exactly one $x$, you have to say that it is true for at least one $x$, and also at most one $x$. So $\exists !xP(x)$ is equivalent to: $$ \exists x P(x) \wedge \forall x\forall y((P(x) \wedge P(y)) \to x = y). $$