1

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.

Fernando Martinez
  • 6,698
  • 19
  • 74
  • 108
  • I don't think this is right. If $P(x)$ happens to be false for all $x$, then $\exists!xP(x)$ is false but your expression would be true, as the left side of the implication would be false for all $x, y$. – 79037662 Nov 26 '21 at 19:30
  • 1
    Your expression seems to be equivalent to "at most one $x$ is such that $P(x)$", rather than exactly one. – 79037662 Nov 26 '21 at 19:32
  • It could be $\exists x(P(x)\wedge\forall y(P(y)\implies y=x))$ or you could take the conjunction of yours and $\exists xP(x)$. Or other options. – plop Nov 26 '21 at 19:33
  • 1
    Yes P(x) and P(y) could be false and implication is still true, so there might not exist an x and y which makes P() true. – Fernando Martinez Nov 26 '21 at 19:37

1 Answers1

2

$\exists !x P(x)$ 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: $$ \exists x P(x) \wedge \forall x\forall y((P(x) \wedge P(y)) \to x = y). $$

Dan Velleman
  • 2,746