How should be interpreted "(∀x)(∀y)(P(x) ∧ P(y) ⇒ x = y)"?

Question

I was reviewing some books of logic and find that the statement "(∃!x)P(x)" is equivalent to "(∃x)P(x) ∧ (∀y)(∀z)(P(y) ∧ P(z) ⇒ y = z)", i understand that "(∃x)P(x)" is read as "there exist a x that, P(x)", therefore "(∀y)(∀z)(P(y) ∧ P(z) ⇒ y = z)" is the part that must indicate that x is unique, but i dont find the way of interpreted it for give it that interpretation, The interpretation that occurs to me is "Every y that P(y), is equal to every z that P(z)", but, why this mean that x is unique?

Answer 1

The formula you provided works, but is a bit redundant. I'd suggest: $$(\exists x) P(x) \wedge (\forall y)(\forall z)(P(y) \wedge P(z) \implies y = z)$$ This essentially just rules out the possibility that there are two different values $y$ and $z$ that both satisfy $P$ but aren't equal.

You could also use: $$(\exists x) P(x) \wedge (\forall y)(P(y) \implies y = x)$$ Where the scope of the $\exists x$ quantification carries over to the second part of the formula. This says that everything satisfying $P$ is $= x$.

Answer 2

The quantifier that you have encountered is called a uniqueness quantifier. The formula $\exists! x \mathop. P(x)$ reads as "there exists a unique $x$ such that $P(x)$ holds".

The first part $\exists x \mathop. P(x)$ means there is at least one $x$ such that $P(x)$ holds as you noted.

The second part $\forall y \forall z \mathop. P(y) \land P(z) \to y = z$ means that there is at most one $x$ such that $P(x)$ holds. The second part by itself does not imply existence.

In prose, it reads, for all entities $y$ and $z$, if $P(y)$ holds and $P(z)$ holds then $y$ is equal to $z$. This fact does hold if there's one thing satisfying $P$ and does hold if there are zero things satisfying $P$, but fails if there are two or more.

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Here's a more complete explanation of where that formula comes from using counting quantifier notation.

$$ \exists_{=1} x \mathop. \varphi(x) \;\;\text{is equivalent to}\;\; (\exists_{\ge 1} x \mathop. \varphi(x)) \land (\exists_{\le 1} x \mathop. \varphi(x)) $$

$\exists! x \mathop. \varphi(x) $ and $\exists_{=1} x \mathop. \varphi(x)$ mean the same thing: there exists exactly one $x$ such that $\varphi(x)$ holds.

The expression $\exists_{\ge 1} x \mathop. \varphi(x)$ will be true precisely when some $x$ satisfying $\varphi(\cdots)$ exists, so that conjunct is just $\exists x \mathop. \varphi(x)$.

The expression $\exists_{\le 1} x \mathop. \varphi(x)$ is harder. We need to remember that quantifying over an empty set is vacuously true. Given that observation in mind, we can note that there's at most one $x$ satisfying $\varphi(\cdots)$ precisely when any two values satisfying $\varphi(\cdots)$ have to be equal.

Using a bounded quantifier in a totally ad hoc way, we can write this as:

$$ \exists_{\le 1} x \mathop. \varphi(x) \;\;\text{is equivalent to}\;\; \forall y \in \varphi \mathop. \forall z \in \varphi \mathop. y = z $$

And then we can paraphrase away the bounded quantifier, i.e. rewriting $\forall x \in \varphi \mathop. \psi(x)$ to $\forall x \mathop. \varphi(x) \to \psi(x)$.

This gives us:

$$ \exists_{\le 1} x \mathop. \varphi(x) \;\;\text{is equivalent to}\;\; \forall y \mathop. \varphi(y) \to (\forall z \mathop. \varphi(z) \to y = z) $$

Or, equivalently

$$ \exists_{\le 1} x \mathop. \varphi(x) \;\;\text{is equivalent to}\;\; \forall y \forall z \mathop. \varphi(y) \land \varphi(z) \to y = z $$

Answer 3

The statement reads: For any objects $y$ and $z$: if both $y$ and $z$ have property $P$, then $y$ and $z$ must in fact be the very same one object.

Would this statement be true in any world where there are two or more objects with property $P$? No, because then one can easily pick two different objects that both have property $P$. So, the statement can only be true if there is at most one objects with property $P$.

Together with the existential that states that there is at least one object with property $P$, you just obtain that there is exactly one object with property $P$