# Axiom of Replacement (Question of notation)

For each formula $\phi$ without $Y$ free, the universal closure of the following is an axiom:

$\forall x\in A \exists !y \phi(x,y) \Longrightarrow \exists Y \forall x\in A \exists y\in Y \phi(x,y)$

My question is about the $!$, does that mean that $y$ is bound in $\phi$? What does the exclamation mark mean? Is it there so that $y\ne \{x: x\not\in x \}$? I am confused.

#### Solutions Collecting From Web of "Axiom of Replacement (Question of notation)"

Rustyn, $\exists !$ is the quantifier “there is a unique”. This is just an abbreviation, as it can be defined in terms of the standard quantifiers: $\exists! x\psi(x)$ is just $\exists x(\psi(x)\land\forall y(\psi(y)\to y=x))$.

Anyway, this version of replacement (where the graph of $\phi$ is a “function”) is equivalent to the more generous version where the $\exists!$ is replaced by the usual $\exists$ (where the graph of $\phi$ is just a “relation”).