Which axiom of set theory does this formula represent ? Why? Which axiom of set theory does the statement below represent? Why?
\begin{align}\exists x\bigg(&\forall y\Big(\neg\exists z\left(z\in y\right)\to y\in x\Big)\\&\land\forall w\Big(w\in x\to\forall u\big(\forall v\big(v\in u\leftrightarrow\left(v=w\lor v\in w\right)\big)\to u\in x\big)\Big)\bigg)\end{align}
 A: This statement is equivalent to the following
$$\exists x : ((\emptyset \in x) \wedge (\forall w \in x, \ \{w\}\cup w \in x))$$
i.e. the Axiom of Infinity.
A: In words, the statement is saying: there exists $x$ such that $\emptyset \in x$ and for all $w$, if $w \in x$, then the successor of $w$ is an element of $x$.
You should be able to take it from here.
A: Break it up into smaller pieces:


*

*$\lnot \exists z (z \in y)$ says, "$y$ is the empty set".

*So $\forall y (\lnot \exists z (z \in y) \to y \in x)$ says, "if $y$ is the empty set, then $y$ is a member of $x$".

*$\forall v (v \in u \leftrightarrow (v = w \lor v \in w))$ says, "the members of $u$ are $w$ and the members of $w$", or more simply, $u = w \cup \{ w \}$.

*So $\forall u (\forall v (v \in u \leftrightarrow (v = w \lor v \in w)) \to u \in x)$ says, "if $u = w \cup \{ w \}$, then $u$ is a member of $x$".

*So $\forall w (w \in x \to \forall u (\forall v (v \in u \leftrightarrow (v = w \lor v \in w)) \to u \in x))$ says, "if $w$ is a member of $x$ and $u = w \cup \{ w \}$, then $u$ is also a member of $x$".


Thus, the whole statement translates to:

There is a set $x$ that satisfies the following conditions:
  
  
*
  
*If $y$ is the empty set, then $y$ is a member of $x$.
  
*If $w$ is a member of $x$ and $u = w \cup \{ w \}$, then $u$ is also a member of $x$.
  

You should now recognise this as the axiom of infinity.
