Prove that for any two sets $A$ and $B$, $A\notin B$ or $B\notin A$
Solution 1:
If $A\in B$ and $B\in A$, then $X=\{A,B\}$ is a set which is:
- Non-empty,
- each of the members of $X$ has a non-empty intersection with $X$. Namely, $A\in B\cap X$ and $B\in A\cap X$.
This is a contradiction to the axiom of regularity, therefore either $X$ is not a set, or $A\in B\land B\in A$ is false. But since pairing ensures that $X$ is a set, we can only conclude that $A\notin B$ or $B\notin A$.