**The exercise:** Use the axiom of regularity to prove that if $\displaystyle a$ is a set such that $\displaystyle a\subseteq a\times a$, then $\displaystyle a=\emptyset$.

**The axiom of regularity:** Every nonempty set $\displaystyle a$ contains an element $\displaystyle b$ such that $\displaystyle a\cap b=\emptyset$.

**Definition of cartesian product:** $\displaystyle a\times b=\{w\in\mathcal{P}(\mathcal{P}(a)):$ $\displaystyle (\exists x)(\exists y)(x\neq y\wedge x\in a\wedge y\in b\wedge (z)(z\in w\leftrightarrow z=\{x\}\vee z=\{x,y\}))$$\displaystyle \vee (\exists x)(x\in a \wedge x\in b \wedge (z)(z\in w\leftrightarrow z=\{x\}))\}$.

More simply, we say $\displaystyle a\times b=\{(x,y):x\in a\wedge y\in b\}$, where ordered pairs are defined as $\displaystyle (x,y)=\{\{x\},\{x,y\}\}$.

I believe this exercise is supposed to be easy, but I'm having a lot of trouble with it. Assistance would be much appreciated!