From axiom of regularity, how can I prove following statement? $$x \notin y \lor y \notin x$$
I’ve been reading about ZF theory. So far, I understood the axiom of regularity as follows:
$$\forall x (x \neq \phi \Rightarrow \exists z. z \in x \land z \cap x = \phi)$$
From this, (and with other axioms in ZF), I should be able to derive the first statement, but have not managed to do so.
Question:
- How can I derive the statement within ZF?
The axiom of regularity states that:
$$\forall x(x\neq \emptyset\rightarrow \exists y(y \in x \wedge y\cap x=\emptyset))$$
Now let $a, b$ be given. Consider $x=\{a, b\}$. By the axiom of regularity, there exists $y \in \{a, b\}$ such that $y\cap \{a, b\}=\emptyset$. Since $y \in \{a, b\}$, either $y=a$ or $y=b$. If $y=a$, then $y\cap \{a, b\}=\emptyset$, so $b \notin y=a$. If $y=b$, then $y\cap \{a, b\}=\emptyset$, so $a\notin y=b$.