Axiom Regularity in $L$ Without $WF$

Question

I know, how I can prove that Axiom regularity holds in $L$ with $WF$, so axiom regularity holds in every transitive subset $WF$.

But how can I prove, that axiom regularity holds in $L$ without using $WF$?

Answer 1

First prove that the axiom of regularity is equivalent to the existence of a rank function. Namely, a function $r$ from sets to ordinals such that when $x\in y$ it follows that $r(x) < r(y)$.

Then use the definition of $L$ to come up with such a rank function.

Answer 2

It's a bit unclear what you're asking for; I think you're asking for a proof which doesn't use the assumption that $V$ is actually well-founded.

The proof amounts to the following:

If $M\models ZF$ and $X$ is a definable transitive subclass of $M$, then $(X, \in^M\upharpoonright X)$ satisfies regularity.

Let $x\in X$. By regularity in $M$, there is some $y\in x$ with $y\cap x=\emptyset$; by transitivity of $X$, $y\in X$. So $X\models \exists a(a\in x, a\cap x=\emptyset)$.