# Has by well-foundedness every non-empty class an $R$-minimal element? Also if axiom REG is not assumed?

Working in $\mathbf{ZF}$ let $R$ be a proper class of ordered pairs that is well-founded. This means that for every non-empty set $a$ there is a set $b\in a$ such that $cRb\implies c\notin a$. Here $cRb$ is a notation for $\langle c,b\rangle\in R$ and $b$ is a so-called $R$-minimal element of $a$. If $R$ is local (i.e. collections $\{x\mid xRb\}$ are all sets) then it can be shown that also non-empty proper classes have $R$-minimal elements.

I encountered a proof that made the condition of being local redundant. It made use of an operation on classes that adds to each class a set that is contained in it (Bottom-operation) but the definition of this operation relied on the regularity axiom.

My question:

Is there a proof that every non-empty class has an $R$-minimal element that does not make use of the regularity axiom?

Another formulation:

If $R$ is a class of ordered pairs that is well-founded, then is it legal to apply $R$-induction if all axioms of $\mathbf{ZF}$ are accepted with exception of the axiom of regularity?

• So the question is really in $\sf ZF-Reg$, without assuming the locality. – Asaf Karagila Jun 20 '15 at 18:14
• @AsafKaragila Yes. If there is locality then it can be proved without REG. That proof makes use of $\overline R$ (the transitive closure wich is also local in that case). – drhab Jun 20 '15 at 18:23
• Now reposted at mathoverflow.net/q/214775 – Emil Jeřábek Aug 20 '15 at 16:20