Working in $\sf ZF-Reg.$ can we define the unary predicate "is an ordinal", denoted by "$\operatorname {od}$", meaning is a von Neumann ordinal, in a recursive manner?
The usual definition of ordinal is as an $\in$-well ordered transitive set, or equivalently as a transitive set of transitive sets that is $\in$-well founded.
Here I'm contemplating the following definition by recursion:
Define: $ \operatorname {od} (x) \iff \forall y \, (y \in x \leftrightarrow \operatorname {od}(y) \land y \subsetneq x )$
In English: an ordinal is a set of all proper subsets of it that are ordinals.
My rationale is that in $\sf ZF$, where we have foundation, then clearly the above definition entails that every ordinal is a transitive set of ordinals, and so a transitive set of transitive sets which is well founded, therefore clearly a von Neumann ordinal. Now suppose that Regularilty fails, then for an argument by negation, suppose that we have an ordinal $\zeta$ that is not well founded, then the set $\operatorname {Ord}^\zeta$ of all von Neumann ordinals in $\zeta$ would be well founded and so a von Neumann ordinal, and therefor we'll have $\operatorname{Ord}^\zeta \in \zeta$, and so $\operatorname{Ord}^\zeta \in \operatorname{Ord}^\zeta$, which violates the well foundedness of $\operatorname{Ord}^\zeta$.