It is well known that the von Neumann universes $V_{\alpha}$ is a model of ZF(C) when $\alpha$ is an inaccessible cardinal. In the following let $V$ be such a model of ZF(C). It is also well known (see corollary 5.3 of Set Theory, The Third Edition, by Thomas Jech) that assuming axiom of choice every successor aleph cardinal, $\aleph_{\beta+1}$, is a regular cardinal for any ordinal number $\beta$. This means that in ZFC (together with the necessary assumption on existence of the inaccessible cardinal necessary to construct $V$) we have:

For any cardinal number $\alpha \in V$ in the universe there is a

regularcardinal $\beta \in V$ in the same universe, such that $\alpha < \beta$.

Does this statement hold in ZF? In other words, is it provable without the axiom of choice that for any given cardinal number in a universe there a strictly larger and regular cardinal *in that universe*?

**Edit:**
The answer to the question above is no as pointed out in the comments by Mohammad Golshani. The question then is what other axiom, weaker than the axiom of choice allows us to prove the statement above?

WISC is independent from ZF(2012) staff.fnwi.uva.nl/b.vandenberg3/papers/WISC.pdf, reworked as Theorem 5.1 in arXiv:1207.0959, where WISC is renamed AMC, unwisely IMHO, due to a name clash). +1 modulo the obvious correction, which I suggest you work into the main question rather than leave as a comment. $\endgroup$ – David Roberts Aug 22 '17 at 7:51