or (contrariwise) that $NGB$ + "There exists a Reinhardt cardinal" is consistent?

The question is partially in the title. $NGB$ is used for the reasons stated in the Hamkins, Kirmayer, and Perlmutter paper, "Generalizations of the Kunen inconsistency", *Annals of Pure and Applied Logic*, 163 (2012) pp. 1873-1876 (Section 1, "A few metamathematical preliminaries").

In order that this question should not be deemed overbroad, answers must be limited in scope to specific mathematical difficulties (eg. the Erdos-Hajnal theorem used in in proving the Kunen inconsistency in $NGBC$ or $NBC$ + $Choice$ has no known proof in $NGB$ or is known not to be provable without choice--however, those who would say that the Erdos-Hajnal theorem cannot be proven without choice must show how this fact allows for a non-trivial elementary embedding $j$: $V$ $\rightarrow$ $V$ and the existence of the requisite critical point $\kappa$ in $NGB$). Those who are so inclined might make mention of Rupert McCallum's recent attempt in showing that "the existence of a non-trivial elementary embedding $j$: $V_{\lambda + 2}$ $\rightarrow$ $V_{\lambda + 2}$ for some ordinal $\lambda$...is not consistent with $ZF$" (quote from McCallum's withdrawn short preprint, *The Choiceless Cardinals are Inconsistent*, arXiv:1712.09768, which is discussed on Joel David Hamlin's blog — I would be especially interested in getting a nice explanation (i.e. analysis) of the gap in this argument and the relavance of the Laver paper which Gabriel Goldberg directed him to).

Though at first glance the answers to these questions might seem "primarily opinion-based", the mathoverflow community should realize the value of informed opinion and (yes) informed speculation regarding this topic, and use this question as an opportunity to provide these 'clues' (think of the proof of the 4-color theorem) to both the mathoverflow community and the mathematical community as a whole.

necessarystumbling block. The best that can be done is to point out the key step which "no known argument can get around," and observe that this step doesn't go through without choice. So I don't understand what you're looking for in an answer, short of a new major result in set theory, if that doesn't satisfy you. $\endgroup$ – Noah Schweber Oct 13 '18 at 23:26definitiveanswer, one might have relevant facts that could lead to a definitive answer. An easy example, there might be, say, in some obscure journal, a proof of the relevant Erdos-Hajnal theorem that doesn't use choice, or a choice principle weaker than $AC$ (say, the Axiom of Dependent Choices). The relevant Erdos-Hajnal theorem has been around a long time so such a theorem might be out there. A proof that the relevant Erdos-Hajnal theorem cannot do without $AC$ may suggest that one might be able to construct a model of $NGB$ that has a $\endgroup$ – Thomas Benjamin Oct 15 '18 at 14:06