TITLE: Does Con(ZF + Reinhardt) really imply Con(ZFC + I0)?
QUESTION [25 upvotes]: The question is: if I assert in ZF that there exists a Reinhardt cardinal, do I really get a theory of higher consistency strength than when I assert in ZFC that there exists an I0 cardinal (the strongest large cardinal not known to be inconsistent with choice, as I understand)? This is implicit in the ordering of things on Cantor's Attic, for example, but I've been unable to find a proof (granted, I don't necessarily have the best nose for where to look!).
One thing that worries me is that when there is a ZFC analog of a ZF statement, many equivalent formulations of the ZFC statement may become inequivalent in ZFC. So we don't have much assurance that the usual definition of a Reinhardt cardinal is "correct" in the absence of choice.
I think it should be clear that Con(ZF + Reinhardt) implies Con(ZF + I0). But again, it's not clear that ZF+I0 is equiconsistent with ZFC+I0.
It's apparently not possible to formulate Reinhardt cardinals in a first-order way, so I should really talk about NBG + Reinhardt, or maybe ZF($j$) + Reinhardt, where ZF($j$) has separation and replacement for formulas involving the function symbol $j$.
EDIT
Since this question has attracted a bounty from Joseph Van Name, maybe it's appropriate to update it a bit. Now, I'm not actually a set theorist, but it's not even clear to me that Con(ZF + Reinhardt) implies Con(ZFC + an inaccessible). So perhaps the question should really be: what large cardinal strength, if any, can we extract from the theory ZF + Reinhardt?

REPLY [2 votes]: Mohammed Golshani's link doesn't work, so I have reconstructed a sketch of a proof. The key fact is this (You can find a proof in most Set Theory textbooks):
Theorem: If $DC_\omega$ holds and $D$ is an $\omega_1$-complete ultrafilter, then the ultraproduct of $N$ by $D$ is well-founded, for any inner model $N$.
Theorem: If $DC_\lambda$ holds and $\kappa$ is $I0$ (With target $\lambda$) if and only if there is a non-principal $\kappa$-complete $L(V_{\lambda+1})$-ultrafilter on $V_{\lambda+1}$.
Proof. Let $D$ be such an ultrafilter. To verify the existence of a such an ultraproduct, we can code elements of $D$ as function $F: X^{\lt\lambda}\rightarrow X'$ for every set $X'=\{\chi_x|x\in V_{\lambda+1}\}$, there is a function $f: \lambda\rightarrow X$, such that $f(\chi_x)\in\chi_x$, and so some $A\in D$, such that $\{\chi_x|x\in A\}$ admits a Choice function. Then if $M\cong Ult_D(L(V_{\lambda+1}))$ is the ultrapower $M\ni V_{\lambda+1}$. $L(V_{\lambda+1})$ inherits a well-order for each element (A well order not necessarily in $L(V_{\lambda+1})$ from $L(V_{\lambda+1})$. Then by condensation $M= L(V_{\lambda+1})$ and $\kappa$ is the critical point of the ultrapower embedding $k_D: L(V_{\lambda+1})\prec L(V_{\lambda+1})$. For the other direction, define an ultrafilter $D=\{X\subseteq V_{\lambda+1}|j\restriction V_\lambda\in j(X)\}$. This satisfies all the necessary properties.◼
Theorem: If $\kappa$ is Reinhardt, indeed even the critical point of $j: V_{\lambda+2}\prec V_{\lambda+2}$, then $\kappa$ is $I0$.
Proof. Let $\kappa$ Reinhardt as witnessed by $j: V\prec V$, and let $\lambda$ be the least fixed point above $\kappa$. Let $D=\{X\subseteq V_{\lambda+1}|j\restriction V_\lambda\in j(X)\}\cap L(V_{\lambda+1})$. Then $D$ satisfies all the necessary properties. The second case is not much trickier, as it uses the same argument.◼