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A. S. Daghighi, M. Golshani, J. D. Hamkins, and E. Jeřábek proved in "The foundation axiom and elementary self-embeddings of the universe" that, working in ZFGC$^{\text{−f}}$+BAFA, there are nontrivial automorphisms and elementary embeddings of the universe $V$ into itself.

Accordingly, Kunen inconsistency is circumscribed for this class of ill-founded theories.

Does it follow that $\text{ZFGC}^{\text{−f}}+\text{BAFA}+\exists\kappa(κ \text{ is Reinhardt})$ is a non inconsistent extension of $\text{ZFGC}^{\text{−f}}+\text{BAFA}$?

If so, is it known which of the large cardinal properties would $\text{ZFGC}^{\text{−f}}+\text{BAFA}+\exists\kappa(κ \text{ is Reinhardt})$ imply?

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  • $\begingroup$ Let me stress that as noted at the end of Section 4, what we write about BAFA is mostly a survey of previously known results. I don’t recall whether it is already given in Boffa’s paper, but the existence of nontrivial elementary embeddings is definitely proved and used in an essential way in the Ballard–Hrbáček paper. $\endgroup$ – Emil Jeřábek Mar 20 '15 at 16:10
  • $\begingroup$ Dear Dr Jeřábek, thanks for the note. I oversimplified because I wasn't sure of which results you've presented in your paper was previously known. $\endgroup$ – Andrea Tito Nespola Mar 20 '15 at 18:11
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I'm glad to hear you're reading our paper, which can be found here: The foundation axiom and elementary self-embeddings of the universe. Click through to the arxiv for a pdf — and I note that the title you mention is from an earlier draft of this article, so you may want to look at the updated version of the article.

In theorem 1 of the article, we prove in $\text{ZFC}^{-f}$ that any $\Sigma_1$-elementary embedding $j:V\to V$ must fix every well-founded set, and in particular, every ordinal. This is the residue of the Kunen inconsistency in this foundation-ness context. It follows, however, that although as you mention BAFA proves the existence of nontrivial elementary embeddings $j:V\to V$, we do not get nontriviality on the ordinals, and so there are no Reinhardt cardinals here to be found. The embeddings provided by BAFA have no critical points.

Further, since the consistency strength of BAFA is no greater than that of ZFC, it follows also that consistency-wise, one cannot provably get any large cardinals from such an embbedding in $\text{ZFC}^{-f}$.

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  • $\begingroup$ Professor Hamkins, thanks for the answer: I'm truly honored. The last paragraph of your answer stresses that $\exists\kappa (\kappa \text{ is Reinhardt})$ cannot be a theorem of a $\text{ZFC}^{-\text{f}}+\text{BAFA}$ theory. Now, $\exists\kappa (\kappa \text{ is Reinhardt})$ cannot even (consistently) be added to $\text{ZFC}^{-\text{f}}+\text{BAFA}$ as an axiom because the existence of a Reinhardt cardinal implies the existence of nontrivial elementary embeddings of the $V$ into itself, whilst it is a consequence of $\text{ZFC}^{-\text{f}}$ that any such embedding is trivial. $\endgroup$ – Andrea Tito Nespola Mar 20 '15 at 17:22
  • $\begingroup$ Edit of the previous comment: "...whilst it is a consequence of $\text{ZFC}^{−\text{f}}$ that any such embedding is trivial on the well-founded part of the universe". In my previous comment I was checking whether I'm getting it right. $\endgroup$ – Andrea Tito Nespola Mar 20 '15 at 17:31
  • $\begingroup$ @JDH: Is there any further weakening of $ZFC^{-f}$ that will keep will keep some $\Sigma_1$-elementary embedding j: $V$$\rightarrow$$V$ from fixing every well-founded set, and in particular, every ordinal so that one can get nontriviality on the ordinals? $\endgroup$ – Thomas Benjamin Mar 24 '15 at 23:59
  • $\begingroup$ If you drop AC, then it is not known whether there can be such an embedding nontrivial on the well-founded sets. This is the situation of Reinhardt cardinals in ZF without AC. $\endgroup$ – Joel David Hamkins Mar 25 '15 at 0:10
  • $\begingroup$ @JDH: Are there any papers that have results regarding nontrivial elementary embeddings j:$V$$\rightarrow$$V$ in $ZF^{-f}$+$BAFA$? Is it possible to prove an analogue to your Thm.1 in this system? Is the consistency strength of $BAFA$ no greater than $ZF$ as well? $\endgroup$ – Thomas Benjamin Mar 25 '15 at 1:30

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