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Working in language $\mathcal L_{\Omega^+,\Omega^+}$ where $\Omega$ is the first strongly inaccessible cardinal. If we add a primitive partial $\Omega$-ary function $F$ and a primitive constant $\varnothing$, then axiomatize:

Exclusion: $(\forall x_i)_{i < \Omega}: F((x_i)_{i < \Omega}) \neq \varnothing$

Commutative:$(\forall x_i)_{i < \Omega} (\forall y_i)_{i < \Omega} : \\F((x_i)_{i < \Omega}) =F((y_i)_{i < \Omega}) \iff \bigwedge_{i < \Omega} (\bigvee_{j< \Omega} (x_i = y_j)) \land \bigwedge_{j < \Omega} (\bigvee_{i < \Omega} (x_i = y_j))$

Existence:$ (\forall x_i)_{i < \Omega}: \\ \bigvee_{1+\kappa < \Omega}[(\exists y_j)_{j < 1+\kappa} (\bigwedge_{i < \Omega} (\bigvee_{j < 1+\kappa} (x_i=y_j)))] \iff \exists y: y=F((x_i)_{i < \Omega })$

Restriction: $\forall x: x=\varnothing \lor (\exists x_i)_{i < \Omega}: y=F((x_i)_{i < \Omega})$

If we define set membership $\in$ as:

Define: $a \in b \iff (\exists x_i)_{i < \Omega}: b=F((x_i) _{i < \Omega}) \land \bigvee_{i < \Omega} (a=x_i)$

Would that interpret $\sf ZFC$-$\sf Reg.$?

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    $\begingroup$ Apart from any claims made about $F$ or axioms, the logic $L_{\Omega^+,\Omega^+}$ just in the language of $=$ can interpret ZFC on any domain of size at least $\Omega$, since the infinitary language allow us to quantify over objects of size less than $\Omega$, and with this we can talk about well-founded extensional relations coding arbitrary elements of $V_{\Omega}$. For example, there is a tuple coding $V_\Omega$, which is a model of ZFC, and so we can interpret ZFC. Indeed, we can interpret KM this way as well. $\endgroup$
    – Joel David Hamkins
    Commented Nov 17 at 21:29
  • $\begingroup$ Out of curiosity, how do we define $L_{\Omega^+,\Omega^+}$ without starting in $ZFC$ plus the existence of inaccessibles to begin with? $\endgroup$
    – Alec Rhea
    Commented Nov 17 at 23:25
  • $\begingroup$ @AlecRhea Apologies for leaving a comment not related to the question, but a while ago I tried to contact you about one of your preprints. Is your gmail account still current? (If this kind of comment is discouraged by MO, I can delete it.) $\endgroup$
    – C7X
    Commented Nov 18 at 0:00
  • $\begingroup$ @C7X It is current, but unfortunately too active — due to me using it for everything I have 10k+ unread emails and frequently miss things. If you let me know what email address you sent it from I’ll sift through and respond in the next few days. $\endgroup$
    – Alec Rhea
    Commented Nov 18 at 7:17
  • $\begingroup$ @AlecRhea, I think ZFC is assumed at the metatheory. But, I don't know if this would seep through to the theory itself? $\endgroup$
    – Zuhair Al-Johar
    Commented Nov 18 at 11:52

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