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Working in "MK-Regularity-Limitation of Size + Replacement for sets", call it the Base theory, let's coin the following axiom:

Axiom of Super-Choice:$$\forall \ relation \ R \ \exists F \subset R \ (F: dom(R) \to rng(R))$$ Where: $$R \text { is a relation } \equiv_{df} \forall r \in R \ \exists x,y \ (r=\langle x,y \rangle)\\dom(R)=\{x| \exists y (\langle x,y \rangle \in R)\} \\rng(R)=\{y|\exists x (\langle x,y \rangle \in R)\}$$

I think this would be equivalent to Global Choice over the Base theory.

Now if we weaken the above to:

Axiom of Proper Class Choice:$$\forall \ relation \ R \ \text{ with proper class rows } \\\exists F \subset R \ (F: dom(R) \to rng(R))$$

where: $$ z \text{ is a row of }R \iff \\ \exists x \in dom(R)[z=\{y| \langle x,y \rangle \in R\}]$$

A relation $R$ is with proper class rows, if every row of it is a proper class.

Would that still be equivalent to Global Choice over the Base theory?

If not, then if we also add Choice over sets [i.e., every set has a choice function], would that result in proving Global Choice?

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  • $\begingroup$ You might find this useful: jdh.hamkins.org/…. The theorem there gives several equivalent formulations of global choice over Gödel-Bernays set theory, which would be a natural base theory. $\endgroup$ – Joel David Hamkins Feb 1 at 17:18
  • $\begingroup$ @JoelDavidHamkins, Thanks for the reference! $\endgroup$ – Zuhair Al-Johar Feb 1 at 17:22
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Yes, this is still equivalent: given any relation $R$, consider the new relation $$R^{bigrows}=\{\langle x,y\rangle: \exists a,b(y=\langle a,b\rangle\wedge \langle x,a\rangle\in R)\}.$$ Basically, $R^{bigrows}$ just "pads out" the rows of $R$ with a dummy coordinate.

But from a choice function $F$ for $R^{bigrows}$ we can extract one for $R$ itself: consider $$x\mapsto \pi_0(F(x))$$ (where $\pi_0$ denotes projection onto the left coordinate).

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  • $\begingroup$ Basically, restricting attention to proper classes is never going to make things substantively easier - the idea here is the same, although the implementation is different, as in my answer to your earlier question. $\endgroup$ – Noah Schweber Feb 1 at 17:28

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