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Let $Frege(X)$ be the set of all equivalence classes of subsets of $X$ under equivalence relation bijection. formally:

$Frege(X)= \{\{Y \subseteq X: |Y|=|Z|\}: Z \subseteq X \}$

Where $||$ is cardinality function defined after Scott's.

My question is about the following statement:

$\forall X: |Frege(X)| \leq |X|+1$

Is it equivalent to the axiom of choice over the rest of axioms of $\sf ZF$?

More precisely can $\sf ZF$ plus the above statement prove the axiom of choice?

Afternote: To spell the above without invoking Scott's cardinality just replace $|Y|=|Z|$ by $\exists f(f: Y \to Z \land f \text{ is a bijection})$, and replace $|Frege(X)| \leq |X| + 1$ by $\exists g (g: Frege(X) \to X \cup \{X\} \land g \text{ is an injection})$

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    $\begingroup$ I think invoking Scott cardinality just makes this messier: you're just asking about the principle "For all $X$ there is an injection from $\{$subsets of $X$ modulo bijectability$\}$ to $X\sqcup\{*\}$," right? $\endgroup$
    – Noah Schweber
    Jun 28, 2020 at 23:48
  • $\begingroup$ @NoahSchweber, yes definitely you can speak about them in terms in injections, etc... but it burns down to the same thing. $\endgroup$
    – Zuhair
    Jun 29, 2020 at 0:41
  • $\begingroup$ Yes, I only mentioned that since some readers might not be familiar with the notion of Scott cardinality. $\endgroup$
    – Noah Schweber
    Jun 29, 2020 at 1:18
  • $\begingroup$ OK, I've edited the post. Thanks! $\endgroup$
    – Zuhair
    Jun 29, 2020 at 9:40

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