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Definable Partition Principle: If $\phi;\psi$ are formulas in which only the symbol $x$ occur free, then: $$A = \{x \mid \phi\} \land B=\{x \mid \psi\} \land B \, ||| \, A \to B \leq A$$ where $|||$ means “is a partition of“; $B \leq A$ means there exists an injection from $B$ to $A$

Is this provable in ZF? If not, being a weaker principle than the full Partition Principle, would it be provable for it to be not equivalent to AC over ZF?

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    $\begingroup$ It is definitely not provable in ZF, since it is consistent $\mathbb R/\mathbb Q$ doesn't embed into $\mathbb R$. On the other hand, it follows from full partition principle, which is not known to imply AC, so I suspect the same holds for your principle. $\endgroup$
    – Wojowu
    Commented Jan 12, 2022 at 10:54
  • $\begingroup$ We don't know if the full PP implies choice. How would a weaker condition do it? $\endgroup$
    – Asaf Karagila ♦
    Commented Jan 12, 2022 at 10:56
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    $\begingroup$ @AsafKaragila, is it known to be NOT equivalent to the axiom of choice? I've edited the question. $\endgroup$
    – Zuhair Al-Johar
    Commented Jan 12, 2022 at 11:19
  • $\begingroup$ If it's not known to imply AC, how would it be known to not imply it? $\endgroup$
    – Asaf Karagila ♦
    Commented Jan 12, 2022 at 13:01
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    $\begingroup$ @AsafKaragila, the full PP is the one not known to imply AC! This principle is a much weaker form, so it might be the case that this weaker version can be easily provable to be not equivalent to AC $\endgroup$
    – Zuhair Al-Johar
    Commented Jan 12, 2022 at 18:32

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