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Let $ * $ range over cardinal relations $ \{<,<>\}$; if we add the following axiom to $\sf ZF$, would that prove a known form of choice?

Parallelism: $ |x| * |y| \leftrightarrow |\mathcal P(x)| * |\mathcal P(y)| $

Where, $<>$ stands for incomparability; that is, absence of an injection in both directions.

Zuhair Al-Johar
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  • I don't understand why you have so many relations with the bidirectional implication and universal quantifier. $=$ and $<$ are enough. – Asaf Karagila Feb 01 '23 at 09:39
  • @AsafKaragila, OK, I've edited it. I kept the bidirectional because it's more in spirit with this assertion. – Zuhair Al-Johar Feb 01 '23 at 10:04
  • I didn't say to remove the bidirectional implication, I just said that with it are $=$ and $\neq$ are equivalent (and with the universal quantifier, $<$ and $>$ are equivalent). – Asaf Karagila Feb 01 '23 at 10:08

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