Re: Every set x equinumerous with a set y disjoint from x?

["MoeBlee"]

Rupert wrote:
> No, David's proof is correct and doesn't use choice.

Then I don't understand the proof, since I can't see how it goes
through without choice. Would you kindly convey it your own manner? I
am very much interested in seeing the theorem proven without choice,
regularity, or replacement.

Thanks,

MoeBlee