[Haskell-cafe] Re: Theory about uncurried functions

[Hans Aberg]

On 5 Mar 2009, at 17:06, Daniel Fischer wrote:

> Cantor-Bernstein doesn't require choice (may be different for  
> intuitionists).
> http://en.wikipedia.org/wiki/Cantor-Bernstein_theorem

Yes, that is right, Mendelson says that. - I find it hard to figure  
out when it is used, as it is so intuitive.

Mendelson says AC is in fact equivalent proving
   all x, y: card x <= card y or card y <= card x

   Hans Aberg