Given any ennumeration of every rational, and there are lots of them, show that it does not ennumerate all of them by finding one that has been left out of the ennumeration.
Absent such a demonstration, WM is just blowing smoke again.
> *Every* rational can be enumerated. But for *every* rational number we > have: There are infinitely many not enumerated numbers remaining. > Briefly: > > forall n : the quantification "forall n" is wrong.
Only in Wolkenmuekenheim can it be wrong, and then only because "A and not A" too often holds in Wolkenmuekenheim. --