On 8 Feb., 23:52, Virgil <vir...@ligriv.com> wrote: > In article > <64c6e6d9-d039-48bf-9cd6-7c614cee3...@j4g2000vby.googlegroups.com>, > > > > > > WM <mueck...@rz.fh-augsburg.de> wrote: > > On 8 Feb., 23:26, William Hughes <wpihug...@gmail.com> wrote: > > > More WM logic > > > > L is a potentially infinite > > > list and d is the potentially infinite > > > anti-diagonal > > > > From > > > i. For every natural number n, d > > > is not the nth line of L > > > correct. > > > > ii. i. implies that there is no > > > natural number m such that > > > d is the mth line of L > > > No such m can be fixed. > > It is "fixed" in the sense of not existing at all! > > > > > > iii. d may or may not be a line of L > > > There is no part of d(potential) that is surpassing every line of a > > suitable list. > > If every member of the list has a last digit but d does not,
That is one side of the medal, but it is not the only side.
It is exactly as if you would prove that the even numnbers are larger than the odd numbers, by showing that for every off number there is a larger even number. Of course the latter is right, but it does not prove the claim.
> then for > every member of the list there will be a first FIS of d surpassing it,
and for every FIS of d there will be a first line of the remaining list surpassing it.
> and following it, a lot more of them following that first one..
and following this first line there a lot more with the same surplus. > > At least outside the idiotic constraints of WMytheology.
There are no constraints. Is every FIS of d surpassed by a line of the list or is there a first FIS that is not surpassed? In mathematics the defender of such a position should be able to either prove it or to show an example.
You have already agreed hat d is not actually infinite, because you are going to show uncountability without actual infinity.
