In article <firstname.lastname@example.org>, email@example.com wrote:
> On Wednesday, 26 June 2013 17:48:25 UTC+2, dull...@sprynet.com wrote: > > > >For all n exists k: >d_1, d_2, d_3, ..., d_n = q_k1, q_k2, q_k3, ..., > > >q_kn. >This theorem it is not less important than Cantor's theorem: For > > >all >k: d =/= q_k. > >Both theorems contradict each other. > > > Erm, no they don't. If your "for every n there exists k such that" was > > instead "there exists k such that for every n" then they would contradict > > each other. > > Try to find a d_n that, together with all its predecessors d_j for 0 < j < n, > is not in a q_k. Then you may boast "erm". Otherwise learn that, in > mathematics, there is no continuation of d beyond every d_n.
But there is a continuation beyond ANY d_n, which is all that is needed to prove WM totally wrong again.
And so WM's argument above is broken by revealing another example of his quantifier dyslexia, --