On Tuesday, 25 June 2013 23:09:28 UTC+2, Virgil wrote: >> For all n exists k: > d_1, d_2, d_3, ..., d_n = q_k1, q_k2, q_k3, ..., q_kn.
> So what? That does not prove that there exists any k for all n.
What is "all n"? You want to play the trick, "who first moves has lost". If I first give k, then you find a better n. No, I find a k for all n that you can specify. That is enough.
> Quantifier dyslexia cripples WM's arguments again! WM's claims above do NOT establish existence of any k for which d_n = q_k_n for ALL n in |N, which is what WM would need to prove his false claim.
If your |N contains more n than anybody can name, then you are right. But then |N belongs to matheology, not to mathematics. If your claim is correct for nameable n, then name an n for which my theorem fails. You will fail.