In article <email@example.com>, WM <firstname.lastname@example.org> wrote:
> On 11 Apr., 09:47, fom <fomJ...@nyms.net> wrote: > > On 4/11/2013 1:49 AM, WM wrote: > > > > > On 10 Apr., 23:04, Virgil <vir...@ligriv.com> wrote: > > > > >>> For all n is not for all n? d has more digits than all? > > >>> My proof is valid for all n. > > > > >> Your argument only holds for finite sequences, > > > > > for *all* finite sequences. > > > > That is exactly what was said. > > > Not in matheology. Seems that Virgil objected to "all n" by only > "finite sequences". > > > >> but any anti-diagonal, by > > >> not being a finite sequence, is exempt. > > > > > So for all n: d_1, ..., d_n covers less digits than for all n: d_n?
Consider the statements A meaning "For every n, d_n is the first n digits of d" and B meaning "For every n, d_1, d_2, ...,d_n are the first n digits of d"
Then B does not imply A since B can be ttrue when A is false. --