In article <e827583d-6246-4dd1-a860-bc7da80bfbcd@r3g2000yqd.googlegroups.com>, WM <mueckenh@rz.fh-augsburg.de> wrote:
> On 15 Feb., 23:27, Virgil <vir...@ligriv.com> wrote: > > In article > > <8847efa6-6663-40e9-a61e-76bba7f34...@dp10g2000vbb.googlegroups.com>, > > > > > > > > > > > > WM <mueck...@rz.fh-augsburg.de> wrote: > > > On 15 Feb., 00:53, Virgil <vir...@ligriv.com> wrote: > > > > > > > > And just this criterion is satisfied for the system > > > > > > > > 1 > > > > > > 12 > > > > > > 123 > > > > > > ... > > > > > > > > For every n all FISs of d are identical with all FISs of line n. > > > > > > For every n there is an (n+1)st fison of d not identical to any FIS of > > > > line n. > > > > > That does not prove d is not in the list, but only that d is not in > > > the first n lines of the list. > > > > For every n. > > For every n there are infinitely many lines following. > You never can conclude having all n. >
Then WM must be willing to give up all induction arguments and proofs by induction, as they are all have the same basis as Cantor diagonal arguments: if something is true for the first natural, and whenever true for a natural 'n' is also true for the natural 'n + 1', then it is true of ALL n.
Thus true for the first natural and if true for n then true for n+1 DOES allow one to conclude having all n.
At least outside of Wolkenmuekenheim.
Does WM really want to give up the proof by induction? --
