```Date: Feb 16, 2013 4:37 PM
Author: Virgil
Subject: Re: Matheology � 222 Back to the roots

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+1DOES allow one to conclude having all n.At least outside of Wolkenmuekenheim.Does WM really want to give up the  proof by induction?--
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