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+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?
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