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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