Date: Mar 26, 2013 5:07 PM
Author: Virgil
Subject: Re: Matheology � 224

In article 
<0a02c905-583f-4aa8-a22d-c5802e718c77@r1g2000yql.googlegroups.com>,
WM <mueckenh@rz.fh-augsburg.de> wrote:

> On 26 Mrz., 21:17, Virgil <vir...@ligriv.com> wrote:
> > In article
> > <2dc8b38d-3376-4a6c-89e4-ad4b059d8...@r1g2000yql.googlegroups.com>,
> >
> >  WM <mueck...@rz.fh-augsburg.de> wrote:

> > > On 25 Mrz., 23:12, Virgil <vir...@ligriv.com> wrote:
> >
> > > > Lets see WM's statement of the inductive principle.
> >
> > > Let P(1)
> > > and let P(x) ==> P(x+1)

> >
> > > Then P(n) at least for every natural number.
> >
> > > Proof: For P(2) follows from P(1), P(3) follows from P(2), and so on.
> >
> > > More is not required.
> >
> > If proof is not required, or even possible, in any system in which
> > induction, or some equivalent, is not assumed.
> >
> > One acceptable form of induction is:
> >
> > There exists a set of objects, N,  and a zero object, 0,  such that
> >    1. 0 is a member of  N.
> >    2. Every member of N has a successor object in N.
> >    3. 0 is not the successor object of any object in N.
> >    4. If the successors of two objects in N are the same,
> >       then the two original objects are the same.
> >    5. If a set, S, contains 0 and the successor object of every
> >       object in S, then S contains N as a subset.

>
> That is a definition of a sequence, not a proof by induction. It is
> not even a definition of the natural numbers, because even the ordered
> set
> N = (0, pi, pi^2, pi^3, ...)
> obeys your five points.


Mathematical Induction does not require use of natural numbers, but only
of a set which is as inductive as (is order-isomorphic to) the set of
natural numbers, and my form satisfies that requirement.

When WM gets his dander up, he objects to everything, even the truth.

And WM seems unable to keep his dander down when reading my posts.
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