Date: Mar 10, 2013 3:49 PM
Author: Virgil
Subject: Re: Matheology � 222 Back to the roots
In article

<d4059f25-4b4f-48c3-b7ad-38b75b46cc52@h9g2000vbk.googlegroups.com>,

WM <mueckenh@rz.fh-augsburg.de> wrote:

> On 10 Mrz., 01:42, Virgil <vir...@ligriv.com> wrote:

> > In article

> > <9af56365-2be8-419b-a9d0-747008f6c...@fn10g2000vbb.googlegroups.com>,

> >

> >

> >

> >

> >

> > WM <mueck...@rz.fh-augsburg.de> wrote:

> > > On 9 Mrz., 21:07, Virgil <vir...@ligriv.com> wrote:

> >

> > > > > Of course the lines

> > > > > from 1 to n are completely irrelevant.

> >

> > > > There is no way to have a line n+1 without having lines 1 to n preceding

> > > > it, so that no line is irrelevant in determining the position of the

> > > > lines following it.

> >

> > > Wrong. Here is a set that starts at line 4

> >

> > > 1,2,3,4

> > > 1,2,3,4,5

> > > 1,2,3,4,5,6

> > > ...

> > > 1,2,3,4,5,6,7,8,9,10,11,12

> >

> > In your example above, the 1st line would be 1,2,3,4.

> >

> > One does not count as if they were there lines which are not there.

>

> If you mean yourself by "one" then you may be right.

> But that is of little importance for myself and the general opinion

> with respect to the list under discussion for about 500 postings,

> namely

> 1

> 1,2

> 1,2,3

> ...

>

> Regards, WM

Actually, in those 500 postings, you are the only one who agrees with

your false picture of things, and the GENERAL opinion opposes it almost

universally.

And where is your alleged proof that some mapping from the set of all

binary sequences to the set of all paths of a CIBT is a linear mapping?

You have several times claimed it but cannot seem to prove it.

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