Date: Dec 28, 2012 10:41 PM
Author: Graham Cooper
Subject: Re: The Diagonal Argument

On Dec 29, 11:26 am, Virgil <vir...@ligriv.com> wrote:
> In article
>  Graham Cooper <grahamcoop...@gmail.com> wrote:
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> > On Dec 28, 7:18 pm, Virgil <vir...@ligriv.com> wrote:
> > > In article
> > >  Graham Cooper <grahamcoop...@gmail.com> wrote:

>
> > > > Then you agree
>
> > > > 0. T(1,2) T(2,1) T(3,3) T(4,4) T(5,5) T(6,6) ...
>
> > > > is potentially ON THE LIST L!
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> > > But, if T(n,n) is as I previously defined it,
> > > 0. T(1,1) T(2,2) T(3,3) T(4,4) T(5,5) T(6,6) ...
> > > is not even potentially on it.

>
> > HERE IS L
>
> > L(x,y)
> >  +---------------->
> >  | 0. 2 3 4 5 6 7 ..
> >  | 0. 9 8 7 6 5 5 ..
> >  | 0. 1 2 3 1 2 3 ..
> >  | 0. 9 8 9 8 9 8 ..
> >  | 0. 6 5 6 5 6 5 ..
> >  | 0. 5 6 5 6 5 6 ..
> >  |
> >  v

>
> > HERE IS T
>
> > T(x,y)
> >  +---------------->
> >  | 0. 6 6 6 6 5 5 ..
> >  | 0. 5 5 5 5 6 6 ..
> >  | 0. 6 6 6 6 6 6 ..
> >  | 0. 5 5 5 5 5 5 ..
> >  | 0. 5 6 5 6 5 6 ..
> >  | 0. 6 5 6 5 6 5 ..
> >  |
> >  v

>
>  That is not my "T".
>
> My "T(x,y)" gives a single row:
>
>     0. 6 5 6 5 5 5 ... ..
>

That's the D I A G O N A L of T !