Date: Feb 26, 2013 12:11 PM
Subject: Re: Matheology § 222 Back to the roots

On 26 Feb., 13:11, William Hughes <> wrote:
> On Feb 26, 12:47 pm, WM <> wrote:
> We both agree
>  There does not exist an m
>  such that the mth line
>  of L is coFIS with the diagonal
>  (here we interpret "There does
>  not exist" to mean "we cannot find").
> So we agree any such m must be an
> unfindable natural number.

It is a variable that can take any natural number.

> I am not interested in arguments
> about whether an unfindable number exists.
> [I still do not understand why
> WM rejects the obvious proof by contradiction

I do not understand why WH rejects the obvious proof by contradiction
that all natural numbers of the list are in one line together.
> Suppose that P is a predicate such that
> for every natural number m, P(m) is true.
>      Assume a natural number, x, such that P(x)
>      is false exists.

For every natural x, the proposition "P(x) = there are more than one
line necessary to contain all natural numbers from 1 to x of the list"
is false.

>      call it k
>      Then P(k) is both true and false.
>      Contradiction,  Thus the original assumption
>      is false and no natural number, x, such
>      that P(x) is false exists)

Why then do you maintain that proposition?

Or do you withdraw it?
> ]
> <snip>

> > Every natural number is findable.
> Which, according to WM does
> not mean that you can prove every
> natural number is findable.

Every natural number is findable. "The number of the last line of the
list" or "the last natural number" are simply variables that can take
natural numbers as values. But these values cannot be fixed, known,

Regards, WM