Date: Apr 11, 2013 10:42 AM
Author: William Hughes
Subject: Re: Matheology § 238
On Apr 11, 4:20 pm, WM <mueck...@rz.fh-augsburg.de> wrote:

> On 11 Apr., 12:49, William Hughes <wpihug...@gmail.com> wrote:

>

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> > On Apr 11, 8:28 am, WM <mueck...@rz.fh-augsburg.de> wrote:

>

> > > On 10 Apr., 22:53, William Hughes <wpihug...@gmail.com> wrote:

> > <snip>

>

> > > > Thus, the fact that there is no line (along with

> > > > all its predecessors) that cannot be removed

> > > > is not a contradiction.

>

> > > It is not a contradiction with mathematics. So far I agree. But it

> > > would be a contradiction in case someone (and there are many here

> > > around) maintained ~P for some d_n if there is a proof of P for all

> > > FISs of d:

>

> > I do not claim this. I claim that the collection of all d_n does

> > not have the property P.

>

> That is not in question.

> My claim is this:

> If we have the propositions (with d_n a digit)

> A = for every n: P(d_n)

> B = for every n: P(d_1, d_2, ..., d_n)

> Then B implies A.

>

> Do you agree?

Indeed, however, B does not imply

P(d_1,d_2,d_3....)

So there is no contradiction is saying that A and B

are true but it it not true that P(d_1,d_2,d_3,...)