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:
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> > > On 10 Apr., 22:53, William Hughes <wpihug...@gmail.com> wrote:
> > <snip>
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> > > > Thus, the fact that there is no line (along with
> > > > all its predecessors) that cannot be removed
> > > > is not a contradiction.

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> > > 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:

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> > 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,...)