Date: Apr 11, 2013 12:00 PM
Author: William Hughes
Subject: Re: Matheology § 238

On Apr 11, 5:28 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> On 11 Apr., 16:42, William Hughes <wpihug...@gmail.com> wrote:
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> > On Apr 11, 4:20 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
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> > > 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.

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> > > 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.

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> > > Do you agree?
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> > Indeed, however, B does not imply
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> > P(d_1,d_2,d_3....)
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> That is not required. It is only required that B implies
> A =  for every n: P(d_n).
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> > 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,...)-

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> So A does not imply P(d_1,d_2,d_3,...) either?


correct.


Now let P be (can remove the collection without changing
the union of the remaining lines). We have there is no
contradiction in saying that for all n, the nth line can
be removed.