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.