```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:>>>>>>>>>> > On Apr 11, 4:20 pm, WM <mueck...@rz.fh-augsburg.de> wrote:>> > > On 11 Apr., 12:49, William Hughes <wpihug...@gmail.com> wrote:>> > > > 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....)>> That is not required. It is only required that B implies> A =  for every n: P(d_n).>>>> > 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,...)->> So A does not imply P(d_1,d_2,d_3,...) either?correct.Now let P be (can remove the collection without changingthe union of the remaining lines).  We have there is nocontradiction in saying that for all n, the nth line canbe removed.
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