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
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,...)