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?
> Since the fact that there is no d_n that > does not have the property P does not mean that the collection of > all d_n has the property P there is no contradiction. > > > For all n: d_1, d_2, ..., d_n have the property P. > > > Matheology requires: The sequence of all d_n constitutes the real > > number. > > yes
Therefore a property P(d_n) for all n holds for all digits of the real number. > > > The sequence of all d_1, ..., d_n does not constitute a real number. > > Nope, no one of the elements of the sequence constitues the real > number
Therefore a property P(d_1, ..., d_n) for all n does not hold for all digits of the real number?