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?
Here is a better presentation:
In a Cantor list the argument can be written: For every n: (d_1, ..., d_n) differs from every entry (qk1, ..., qkn) with k =< n, i.e., i.e., every entry of the first n lines. That is exactly Cantor's argument, not more and not less.
In a list containing all rational numbers, the counter-argument can be written: For every n: (d_1, ..., d_n) does not differ from infinitely many entries (qk1, ..., qkn) with k > n.
Why should the "for all n" only in one case be exhaustive?