In article <email@example.com>, WM <firstname.lastname@example.org> wrote:
> On 21 Apr., 00:03, Virgil <vir...@ligriv.com> wrote: > > In article > > <ba068c96-ce1a-4a80-b84f-be765c029...@a14g2000vbm.googlegroups.com>, > > > > WM <mueck...@rz.fh-augsburg.de> wrote: > > > There is no term s_n of S that contains all natural numbers. This > > > condition requires > > > (2) exist j, k, m, n : m e s_j & ~(m e s_k) & ~(n e s_j) & n e s_k. > > > > Nothing outside of Wolkenmuekenheim implies > > "(2) exist j, k, m, n : m e s_j & ~(m e s_k) & ~(n e s_j) & n e s_k" > > though it is implied often enough enough inside Wolkenmuekenheim. > > Otherwise all naturals in s_k are in in s_j for k < j. This implies > that all numbers in S are in one and the same s.
Not outside of Wolkenmuekenheim, it doesn't! > > Consider the representation as a table > > 1 > 2, 1 > 3, 2, 1 > ... > n, ..., 2, 1 > ... > > All initial segments of |N (including |N itself) are in the first > column, but not in the lines of the table?
Only in Wolkenmuekenheim!
outside of Wolkenmuekenheim, each n in |N is named in all columns of such a table, at least when the table is completed and in all but finitely many of its rows then.
> > > > if (2) were true outside of > > Wolkenmuekenheim, one would have both j < k and k < j simultaneously > > true as well. > > Yes that's just the contradiction.
But that only holds INSIDE Wolkenmuekenheim, where WM can claim at least partially to control truth, not outside it, where he is powerless. --