Date: Apr 20, 2013 4:16 AM
Author: mueckenh@rz.fh-augsburg.de
Subject: Matheology § 255

Matheology § 255

Let S = (1), (1, 2), (1, 2, 3), ... be a sequence of all finite
initial sets s_n = (1, 2, 3, ..., n) of natural numbers n.

Every natural number is in some term of S.
U s_n = |N
forall n exists i : n e s_i.

S is constructed by adding s_(i+1) after s_(i). So we have
(1) forall n, forall i : (n < i <==> ~(n e s_i)) & (n >= i <==> n e
s_i).

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.

(2) is in contradiction with (1).

A better readable version is in § 255 of
http://www.hs-augsburg.de/~mueckenh/KB/Matheology.pdf

Regards, WM