On 20 Apr., 23:36, Virgil <vir...@ligriv.com> wrote:
> > > Thus for every s_i there is i+1 in |N not in s_i, but Union(S) = |N. > > > That is trivial and requires no explanation. > > > U s_i = |N. > > > But either there are all naturals in one s or not. > > That is simple TND. > > And since for EVERY such s, max(s) + 1 is NOT a member of s, there > cannot be all natruals in any s.
Then for all j: there is an n not in s_j. Does the union contain more naturals than the complete sequence?
> > > If *not* all naturals in one s, then > > exist j, k, m, n : m e s_j & ~(m e s_k) & ~(n e s_j) & n e s_k. > > Once before claimed and also once beforE DEBUNKED. > > NOTE that a e s_b iff a <= b and ~( a e s_b) iff a > b, so that > m e s_j & ~(m e s_k) & ~(n e s_j) & n e s_k > iff > (m <= j) & (m > k) & (n >j) & ( n <= k) > But if (m <= j) & (n >j) then m < n > And if (m > k) & (n <= k) then n < m > So it follows that m > m and n > n, at least
if all naturals are in the terms s_i of S but not all are in one s_i. Yes, that's the contradiction.
Otherwise there are more naturals in the union than in the terms of the sequence. That would be another contradiction.