Virgil
Posts:
8,833
Registered:
1/6/11


Re: Matheology � 253
Posted:
Apr 21, 2013 4:25 PM


In article <f1bd3d7d8f124b9e9063c240a5f93d71@r4g2000vbf.googlegroups.com>, WM <mueckenh@rz.fhaugsburg.de> wrote:
> On 19 Apr., 22:50, Virgil <vir...@ligriv.com> wrote: > > > > There exist m, n, a, b in N, > > > such that m is in FISON(a) and not in FISON(b) and n is in FISON(b) > > > and not in FISON(a). > > > > For each natural n in N, FISON(n) is defined by WM to be the set of all > > naturals less than or equal to the natural number n. > > Thus for natural numbers m ands n, > > n is in FISON(m) if and only if n <= m, and > > similarly n is NOT in FISON(m) if and only if m < n. > > Thus if m is in FISON(a) and not in FISON(b), then b < m <= a > > and if n is in FISON(b) and not in FISON(a), then a < n <= b > > > > Thus in WM's world one must have b < a and a < b simultaneously. > > Explain how N elements (n) can be distributed, with repetition, among > M sets (s_k) such that there are all elements n represented at least > once, all s_k are used too, but not all elements n are in in one set > s_k.
WM himself has shown us how to do it:
FISON_1 = s_1 = { 1 } FISON_2 = s_2 _ { 1 2 } = { 2 1 } FISON_3 = s_3 = { 1 2 3 } = { 3 2 1 } ...
So FISONs do it very neatly!
> Is it possible to avoid the condition: > exist j, k, m, n : m e s_j & ~(m e s_k) & ~(n e s_j) & n e s_k"?
Yes! Trivially! it is, in fact, impossible for any 4 naturals j, k, m, n not to avoid it. > > How? How not?
See above. If WM claims such naturals can exist, let him show us an example of specific values which do satisfy that false claim.
Also. from a previous post:
WM's claim: > 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 in WM's world.
But OUTSIDE WM's world, NOT all naturals can be in an one FISON, s_n > > If only ~(n e s_j) & n e s_k, then also m e s_k is possible, and then > n and m are in s_k. No evidence that not all naturals in one s.
Maybe no such evidence in Wolkenmuekenheim, but there is enough outside Wolkenmuekenheim, to satisfy everyone other than WM.
For every i in N, max(s_i) = i and i+1 is not in s_i.
Which is sufficient evidence OUTSIDE WOLKENMUEKENHEIM to show that not all naturals are in any one of the s_i, at least not OUTSIDE WOLKENMUEKENHEIM . 

