## Friday, May 18, 2018

### Diamond Bracket Forms and How to Count to Two; 5 of 10

From "Cybernetics & Human Knowing", Vol. 24 (2017), No. 3-4, pp. 161-188
4. Fixedpoint Lattices

Any diamond-logic function F(x) has the extreme fixedpoints: Fn(i), Fn(j), the leftmost and rightmost fixedpoints. But often this is not all.
In general, F has an entire lattice of fixedpoints.

Lemma: For any diamond-logic f;
f(x) min f(y)  >  f(x min y)
f(x) max f(y)  <  f(x max y)
Proof is an exercise for the student.

Theorem: If F is diamond-logic, and has fixedpoints a and b, then these fixedpoints exist:
a minF b   =  the rightmost fixedpoint left of both a and b  =  F2n(a min b)
a maxF b  =  the leftmost fixedpoint right of both a and b  =  F2n(a max b)

Proof. Let a and b be fixedpoints, and let c be a fixedpoint left of both a and  b. Then  (a min b) > c ;  so (a min b) =  F(a) min F(b)  > F(a min b) >  F(c)  =  c
Ergo (a min b) is a left seed greater than c:
(a min b)  >   F(a min b)   >   F2(a min b)   >   ...
>   F2n(a min b)   =   F(F2n(a min b))   >    c
Therefore F2n(a min b) is a fixedpoint left of a and of b, and is also the rightmost such fixedpoint. Thus,  F2n(a min b)   =    a minF b .        QED.

Similarly,     F2n(a max b)   =    a maxF b .        QED.
For instance, consider the following Brownian form:
______________________________d
_________________c   |
|    _____a _____b |    |
|        |      |  |    |
|      | |____| |  |    |
|      |________|  |    |
|_______________________|

This is equivalent to this bracket-form system:
a = [b]   ;     b = [a]   ;     c = [ab]  ;   d = [cd]  .
In the standard interpretation, (a,b,c,d) is a fixedpoint for:
F(a,b,c,d)  =  ( ~ b,  ~ a,  ~(a Ú b),  ~(c Ú d) )
In the nand-gate interpretation:
d     =    ~(d Ù c)    =    ~d Ú ~c    =    (d Þ da )
Sentence d says “If I’m not mistaken, then A is both true and false”.
In the nor-gate interpretation:
d        =       Da  -  d   ;
Sentence d says “A is true or false, and I am a liar.”
F has this fixedpoint lattice:

tffi ----- tffj
/               \
/                 \
iiii                   jjjj
\                 /
\               /
ftfi ----- ftfj

To find minF of tffj and ftfj, first take their minimum, then apply F three times:                    (tffj min ftfj) = iifj   è   iiij   è   iiit   è   iiii