Chapter 9 :
Trinary Logics
Majorities
Octohedral distribution
The Hexagram
9A. Majorities
Define
three majority operators:
MI(x,y,z) = (x &FT y) &TF (y &FT
z) &TF (z &FT x)
MT(x,y,z) = (x
&IF y) &FI
(y &IF z) &FI (z &IF x)
MF(x,y,z) = (x
&IT y) &TI
(y &IT z) &TI (z &IT x)
The
subscripted majority operator has these identities:
Symmetry: Ma(x,y,z) = Ma(y,z,x) = Ma(z,x,y) = Ma(x,z,y)
= Ma(z,y,x) = Ma(y,x,z)
Coalition: Ma(x,x,y) = Ma(x,x,x) = x
Cancellation: Ma#b(x, a, b) = x
Mediocrity: Mf(f,i,t) =
f ; Mi(f,i,t) =
i ; Mt(f,i,t) =
t
The
subscript ‘a’ is the “Chairman’s Subscript”: it decides three-way ties. These
axioms suffice to define all values of Ma(x,y,z); for either {x,y,z} has three distinct
elements (and then Symmetry plus Mediocrity applies) or {x,y,z} has
coincidental elements (so Symmetry and Coalition apply).
Permutations: for any permutation P, P(Ma(x,y,z)) = MP(a)(P(x),P(y),P(z))
Subscripts: x &ab
y =
Ma#b(x, y, a)
Exercise
for the student:
Prove that &fi,
&ft, &tf, &ti all distribute over
Mi; but not &if or &it.
Prove that if a, b and c
are three different forms, then
&ac, &ab, &ba,
&bc all distribute over Mc; but not &ca or
&cb.
9B. Octohedral Distribution
Say that “&ab
distributes over &de” if, for every x, y and z from among the
three forms:
x &ab ( y &de
z ) = (x &ab
y) &de ( x &ab z
)
There
are three cases: {x,y,z} has 1 element, or 2, or 3.
Case 1:
{x,y,z} has only 1 element.
Then the
equation follows by Recall for each operator:
x&pqx = x .
Case 2:
{x,y,z} has only 2 elements - say, {F,I}.
Then the
operators &ab
and &de would be min or max operators on the 2-element
lattice F<I; these distribute over each other and themselves, therefore &ab would distribute over &de.
Case 3:
{x,y,z} = {F,I,T}.
There
are three distinct values for x, y and z. Call them a, b, and c; so there are
three cases for x; it equals a, or b, or c.
If X = a,
then:
x &ab (y &de z) =
a &ab (y &de z) = a
and
(x &ab
y) &de (x &ab z) = (a &ab
y)&de(a &ab z) = a&dea = a
If X = b,
then:
x &ab (y &de
z) = b &ab (y &de
z) =
(y &de z)
and
(x &ab
y) &de (x &ab z) = (b &ab
y) &de (b &ab z) = (y
&de z)
If X = c,
then:
x &ab (y &de
z) = c &ab (y &de
z) =
c &ab (a &de b)
and
( x &ab
y) &de ( x &ab z ) = (c &ab
a) &de (c &ab b)
= a &de
c
-
because
y and z are the other two values a and b.
So now the question is: for what d and e is the following
true?
c &ab (a &de
b) =
a &de c
Just check
all six possibilities for d and e:
d e c&ab(a&deb) a &de c check?
-------------------------------------------------------------------------------
a b c&ab(a&abb) = a a &ab c = a yes
a c c&ab(a&acb) = a a
&ac c = a yes
b a c&ab(a&bab) = c a &ba c = c yes
b c c&ab(a&bcb) = c a
&bc c = a no!
c a c&ab(a&cab) = c a &ca c = c yes
c b c&ab(a&cbb) = a a
&cb c = c no!
Note the
two exceptions:
c &ab (a&bcb) = c ; a &bc c = a
c &ab (a&cbb) = a ; a &cb c = c
They
are d=b and e=c, or d=c and e=b; in neither case is either d or e equal to a,
whereas one of them equals a in every other case.
Therefore this Theorem:
Octohedral Distribution
&ab
distributes over &cd
if and only if
a=c or a=d
9C. The Hexagram
Octohedral distribution is
illustrated by the following “distribution octohedron”, a.k.a. the “Hexagram”.
Lines join mutually distributing operators, lines with arrows join one-way
distributing operators. The pivot a#x is here denoted ~ax; the
double-crosses d#(e#x) are denoted d#(e#.
The triples (&AC,
&BA, &CB) and (&CA,
&BC, &AB) display cyclic distribution:
The first distributes over the second;
The second distributes over the third;
The third distributes over the first;
But not the other way around!
Here is the Hexagram in terms of Kleenean logic:
Here it is in terms of triple forms:
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