Chapter 9 :
Trinary Logics
Majorities
Octohedral distribution
The Hexagram
9A. Majorities
Define
three majority operators:
M_{I}(x,y,z) = (x &_{FT} y) &_{TF} (y &_{FT}
z) &_{TF} (z &_{FT} x)
M_{T}(x,y,z) = (x
&_{IF} y) &_{FI}
(y &_{IF} z) &_{FI} (z &_{IF} x)
M_{F}(x,y,z) = (x
&_{IT} y) &_{TI}
(y &_{IT} z) &_{TI} (z &_{IT} x)
The
subscripted majority operator has these identities:
Symmetry: M_{a}(x,y,z) = M_{a}(y,z,x) = M_{a}(z,x,y) = M_{a}(x,z,y)
= M_{a}(z,y,x) = M_{a}(y,x,z)
Coalition: M_{a}(x,x,y) = M_{a}(x,x,x) = x
Cancellation: M_{a#b}(x, a, b) = x
Mediocrity: M_{f}(f,i,t) =
f ; M_{i}(f,i,t) =
i ; M_{t}(f,i,t) =
t
The
subscript ‘a’ is the “Chairman’s Subscript”: it decides threeway ties. These
axioms suffice to define all values of M_{a}(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(M_{a}(x,y,z)) = M_{P(a)}(P(x),P(y),P(z))
Subscripts: x &_{ab}
y =
M_{a#b}(x, y, a)
Exercise
for the student:
Prove that &_{fi},
&_{ft}, &_{tf}, &_{ti} all distribute over
M_{i}; but not &_{if }or &_{it}.
Prove that if a, b and c
are three different forms, then
&_{ac}, &_{ab}, &_{ba},
&_{bc} all distribute over M_{c}; 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&_{pq}x = 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 2element
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&_{de}a = 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&_{de}b) a &_{de} c check?

a b c&_{ab}(a&_{ab}b) = a a &_{ab} c = a yes
a c c&_{ab}(a&_{ac}b) = a a
&_{ac} c = a yes
b a c&_{ab}(a&_{ba}b) = c a &_{ba} c = c yes
b c c&_{ab}(a&_{bc}b) = c a
&_{bc} c = a no!
c a c&_{ab}(a&_{ca}b) = c a &_{ca} c = c yes
c b c&_{ab}(a&_{cb}b) = a a
&_{cb} c = c no!
Note the
two exceptions:
c &_{ab }(a&_{bc}b) = c ; a &_{bc }c = a
c &_{ab }(a&_{cb}b) = 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 oneway
distributing operators. The pivot a#x is here denoted ~_{a}x; the
doublecrosses 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:
No comments:
Post a Comment