Laws of Triple Form: 6 of 12

     Chapter 5 : Completeness Theorems

          Anchored normal forms

          Complete deduction

          Complete re-entrance

        5A. Anchored normal forms

First note this lemma about minima:

x min (y min z)  =  (xVI) & (yVI) & (zVI) & (xVyVz)

Now consider the following Kleenean Anchor Forms:

A₀(a,b,c,x)     =   (a & x) V (b & ~x) V (c & dx) V  (a&b&c)

A₁(a,b,c,x)     =   (a V ~x) & (b V x) & (c V Dx) & (aVbVc)

I call these ‘anchor’ forms because of the abc terms.

The Absorption axiom and the mimima lemma imply these equations:

A₀(a,b,c,F)   =     b V  (a&b&c)                                   =             b

A₀(a,b,c,I)    =   (a&I) V (b&I) V (c&I) V (a&b&c)  =   a min b min c

A₀(a,b,c,T)   =     a V  (a&b&c)                                   =             a

Exercise for the Student: prove A₀(a,b,c,x) = A₁(a,b,c,x) . Call them both A.

If f(x) is Kleenean, then it preserves inner order;

so      f(I) min f(T) min f(F)  =  f(I) ;  therefore;    

f(x)    =       A( f(T), f(F), f(I), x )

=    (f(T) & x)  V  (f(F) & ~x)  V  (f(I) & dx)  V   (f(T)& f(F)& f(I))

=    (f(T) V ~x)  &  (f(F) V x)  &  (f(I) V Dx)  &  (f(T)V f(F)V f(I))

          =    [ [f([])] [x] ]   [ [f([[]])] x ]  [ [f(6)] x[x] ]  [ [f([])] [f([[]])] [f(6)] ]

          =    [  [ f([]) [x] ]  [ f([[]]) x ]   [ f(6) x[x] ]   [ f([]) f([[]]) f(6) ]  ]

This is the Anchored Normal Form. 

          5B. Complete deduction

          If the triple-form form equation

                   F       =       G

          is necessarily true, then the equation is provable from the bracket axioms.

          Proof is by induction on the number of variables.

          Case 0. If  F and G have no variables, then the equation F=G is a bracket-arithmetic equation; these can be calculated from the tables, which are a consequence of the bracket axioms. Therefore F=G is provable from the axioms.

          Case N implies Case N+1. Suppose that all N-variable identities are provable from the bracket axioms. Now suppose that F=G contains variable N+1; call it x. Then these equations are provable:

          F(0)   =       G(0)

          F(1)   =       G(1)

          F(6)   =       G(6)

since these have only N variables. Therefore:

          F(x)   =       A( f(T), f(F), f(I), x )       by anchored normal form

                   =       A( f(T), f(F), f(I), x )       by above equations

                    =       G(x)                               by anchored normal form

          Case 0 holds; recurrence holds; therefore the theorem is true, by induction.

          Like with Brownian forms, any identity-proof constructed this way is equivalent to table look-up, with 3^n cases, taking 3^n steps. And as with Brownian forms, finding exceptions to equations is an NP-complete problem.

          5C. Complete re-entrance

Given a re-entrant form, can we evaluate them in Kleenean logic? And if so, how? It turns out that inner order permits us to do so in general. For any Kleenean function F(x), we have the following:

The Self-Reference Theorem:

Any self-referential Kleenean system has a fixedpoint:

F(x) = x

Proof: Recall that all Kleenean functions preserve inner order.

i is the leftmost set of values, hence this holds:

i     <    F(i)

Therefore, i is a right seed for F:

i   <  F(i)  <  F²(i)  <  F³(i)  < ... Fⁿ(i)  =   F(Fⁿ(i))

i generates the “leftmost” fixedpoint. 

QED.

I call this process “productio ex absurdo”, meaning production from the absurd; in contrast to “reduction to the absurd”, boolean logic’s refutation method. Paradox logic begins where boolean logic ends. To see productio ex absurdo in action, consider this system:

          A       =       [ B C [] ]

          B       =       [ A B C ]

          C       =       [ [A] B ]

Iterate this system from 6:

     

                  |                    |       

A  =   [ B C [] ] |  B  =  [ A  B  C ] | C  =  [[ A ] B ]

__________________|____________________|___________________

                  |                    |

               6  |                  6 |                6

__________________|____________________|__________________ 

                  |                    |

 [ 6 6[]]  = [[]] |  [ 6 6 6 ]    =  6 |  [[6]6]    =   6

__________________|____________________|__________________ 

                  |                    |

 [ 6 6[]]  = [[]] |  [[[]]6 6 ]   =  6 | [[[[]]]6]  =  [[]]

__________________|____________________|__________________ 

                  |                    |

 [[[]]6[]] = [[]] |  [[[]]6[[]] ] =  6 | [[[[]]]6]  =  [[]]

__________________|____________________|__________________ 

After three cycles it stops. The leftmost fixedpoint is: A = void, B = curl, C = void. All fixedpoints are right of the leftmost fixedpoint; therefore A = C = [[]]; therefore B = [B];  therefore B = 6. This fixedpoint is the only one for this system.

Other systems may have more than one fixedpoint. The “toggle” system:

A       =       [B]

B       =       [A]

has three fixedpoints:    (1,0), (0,1), (6,6)

The “triplet” system: 
                   A       =       [ B C ]

                   B       =       [ C A ]

                   C       =       [ A B ]

          has four fixedpoints:  (1,0,0), (0,1,0), (0,0,1), (6,6,6).