Tuesday, July 10, 2018

Laws of Triple Form; 2 of 12

          Chapter 1:  Brownian Forms

          Form arithmetic
          Form algebra
          Isomorphic twice to Boolean logic
          Normal forms
          Complete deduction
          Incomplete re-entrance

          1A . Form Arithmetic

Let empty space on a page be called a form:

Let us be able to mark such a space by placing brackets in it:

[ ]

Let the first such form be called void, and the second be called mark.
For any forms, let us be able to make these new forms:

Bracket:                          [ x ]
Juxtaposition:                x y

The brackets distinguish only inside from out, not left from right; so juxtaposition is commutative:                 
x  y    =   y   x

-      where “=” means “is confused with”.

Let two marks juxtaposed denote the same mark:

[ ]  [ ]    =   [ ]

Call this the Law of Calling:
To call twice is to call.

Let a bracket within a bracket denote crossing a boundary twice, and hence not at all:

[ [ ] ]  =

Call this the Law of Crossing:
To cross twice is not to cross.

Let there be a ‘void mark’, or ‘parenthesis’, or ‘doublecross’, defined as:
( x )  =  [ [ x ] ]   =  x
With the double-cross (x), we can say that juxtaposition is associative:

          x(yz) = (xy)z

To avoid confusing empty spaces on the page, let us use these symbols for void and mark:

0       =   ()  =  [[]]  =  
1       =   [] 

That implies these tables for bracket and juxtaposition:

        x  0          1
y    -|------------------
0     |     0          1
1     |     1          1

                        [x]   |     1          0

          Define the operation ‘rejuxtaposition’:

                   [ [x] [y] ]

It has this table:

        x  0          1
y    -|------------------
0     |     0          0
1     |     0          1

This is the Form Arithmetic of G. Spencer-Brown, in his “Laws of Form”.

Exercises for the student:

Prove that any form made from only doublecross, void and juxtaposition equals void.

Prove Spencer-Brown’s Arithmetic Theorem:
Any form made from void, bracket and juxtaposition equals mark or void.

1B. Form Algebra

These are the arithmetic axioms for brackets:

Calling:      [] []  =  []
Crossing:    [[]]   =  

Exercise for the student:
Prove the validity of these bracket axioms:

Position:                         [[x]x]          =      
Transposition:              [[x][y]] z     =       [[xz][yz]]

          Exercises for the student:
From the bracket axioms, prove these identities:

          Doublecross:        [[a]]            =       a
          Generation:         [ab] b         =       [a] b
          Attractor:             [] a              =       []
          Occultation:         [[a]b]a        =       a
          Recall:                  a a              =       a
          Extension:    [[a][b]] [[a]b]     =       a
          Echelon:               [[[a]b]c]      =       [ac] [[b]c]
[ax][bx]      =       [ [[a][b]] x ]
          General Retransposition:
      [a1x][a2x]…[anx]      =       [ [[a1][a2]…[an]] x ]
[ax] [b[x]]   =       [  [[a]x]  [[b][x]]  ]

            1C. Isomorphic twice to Boolean logic

          Consider these tables for Boolean logic, where T means ‘true’, F means ‘false’, ~ means ‘not’, & means ‘and’, and V means ‘or’;

V     x   F          T
y    -|------------------
F     |    F          T
T     |     T          T

&      x  F          T
y    -|------------------
F     |    F          F
T     |     F          T

                        ~x   |     T          F

          Note the similarity to the form tables previously. In fact these systems are isomorphic twice:

          (   [[]], [], [x], xy, [[x][y]]  )  is isomorphic to:

          (   F, T, ~x, xVy, x&y  )   ,    and to:

          (   T, F, ~x, x&y, xVy  ) 

          Boolean logic is isomorphic to itself. Its automorphism group is S2; the symmetries of two objects. 

1D. Normal forms

          Recall Echelon, General Retransposition and Crosstransposition:

        Echelon:                                                       [[[a]b]c]          =          [ac] [[b]c]
            General Retransposition:             [a1x][a2x]…[anx]       =          [ [[a1][a2]…[an]] x ]
            Crosstransposition:                                  [ax] [b[x]]      =          [  [[a]x]  [[b][x]]  ]

          These results imply a normal form for all bracket expression, in five steps:

          Step 1: Any bracket expression F is provably equal to one no more than two brackets deep.
          Proof: use Echelon enough times.

          Step 2: Any bracket expression, containing variable x, is provably equal to the bracket expression:
          [ A x ]  [ B [x] ]  C
          where A, B and C do not contain the variable x.
          Proof: use step 1, then use General Retransposition to collect all the x terms in one bracket, and all the [x] terms in another bracket.

          Step 3: Such a bracket expression equals, by Crosstransposition:
          [  [ [A] x ]  [ [B] [x] ]  ]  C

          Step 4: It therefore equals, by Transposition:
          [  [ [A] C x ]  [ [B] C [x] ]  ]

          Step 5: Let F(x) denote that expression, then substitution yields
          F(0) = [A]C
          F(1) = [B]C

          F(x)   =   [  [ F(0) x ]  [ F(1) [x] ]  ]
          Call this the standard normal form.

1E. Complete deduction

          Complete Deduction Theorem:
          If the bracket 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)           
since these have only N variables.
          F(x)   =   [  [ F(0) x ]  [ F(1) [x] ]  ]       by standard normal form
                    =   [  [ G(0) x ]  [ G(1) [x] ]  ]      by the preceding equations
                    =   G(x)                                        by standard normal form

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

          Note that any proof thus constructed will be equivalent to table look-up, as it will check all 2^n cases, for n = number of variables; and thus take 2^n steps.

In complexity theory, finding exceptions is an NP-complete problem.

1F. Incomplete re-entrance

          Bracket forms are built of brackets and juxtaposition of other bracket forms; a chain ultimately ending in the void. Or does it? Could there be a form founded upon itself? Or even systems of forms founded upon each other?

Consider the forms
          A       =       [B]
          B       =       [A]

          This system, a ‘toggle’, has two solutions:
          A=0, B=1
          A=1, B=0
          Both make sense, but which is it?

          Now consider a worse case; the form
          L       =       [L]
          L can equal neither 0 or 1.

          Therefore Brownian bracket forms are incomplete for re-entrance. To solve the equation, you need a third form.

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