## 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]
Retransposition:
[ax][bx]      =       [ [[a][b]] x ]
General Retransposition:
[a1x][a2x]…[anx]      =       [ [[a1][a2]…[an]] x ]
Crosstransposition:
[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

Therefore
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.
Therefore:
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.