Kindergarten and Differences/Handouts

Materials for a better exploration of the different worlds of mathematics

Rudolf Kaehr Dr. phil@

Copyright ThinkArt Lab ISSN 2041-4358

 

Abstract

Handouts are for free. These handouts are intended to help teachers and parents to avoid abusive mental suppression of their children by indoctrinating them against their intuition and will to become parrots, fit for the use of their early bank accounts. And it’s free! No charge, like for the Soft Start program. But there is also no guarantee included for the mental health of the applicants. They might easily become alienated.
(work in progress, vers. 0.3, Nov. 2013)

1.  Counting in 5 different worlds

1.1.  What’s about?

1.1.1.  How did it start?

All started with the insight that the innocent question of a teacher: “How much is 2 + 2?” isn’t as trivial as he thought. Before the child answered this simple question it returned it on another level with its own question: “Am I selling or am I buying?”

Everybody knows the games of partitions, permutations and prolongations of sequences for forms, played with shapes of different colors.

Here, 5 different ways of playing such games are introduced.

I call them the Leibniz, the Pascal, the Brown, the Mersenne and the Stirling games.

The differentiation of the games are defined by the different rule sets of the games.

1.1.2.  The Leibniz game

The Leibniz game is defined by some strict axioms.

Classical rules

typeset structure

Wording

Two elements are not equal one element.
Different elements are different and not equal.

Little task

Given 2 elements and 3 places, how many different constellations of the two different elements on the 3 places are possible in the Leibniz world?

Leibniz(3,2) = 8:

  typeset structure

Answer: The Leibnizian order for 2 elements and 3 places has 8 constellations.

Symmetry

There is also a nice symmetry between the first and the second half of the Leibniz constellations.

typeset structuretypeset structure

Successors

Alphabet ∑ = {}
succ() = • •

Alphabet ∑ = {•, }
succ() = {• •, • }
succ() = { •, • }

These successors are defining a binary tree. With 3 elements the successors are defining a ternary tree.

Binary tree for Leibniz
typeset structure

Reversion
As easy as successions are reversions of patterns with 4 elements.
(O ▲) = pattern
rev(O ▲) = ( O) : reversion of the pattern.
Hence, rev(O ▲) != (O ▲).

1.1.3.  The Pascal game

Between the Leibniz and the Brownian game with its fundamental commutativity of terms, the realm of Pascal partitions has to be placed. The Pascal game is also defined in the general system of graphematics as a deutero-structure.

The Pascal game is defined by some strict axioms:

Pascal rules

typeset structure

Wording

Different elements are not equal.

Partitions for 3 elements and 3 positions

When we look at parenthesis with order 3 there are 5 possibilities.

typeset structure

This bracket notation corresponds to the tuple notation:
[1,1,1,1], [2,1,1], [2,2], [1,3], [4].

Partitions for 4 elements and 4 positions
1+1+1+1: ()()()():  • •    : aaaa  [1,1,1,1]
1+1+2:     ()() (1):  O •    : aabc [2,1,1]
(2)(2):       (1)(1):   O O     : aabb [2,2]
1+3:           ()(2):    O •    : abbb [3,1]
4:               (3):       O      : abcd [4]

Combinatorics

In contrast, the Brownian distribution counts 4, and not 5 possibilities for n = 3 according to the immanent commutativity rule: (())() = ()(()).

() () () ; () (()) ; (() ()) ; ((())) .

The general number of possibilities is calculated with the partitions of numbers P(n).

typeset structure

typeset structure   (sequence A000041 in OEIS).

[Graphics:HTMLFiles/Kindergarten and Differences-Handouts_11.gif]

typeset structure
typeset structure

typeset structure

Little task
Given 4 elements and 4 places, how many different constellations of the four different elements on the 4 places are possible in the Pascal world?

Pascal(4,4) = 5:

  typeset structure

Answer: The Pascal partition order for 4 elements and 4 places has 5 constellations.

Symmetry
There is no obvious symmetry for this partition system. Symmetry is given internally for typeset structure), (typeset structure) and (typeset structure).

1.1.4.  The Spencer-Brown game

The basic rules for the Brownian distinction calculus

typeset structure

Rule 1. () () = ()
Rule 2. (()) = ⌀
3. Substitution rules

Wording
Rule1: A distinction of 2 distinctions is a distinction.
Rule2: A distinction of a distinction is no distinction.

In colors
Rule1.  • • =
Rule2.   typeset structure =

Especially:
(( )) ( )  = ( ) (( ))  :  typeset structure = typeset structure .
Red in red kills red,  typeset structure=⌀ and red saves red
equal
red with red in red typeset structure kills red typeset structure=⌀ and saves red .
Hence,  typeset structure = typeset structure = •.

Superpositions
(() (((()))) =  typeset structure
(() ())     =     typeset structure
(())         =     typeset structure
(())         =     ⌀

In words:
typeset structure : Red with red in red in red kills red and red saves red: typeset structure .
typeset structure   : Red with red in red saves red : typeset structure .
typeset structure       : Red in red kills red  ⌀ .

Little task
Given 2 elements and 3 places, how many different constellations of the two different elements on the 3 places are possible in the Brownian world?

Brown(3,2) = 4
  typeset structure

Answer: The Brownian order for 2 elements and 3 positions has 4 constellations.

The number of forms (not possibilities) of degree 3 is 4 and not 5 as for the Pascal partitions. The forms are:

typeset structure

Normal form

The written table is accepting the normal form for Brownian constellations. Hence, the two constellations typeset structure, typeset structure typeset structureare represented by the single constellation typeset structure in Brownian normal form.

Symmetry

In contrast to the Pascal world, there is a nice symmetry between the first and the second half of the patterns based.on the commutativity of the forms.

typeset structure   typeset structure

Successor

Alphabet ∑ = {•, }
succ() = {(• •), (), ( • ) }  

Addition Sum

sum(, ⌀) =
sum(• •, ) = {• • •, • • }.
sum(, ) = { }
sum( , ) = {• • •}.

Reversion for Brownian patterns

rev() = ( ) and () = ( ).

1.1.5.  The Mersenne game

The basic rules of the calculus of differentiations

typeset structure

Rule 1. () () = ⌀
Rule 2. (()) = ()
3. Substitution rules

Wording

Rule1: A differentation between 2 differentiations is an absence of a differentiation.
Rule2: A differentiation of a differentiation is a differentiation.

In colors
Rule1.   = ⌀
Rule2. typeset structure =

Other wording

Blue with blue kills blue.
Blue in blue saves blue.
The rules are also well understood as oriented actions.
Rule1.    = ⌀  is an equational notation for to corresponding actions:
Rule1a. ==> ⌀ and
Rule1b. <== ⌀
Rule2.   typeset structure = this also holds for Rule2
Rule2a   typeset structure ==>
Rule2b. typeset structure <==

Some examples

1. ()()() = (()(())(()) : the same are the same, thus there is no differentiation.
= typeset structure typeset structure typeset structure
=
typeset structure =
Thus,   = typeset structure typeset structure typeset structure.

2. ( ) ( ) () = () : rule1         :   =
(( )) ( )  = ⌀  : rule2, rule1   : typeset structure •  =
((()))     = () : rule2             : typeset structure = •

Especially
3.
(( )) = ( ) () ()                 : typeset structure = • • •
Proof of typeset structure = • • •
[• •] •  : brackets
[] •     : rule1
•           : rule1.
typeset structure         
•           : rule2

Wording

In a Mersenne universe, the order of 2 different elements is relevant. In contrast to the Brownian universe, they are therefore different.
But a constellation of two same elements is equal to another constellation of two same elements.

Alternative wording

Red and green together are safe.
Two greens together are killed by two reds.

Little task

Given 2 elements and 3 places, how many different situations of the two different elements on the 3 places are possible in the Mersennian world?

Mersenne(3,2) = 7

  typeset structure

Answer: The Mersenne order for 2 elements and 3 positions includes 7 situations.

Symmetry
The nice symmetry of the whole set as we have seen for Leibniz and Brown is broken.

typeset structure typeset structure  typeset structure

Successor

Alphabet ∑ = {•, }
succ(typeset structure) = {(typeset structuretypeset structure), (typeset structure), (typeset structure)}.

Addition Sum

sum(typeset structure, ⌀) = typeset structure
sum(typeset structure typeset structure, typeset structure) = {typeset structure typeset structure typeset structure, typeset structure typeset structure typeset structure, typeset structure typeset structure typeset structure}
sum(typeset structure, typeset structure) = {typeset structure typeset structure, typeset structure}
sum(typeset structure, typeset structure) = {typeset structuretypeset structure, typeset structure typeset structure typeset structure}.

Reversion for Mersenne

rev(ab) = (ba) and (ab) != (ba)
rev(typeset structure) = (typeset structure) and (typeset structure) !=  (typeset structure).

Comparison Brown and Mersenne

Interestingly, there are some coincidences between both calculi. Both are deducing form the 3 brackets one resulting bracket: ( ) ( ) () = ().
But the way they are doing it is differently organized according to the 2 different rule sets.
It is a common failure to not to recognize this crucial difference.

Mersenne : ( ) ( ) () = () :
by rule1 :  
= •  :
( ) = (⌀) =
( ( )  = (⌀) =
Hence, • = • .

Brown: ( ) ( ) () = () :
by rule1 :
( ) = () =
(• •) = ()=
Hence, • • = .

In contrast:
typeset structure = • • •
typeset structure =
• • • =

Hence, typeset structure = • • •.
typeset structure != • •
typeset structure = ⌀
• • =
Thus, ⌀ != .

1.1.6.  The Stirling game

A Stirling blend

For a Stirling approach, the fact that the concept of patterns, i.e. ordered strings or configurations of identity-free elements, is crucial, leads to the following rules.

Those rules shall be understood as a blend of Brownian, Rule3, and Mersennian, Rule2, rules. A blend always produces also something new: Rule1 and Rule4.

Rule1. () = (())
Rule2. () () = (()) (())
Rule3. () (()) = (()) ()
Rule4. ()()(()) != ()(())() != ()(())(()) != ()(())((())).

In colors
Rule1. typeset structure
Rule2. typeset structure typeset structure = typeset structure typeset structure
Rule3. typeset structure = typeset structure
Rule4. typeset structure != typeset structure != typeset structure != typeset structure.

Another setting:
Rule1. typeset structure
Rule2. typeset structure typeset structure = typeset structure typeset structure
Rule3. typeset structure = typeset structure
Rule4. typeset structure != typeset structure != typeset structure != typeset structure.

Wordings of constellations

For a Stirlingian game with 3 elements, some typical situations occur.
1.  • • typeset structure typeset structuretypeset structuretypeset structure,
     typeset structuretypeset structure typeset structure typeset structuretypeset structuretypeset structuretypeset structure
     etcetera
2. typeset structure ≡ rev(typeset structure)  : reversion
3. typeset structure ≡ rev( typeset structure) : self-symmetry
    typeset structure ≡ rev(typeset structure)|
   typeset structure ≡ rev(typeset structure)

Wordings of rules for Stirling(3,3)

typeset structure:        Blue kills red.
typeset structuretypeset structure:  Blue together with red kills red together with blue.
typeset structure:        Two blue together with one red, and
typeset structure:        one blue together with one red and one blue, and
typeset structure:        one blue with two reds, are safe in the Stirlingian world.
typeset structure:        As well as blue and red and green together.
This constitutes a kind of safety in groups.

Little task

Given 3 elements and 3 places, how many different patterns (morphograms) of the three different elements on the 3 places are possible in the Stirlingian world?

            Sn (3,3) = 5
numeric   symbolic alphabetic
[13]:           • • : aaa
[122typeset structure:       • • •  : aab
[112111]:     : aba
[1122]:        : abb
[112131]:   O : abc

Answer: The Stirling order for 3 elements and 3 positions for distribution is 5.

Hence, there are 5 different morphograms for 3 elements and 3 positions. The choice of the color of the elements, here as blue, red and green is arbitrary and ruled by its normal form.

Further tasks for more complex situations

Sn (3,2) = 4  Sn (3,3) = 5   Sn (4,2) = 8  Sn (4,3) = 14   Sn (4,4) = 15
  typeset structure       typeset structure         typeset structure     typeset structure     typeset structure       

Symmetry

Here, again, the symmetry of the set of the basic patterns is broken.
But there are some nice internal symmetries left.
rev(typeset structure) = (typeset structure), that is rev(typeset structure) = (typeset structuretypeset structuretypeset structure) but (typeset structuretypeset structuretypeset structure) = (typeset structure).
Self-symmetric patterns: (typeset structure), (typeset structure), (typeset structure).

typeset structure

The function  ϕ is iterative if it repeats a given element, and accretive if it adds a new element. There is no recurs to a pre-given alphabet necessary. The successor operation is recurring retrograde to the predecessor elements and iterates the produced elements iteratively and adds accretively a new element to the system.

Resulting in the production of the 5 trito-patterns with 3 elements:
[aaa], [aab], [aba], [abb], [abc].

                &nb ... sp;     [b]    [a] [b]   [c]    <br />

Successors in colors

<br /> succ (•) = ((• •)/(•   •)) <br /> <br /> succ    ... 226;   •   O   • <br /> •   •   O   O)/(•   •   O   ◆))

Left successor

Example l - succ (•   •   • ) = ((•   •   •   •   <b ... O   O  )) succ (•   •   • )   !=   l - succ (•   •   • ) Null

left - succ                             pattern                                 right - succ   ... 226;   O   •         -                                       •   •   •   O

<br /> Addition

add (• •, •   •) = ((•   •   •   • <br /> •   •   •   •)/(•   •   •   O))

Multiplication

kmul ([•   •] [•   •]) = ((•   •   •   • <br / ...  •   O   • <br /> •   •   •   O)/(•   •   O   ◆))

Difference notation for morphograms

Difference notation with ν=non-equal and ε=equal
The fact that the presentation of the morphograms by specific elements is arbitrary has to be considered as crucial. Therefore, not the elements are determining the morphic patterns but the differences between the elements.
This is well depicted for the example [typeset structure].

typeset structure  

A useful notation is given with the matrix of the ε/ν-structures.

•      • •     ••     • •    O
typeset structure     typeset structure     typeset structure    typeset structure    typeset structure     

Comparison for systems with 2 elements an 3 places.

Little task in different worlds

Teacher: Given 2 elements and 3 places, how many different partitions of the three different elements on the 3 places are possible?

Child: In which world should the partition happen? In the Leibniz, the Pascal, the Brown, the Mersenne or the Stirling world? Or do you offer some others too?

The teachers task can be correctly answered by at least 5 different solutions.

   Leibniz        Brown      Mersenne     Stirling     Pascal        Proto
  typeset structure     typeset structure     typeset structure     typeset structure   typeset structure  typeset structure

There is a coincidence in the numbers of Brownian and Stirlingian tables for Sys(3,2). Also Pascal and proto-structures are coinciding on this level, i.e. Sys(3,2) and Sys(3,3).

   Leibniz          Brown       Mersenne      Stirling       Pascal       Proto
  typeset structure     typeset structure     typeset structure     typeset structure   typeset structure   typeset structure

Crucial differences

Not just the answers to the little tasks depends on the world model used but also the structure of the simples operations, like succession, addition, reversion, etcetera, differ significantly.

Leibniz world

The main model for succession is given by the Stroke calculus as it is fundamental for the Leibniz world.

Stroke calculus

Rule1.  ==> |
Rule2. n ==> n |
Meta-Rule3.  n ∈ Var, repetition of Rule2.

The main feature is the abstract concatenation of an atomic element to the just produced strokes, represented by n. Thus, Rule2. n ==> n |.

typeset structure

This simple feature differs depending on the world model.

Brownian world

For the Brownian model, the successor operation is still in the spirit of the Leibnizian successor but modified by the specific Brownian features of commutativity.

               &nbs ... A0;^3} , {2 ^4}     <br />         

     Numeric indicational rules <br /> <br />     R1 : &nbs ... t; {1 ^(n + 1) 2 ^n} | {1 ^n 2 ^(n + 1)}     <br />

       typeset structure

Mersennian world

For the Mersennian model, the successor operation is still in the spirit of the Leibnizian successor but modified by the specific Mersennian features of non-commutativity.

             Numeric Mersenne rules <br ...  {2 ^n 1 ^n 2 ^1}     <br />      

             typeset structure

Stirlingian world

For the Stirlingian model, the successor operation is not anymore in the spirit of the Leibnizian successor.

The Stirlingian successor operation is defined by the feature of “retro-grade” recursivity.

Hence, the retrograde recursion gets a meta-rule which controls the successor-procedure in respect of the structure of the added morphogram.

                Kalu Overscript[z, ... bsp;        Retrograde recursion example of a Stirling addition

             <br />    & ... ;  ==>   collect (stop ) . <br />   {[a], [aa], [ab], [aaa], [aab], [aba], [abb], [abc]}

1.2.  Palindromes all over the worlds

1.2.1.  Leibniz’ palindromes

The rules for building classical palindromes are easy to understand. If we add to a given element the same additional element on the right and on the left side, we get a palindrome:

For an alphabet ∑ = {•, } we get:
==> • • •,
==> • • •, •.

Leibniz palindromes are standard and well studied.

Instead of demonstrating many examples, the production rules are defining how to produce palindromes. They are given with this little grammar, consiting on an alphabet and a set of production rules.    

   typeset structure

Odd and even palindromes

Odd palindrome

With rule 3 we introduce a start token, say typeset structure. Now S is typeset structure, and typeset structure is palindrome.
Apply rule1 to S: S --> typeset structure S typeset structure. Now S is typeset structure typeset structure typeset structure, is palindrome,
Apply rule2 to S: S --> typeset structure(typeset structure S typeset structure) typeset structure. Now S is typeset structuretypeset structure typeset structure typeset structuretypeset structure, is palindrome.
Apply rule1 to S: S  --> typeset structure S typeset structure. Now S is   typeset structure typeset structuretypeset structure typeset structure typeset structuretypeset structure typeset structure, is palindrome.
And so on.

The order of the application of the rules rule1 and rule2 is free. The result is always symmetric, and therefore a palindrome. There are no surprises included in this parcel.

Even palindrome

A more interesting example is given with ( O O).
The alphabet is: typeset structure and a new
rule4: S --> S ▲,
rule5: S --> O S O.

With rule 3 we introduce a start with the empty token typeset structure. Now S is typeset structure, typeset structure is a nil- palindrome.
rule4: S --> ▲ S ,
rule1: S --> typeset structure ( S ) typeset structure,
rule2:  typeset structure S typeset structure --> (typeset structure S typeset structure ) • ,
rule5: (typeset structure S typeset structure ) --> O ( (typeset structure S typeset structure ) • ) O.

With rule3 we replace S by typeset structure: thus we get the palindrome: O O.
The same holds here. Free application of the rules, and no surprise in the box.

A simple palindromy checker for Leibniz palindromes

The little palindrome production rules are producing palindromes. The same rules applied  backwards to an arbitray string lets easily decide if the string is a palindrome or not.

This is easily realized with two palyers, the head (H) and the tail (T) manager, deciding the equality or non-equality of their states in respect to the positions.

     typeset structure

1.2.2.  George Spencer Brown’s palindromes

Partition based palindromes

Brownian constellations are order-free, i.e. their elements are commutative, and are allowed to change position. Hence, Brownian palindromes are free under permutation.
This commutativity poses a problem for a proper definition of Brownian palindromes.
Brownian constellations are mathematically heap, i.e. multi-sets, and not linear sequences.

Therefore, can a multi-set be palindromic?

A solution is given with the application of the Brownian standard normal form, bnf, of constellations.

rev(typeset structure) = (typeset structure) and (typeset structure) = (typeset structure).

The constellation (typeset structure) is a palindrome because of its commutatiity.

Is the accepted Brownian constellation [ ] a palindrome?

Because of the standard normal form convention for Brownian constellations, bnf, we know that [ ] =Brown [ ]. But, the pattern [ ] is palindromic.

That is, the standard form constellation [ ] represents the set of equivalent constellations
{[ ], [ ]}.

typeset structure

Brownian palindromes

PalBrown (4) = {(aa), (bb); (aaa), (bbb); (aaaa), (bbbb)}.

1.2.3.  Mersenne’s palindromes

Mersenne situations are ordered sequences. They can be read forwards and backwards, hence they might be palindromic. Mersenne’s palindromes are similar to Leibniz palindromes with the crucial difference that homogeneous sequences are equal, i.e. = typeset structure typeset structure.
rev(typeset structure) = (typeset structure) and (typeset structure) !=  (typeset structure).

typeset structure

Mersennian palindromes

PalMers (4,2) = {(aa); (aaa), (aba), (bab); (aaaa), (abba), (baab)}

1.2.4.  Stirling’s palindromes

Stirling palindromes had been well studied recently.

           typeset structure

Production examples for even palindromes   

P: w1!=w2: [w1=typeset structure, w2=typeset structure]: P = [typeset structure,typeset structure]
P: w1=w2: [w1=typeset structure, w2=typeset structure]: P = [typeset structure,typeset structure].

                                   PalStirling(4):
                    rules        results           
P = [typeset structure,typeset structure]:   w1Pw2  : [typeset structure,typeset structure,typeset structure,typeset structure]  ; rule1(=rule2)
                   w3Pw3  : [typeset structure,typeset structure,typeset structure,typeset structure]   ; rule3
                   w3Pw4  : [typeset structure,typeset structure,typeset structure,typeset structure]   ; rule4
P = [typeset structure,typeset structure]:   w1Pw2  : [typeset structure,typeset structure,typeset structure,typeset structure]   ; rule1
                   w2Pw1  : [typeset structure,typeset structure,typeset structure,typeset structure]   ; rule2
                   w3Pw3  : [typeset structure,typeset structure,typeset structure,typeset structure]   ; rule3
                   w3Pw4  : [typeset structure,typeset structure,typeset structure,typeset structure]   ; rule4

Quite obviously, a pattern like [typeset structuretypeset structuretypeset structuretypeset structure] doesn’t read forwards and backwards the same in a Leibniz world. But read as a deep-structural pattern of differences it does. Hence, the pattern is a Stirlingian palindrome.

Test
The difference-structure of [typeset structure typeset structure typeset structuretypeset structure] is:

typeset structure

This matrix is obviously symmetric. Thus, it represents a palindrome.
The same holds for the next example [typeset structure typeset structure typeset structure typeset structure]:

typeset structure .

Test
- ENstructureEN[1,2,3,1];
val it = [[],[N],[N,N],[E,N,N]] : EN list list
- ENstructureEN[1,3,2,1];
val it = [[],[N],[N,N],[E,N,N]] : EN list list

Matrix comparison

[1,2,3,1]   [1,3,2,1]
typeset structure   typeset structure

Therefore, the pattern (typeset structure) is palindromic.

Bisymmetric examples

Bisymmetric examples
Is the pattern  [O O] a palindrome?

Reversion method

The naive method deals with the pattern as they are perceived and not with the differences that are not perceived but recognized by analysis. Hence the inversion of the pattern  [O O] is the pattern [O O • ]. Both are differential symmetric and the matrix of the differences are equal. Hence the patterns are palindromic.

ENstructureEN([O O]) = ENstructureEN(rev([O O])).

But with this approach we are not dealing with the differences as our primary objects but with the patterns with their arbitrary elements.

(a) =[O O]    rev(a) = [O O • ]
typeset structure           typeset structure

Bisymmetry method

Morphograms are not dealing with the identity of their elements, but with the pattern defined by the differences between the elements only. Therefore we have to apply a different method. This method is focusing on the differences they are notified with the matrix only.

The method is called bilateral symmetry, in short: bisymmetry.

ENstructureEN([O O]) =bisym rev(ENstructureEN([O O])).


typeset structure

Examples for Stirling palindromes

Palindromes pal(7,7):
   [1,1,1,1,1,1,1],[1,1,1,2,1,1,1],[1,1,1,2,3,3,3],[1,1,2,1,2,1,1],
   [1,1,2,1,3,1,1],[1,1,2,2,2,1,1],[1,1,2,2,2,3,3],[1,1,2,3,1,2,2],
   [1,1,2,3,2,1,1],[1,1,2,3,2,4,4],[1,1,2,3,4,1,1],[1,1,2,3,4,5,5],
   [1,2,1,1,1,2,1],[1,2,1,1,1,3,1],[1,2,1,2,1,2,1],[1,2,1,2,3,2,3],
   [1,2,1,3,1,2,1],[1,2,1,3,1,4,1],[1,2,1,3,2,1,2],[1,2,1,3,4,2,4],
   [1,2,1,3,4,5,4],[1,2,2,1,2,2,1],[1,2,2,1,3,3,1],[1,2,2,2,2,2,1],
   [1,2,2,2,2,2,3],[1,2,2,3,1,1,2],[1,2,2,3,2,2,1],[1,2,2,3,2,2,4],
   [1,2,2,3,4,4,1],[1,2,2,3,4,4,5],[1,2,3,1,2,3,1],[1,2,3,1,3,2,1],
   [1,2,3,1,3,4,1],[1,2,3,1,4,2,1],[1,2,3,1,4,5,1],[1,2,3,2,1,2,3],
   [1,2,3,2,3,2,1],[1,2,3,2,3,2,4],[1,2,3,2,4,2,1],[1,2,3,2,4,2,5],
   [1,2,3,3,3,1,2],[1,2,3,3,3,2,1],[1,2,3,3,3,2,4],[1,2,3,3,3,4,1],
   [1,2,3,3,3,4,5],[1,2,3,4,1,2,3],[1,2,3,4,1,5,3],[1,2,3,4,2,3,1],
   [1,2,3,4,2,3,5],[1,2,3,4,3,1,2],[1,2,3,4,3,2,1],[1,2,3,4,3,2,5],
   [1,2,3,4,3,5,1],[1,2,3,4,3,5,6],[1,2,3,4,5,1,2],[1,2,3,4,5,2,1],
   [1,2,3,4,5,2,6],[1,2,3,4,5,6,1],[1,2,3,4,5,6,7].

1.3.  Logics in different worlds

1.3.1.  Logic in a Brownian world

typeset structure

Classical interpretation:
Log(CI) = {true ≡typeset structure}

"Here is how we shall model elementary logic using Laws of Form. We shall take the marked state for the value T (true) and the unmarked state for the value F(false). We take NOT as the operation of enclosure by the mark.” (l. Kauffman)

Log(CI) = {true ≡typeset structure}

      typeset structure

Graphematical interpretation

The graphematical interpretation of the Brownian calculus is emphasizing the crucial fact of its intrinsic commutativity.

      typeset structuretypeset structure       

    Iterative indicational graph   <br />       & ... f              fff <br />   

    Accretive indicational graph    <br />     &n ... nbsp;         (abc) <br />       

Semantics of the indicational domain
val({aa, ab, bb}) = {tt, tf, ff}

val(aa) = (tt)
val(ab) = (tf)
val(bb) = (ff).

Negation
non(tt) = ff
non(tf) = tf, because (ab) =
typeset structure(ba)

typeset structure

Statement: Negation in Brown is inversion (negation).

Hence, the conjunction of a true Brownian statement and its negation is a contradictional statement.

Numerical truth-values
num(tt) = (1)
num(tf) = (2)
num(ff) = (3).

non(1, 2, 3) = (3, 2, 1)

Conjunction
(tt) (tt) --> (tt)
(tt) (tf) --> (tf)
(tt) (ff) --> (ff)

(tf) (tt) --> (tf)
(tf) (tf) --> (tf)
(tf) (ff) --> (ff)

(ff) (tt) --> (ff)
(ff) (tf) --> (ff)
(ff) (ff) --> (ff)

Truth-tables:
typeset structure

1.3.2.  Logic in a Mersennian world

The distinctional domain of the Mersenne calculus is trichotomic too:

Mers = {tt, tf, ft}.

typeset structure

Negation

  typeset structure

Statement: Negation in Mers is permutation.

Hence, the conjunction of a true Mersenne statement and its negation remains a true statement.

Conjunctiom

num({tt, tf, ft}) = (1,2,3)

non(1,2,3) = (1, 3, 2)

typeset structure    typeset structure

1.3.3.  Comparisons

typeset structure
typeset structure

red:    Brown ∩ Mersenne
blue:   Mersenne \ {red, green}
green: Brown \ {red, blue}
Semiotics: Brown ∪ Mersenne

Comparision of truth-tables

typeset structure        typeset structure

standardized:

typeset structure        typeset structure

Proof-theory and tableaux calculi

For Mersenne calculi a branch of the tableau terminates if the same formulas contains the signatures (tf) and (ft).
For Brown calculi a branch of the tableau terminates if the same formula contains the signatures (tt) and (ff).

1.4.  Cellular automata in different worlds

1.4.1.  CAs in Brownian worlds

A further understanding of the indicational calculus is offered by the study of its dynamism, sketched as cellular automata.

typeset structure

typeset structure

Example
indCA, r = typeset structure
typeset structure

1.4.2.  CAs in Mersenne worlds



1.4.3.  CAs in Stirling worlds

typeset structure

Homogeneous kenoCAtypeset structure applications

typeset structure


typeset structure

Heterogeneous kenoCAtypeset structure compositions

typeset structure

Formal interpretation

typeset structure

2.  How are the 4 games inter-related?

2.1.  Systematic Diagrams

               Systems   ...                                                                                              a   b

       typeset structure   

      Connection table in normal form for sys(2,3)

      typeset structure

      Connection table in normal form for sys(2,2)

      typeset structure

2.2.  Systematic graphs


            typeset structure

 Alphabet :                ... bsp;     •   •        <br />  <br />

             <br />    & ... p;          Leibniz     <br /> <br />

2.3.  Polyfunctorial mediations

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