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 S*oft 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)

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.

The Leibniz game is defined by some strict axioms.

**Classical rules**

**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:

*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.

**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**

**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 • • ▲ ▲).

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**

**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.

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).

(sequence A000041 in OEIS).

**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:

*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 ), () and ().

**The basic rules for the Brownian distinction calculus**

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. = ⌀

**Especially**:

(( )) ( ) = ( ) (( )) : • = • .

Red in red kills red, =⌀ and red • saves red •

equal

red • with red in red kills red =⌀ and saves red •.

Hence, • = • = •.

**Superpositions**

(() (((()))) =

(() ()) =

(()) =

(()) = ⌀

**In words:**

: Red with red in red in red kills red and red saves red: .

: Red with red in red saves red : .

: 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

*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:

**Normal form**

The written table is accepting the normal form for Brownian constellations. Hence, the two constellations , are represented by the single constellation 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.

**Successor**

Alphabet ∑ = {•, •}

succ(•) = {(• •), (• •), (• • ) }

**Addition Sum**

sum(•, ⌀) = •

sum(• •, •) = {• • •, • • •}.

sum(• •, •) = {• • •}

sum(• •, •) = {• • •}.

**Reversion for Brownian patterns**

rev(• •) = (• •) and (• •) = (• •).

**The basic rules of the calculus of differentiations**

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. = •

**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. = • this also holds for Rule2

Rule2a ==> •

Rule2b. <== •

**Some examples**

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

• • • =

• • • = •

= •

Thus, • • • = .

2. ( ) ( ) () = () : rule1 : • • • = •

(( )) ( ) = ⌀ : rule2, rule1 : • = ⌀

((())) = () : rule2 : = •

**Especially3. **(( )) = ( ) () () : = • • •

Proof of = • • •

[• •] • : brackets

[⌀] • : rule1

• : rule1.

• : 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

*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.

**Successor**

Alphabet ∑ = {•, •}

succ() = {(), (), ()}.

**Addition Sum**

sum(, ⌀) =

sum( , ) = { , , }

sum(, ) = { , }

sum(, ) = {, }.

**Reversion for Mersenne**

rev(ab) = (ba) and (ab) != (ba)

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

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**:

= • • •

= •

• • • = •

Hence, = • • •.

!= • • •

= ⌀

• • • = •

Thus, ⌀ != •.

**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.

Rule2. =

Rule3. =

Rule4. != != != .

**Another setting:**

Rule1.

Rule2. =

Rule3. =

Rule4. != != != .

**Wordings of constellations**

For a Stirlingian game with 3 elements, some typical situations occur.

1. • • • ≡ ≡ ,

≡ ≡

etcetera

2. ≡ rev() : reversion

3. ≡ rev( ) : self-symmetry

≡ rev()|

≡ rev()

**Wordings of rules for Stirling(3,3)**

: Blue kills red.

≡ : Blue together with red kills red together with blue.

: Two blue together with one red, and

: one blue together with one red and one blue, and

: one blue with two reds, are safe in the Stirlingian world.

: 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*

[1^{3}]: • • • : aaa

[1^{2}2: • • • : aab

[1^{1}2^{1}1^{1}]: • • • : aba

[1^{1}2^{2}]: • • • : abb

[1^{1}2^{1}3^{1}]: • • 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

**Symmetry**

Here, again, the symmetry of the set of the basic patterns is broken.

But there are some nice internal symmetries left.

rev() = (), that is rev() = () but () = ().

Self-symmetric patterns: (), (), ().

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].

**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 [].

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

• • • • • • • • • • • • • • O

**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

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

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 |.

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.

**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.

**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.

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.

**Odd and even palindromes**

**Odd palindrome**

With rule 3 we introduce a start token, say . Now S is , and is palindrome.

Apply rule1 to S: S --> S . Now S is , is palindrome,

Apply rule2 to S: S --> ( S ) . Now S is , is palindrome.

Apply rule1 to S: S --> S . Now S is , 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: and a new

rule4: S --> ▲ S ▲,

rule5: S --> O S O.

With rule 3 we introduce a start with the empty token . Now S is , is a nil- palindrome.

rule4: S --> ▲ S ▲ ,

rule1: ▲ S ▲ --> (▲ S ▲) ,

rule2: ▲ S ▲ --> • ( ▲ S ▲ ) • ,

rule5: • ( ▲ S ▲ ) • --> O (• ( ▲ S ▲ ) • ) O.

With rule3 we replace S by : 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.

**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() = () and () = ().

The constellation () 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

{[• • •], [• • •]}.

**Brownian palindromes**

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

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. • • = .

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

**Mersennian palindromes**

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

Stirling palindromes had been well studied recently.

**Production examples for even palindromes **

P: w1!=w2: [w1=, w2=]: P = [,]

P: w1=w2: [w1=, w2=]: P = [,].

Pal_{Stirling}(4):

**rules results ** ** **

P = [,]: w1Pw2 : [,**,**,] ; rule1(=rule2)

w3Pw3 : [,,,] ; rule3

w3Pw4 : [,**,**,] ; rule4

P = [,]: w1Pw2 : [,**,**,] ; rule1

w2Pw1 : [,**,**,] ; rule2

w3Pw3 : [,,,] ; rule3

w3Pw4 : [,**,**,] ; rule4

Quite obviously, a pattern like [] 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 [ ** **] is:

This matrix is obviously symmetric. Thus, it represents a palindrome.

The same holds for the next example [ ]:

.

**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]

Therefore, the pattern () is palindromic.

**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 • • • ]

**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])).

**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**].

**Classical interpretation**:

Log(CI) = {true ≡}

*"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 ≡}

**Graphematical interpretation**

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

**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) =(ba)

*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:**

The *distinctional* domain of the Mersenne calculus is trichotomic too:

*Mers* = {tt, tf, ft}.

**Negation**

*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)

red: Brown ∩ Mersenne

blue: Mersenne \ {red, green}

green: Brown \ {red, blue}

Semiotics: Brown ∪ Mersenne

**Comparision of truth-tables**

standardized:

**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).

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

**Example**

indCA, r =

**Homogeneous kenoCA**** applications**

**Heterogeneous kenoCA**** compositions**

**Formal interpretation**

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

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