Supposed I want to read some passages of a text to an audience, but only the important ones. For that it's necessary to

For simplification I suppose that on a given piece of paper (represented by a frame which corresponds to the "unwritten cross") are written exactly two paragraphs, of which the first contains all important and the second all unimportant sentences.

Furthermore for an easy representation the word "IMPORTANT" shall represent the important, the word "UNIMPORTANT" the unimportant paragraph.

Thereby there are three possibilities to distinguish these paragraphs from another :

1. I mark the important paragraph by one and the unimportant by another way, for example :

IMPORTANT | |

IMPORTANT | UNIMPORTANT |

IMPORTANT |

3.a :

IMPORTANT | UNIMPORTANT |

IMPORTANT UNIMPORTANT | |

Furthermore will be considered what happens if I mark twice. There are two possibilities doing this :

1. Marking the same paragraph twice, for example :

PARAGRAPH 1 || PARAGRAPH 2 |

PARAGRAPH 1 | PARAGRAPH 2 | |

Back to the marks :

Now I will consider if marks are equal

Doing this I only consider

Doing this I realize :

If I mark the same paragraph twice I do not distinguish them more clear than marking once. Therefore :

PARAGRAPH 1 || PARAGRAPH 2 |

PARAGRAPH 1 | PARAGRAPH 2 |

Furthermore:

If I mark both paragraphs the difference between them vanishes,therefore :

PARAGRAPH 1 | PARAGRAPH 2 | |

PARAGRAPH 1 PARAGRAPH 2 |

| | |
= |

And these are (nearly) the initial equations of the primary arithmetic of the laws of form.

They arise by keeping the marks and leaving out the text. In the end it's not essential what it is I mark, texts or something else, only the marks matter.

But George Spencer Brown uses other kinds of marks, closed curves, for example circles, which divide the plane (the piece of paper) into exactly two totally severed parts, the inner and the outer side, corresponding to the distinction of two completely different states. But doing this another complication arises. Because of this I first considered the easyer case of marking by strokes. For example the vertical stroke is a symetrical mark. It clearly divides two sides from another but they are not different in any other way. It doesn't matter if the marked text stands left or right beside the stroke. Not so in case of a circle ! It's an asymetric cleavage of the space and it's not the same if a value is corresponded to the inside or the outside. In case of a stroke I only have to decide which value it should mark, in case of a circle I have further to decide if it should mark the inner or the outer side. This is not a matter of interpretation but of the calculus itself. The following consideration shall show that it makes a difference :

I choose the brackets ( ) to represent the mark, even they have an inner and an outer side.

1.The inside shall be marked, thereby :

(PARAGRAPH 1) PARAGRAPH 2 |

1.a :

((PARAGRAPH 1)) PARAGRAPH 2 |

1.b :

(PARAGRAPH 1) (PARAGRAPH 2) |

Different from the strokes theres no need to pay attention if the marks are standing side by side or one upon another. (( )) and ( )( ) cannot be confused. Using asymetric marks the notation becomes topological invariant.

2. Now the mark shall indicate its outside:

PARAGRAPH 1 (PARAGRAPH 2) |

2.a :

PARAGRAPH 1 ( ) (PARAGRAPH 2) |

Therefore :

I1. ( )( ) = ( ).

Now I mark both paragraphs :

2.b :

PARAGRAPH 1 (( )PARAGRAPH 2) |

I2. (( ))=

Because the first mark (counted from the outside to the inside) completely separates its inner space from its outer space,paragraph 2 can only be indicated by a second mark inside the first mark.

The comparison shows that only case 2 (the mark indicates its outside) leads to equations similar to the equations given by using strokes as marks. But even the commitment 1 is possible (Felix Lau "Die Form der Paradoxie" Kapitel 1:"Der Indikationenkalkül"), even this is consistent to the intention of marking, but would lead to a different calculus.

Another reason (additional to the intention to reach correspondence to the use of strokes) for choosing case 2 is given by considering the calculus in reference to logic.

A1. TT = T A2. FT = T A3. TF = T A4. FF = F A5. T = (F) A6. F = (T) |
The constants represent the values of propositions. Thereby for example TF means that of two with "or" connected propositiones the first is true and the second is false. Then TF =T means, that the proposition composed out of two single propositiones is true. |

The fact that it is not usual to do so, doesn't mean that it is impossible to do so. It is possible and that means that it is possible to make calculations by using the equations A1 up to A6 and thereby in fact we have an arithmetic. Now, using variables, I can propound equations, which are propositions about this arithmetic. For example the equations A1 up to A4 show, that a composited expression always has the value T if it contains at least once the value T, that means : aT = T (stricly spoke it has to be proved that this is true in general for any long expressions represented by a, but in this quite nonsystematical introduction this shall be omitted).

Furthermore :

aF = F,if a = F;

aF = T,if a = T.

As a whole aF has alway the same value as a itself, therefore : aF = a.

Using A6 by substitution of F by (T) equation A5 T = (F) gives : T = ((T)). In the same way I get F =((F)).

Thereby : a = ((a)).

Untill we get the following algebraic equations :

A7. aT = T

A8. aF = a

These equations contains even constant as variable parts and connect the algebra and the arithmetic.

A9. a = ((a))

this equation is completely algebraical.

but the equations A7 up to A9 are **no consequences** of the arithmetic, because they

If two expressions are equal in both values of any variable, they are equal. This is theorem 16 in the*Laws of Form*.But it's not only true in the LoF. To realize this I tabulate the values of two expressions,
for example :

a b ((a)((a)b)) ((a) (b)) T T T T T F F F F T F F F F F FNow I consider any variable, for example a. If both expressions are equal in every row, in which a has the value T, they are equal in half of all rows. If they are even equal in every row with a = F, they are equal in the other half of all rows. Thereby the whole table is covered ! Thereby both expressions are equal in every case and there is no need to consider the values of all other variables and thereby there is no need to tabulate the values every time.

George Spencer Brown proves this theorem especially in reference to his calculus and therefore completely different.

Example :

( (a)( (a)b ) ) = ( (a)(b) )

1. a = TBoth expressions are equal in both values of a, therefore they are equal in general. Especially case 1 shows, that there is no need to know, what's the value of b. Both expressions equal b, therefore they are equal, no matter what's the value of b. The same example shall be considered again in section 2.b .

( (a)( (a)b ) ) = ( (T)( (T) b) ) ( (a)(b) ) = ( (T)(b) ) A6 = ( F ( F b) ) A6 = ( F (b) ) A8 = ( F ( b) ) A8 = ( (b) ) A8 = ( ( b) ) A9 = b A9 = b Therefore in this case ( (a)( (a)b ) ) = ( (a)(b) ) 2. a = F ( (a)( (a)b ) ) = ( (F)( (F)b ) ) ( (a)(b) ) = ( (F)(b) ) A5 = ( T ( T b ) ) A5 = ( T (b) ) A7 = ( T ( T ) ) A7 = ( T ) A6 = ( T F ) A6 = F A3 = ( T ) A6 = F

Untill all this has nothing to do with the

I ask : how do I have to manipulate the left side of this equation to get the right side, with other words : by what I have to substitute F on the left side to get the right side ?

Answer : by nothing! In other words : if I assume F =

then even aF = a. Thereby I have reduced aF = a to the more fundamental equation F =

Now I can assume this new equation and see what happens, if the calculus works in this way.

First : T = (F) , F =

Therefore : T = ( ). Thereby the constant T get substituted by the same sign representing the negator. Thereby the difference vanishes between the object on which the operator operates and the operator itself. Thereby the new calculus gets even notional more fundamental than the previous one. It contains only one constant (instead of two of them previously), which is the same as the operator and if I undo the substitution of F by

With W = ( ) and F =

the arithmetical equations A1 up to A6 become :

1'. ( )( ) = ( ) 2'. ( ) = ( ) 3'. ( ) = ( ) 4'. = 5'. ( ) = ( ) 6'. = (( ))Equations A2' up to A5' are trivial and only say, that every constant equals itself. As essential remain A1'and A6'. These are exactly the same equations as I1 and I2, which raised in section 1 "Marks" by choosing the outside as indicated space. This is the announced second reason for this choose.

Using these new arithmetical equations for initials it's possible to derive longer expressions from shorter ones or to simplify composited expressions in the same way as before and thereby to build up the

( ) = ( )( ) = ( ) ( (( )) ) = ( ) ( (( )( )) ) etc.

Belonging to this arithmetic it's possible to build up the primary algebra, suitable for calculations like in section 2.a.Now the same example shall be considered again :

a | b | ((a)((a)b)) | ((a) (b)) |
---|---|---|---|

( ) | ( ) | ( ) | ( ) |

( ) | |||

( ) | |||

1. : a = ( ) ( (a)( (a)b ) ) = ( (( )) ( (( ))b ) ) ((a)(b)) = ((())(b)) 2-mal I2 = ( ( b ) ) I2 = ( (b)) I2 = b I2 = b 2. : a = ( (a)( (a)b ) ) = ( ( )( ( )b ) ) ((a)(b)) = (( )(b)) T2 = ( ( ) ) T2 = (( ) ) I2 = I2 =Theorem T2 (so called in LoF) arises with aW = W and W = ( ) as a( ) = ( ). Even in the new algebra both expressions are equal. Obviously in the primary algebra with its new constants calculations are much easyer and shorter as before.

On both ways, from the beginning and from the end, arise the same equations I1 and I2.
Thereby bot ways to the *Laws of Form* has come together and this first introduction has come to its end.

Final words :

This reconstruction should show how *Laws of Form* could have been discovered if they weren't discovered at all and should make it more plausible why they are how they are.
How George Spencer Brown himself truely discovered them, I don't know. Even that my reconstruction is not exact is obvious. Who want's to have a mathematical exact representation shall read *Laws of Form* itself.

Questions, suggestions, anything else ?