There is no Redundancy in the Set of Initial Equations
On this page, we prove that there is no redundancy in the set of three initial equations which generates the calculus of circles and letters. That is to say, no one of the three equations can be arrived at as a consequence of the other two. We consider the various possibilities in turn.
I. Could 'wrap' be a consequence of 'delete' and 'copy'?
Kauffman[1] has provided a neat way of settling this question. Consider the following variation on the arithmetic of the mark (and let's call this variation Arith1). Besides the circle (with its capacity for multiplying and nesting) let there be a second mark, θ (the Greek letter theta). Expressions are made out of O and θ; for example
The following two rules will enable us to evaluate any such expression, by working up from the deepest space(s) (as in Valerie's Mark-up Method).
Rule 1: Value of a composite expression
Suppose E is a product of prime factors
E = ABCD..
whose values are already known; and that in each case the value is _ , O or θ. Note that each prime factor is either a circle surrounding another expression F, or is the new mark, θ. Then (1) if any of the factors has value O, the value of E is O; (2) otherwise, if any of the factors has value θ, the value of E is θ; (3) otherwise (in case all the factors have value _ ), the value of E is _ .
Thus θ is intermediate between the dominant O and the recessive _ .
Rule 2: Value of a circle containing another expression
Suppose F is contained in a circle, and the value of F is known (and is _ , θ or O). Then the value of (F) is _ if the value of F is O ; and O in both other cases.
The evaluation procedure may be regarded as determined by the following arithmetical rules:
O O = O
(O) = _
θ O = O
θ θ = θ
(θ) = O
The rules enable any expression in Arith1 to be reduced to one of the three values _ , θ and O.
As usual, we may write an equals sign between two expressions if (and only if) they reduce to the same value.
We may now verify that
I2: A O = O
and
I3: A(B) = A(AB)
are true as general rules in Arith1. The first says that the value of any expression of the form A O is O, which is an easy consequence of the 'dominance' of O; the second we must verify by considering the different possible cases:
(i) A = _ ; I3 reduces to (B)=(B) which is obviously true.
(ii) A = θ. Now B differs in value from θ B only in case B = _ ; but even in this case (B) = O has the same value as (θB) = (θ) = O.
(iii) A = O. Both sides reduce to O and again the equation is satisfied.
Thus, in Arith1,
A O ↔ O
A(B) ↔ A(AB)
are value-preserving moves; as is any embedding or combination of these moves.
Now, change focus back to the CL-calculus. Suppose that there was some combination of the (un)delete and (un)copy moves which turned the expression
a
into the expression
((a)).
Then one could perform the very same sequence of moves in Arith1, but starting with the expression
θ,
and turning it into
((θ)).
(Note: if one or more of the moves in the CL-calculus is undelete, then such a move introduces a new expression into the space that already contains an empty circle; and the new expression may well contain new letters b, c, d and so on. So there is a question, how to translate such a move into a move in Arith1. However, the new letters must disappear again by the end of the sequence of moves, since the end-product of the moves is just ((a)); therefore, any of the three values may be substituted for each of the new letters. It will make no difference in the end.)
But the value of ((θ)) is _ , not θ: this is absurd, because each individual move preserves value. Therefore, our supposition must be false: there can be no combination of the moves delete and copy which turns a into ((a)) in the CL-calculus.
The same argument, rephrased
Suppose we introduce a valuation on all algebraic expressions. We calculate the value of an expression E as follows: replace each letter in E by the symbol θ; then evaluate the expression according to the procedures of Arith1. Then, delete and copy are value-preserving moves, but wrap is not. Therefore no combination of delete and copy can result in wrap.
II. 'Delete' cannot be a consequence of 'wrap' and 'copy'
For this we invent a new three-valued arithmetic called Arith2. This time we use the symbol φ (Greek letter phi) for the third value, and for the rules of the arithmetic we take the following:
O O = O
(O) = O
(φ) = φ
φ O = φ
φ φ = φ .
As ((φ)) = (φ) = φ, the wrap equation is satisfied. delete is clearly not satisfied, as φ O≠ O. As for copy, we need to check cases:
(i) Let A = _ ; then A(B) = (B) = A(AB).
(ii) Let A = φ; then A(B) = φ, whatever B is, and the same goes for A(AB).
(iii) Let A = O; then certainly O(B) = O (OB) when B = _ or O. But when B = φ then B = OB; and again the equation is satisfied.
Thus copy is valid in Arith2.
We may thus create a new valuation on all algebraic expressions, by replacing all letters with the symbol φ and reducing the new expression using the rules of Arith2. Then wrap and copy are value-preserving, but delete is not. Therefore no combination of wrap and copy can result in delete.
III. 'Copy' cannot be a consequence of 'wrap' and 'delete'
For this we employ yet another arithmetic, Arith3, whose third value χ (Greek letter chi) obeys the following rules:
O O = O
(O) = O
(χ) = χ
χ O = O
χ χ = χ .
It is easy to see that wrap and delete are satisfied; but copy fails in the following case: A = χ, B = _ . For then
A(B) = χ( ) = O
but
A(AB) = χ(χ) = χ χ = χ.
Now, if copy were a consequence of wrap and delete in the CL-calculus, then so would be the general equation
A(A) = O
(obtained from copy by substituting B = _ ; then applying delete). Now set up a valuation on algebraic expressions in the usual way. Then wrap and delete are value-preserving, but
a(a) → O
is not; this last cannot therefore arise as a combination of wrap and delete.
[1] Louis H. Kauffman, Laws of Form - an Exploration, pages 31-32, see also page 35. Kauffman proves a slightly more difficult proposition than ours, namely, that wrap is not a consequence of copy and complements.
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