### An Application of the CL-Calculus (Set Theory)

In our development of the circle-and-letter calulus (CL-calculus) we have taken care never to make any assumptions about what the mysterious letters, occurring in CL expressions, actually stand for, if anything. Thanks to the canon of substitution, it is true, there is a sense in which any letter may 'stand for' any CL expression; but since most such expressions themselves contain more letters, this is not much of an explanation of what the letters 'really' stand for. It would seem, however, that were the letters to stand for something, then any CL expression ought to stand for the same sort of thing.

Now, there is of course one interpretation or application of the calculus that we already know about. In the course of our story, the calculus was as it were made in the image of the Algebra of the Mark; that is, the system of General Rules discovered during our investigation of the Arithmetic of the Mark (see Algernon the Algebraist). The CL-calculus was 'taken out' of the arithmetic (to borrow Spencer Brown's phrase). Going backwards from the calculus, if we say that a letter, such as x, may stand for any arithmetic expression, we arrive at the interpretation of the calculus as the algebra of the mark, that is, as a system of General Rules governing the mark's arithmetic.

One outcome of the last section (Consequence No.4 of the Standard Form Theorem) was that the calculus, when interpreted in this way, furnishes a complete description of all the valid general rules of the arithmetic.

However, other interpretations of the calculus are possible. For example, a possible new interpretation arises as follows. Let U (for 'universal') be a set. Suppose that any letter, such as x, stands for a subset of U. In view of our remarks above, our first task must now be to show how any CL-expression may be interpreted as a subset of U (given that each of the letters occurring in the expression stands for such a subset). Then we must verify that the axioms of the CL-calculus (i.e. the three initial equations) are consistent with the interpretation.

Let us work our way towards the interpretation of an arbitrary CL-expression in easy stages. First of all, if a stands for a subset of U, b for another, we interpret the expression ab as the intersection of a with b. Secondly, (a), or a encircled, is to be interpreted as the complement of a in U. That is to say, the elements of (a) are all the elements of U which are not in a.

How is the empty expression to be interpreted? Since

a _ = a

for all a, the intersection of a with _ - or rather, with our interpretation of _ , whatever that may turn out to be - is always a. The only subset of U which satisfies this condition is U itself: therefore we are forced to interpret _ as U. The empty circle, O must then be the complement of U, that is, all the elements of U that are not in U; in other words, none of them. Therefore O is the subset with no members, the empty set. It's time to check the axioms:

1. ((a)) = a    (the complement of the complement of a is a itself)
2. a O = O    (the intersection of a with the empty set is the empty set)
3. a(b) = a(ab)

The first two are easy to verify; the third is more obscure, but easy to verify with the help of a Venn diagram. a(b) is the intersection of a with not-b; in other words, it is a less the intersection of a with b. But the interpretation of the right-hand side is just that: the intersection of a with the complement of a-intersect-b. Thus the three axioms are consistent with the interpretation, as are the canons of embedding, substitution and concatenation. Therefore, any equivalence of expressions that is provable in the CL-calculus may be interpreted as equality of the corresponding subsets.

Exercise for the reader: Check that the expression

((a)(b))

represents the union of the sets a and b.

#### The subset relation

To assert a(b) = O is to assert that the intersection of a with not-b is empty; in other words, there is no element of a that is not also an element of b; or, a is a subset of b.

The mutual dominance lemma tells us that if a is a subset of b, and b is a subset of a, then a equals b. We may also prove the following: if a is a subset of b, and b is a subset of c, than a is a subset of c. For,

a(c) = a(a(b))(c) = a((b))(c) = ab(c) = aO = O

#### Single-element subsets

Suppose that σ is an element of U, and that s is the subset whose only element is σ. Let a be a subset of U, and consider the expression sa. If σ happens to belong to a, the product will be s; otherwise it will be the empty set O.

Now let E be a CL expression, involving various letters a, b, c... ; and suppose that we want to know whether or not σ belongs to the set described by E. To this end we consider the product sE. Now, by the copy axiom, s may be copied into the expression, so that each letter (for example a) is replaced by the product of that letter with s (for example sa). Now sa will be either s or O, depending on whether or not σ belongs to a. In the first case, s may be uncopied, so that in effect a will have been replaced by _ . Thus a may be replaced by _ if σ belongs to a, O if it does not. The same applies to all the other letters in the expression.

Thus, in the product sE, E may be replaced by an arithmetic expression, obtained by substituting a primitive value for each letter occurring in E, namely, _ if σ belongs to the subset in question (i.e. the subset represented by the letter); O, if it does not.

The arithmetic expression may then be reduced to a primitive value. If the value is _ , then sE reduces to s; if O then sE reduces to O. In the first case, σ does indeed belong to the subset represented by E; in the second case, not.

We see that the reduction of the arithmetic expression corresponds to a sort of logical argument; and we begin to see that the interpretation of the CL-calculus in terms of sets brings with it, as a special case, an interpretation in terms of Logic. Of course, it has been evident from the very beginning of our story that the calculus of the mark has something to do with logical thought; but this is the point at which the connection comes clearly into the light of day.

When we wish to ascertain whether σ belongs to E, we replace each letter in E by _ if σ belongs to the subset represented by the letter; by O if σ does not belong to the subset. Now suppose that a and b are two subsets satisfying the relation

a(b) = O

so a is a subset of b. We may also write:

(a(b)) = _ ,

implying, as we learnt in the chapter on dominance, that a dominates b, or

ab = a .

In the present interpretation, 'a dominates b' corresponds to the fact that the statement 'σ is in a' is a stronger statement than 'σ is in b'. That is how we could interpret the symbolic statement

a≥b.

As subsets, however, a is inside b and therefore smaller than b; for this we use the notation

a⊂b.

There is here a discrepancy, hiding a paradox, first attended to (so far as I know) by Leibniz. He said (in effect) that from the point of view of extension, b is greater than a; but from the point of view of intension, a is greater than b.

Exercise: (1) Let a,b,c,d be distinct single-element sets (so ab=O, ac=O, etc.). Prove (in the CL-calculus) the following:

((a)(b)) ((c)(d)) = O.

(2) Suppose that e and f are also single-element sets, distinct from a,b,c,d and also from each other. Prove that

((a)(b)(c)(d)) ((c)(d)(e)(f)) = ((c)(d)).

#### Duality

A fact that I find slightly irritating (which probably indicates that I haven't yet got to the bottom of it) is that the interpretation of the CL-calculus in terms of subsets, given above, is not the only one possible. Alternatively, we may interpret the 'product expression' ab not as the intersection of the two sets, but as their union. Then _ must be interpreted as the set which makes no difference when it is united with another set. This is certainly not U, because U absorbs any subset when united with it; rather, _ must be interpreted as the empty set, and it is O that must be interpreted as the universal set.

To put the set with a single element to work, we need to consider not sa but (s)a. If σ belongs to a, then (s), which consists of everything except the element σ is completed by union with a so that it becomes U, represented by O. If σ does not belong to a, then a is no use; in fact a is already contained in (s) and so the union of (s) with a is just (s).

To decide whether σ belongs to e, we consider the product (s)e. We then copy (s) into the expression so that each letter, such as a, has (s) beside it. If σ belongs to a we can replace a by O ; if not, a can be replaced by (s), which can then be uncopied leaving _ . Thus the rule is, replace a letter by O if σ belongs to the corresponding subset; by _ if it does not. Evaluate the resulting arithmetic expression, and combine the value with (s). If the value is O the result of the whole calculation is O; otherwise, it is (s).

It is just as before, except that O now means 'Yes, σ does belong to a', while _ now means 'No, σ does not belong to a'.

Which of the two interpretations is preferable? My own preference is that the void, _ , should indicate Yes, the truth, which is true whether or not it is spoken; while the mark, O, falsehood, does not exist unless it is spoken. You may not agree - perhaps the choice is best left as a matter of taste.