### Peircean Illative Transformations

We here describe a rule, due essentially to C.S.Peirce (1839-1914), for converting circle-and-letter expressions, not into their equivalents – as has been our concern up until now – but into lesser expressions; by which we mean that in each instance the transformed expression is dominated by the original expression. So if Y is converted into Z,

YZ = Y .

Thus Z, the lesser light, is eclipsed by Y, the greater light.

Lemma (‘Illation’)[1]: Let Y be any expression; let Z be obtained from Y either (a) by removing a sub-expression X from any space in Y whose depth is an even number (including zero) or (b) by inserting a sub-expression X into any space in Y whose depth is an odd number. Then Z is dominated by Y.

Proof: We proceed by first proving the following sub-lemma:

Sub-Lemma: Let A and B be CL expressions, and suppose that P ≥ Q ;
then (AP)B ≤ (AQ)B.

Proof of sub-lemma: We are given that PQ=P, therefore

(AP)B (AQ)B = (AP)(AQ)B [uncopy]
= (APQ)(AQ)B [premise]
= (P((AQ)))(AQ)B [wrap]
= (P( )) (AQ) B [uncopy]
= (()) (AQ) B [delete]
= (AQ)B [unwrap],

thus (AQ)B dominates (AP)B, which is what needed to be shown ⋄

Applying the sub-lemma to its own outcome, from

(AP)B ≤ (AQ)B

we deduce

((AP)B)C ≥ ((AQ)B)C

and thence

(((AP)B)C)D ≤ (((AQ)B)C)D,

and so on; with the relation changing from ≤ to ≥, and vice versa, each time another layer is added to the onion. Returning to the main lemma, if Y contains the expression X at some even level, then expressions A, B, C ...H may be chosen so that Y may be written in the form

Y = (...((AX)B)C...)H,

where the number of circles is even. Now X certainly dominates the void expression _ , so applying the sub-lemma as many times as required (with P=X, Q=_) we obtain

Y = (...((AX)B)C...)H ≥ (...((A)B)C...)H =Z.

Similarly, if Z contains X at an odd level then

Z = ((...((AX)B)C...)H)J ≤ ((...((A)B)C...)H)J = Y ;

so Z, whether it is obtained by removing the expression X from an even level, or adding X in at an odd level, is dominated by Y ⋄

Exercise: Prove that ((ax)b)c dominates ((a)b)c by taking legal steps in the Play Area. Then do the same with deeper expressions... (This is quite fun!)

(Hint: to prove that E dominates F, start with the expression EF and show that it may be transformed into expression E.)

##### Note

[1] The unfamiliar words 'illative' and 'illation' were used by Peirce, roughly as synonyms of 'inferential' and 'inference'. In Latin, illatus is the past participle of infero, so illatus est would mean 'it has been inferred'. For Peirce, but not for us, illative transformations have to do with logical inference. Our discussion is in terms of one expression 'dominating' another; the interpretation in terms of propositional logic is something that can be bolted on later, if desired.

Peirce's account of these transformations can be found in his Collected Papers, Volume 4, paragraphs 372-417 (available online at existentialgraphs.com). Note that his operations of 'iterate' and 'deiterate' anticipate our 'copy' and 'uncopy'; and his Rule 3, that anything can have double enclosures added or taken away, anticipates our 'wrap' and 'unwrap'.

The Illation Lemma is an invaluable extra weapon in the armoury of the circle-and-letter calculator. For example, by removing first b and then c from the second level in the expression ((ab)(ac)) we obtain:

((ab)(ac)) ≥ ((a)(a)) = ((a)) = a   [illation, uncopy, unwrap].

Therefore

((ab)(ac)) a = ((ab)(ac)).

This gives us a very quick proof of the law of distribution:

a((b)(c)) = a ((ab)(ac))  [copy]
= ((ab)(ac))  [by the result just obtained].

In a way, this is an absurdly roundabout way of proving distribute, involving as it does first the development of the notion of dominance, then the proof of the Illation Lemma. On the other hand, all the steps in the proof have a certain remarkable naturalness, following from the Bricken initials in a seemingly inevitable fashion. The direct proof, on the other hand, requires a somewhat 'artificial' move to get rid of the factor a in the expression

a((ab)(ac))

(one has to first wrap up the a, then copy the second factor into the space alongside (a) in the expression ((a)) ). In the end, it's a matter of taste.

It is of interest, perhaps, to see how the Illation Lemma may be justified from the perspective of the Test Theorem (see Consequences of the Standard Form Lemma). This says that two CL expressions are CL-equivalent if and only if they turn into arithmetic expressions of equal value whenever primitive values ( _ or O) are substituted for the letters in the expressions. A corollary is that A dominates B if and only if the expression (A(B)) reduces to _ under any substitution of values for the letters in A and B.

Let us consider, then, a pair of expression Y and Z, related to one another in the manner described in the statement of the Illation Lemma (above); that is to say, Z is obtained from Y, either (a) by removing an expression X from an even-depth space of Y, or (b) adding X to an odd-level space of Y. Considering first case (a), we have

Y = ((...((AX)B)C...)H)J    [with X nested in an even number of circles]
Z = ((...((A)B)C...)H)J .

Now consider the difference in value between Y and Z, when primitive values are substituted for the letters occurring in the sub-expressions A, B, ... J and X. A difference in value can only arise from the presence of X in Y, but not in Z. But the value of X can only make a difference on the rare occasions when all of A, B, C, ... J have value _ . Otherwise, X’s ‘signal’ gets drowned out. On those occasions, the value of Y is

((...((X))...)) = X,

thanks to the even number of enclosures; while the value of Z is just _ . So, on these occasions, (Y(Z)) takes the same value as (X( )), namely _ ; and on all the other occasions, since Y and Z take the same value, the ‘dominance expression’ (Y(Z)) also takes the value _ . Therefore, by the Test Theorem, (Y(Z)) is CL-equivalent to the void; therefore Y dominates Z.

Case (b) is similar. This time

Y = (...((A)B)C...)H
Z = (...((AX)B)C...)H   [with X nested in an odd number of circles].

Again, Y and Z can only differ in value if all of A, B, ... H take the value _ , in which case Y will take the value O, while Z will take the same value as (X). Again, the dominance expression will take the same value as the expression (O(X)), namely _ ; so again Y must dominate Z ⋄