# Boole's use of supplementary classes

Boole could have represented "$$x$$ is a subclass of $$y$$" via the formula $$x=xy$$ (where $$xy$$ signifies the intersection of the class $$x$$ and the class $$y$$), but according to Daniel Bonevac and Josh Dever's "A History of the Connectives", Boole instead used $$x=vy$$ where $$v$$ is “a class indefinite in every respect”. (The quoted phrase is Boole's; see page 218 of Bonevac and Dever.) The use of supplementary classes complicates the algebraic proofs of syllogisms, so why did Boole go this route?

Additionally, I'm puzzled by the fact that Boole re-used supplementary classes, writing (for instance) "$$x=vy$$ and $$y=vz$$" where I would have guessed he'd want to write "$$x=vy$$ and $$y=wz$$" (why use the same indefinite class twice?).

I'd be grateful if someone explained what I'm missing here.

Daniel Bonevac and Josh Dever, "A History of the Connectives", in "The many valued and non-monotonic turn in logic", Handbook of the History of Logic, volume 8, edited by Dov M. Gabbay and John Woods, Elsevier, 2007.

• Perhaps Boole feared that $x=xy$ would be mistaken for a circular definition of $x$? Commented Jul 17 at 17:06
• Maybe useful the annotated version of Boole's 1848 The Calculus of Logic. I mostly agree with your comment: Boole want to preserve the possibility to manipulate algebraically the "equations". Commented Jul 18 at 7:30
• Here's a proposed algebraic proof that if Socrates is human and all human are mortal then Socrates is mortal: letting $x$, $y$, and $z$ denote the class consisting of just Socrates, the class of humans, and the class of mortals, our assumptions translate into $xy=x$ and $yz=y$, so we have $xz=(xy)z=x(yz)=xy=x$, with no need for supplementary classes. Did Boole do it this way? If not, why not? Commented Jul 19 at 17:15
• I have an alternative hypothesis. Boole regarded individual proofs as illustrations of a general method of proof. Perhaps in the interest of uniformity of method, he sometimes opted for a lengthier proof than necessary when the lengthier proof employed a more general method. More specifically, since his proofs of I-propositions required supplementary classes, he could have chosen to use supplementary classes in his proofs of A-propositions for consistency's sake. But that's just a guess. Commented Jul 19 at 23:31

Disclaim: I've no direct access to manuscripts and other "first-hand" Boole's sources: thus, it is hard to be more specific wrt historians writing "It is somewhat of a mystery why Boole..."

We can read the relevant passage of The Mathematical Analysis of Logic (1847) where the algbraic symbolism is used by Boole to analyze propositions:

To express the Proposition, All Xs are Ys. As all the Xs which exist are found in the class Y, it is obvious that to select out of the Universe all Ys, and from these to select all Xs, is the same as to select at once from the Universe all Xs. Hence $$xy = x$$, or $$x (1 - y) = 0$$.

To express the Proposition, Some Xs are Ys. If some Xs are Ys, there are some terms common to the classes X and Y. Let those terms constitute a separate class V, to which there shall correspond a separate elective symbol $$v$$, then $$v = xy$$. And as $$v$$ includes all terms common to the classes X and Y, we can indifferently interpret it, as Some Xs, or Some Ys.

Thus, we have the "elective symbols" $$v$$ from the start.

We can see Dale Jacquette, Boole’s Logic in: Dov Gabbay & John Woods (editors), Handbook of the History of Logic. 4 British Logic in the Nineteenth Century (2008), page 345:

The interpretation of I-type categorical propositions ["Some X’s are Y ’s"as $$v = xy$$] involves the introduction of a special elective operator $$v$$ for the class $$V$$ consisting of the objects (Boole says ‘terms’) common to classes $$X$$ and $$Y$$. The otherwise unauthorized appearance of this new elective operator is not entirely satisfactory, and has been the subject of complaint by many even of Boole’s most sympathetic commentators. The objection is that $$V$$ and $$v$$ in Boole’s translation merely conjure up a name to stand for what would better be represented as the specific operations by virtue of which a set of objects common to classes $$X$$ and $$Y$$ are logically related to the membership of $$X$$ and $$Y$$. [...]

It is somewhat of a mystery why Boole does not simply avail himself of propositional negation or nonidentity in his algebra, but relies instead on a nominal subterfuge, burying away the logic of ‘some’ in the unsymbolized definition of class $$V$$ and its elective operator $$v$$ for I-type and O-type categorical propositions. It is significant that Boole never uses the nonidentity sign, ‘$$\ne$$’, anywhere in his system, even though it is commonly found in arithmetical algebra. Boole, as previously mentioned, moreover, explicitly denies that any proposition in his logic can be negated or ‘negatived by a legitimate operation’ [Mathematical Analysis of Logic, 18–19]. It would be interesting to know why Boole, who evidently considered and deliberately rejected the possibility, decided to exclude propositional negation from his logic. What specific reasons did he have? The question is philosophically important because the absence of negation is the chief obstacle to regarding Boole’s analysis especially of hypothetical or secondary propositions and hypothetical and other related inferences as providing all the essentials of a proto-propositional (and more generally, proto-first-order propositional and predicate-quantificational symbolic) logic.

And page 349:

Boole’s use of elective symbol $$v$$ nevertheless raises philosophical difficulties, both in Mathematical Analysis of Logic, and in its expanded application in Laws of Thought. Symbol $$v$$ functions properly only by virtue of its special interpretation, and in other ways does not have the same symbolic role as other class terms. This is seen among other ways in the fact that from $$x = y$$ and $$y = z$$, it follows logically in Boole’s system that $$x = z$$. From the fact that $$x = v$$ and $$y = v$$, meaning that class $$x$$ and class $$y$$ are not empty, but contain some indefinite number of objects, however, it does not at all follow logically that therefore $$x = y$$. If Boole tries to block the unwanted inference either by restricting identities to class terms, so that $$x = v$$ and $$y = v$$ cannot be written in the algebra, the restriction would only serve to further underscore the peculiar nature of $$v$$. Moreover, the parallel problem would still remain for the canonical Boolean identities, $$x = vy$$ and $$z = vy$$, to symbolize the facts that all x’s are (some) y’s and that all z’s are (some) y’s — say, that all whales are (some) mammals and all elephants are (some) mammals — from which it certainly does not follow that all and only whales are elephants, nor that the class of all whales is identical to the class of all elephants. To the extent that $$v$$ looks like but does not function symbolically as a class term according to general algebraic conversion and transformation rules, Boole fails to satisfy his own requirement that all of logic be interpreted by means of purely algebraic syntactical operations.

We can see also the Introducion to Ivor Grattan-Guinness & Gerard Bornet (editors), George Boole: Selected Manuscripts on Logic and its Philosophy (Springer, 1997), page 30:

Boole realised that a given set of premises may not have any solutions, or maybe more than one; and to accommodate the latter possibility he introduced "V" to symbolise an indefinite proportion of members of a class, from none through "some" to all of them; the corresponding elective symbol was "$$v$$". This notion has often perplexed readers, but it allows him to handle the lack of a cancellation law in his algebra: namely that if $$xy = xz$$, then it is not necessarily the case that $$y = z$$. In addition, it enabled him to express many propositions in terms of equations rather than inequalities, which would have complicated the algebra considerably. However, he offered no laws which "$$v$$" should satisfy, and could not always distinguish between traditional forms of proposition and those involved in the quantification of the predicate (which he did not analyse explicitly); for example, "$$vx =vy$$" could cover both "Some Xs are Ys" and "Some Xs are some Ys".

Further, contrary to Boole's apparent belief, the solutions found by his methods were not always complete. For example, for the universal affirmative proposition "All Ys are Xs", symbolised as $$(1-x)y = 0$$, he put forward $$y = vx$$ as "the most general solution"; but he should have noticed that $$x = 0$$ was missing from it, and that it did not hold if $$x = 0$$ and $$v$$ was a class such that $$vy \ne 0$$ [Corcoran & Wood]. It is curious that a logician should seem to slip over the difference between necessary and sufficient conditions, or that a mathematician with a strong interest in singular solutions of differential equations did not notice analogous problems in his logic. In fact, in this case Boole realised that $$v = xy$$ was a better version than $$y = vx$$ (so did the Irish mathematician Charles Graves after reading MAL, and he told the Royal Irish Academy in a note (1850).

The process seems to be: from the for A "All Xs are Ys": $$(1-y)x = 0$$,and E "No Xs are Ys": $$yx = 0$$, Boole derives the "solutions": $$x = vy$$ and $$x = v(1-y)$$.

We can check the first one substituting $$x=vy$$ into $$(1-y)x = 0$$ and verifying that $$(1-y)vy=vy-vyy=0$$, that holds irrespective of the nature of $$v$$. The same with the second pair of equations.

The first solution may be interpreted as "Some Ys are Xs" and the second with "Some not-YFs are Xs."

As noted by Corcoran & Wood (1989): There should be no mistake about our claim that Boole's sole ground for asserting that $$x = vy$$ follows from $$(1 - y)x = 0$$ is given above. This is so despite the fact that it should have been obvious to Boole that $$x = 0$$ is also a solution and that if $$x = 0$$ and $$y$$ is any nonempty set then the equation is true but the solution $$x = vy$$ is false (for $$v$$ any set not disjoint with $$y$$). One should also be clear about the fact that $$x = vy$$ is not deducible from $$(1 - y)x = 0$$ using the rules that Boole explicitly states. The reason is simply that Boole's explicitly stated rules are all satisfied when interpreted as in ordinary class algebra.

The "unpleasan" use of the mysterious symbol $$v$$ was noted quite early. See

"Some X's are Y's" is written as $$xy \ne 0$$ and "All Y's are X's" as $$x'y=0$$.

• I appreciate the information about Boole's treatment of I-type propositions, but I don't see how it sheds light on my question about A-type propositions. ($v=xy$ is very different from $x=vy$.) Have I misinterpreted Bonevac and Dever? Or is their account of Boole's system incorrect? Commented Jul 18 at 23:08
• The comment I made on July 18 no longer applies, thanks to Mauro's more detailed explanation of Boole's treatment of A-type and I-type propositions. But I am left wondering whether Boole (or any of his successors, for that matter) ever presented the 𝑥𝑧=(𝑥𝑦)𝑧=𝑥(𝑦𝑧)=𝑥𝑦=𝑥 style of proof for Barbara syllogisms. For that matter, once Boole's successors extended + to allow unrestricted addition of classes, the alternative proof 𝑥+𝑧=𝑥+(𝑦+𝑧)=(𝑥+𝑦)+𝑧=𝑦+𝑧=𝑧 became available. This proof is natural from the viewpoint of modern Boolean lattice theory, but who first proposed it? Commented Jul 25 at 14:48