the numerical conditions of logical classes. In a paper published among the Memoirs of the Manchester Literary and Philosophical Society, Third Series, vol. IV. p. 330 (Session 1869–70), I have pointed out that we can apply arithmetical calculation to the Logical Alphabet. Having given certain logical conditions and the numbers of objects in certain classes, we can either determine the numbers of objects in other classes governed by those conditions, or can show what further data are required to determine them. As an example of the kind of questions treated in numerical logic, and the mode of treatment, I give the following problem suggested by De Morgan, with my mode of
“For every man in the house there is a person who is aged; some of the men are not aged. It follows that some persons in the house are not men.”6
| Now let | A = person in house, |
|---|---|
| B = male, | |
| C = aged. |
By enclosing a logical symbol in brackets, let us denote the number of objects belonging to the class indicated by the symbol. Thus let
| (A) = | number of persons in house, |
|---|---|
| (AB) = | number of male persons in house, |
| (ABC) = | number of aged male persons in house, |
and so on. Now if we use w and w′ to denote unknown numbers, the conditions of the problem may be thus stated according to my interpretation of the words—
that is to say, the number of persons in the house who are aged is at least equal to, and may exceed, the number of male persons in the house;
that is to say, the number of male persons in the house who are not aged is some unknown positive quantity.
If we develop the terms in (1) by the Law of Duality (pp. 74, 81, 89), we obtain
Subtracting the common term (ABC) from each side and substituting for (ABc) its value as given in (2), we get at once
and adding (Abc) to each side, we have
The meaning of this result is that the number of persons in the house who are not men is at least equal to w + w′, and exceeds it by the number of persons in the house who are neither men nor aged (Abc).
It should be understood that this solution applies only to the terms of the example quoted above, and not to the general problem for which De Morgan intended it to serve as an illustration.
As a second instance, let us take the following question:—The whole number of voters in a borough is a; the number against whom objections have been lodged by liberals is b; and the number against whom objections have been lodged by conservatives is c; required the number, if any, who have been objected to on both sides. Taking
then we require the value of (ABC). Now the following equation is identically true—