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
A = voter,
B = objected to by liberals,
C = objected to by conservatives,
then we require the value of (ABC). Now the following equation is identically true—
(ABC) = (AB) + (AC) + (Abc) - (A). (1)
For if we develop all the terms on the second side we obtain
(ABC) = (ABC) + (ABc) + (ABC) + (AbC) + (Abc)
- (ABC) - (ABc) - (AbC) - (Abc);
and striking out the corresponding positive and negative terms, we have left only (ABC) = (ABC). Since then (1) is necessarily true, we have only to insert the known values, and we have
(ABC) = b + c - a + (Abc).
Hence the number who have received objections from both sides is equal to the excess, if any, of the whole number of objections over the number of voters together with the number of voters who have received no objection (Abc).