How to do Categorical Statements using set algebra.

[BaalChatzaf]

The categorical statement forms are the basis of Aristotle's categorical logic (as defined in -Prior Analystics). There are four categorical forms:

A All A are B

E No A are B

I Some A is B

O Some A is not B.

I will do this using set algebra. First a notational definition. 0 stands for the empty set. ^ stands for set intersection, v stands for set union (the or'ing of two sets) and != means not equal and - means complement. -A is the set of all things not in A. In the special case -0 is the set of everything, the so called universal set.

A is rendered as A^B = A

E is rendered as A^B = 0 (or A^-B = A)

I is rendered as A^B != 0

O is rendered as A^-B != 0

This puts all the machinery of set algebra at the disposal of categorical logic. There is a caveat. From the equality A^B = A one cannot infer A != 0 since the equality A^B = A holds when A = 0. And this is the main difference between the modern treatment of categorical propositions and the classical (Aristotelian) treatment. This is the removal of existential import and it is done so that empty sets can be included in the logic (as it is included in the algebra of sets). We get set union by the equation -(A^B) = -Av-B hence

AvB = -(-A^-B). The empty set 0 and its complement 1 (here 1 means the universal set) play roles similar to numerical 0 and 1. So A^1 = A. A^0 = 0 Av0 = A There are differences however;

Av1 = 1, A^-A = 0 and A v -A = 1.

I can't leave without proving BaRBaRa which is the primary syllogism of classical logic.

Classically it says:

all A are B and all B are C imply all A are C. That is A^B = A and B^C = B imply A^C = A

Here goes:

A^B = A and B^C =B : givens

substitute B^C for B in A^B = A and get A^(B^C) = A.

By the associative law for ^ (take my word for this), we get

(A^B)^C = A. But A^B = A (given) so substituting we get A^C = A which is to say all A are C.

q.e.d.

Slicker than beets!

The inclusion property does NOT depend on the existence of an element in the smaller set. It holds for ALL sets, not just non-empty sets. Now you know why the modern treatment is more general and uniform than the classical treatment.

The other categorical syllogisms are almost as easy and I will leave them as exercises for the interested reader.

Ba'al Chatzaf

Edited May 6, 2009 by BaalChatzaf