Talk:Simple theorems in the algebra of sets

Proof of Proposition 3 is below:

Proof: We shall prove (a) and leave (b) as an exercise. Each side of equation (a) defines a set and we wish to prove that these sets are equal. By Proposition 2, a possible strategy is to show that each side is a subset of the other.
 * 1) Pick any element x of the left-hand side (LHS). Then, by definition of &cap;, x is in A and x is in B &cup; C; that is, x is in A and either x is in B or x is in C (or both). In the first case, x is both in A and in B, so it's in A &cap; B and a fortiori in (A &cap; B) &cup; (A &cap; C). In the second case, x is both in A and in C and so again it's in (A &cap; B) &cup; (A &cap; C). Thus in either case, x is in (A &cap; B) &cup; (A &cap; C). We have shown that every element of the LHS is automatically in the RHS. But this is precisely what we mean by saying that the LHS is a subset of the RHS.
 * 2) Pick any element x of the RHS. Then x is in A &cap; B or x is in A &cap; C (or both). In the first case, x is in A and x is in B; in the second, x is in A and x is in C. In either case, x is in A. Also in the first case x is in B and hence in B &cup; C; in the second case, x is in C and thus again in B &cup; C. We have proved that whatever x is, if it is a member of the RHS, then it is both in A and in B &cup; C and hence by definition is in A &cap; (B &cup; C). We have proved that the RHS is a subset of the LHS.

By Proposition 2, (1) and (2) together prove that LHS = RHS, as required.

Restructure
Cant this be somehow restructured? It sounds like some textbook entry. And the talk page proof seems like it has been taken from some book. --Soumyasch 06:14, 22 March 2006 (UTC)

A \ B
According to the set article, "A \ B" and "A - B" mean the same thing (alternate notation). —Preceding unsigned comment added by Special:Contributions/ (talk)
 * Yes. So? Paul August &#9742; 20:35, 21 May 2006 (UTC)

A ⊆ B if and only if B ⊆ A;
For this to be true, wouldn't A have to be the same as B?

Could someone clarify this one for me? It doesn't make sense to my in it's current form in the article. 121.127.202.6 (talk) 13:58, 14 August 2008 (UTC)