epdf.pub_the-nuts-and-bolts-of-proofs-third-edition-an-intr

144 The Nuts and Bolts of Proof, Third Edition
The following is a representation of the set (AU B)n C.
The equaUty does not seem to be true in general, as it contradicts
the distributive law of union with respect to intersection. Let us
look for a counterexample. If A = {1}, B = {2}, and C = {2,3}, then
AU{BnC)= AU{2} = {12} and (AU B)nC = {12} DC = {2}.
Therefore, the two sets are not equal in general.
b. This equaUty seems to be an extension of one of De Morgan's laws.
Let's try to prove it. The element x belongs to (AH BO C)' if and
only if X^ (^ n J5 n C). This happens if and only {{x^Aorx^Bov
x^ C. This is equivalent to saying that X e A' or x e B' ov x e CThis
happens if and only lixe A' \JB' \J C.
The element (xo,yo) belongs to A if and only if yo
equaUty is equivalent to the equaUty:
- v2. 1. This
yo = (^0 -1) ^0 + 1
X2 + 1
because x^ + 1 7^ 0. Therefore, yQ:=^xl-\ if and only i{{yo — (x\ - 1)/
(XQ + 1)). This means that (xo,yo) e ^ if and only if (xo,}^o) ^ ^- Thus,
the two sets are equal.
a. The base case has been proved in Example 5. Let A — A\ and
B^Ai,
b. Inductive hypothesis: Assume that for some n > 3 the equaUty
(^1 n yl2 n ... n An)' = ^; U ^'2 ^ ... U A; holds true.
c. We have to prove that (Ai fi ^2 H ... n A„ n An+i)' = A; U ^^ U ...
UA^^UA'^_^^. Using the associative property of intersection (see
Exercise 4 in this section), we have:
(yii n ^2 n... n A„ n An+i) = [(^1 n ^2 n... n ^„) n A„+i].