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

Special Kinds of Theorems 75
both A and B, then it would be an element of their intersection, but we
cannot exclude that x belongs to one of the two sets.)
Thus, by the definition of the complement of a set, either x e A' or x e B.
This implies that x e A'D B\
Part 2. A'UB' QiAnsy.
Let X e A' U B\ Then either x e A' or x e B'\ that is, by the definition of
complement of a set, either x^A or x^B. This implies that x is not a
common element of A and 5; that is, x^{Af\ B). Thus, we can conclude that
xe{Af\ By.
As both inclusions are true, the two sets are equal. •
Sometimes the two inclusions can be proved at the same time. We could
have proved the statement in Example 5 as follows:
X e(An By if and only if x^ (A n B) if and only if x^A or x^ B
if and only if x e A' or x e B if and only ii x G A' U B'.
While this kind of proof is clearly shorter than the one presented in
Example 5, it can be trickier because there are fewer separate steps and it is
less explicit. Therefore, proofs of this type can be more difficult to analyze
and it becomes easier to overlook important details and make mistakes.
(See Theorem 9 in the section Collection of Proofs section)
In order to prove that two sets, A and B, are not equal, it is sufficient to
prove that at least one of the two inequalities {A C.B or B c^A) does not hold.
This means that it is enough to show that there is at least one element in one
set that does not belong to the other.
EXAMPLE 6. Let A — {all odd counting numbers larger than 2}
B = {all prime numbers larger than 2}. These two sets are not equal.
and
Proof: We have already seen that all prime numbers larger than 2 are
odd. Therefore, B <zA.