Page 53 - HOW TO PROVE IT: A Structured Approach, Second Edition

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Operations on Sets 39

Figure 6

were two distributive laws for the logical connectives, with ∧ and ∨ playing
opposite roles in the two laws, there might be two distributive laws for set
theory operations too. The second distributive law for sets should say that for
any sets A, B, and C, A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C). You can verify this
for yourself by writing out the statements x ∈ A ∪ (B ∩ C) and x ∈ (A ∪ B) ∩
(A ∪ C) using logical connectives and verifying that they are equivalent, using
the second distributive law for the logical connectives ∧ and ∨. Another way
to see it is to make a Venn diagram.
We can derive another set theory identity by finding a statement equivalent
to the statement we ended up with in part 2 of Example 1.4.4:
x ∈ A \ (B ∩ C)
is equivalent to x ∈ A ∧¬(x ∈ B ∧ x ∈ C) (Example 1.4.4),
which is equivalent to x ∈ A ∧ (x ∈ B ∨ x ∈ C) (DeMorgan’s law),
which is equivalent to (x ∈ A ∧ x ∈ B) ∨ (x ∈ A ∧ x ∈ C)
(distributive law),
which is equivalent to (x ∈ A \ B) ∨ (x ∈ A \ C) (definition of \),
which is equivalent to x ∈ (A \ B) ∪ (A \ C) (definition of ∪).
Thus, we have shown that for any sets A, B, and C, A \(B ∩ C) = (A \ B) ∪
(A \ C). Once again, you can verify this with a Venn diagram as well.
Earlier we promised an alternative way to check the identity (A ∪ B) \
(A ∩ B) = (A \ B) ∪ (B \ A). You should see now how this can be done. First,
we write out the logical forms of the statements x ∈ (A ∪ B) \ (A ∩ B) and
x ∈ (A \ B) ∪ (B \ A):

x ∈ (A ∪ B) \ (A ∩ B) means (x ∈ A ∨ x ∈ B) ∧¬(x ∈ A ∧ x ∈ B);
x ∈ (A \ B) ∪ (B \ A) means (x ∈ A ∧ x ∈ B) ∨ (x ∈ B ∧ x ∈ A).