Page 52 - HOW TO PROVE IT: A Structured Approach, Second Edition

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38 Sentential Logic
intersections, unions, and differences of sets. For example, according to the
definition of intersection, to say that x ∈ A ∩ B means that x ∈ A ∧ x ∈ B.
Similarly, to say that x ∈ A ∪ B means that x ∈ A ∨ x ∈ B, and x ∈ A \ B
means x ∈ A ∧ x ∈ B, or in other words x ∈ A ∧¬(x ∈ B). We can combine
these rules when analyzing statements about more complex sets.

Example 1.4.4. Analyze the logical forms of the following statements:
1. x ∈ A ∩ (B ∪ C).
2. x ∈ A \ (B ∩ C).
3. x ∈ (A ∩ B) ∪ (A ∩ C).

Solutions
1. x ∈ A ∩ (B ∪ C)
is equivalent to x ∈ A ∧ x ∈ (B ∪ C) (definition of ∩),
which is equivalent to x ∈ A ∧ (x ∈ B ∨ x ∈ C) (definition of ∪).
2. x ∈ A \ (B ∩ C)
is equivalent to x ∈ A ∧¬(x ∈ B ∩ C) (definition of \),
which is equivalent to x ∈ A ∧¬(x ∈ B ∧ x ∈ C) (definition of ∩).
3. x ∈ (A ∩ B) ∪ (A ∩ C)
is equivalent to x ∈ (A ∩ B) ∨ x ∈ (A ∩ C) (definition of ∪),
which is equivalent to (x ∈ A ∧ x ∈ B) ∨ (x ∈ A ∧ x ∈ C)
(definition of ∩).

Look again at the solutions to parts 1 and 3 of Example 1.4.4. You should rec-
ognize that the statements we ended up with in these two parts are equivalent. (If
you don’t, look back at the distributive laws in Section 1.2.) This equivalence
means that the statements x ∈ A ∩ (B ∪ C) and x ∈ (A ∩ B) ∪ (A ∩ C) are
equivalent. In other words, the objects that are elements of the set A ∩ (B ∪ C)
will be precisely the same as the objects that are elements of (A ∩ B) ∪
(A ∩ C), no matter what the sets A, B, and C are. But recall that sets with
the same elements are equal, so it follows that for any sets A, B, and C, A ∩
(B ∪ C) = (A ∩ B) ∪ (A ∩ C). Another way to see this is with the Venn di-
agram in Figure 6. Our earlier Venn diagrams had two circles, because in
previous examples only two sets were being combined. This Venn diagram has
three circles, which represent the three sets A, B, and C that are being combined
in this case. Although it is possible to create Venn diagrams for more than three
sets, it is rarely done, because it cannot be done with overlapping circles. For
more on Venn diagrams for more than three sets, see exercise 10.
Thus, we see that a distributive law for logical connectives has led to a
distributive law for set theory operations. You might guess that because there