# Let ∩ and ∪ stand for the words “intersection” and “union,” respectively. Fill in the blanks in the following proof that for all sets A, B, and C, A ∩ ( B ∪ C ) = ( A ∩ B ) ∪ ( A ∩ C ) . Proof: Suppose A , B and C are any sets. (1) Proof that A ∩ ( B ∩ C ) ⊆ ( A ∩ B ) ∩ ( A ∩ C ) : Let x ∈ A ∩ ( B ∪ C ) . [We must show that x ∈ (a) ]. By definition of ∩ , x ∈ (b) and x ∈ B ∪ C . Thus x ∈ A and, by definition of ∪ , x ∈ B or (c) . Case 1 ( x ∈ A and x ∈ B ): In this case, x ∈ A ∩ B by definition of ∪ . Case 2 ( x ∈ A and x ∈ C ): In this case, x ∈ A ∩ C by definition of ∪ . By cases 1 and 2, x ∈ A ∩ B or x ∈ A ∩ C , and so, by definition of ∪ . (d) . [So A ∩ ( B ∪ C ) ⊆ ( A ∩ B ) ∪ ( A ∩ C ) by definition of subset.] (2) Proof that ( A ∩ B ) ∪ ( A ∩ C ) ⊆ A ∩ ( B ∪ C ) : Let x ∈ ( A ∩ B ) ∪ ( A ∩ C ) . [We must show that x ∈ A ∩ ( B ∪ C ) .] By definition of ∪ , x ∈ A ∩ B (a) x ∈ A ∩ C . Case 1 ( x ∈ A ∩ B ) : In this case, by definition of ∪ , x ∈ A and x ∈ B . Since x ∈ B , then x ∈ B ∪ C by definition of ∪ . Case 2 ( x ∈ A ∩ C ) : In this case, by definition of ∪ , x ∈ A (b) x ∈ C . Since x ∈ C , then x ∈ B ∪ C by definition of ∪ . In both cases x ∈ A and x ∈ B ∪ C , and so, by definition of ∪ , (c) . [So ( A ∩ B ) ∪ ( A ∩ C ) ⊆ A ∩ ( B ∩ C ) by definition of (d) .] (3) Conclusion: [Since both subset relations have been proved, it follows, by definition of set equality, that (a).]

### Discrete Mathematics With Applicat...

5th Edition
EPP + 1 other
Publisher: Cengage Learning,
ISBN: 9781337694193

### Discrete Mathematics With Applicat...

5th Edition
EPP + 1 other
Publisher: Cengage Learning,
ISBN: 9781337694193

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Chapter
Section
Chapter 6.2, Problem 6ES
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