The following is a proof that for any sets A, B, and C, A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C). Fill in the blanks.
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 intersection, x ∈ (b) and x ∈ (c). Thus x ∈ A and, by definition of union, x ∈ B or (d).
Case 1 (x ∈ A and x∈ B): In this case, by definition of intersection, x ∈ (e), and so, by definition of union, x ∈ (A ∩ B) ∪ (A ∩ C).
Case 2(x∈ A and x∈ C): In this case. (f). Hence in either case, x ∈ (A ∩ B) ∪ (A ∩ C) [as was to be shown].
[So A ∩ (B ∪ C) ⊆ (A ∩ B) ∪ (A ∩ C) by definition of subset.]
(2) (A ∩ B) ∪ (A ∩ C) ⊆ A ∩ (B ∪ C):
Let x ∈ (A ∩ B) ∪ (A ∩ C). [We must show that (a).] By definition of union, x ∈ A ∩ B (b) x A ∩ C.
Case 1 (x ∈ A ∩ B): In this case, by definition of intersection, x ∈ A (c) x ∈ B. Since x ∈ B, then by definition of union, x ∈ B ∪ C. Hence x ∈ A and x ∈ B ∪ C, and so, by definition of intersection, x ∈ (d).
Case 2 (x ∈ A ∩ C): In this case, (e).
In either case, x ∈ A ∩ (B ∪ C) [as was to be shown]. [Thus (A ∩ B) ∪ (A ∩ C) ⊆ A ∩ (B ∪ C) by definition of subset.]
(3) Conclusion: [Since both subset relations have been proved, it follows, by definition of set equality, that (a).]