The following is a proof that for any sets A, B, andC,

Chapter 6, Problem 6.59

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