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Properties of Sets Lecture 26 Section 5.2 Tue, Mar 6, 2007
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Proving Basic Properties Theorem: Let A and B be sets. Then A B A. Proof: Let x A B. Then x A and x B. Therefore, x A. So A B A.
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Comments The proof uses the logic that S T if and only if x S x T. A Venn diagram alone does not constitute a proof. This theorem is suggestive of the “specialization” argument form (p. 40) p q p What is the connection?
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Identity Laws Theorem: Let A be any set. Then A U = A, A = A.
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DeMorgan’s Laws Theorem: Let A and B be any two sets. Then (A B) c = A c B c, (A B) c = A c B c.
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Set-Theoretic Proofs Theorem: Let A, B, and C be sets. Then (A – B) (C – B) = (A C) – B. Proof: Let x (A – B) (C – B). Then x A – B and x C – B, x A, and x C, and x B, x A C and x B, x (A C) – B.
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Set-Theoretic Proof It then follows that (A – B) (C – B) (A C) – B. The second half of the proof will show that (A C) – B (A – B) (C – B). However, the logic is exactly the reverse of the first half. We may handle that by saying “and conversely” at the end of the first half.
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Comment The preceding theorem is equivalent to the logical equivalence (p q) (r q) = (p r) q which is not hard (at all!) to prove. Proof: (p q) (r q) = p q r q = p r q = (p r) q.
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Question Why was that so much easier than the original proof? Because we know a lot about the operators , , and . We could use the “algebra” of , , and . Is there an algebra of , , and complement?
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Algebraic Properties of Sets See Theorem 1.1.1, p. 14. Commutative Laws: A B = B A. A B = B A. Associative Laws: (A B) C = A (B C). (A B) C = A (B C).
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Algebraic Properties of Sets Distributive Laws: A (B C) = (A B) (A C). A (B C) = (A B) (A C). Identity Laws: A U = A. A = A.
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Algebraic Properties of Sets Complement Laws: A A c = U. A A c = . Double Complement Law: (A c ) c = A. Idempotent Laws: A A = A. A A = A.
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Algebraic Properties of Sets Universal Bound Laws: A U = U. A = . DeMorgan’s Laws: (A B) c = A c B c. (A B) c = A c B c.
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Algebraic Properties of Sets Absorption Laws: A (A B) = A. A (A B) = A. Complements of U and . U c = . c = U.
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The Proof Revisited Theorem: Let A, B, and C be sets. Then (A – B) (C – B) = (A C) – B. Proof: (A – B) (C – B) = (A B c ) (C B c ) = (A C) (B c B c ) = (A C) B c = (A C) – B.
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The Proof Revisited Theorem: Let A, B, and C be sets. Then (A – B) (C – B) = (A C) – B. Proof:
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