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Published byAusten Norris Modified over 4 years ago

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Rosen 1.6

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Approaches to Proofs Membership tables (similar to truth tables) Convert to a problem in propositional logic, prove, then convert back Use set identities for a tabular proof (similar to what we did for the propositional logic examples but using set identities) Do a logical (sentence-type) argument (similar to what we did for the number theory examples)

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Prove (A B) (A B) = B AB (A B)(A B) (A B) (A B) 11 1 01 10 0 00 01 0 11 00 0 00

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Prove (A B) (A B) = B (A B) (A B) = {x | x (A B) (A B)} Set builder notation = {x | x (A B) x (A B)} Def of = {x | (x A x B) (x A x B)} Def of x2 and Def of complement = {x | (x B x A ) (x B x A )} Commutative x2 = {x | (x B (x A x A )} Distributive ={x | (x B T } Or tautology ={x | (x B } Identity = BSet Builder notation

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Set Identities (Rosen, p. 89) A Ø = A A U = A A U = U A Ø = Ø A A = A A A = A (A) = A Identity Laws Domination Laws Idempotent Laws Complementation Law

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Set Identities (cont.) A B = B A A B = B A A (B C) = (A B) C A (B C) = (A B) C A (B C)=(A B) (A C) A (B C)=(A B) (A C) A B = A B A B = A B Commutative Laws Associative Laws Distributive Laws De Morgan’s Laws

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Prove (A B) (A B) = B (A B) (A B) = (B A) (B A) Commutative Law x2 =B (A A) Distributive Law =B U Definition of U =BIdentity Law

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Prove (A B) (A B) = B Proof: We must show that (A B) (A B) B and that B (A B) (A B). First we will show that (A B) (A B) B. Let e be an arbitrary element of (A B) (A B). Then either e (A B) or e (A B). If e (A B), then e B and e A. If e (A B), then e B and e A. In either case e B.

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Prove (A B) (A B) = B Now we will show that B (A B) (A B). Let e be an arbitrary element of B. Then either e A B or e A B. Since e is in one or the other, then e (A B) (A B).

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Prove A B A Proof: We must show that any element in A B is also in A. Let e be an element in A B. Since e is in the intersection of A and B, then e must be an element of A and e must be an element of B. Therefore e is in A.

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Prove A A = A A = (A A) - (A A) = (A) - (A) = Ø Definition of Idempotent Laws Definition of -

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Prove A B = A B Proof: To show that A B = A B we must show that A B A B and A B A B. First we will show that A B A B. Let e A B. We must show that e is also A B. Since e A B, then e A B. So either e A or e B. If e A then e A. If e B then e B. In either case e A B

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DeMorgan Proof (cont.) Next we will show that A B A B. Let e A B. Then e A or e B. Therefore, by definition e A or e B. Therefore e (A B) which implies that e A B Since A B A B and A B A B then A B = A B.

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Prove A (B C)=(A B) (A C) Proof: To show that A (B C)=(A B) (A C) we must show that A (B C) (A B) (A C) and (A B) (A C) A (B C).

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Distributive Proof (cont.) First we will show that A (B C) (A B) (A C). Let e be an arbitrary element of A (B C). Then e A and e (B C). Since e (B C), then either e B or e C or e is an element of both. Since e is in A and must be in at least one of B or C then e is an element of at least one of (A B) or (A C). Therefore e must be in the union of (A B) and (A C).

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Distributive Proof (cont.) Next we will show that (A B) (A C) A (B C). Let x (A B) (A C). Then either x (A B) or x (A C) or x is in both. If x is in (A B), then x A and x B If x B, then x (B C). Therefore x A (B C). By a similar argument if x (A C) then, again, x A (B C). Since A (B C) (A B) (A C) and (A B) (A C) A (B C), then A (B C) = (A B) (A C).

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Prove: [A B A B] [A = B] Proof: We must show that when A B A B is true then A=B is true. (Proof by contradiction) Assume that A B A B is true but A B. If A B then this means that either x A but x B, or x B but x A. If x A but x B, then x A B but x A B so A B is not a subset of A B and we have a contradiction to our original assumption. By a similar argument A B is not a subset of A B if x B but x A. Therefore [A B A B] [A = B].

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Prove or Disprove [A B=A C] [B=C] False! A= Ø, B={a}, C={b} [A B=A C] [B=C] False! A={a}, B= Ø, C={a}

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Prove: A (B-A) = A B Proof: We must show that A (B-A) A B and A B A (B-A). First we will show that A (B-A) A B. Let e A (B-A). Then either e A or e (B-A). If e A, then e A B. Note that e cannot be an element of both by the definition of (B-A). If e (B-A), then e B and e A by the definition of (B-A). In this case, too, e A B. Thus A (B-A) A B.

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Prove: A (B-A) = A B (cont.) Next we will show that A B A (B-A). Let e A B. Then either e A or e B or both. If e A or both, then e A (B-A). The other case is e B, e A. In this case e (B-A) by the definition of (B-A). Again, this means that e A (B-A). Thus A B A (B-A). Therefore A (B-A) = A B.

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