Lec0201sol - (Absorption Law = A B C A U(Identity Law = A B C A(B B(Inverse Law = A B C A B C A B(Distributive Law = A B C A B C A B(Commutative

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Proof of the following set equivalence using a Set Table. ¬ A ¬ B (A B ¬ C) = ¬ (A B C) A B C ¬ (A B C) 0 0 0 1 0 0 1 1 0 1 0 1 0 1 1 1 1 0 0 1 1 0 1 1 1 1 0 1 1 1 1 0 A B C ¬ A ¬ B A B ¬ C ¬ A ¬ B (A B ¬ C) 0 0 0 1 1 0 1 0 0 1 1 1 0 1 0 1 0 1 0 0 1 0 1 1 1 0 0 1 1 0 0 0 1 0 1 1 0 1 0 1 0 1 1 1 0 0 0 1 1 1 1 1 0 0 0 0 Thus, the two sets are equivalent. Using the Set Laws: ¬ (A B C) = ¬ (A B) ¬ C (De Morgan’s) = ( ¬ A ¬ B) ¬ C (De Morgan’s) = ¬ A ¬ B ( ¬ C U) (Identity Law) = ¬ A ¬ B ( ¬ C (A ¬ A) ) (Inverse Law) = ¬ A ¬ B ( ¬ C A) ( ¬ C ¬ A) (Distributive Law) = ( ¬ A ( ¬ A ¬ C)) ¬ B ( ¬ C A) = ¬ A ¬ B ( ¬ C A)
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Unformatted text preview: (Absorption Law) = A B (( C A) U) (Identity Law) = A B (( C A) (B B)) (Inverse Law) = A B ( C A B) ( C A B) (Distributive Law) = A B ( C A B) ( C A B) (Commutative Law) = A B ( C A B) (Absorption Law)...
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This document was uploaded on 07/14/2011.

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