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lec10 (1) - CSE 20 Discrete Math Lecture 10 CK Cheng 1...

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1 CSE 20 – Discrete Math Lecture 10 CK Cheng 4/28/2010

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2 Boolean Algebra Theorem: (Associative Laws) For elements a, b, c in B, we have a+(b+c)=(a+b)+c; a*(b*c)=(a*b)*c. Proof: Denote x = a+(b+c), y = (a+b)+c. We want to show that (1) ax = ay, (2) a’x = a’y. Then, we have x= 1*x=(a+a’)x= ax+a’x = ay+a’y= (a+a’)y= 1*y=y Shannon’s Expansion: (divide and conquer) We use a variable a to divide the term x into
3 Proof of (1) ax = ay ax = a (a+(b+c)) = a a + a (b+c) (Distributive) = a + a (b+c) (Idempotence) = a (Absorption) ay = a ((a+b)+c)) = a (a+b) + a c (Distributive) = (aa + ab) + ac (Distributive) = (a + ab) + ac (Idempotence) = a + ac (Absorption) = a (Absorption)

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4 Proof of (2) a’x = a’y a’x = a’ (a+(b+c)) = a’ a + a’ (b+c) (Distributive) = 0+ a’(b+c) (Complementary) = a’(b+c) (Identity) a’y = a’ ((a+b)+c) = a’ (a+b) + a’ c (Distributive) = a’b+a’c (Theorem 8) = a’(b+c) (Distributive) Therefore: a’x = a’(b +c) = a’y
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lec10 (1) - CSE 20 Discrete Math Lecture 10 CK Cheng 1...

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