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CSE20 Lecture 9

CSE20 Lecture 9 - CSE 20 Lecture 9 Boolean Algebra CK Cheng...

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CSE 20 Lecture 9 Boolean Algebra CK Cheng Feb 11, 2010 Lecture notes 1
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Theorems & Proofs P1: a+b = b+a, ab=ba P2: a+bc = (a+b)(a+c) a(b+c) = ab + ac P3: a + 0 = a, a1 = a P4: a + a’= 1, a a’= 0 2
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Theorem 6: for every a in B, (a')' = a Proof: A is complement of a'. The complement of a‘ is unique Thus a = (a')' Theorem 7: (Absorption Law) For every pair a,b in B, a·(a+b) = a, a + a·b = a Proof: a(a+b) a+a·b = (a+0)(a+b) (P3) = a·1 + a·b (P3) = a+0·b (P2) = a(1+b) (P2) = a + 0 (P3) = a·1 (P3) = a (P3) = a (P3) 3
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Theorem 8 For every pair a, b in B a + a’*b = a + b,a*(a’+ b) = a*b Proof: a + a’*b = (a + a’)(a + b) by P2 = (1)(a + b) by P4 = (a + b) by P3 a*(a’+ b) = a*a’+ a*b (by P3) = 0 + a*b (by P4) = a*b (by P3) 4
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Theorem 9: De Morgan’s Law Theorem: For every pair a, b in set B: (a+b)’= a’b’, and (ab)’= a’+b’. Proof: We show that a+b and a’b’are complementary. According to P4, both of the following have to be true: (a+b) +a’b’= 1, (a+b)(a’b’) = 0 5
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Theorem 9: De Morgan’s Law (cont.) statement justification (a+b)+a’b’ = 1 given (a+b+a’)(a+b+b’) = 1
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