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# H-SC MATH 262 - Lecture 27 - Boolean Algebras

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Boolean AlgebrasSlide 2Properties of a Boolean AlgebraSlide 4Slide 5Set-Theoretic InterpretationLogic InterpretationBinary InterpretationOther InterpretationsSlide 10ConnectionsSlide 12DualityOther PropertiesSlide 15Slide 16The Idempotent LawsSlide 18The Laws of Universal BoundsSlide 20A Very Handy LemmaThe Lemma AppliedDeMorgan’s LawsSlide 24Slide 25The Other DeMorgan’s LawApplicationsBoolean AlgebrasLecture 27Section 5.3Wed, Mar 7, 2007Boolean AlgebrasIn a Boolean algebra, we abstract the basic properties of sets and logic and make them the defining properties.A Boolean algebra has three operators+ Addition (binary) Multiplication (binary)— Complement (unary)Properties of a Boolean AlgebraCommutative Lawsa + b = b + aa  b = b  aAssociative Laws(a + b) + c = a + (b + c)(a  b)  c = a  (b  c)Properties of a Boolean AlgebraDistributive Lawsa + (b  c) = (a + b)  (a + c)a  (b + c) = (a  b) + (a  c)Identity Laws: There exist elements, which we will label 0 and 1, that have the propertiesa + 0 = aa  1 = aProperties of a Boolean AlgebraComplement Lawsa +a = 1a a = 0Set-Theoretic InterpretationLet B be the power set of a universal set U.Interpret + to be ,  to be , and — to be complementation.Then what are the interpretations of 0 and 1?Look at the identity and complement laws:A  0 = A, A  1 = AA  Ac = 1, A  Ac = 0Logic InterpretationLet B be a collection of statements.Interpret + to be ,  to be , and — to be .Then what are the interpretations of 0 and 1?Look at the identity and complement laws:p  0 = p, p  1 = pp  p = 1, p  p = 0Binary InterpretationLet B be the set of all binary strings of length n.Interpret + to be bitwise “or,”  to be bitwise “and,” and — to be bitwise complement.Then what are the interpretations of 0 and 1?Look at the identity and complement laws:x | 0 = x, x & 1 = xx | x = 1, x & x = 0Other InterpretationsLet n be any positive integer that is the product of distinct primes. (E.g., n = 30.)Let B be the set of divisors of n.Interpret + to be gcd,  to be lcm, and — to be division into n.For example, if n = 30, then a + b = gcd(a, b)a  b = lcm(a, b)a = 30/a.Other InterpretationsThen what are the interpretations of “0” and “1”?Look at the identity and complement laws.a + “0” = gcd(a, “0”) = a, a  “1” = lcm(a, “1”) = a,a +a = gcd(a, 30/a) = “1”, a a = lcm(a, 30/a) = “0”.ConnectionsHow are all of these interpretations connected?Hint: The binary example is the most basic.Set-Theoretic InterpretationLet B be the power set of a universal set U.Reverse the meaning of + and  :+ means ,  means .Then what are the interpretations of 0 and 1?Look at the identity and complement laws:A  0 = A, A  1 = AA  Ac = 1, A  Ac = 0DualityOne can show that in each of the preceding examples, if we Reverse the interpretation of + and Reverse the interpretations of 0 and 1the result will again be a Boolean algebra.This is called the Principle of Duality.Other PropertiesThe other properties appearing in Theorem 1.1.1 on p. 14 can be derived as theorems.Double Negation LawThe complement ofa is a.Idempotent Lawsa + a = aa  a = aOther PropertiesUniversal Bounds Lawsa + 1 = 1a  0 = 0DeMorgan’s Lawsbabababa)()(Other PropertiesAbsorption Lawsa + (a  b) = aa  (a + b) = aComplements of 0 and 10 = 11 = 0The Idempotent LawsTheorem: Let B be a boolean algebra. For all a  B, a + a = a.Proof:a  a = a  a + 0= a  a + a a= a  (a +a)= a  1= a.The Idempotent LawsProve the other idempotent lawa a = a.The Laws of Universal BoundsTheorem: Let B be a boolean algebra. For all a  B, a + 1 = 1.Proof: a + 1 = a + (a +a)= (a + a) +a= a +a= 1.The Laws of Universal BoundsProve the other law of universal bounds:a 0 = 0.A Very Handy LemmaLemma: Let B be a boolean algebra and let a, b  B. If a + b = 1 and a  b = 0, then b =a.Proof:The Lemma AppliedCorollary: Let p and q be propositions. If p  q = T and p  q = F, then q = p. Corollary: Let A and B be sets. If A  B = U and A  B = , then B = Ac.Corollary: Let x and y be ints. If x | y == 1 and x & y == 0, then y == x.DeMorgan’s LawsTheorem: Let B be a boolean algebra. For all a, b  B, the complement of (a + b) equalsa b.Proof: We show that (a + b) + (a b) = 1 and that (a + b)  (a b) = 0.It will follow from the Lemma thata b is the complement of a + b.DeMorgan’s Laws(a + b) + (a b) = (a + b + a’).(a + b + b’)= (1 + b).(1 + a)= 1.1= 1.(a + b).(a’.b’) = a. a’.b’ + b. a’.b’= 0.b’ + 0.a’= 0 + 0= 0.DeMorgan’s LawsTherefore,a b is the complement of a + b.The Other DeMorgan’s LawProve the law that a +b is the complement of a  b.Prove the law of double negation, that the complement ofa is a.ApplicationsThese laws are true for any interpretation of a Boolean algebra.For example, if a and b are integers, then gcd(a, lcm(a, b)) = alcm(a, gcd(a, b)) = aIf x and y are ints, thenx | (x & y) == xx & (x | y) ==

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