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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 AlgebrasIn 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 AlgebraCommutative Lawsa + b = b + aa b = b aAssociative Laws(a + b) + c = a + (b + c)(a b) c = a (b c)Properties of a Boolean AlgebraDistributive Lawsa + (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 propertiesa + 0 = aa 1 = aProperties of a Boolean AlgebraComplement Lawsa +a = 1a a = 0Set-Theoretic InterpretationLet 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 = AA Ac = 1, A Ac = 0Logic InterpretationLet 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 = pp p = 1, p p = 0Binary InterpretationLet 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 = xx | x = 1, x & x = 0Other InterpretationsLet 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 InterpretationsThen 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”.ConnectionsHow are all of these interpretations connected?Hint: The binary example is the most basic.Set-Theoretic InterpretationLet 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 = AA Ac = 1, A Ac = 0DualityOne 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 PropertiesThe other properties appearing in Theorem 1.1.1 on p. 14 can be derived as theorems.Double Negation LawThe complement ofa is a.Idempotent Lawsa + a = aa a = aOther PropertiesUniversal Bounds Lawsa + 1 = 1a 0 = 0DeMorgan’s Lawsbabababa)()(Other PropertiesAbsorption Lawsa + (a b) = aa (a + b) = aComplements of 0 and 10 = 11 = 0The Idempotent LawsTheorem: 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 LawsProve the other idempotent lawa a = a.The Laws of Universal BoundsTheorem: 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 BoundsProve the other law of universal bounds:a 0 = 0.A Very Handy LemmaLemma: 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 AppliedCorollary: 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 LawsTheorem: Let B be a boolean algebra. For all a, b B, the complement of (a + b) equalsa b.Proof: We show that (a + b) + (a b) = 1 and that (a + b) (a b) = 0.It will follow from the Lemma thata 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 LawsTherefore,a b is the complement of a + b.The Other DeMorgan’s LawProve the law that a +b is the complement of a b.Prove the law of double negation, that the complement ofa is a.ApplicationsThese laws are true for any interpretation of a Boolean algebra.For example, if a and b are integers, then gcd(a, lcm(a, b)) = alcm(a, gcd(a, b)) = aIf x and y are ints, thenx | (x & y) == xx & (x | y) ==

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