Boolean AlgebrasSlide 2Properties of a Boolean AlgebraSlide 4Slide 5Set-Theoretic InterpretationLogic InterpretationBinary InterpretationOther InterpretationsSlide 10ConnectionsDualityOther PropertiesSlide 14Slide 15ApplicationsBoolean AlgebrasLecture 24Section 5.3Wed, Mar 22, 2006Boolean 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?The identity and complement laws would be interpreted as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?Again, 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.Duality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.Other PropertiesThe other properties appearing in Theorem 1.1.1 on p. 14 can be derived as theorems.Double Negation Law  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 = 0Applications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) ==