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Boolean Algebras. Lecture 27 Section 5.3 Wed, Mar 7, 2007. Boolean 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)

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Boolean algebras l.jpg

Boolean Algebras

Lecture 27

Section 5.3

Wed, Mar 7, 2007


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Boolean 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)


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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)


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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 = a


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Properties of a Boolean Algebra

  • Complement Laws

    • a +a = 1

    • a a = 0


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Set-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 = 0


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Logic 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 = 0


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Binary 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 = 0


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Other 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.


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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”.


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Connections

  • How are all of these interpretations connected?

  • Hint: The binary example is the most basic.


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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 = 0


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Duality

  • One can show that in each of the preceding examples, if we

    • Reverse the interpretation of + and 

    • Reverse the interpretations of 0 and 1

      the result will again be a Boolean algebra.

  • This is called the Principle of Duality.


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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 = a


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Other Properties

  • Universal Bounds Laws

    • a + 1 = 1

    • a 0 = 0

  • DeMorgan’s Laws


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Other Properties

  • Absorption Laws

    • a + (a b) = a

    • a  (a + b) = a

  • Complements of 0 and 1

    • 0 = 1

    • 1 = 0


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The 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.


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The Idempotent Laws

  • Prove the other idempotent law

    a a = a.


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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.


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The Laws of Universal Bounds

  • Prove the other law of universal bounds:

    a 0 = 0.


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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:


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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.


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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.


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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.


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DeMorgan’s Laws

  • Therefore,a b is the complement of a + b.


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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.


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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) == x


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