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# boolean algebras - PowerPoint PPT Presentation

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

Lecture 27

Section 5.3

Wed, Mar 7, 2007

• 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

•  Multiplication (binary)

• — Complement (unary)

• Commutative Laws

• a + b = b + a

• a  b = b  a

• Associative Laws

• (a + b) + c = a + (b + c)

• (a  b)  c = a  (b  c)

• 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

• Complement Laws

• a +a = 1

• a a = 0

• 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

• 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

• 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

• 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.

• 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”.

• How are all of these interpretations connected?

• Hint: The binary example is the most basic.

• 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

• 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.

• 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

• Universal Bounds Laws

• a + 1 = 1

• a 0 = 0

• DeMorgan’s Laws

• Absorption Laws

• a + (a b) = a

• a  (a + b) = a

• Complements of 0 and 1

• 0 = 1

• 1 = 0

• 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.

• Prove the other idempotent law

a a = a.

• 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.

• Prove the other law of universal bounds:

a 0 = 0.

• Lemma: Let B be a boolean algebra and let a, bB. If a + b = 1 and ab = 0, then b =a.

• Proof:

• 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.

• 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.

• (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.

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

• 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.

• 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