boolean algebras lecture 27 section 5.3 wed, mar 7, 2007

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)— Complement (unary)

Properties of a Boolean Algebra

Commutative Lawsa + b = b + aa b = b a

Associative Laws(a + b) + c = a + (b + c)(a b) c = a (b c)

Properties of a Boolean Algebra

Distributive 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 = a

Properties of a Boolean Algebra

Complement Lawsa +a = 1a a = 0

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 = AA Ac = 1, A Ac = 0

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 = pp p = 1, p p = 0

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 = xx | x = 1, x & x = 0

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.

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 = AA Ac = 1, A Ac = 0

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.

Other Properties

The 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 = a

Other Properties

Universal Bounds Lawsa + 1 = 1a 0 = 0

DeMorgan’s Laws

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Other Properties

Absorption Lawsa + (a b) = aa (a + b) = a

Complements of 0 and 10 = 11 = 0

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.

The Idempotent Laws

Prove the other idempotent law

a 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) 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 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 ofa 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)) = alcm(a, gcd(a, b)) = a

If x and y are ints, thenx | (x & y) == xx & (x | y) == x