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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

- 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 = a

Properties of a Boolean Algebra

- Complement Laws
- a +a = 1
- a 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 = A
- A 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 = p
- p 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 = x
- x | 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 = A
- A 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 Law
- The complement ofa is a.

- Idempotent Laws
- a + a = a
- a a = a

Other Properties

- Universal Bounds Laws
- a + 1 = 1
- a 0 = 0

- DeMorgan’s Laws

Other Properties

- Absorption Laws
- a + (a b) = a
- a (a + b) = a

- Complements of 0 and 1
- 0 = 1
- 1 = 0

The Idempotent Laws

- Theorem: Let B be a boolean algebra. For all aB, a + a = a.
- Proof:
- aa = aa + 0
= aa + aa

= a (a +a)

= a 1

= a.

- aa = aa + 0

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 aB, a + 1 = 1.
- Proof:
- a + 1 = a + (a +a)
= (a + a) +a

= a +a

= 1.

- a + 1 = a + (a +a)

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, bB. If a + b = 1 and ab = 0, then b =a.
- Proof:

The Lemma Applied

- Corollary: Let p and q be propositions. If pq = T and p q = F, then q = p.
- Corollary: Let A and B be sets. If AB = 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, bB, 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 thata +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)) = 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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