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