Boolean Algebras

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

2. 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)

3. 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)

4. 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

5. Properties of a Boolean Algebra • Complement Laws • a +a = 1 • a a = 0

6. 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

7. 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

8. 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

9. 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.

10. 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”.

11. Connections • How are all of these interpretations connected? • Hint: The binary example is the most basic.

12. 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

13. 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.

14. 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

15. Other Properties • Universal Bounds Laws • a + 1 = 1 • a 0 = 0 • DeMorgan’s Laws

16. Other Properties • Absorption Laws • a + (a b) = a • a  (a + b) = a • Complements of 0 and 1 • 0 = 1 • 1 = 0

17. 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.

18. The Idempotent Laws • Prove the other idempotent law a a = a.

19. 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.

20. The Laws of Universal Bounds • Prove the other law of universal bounds: a 0 = 0.

21. 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:

22. 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.

23. 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.

24. 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.

25. DeMorgan’s Laws • Therefore,a b is the complement of a + b.

26. 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.

27. 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