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  A c = 1, A  A c = 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  A c = 1, A  A c = 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 = A c. 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