1. Boolean Algebras Lecture 24 Section 5.3 Wed, Mar 22, 2006

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? The identity and complement laws would be interpreted as 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? Again, 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. 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.

13. Other Properties The other properties appearing in Theorem 1.1.1 on p. 14 can be derived as theorems. Double Negation Law   a Idempotent Laws a + a = a a  a = a

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

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

16. 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