1 / 27

Boolean Algebras

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)

Olivia
Télécharger la présentation

Boolean Algebras

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related