1 / 10

Discrete Structures

Discrete Structures. Chapter 6: Set Theory 6.4 Boolean Algebras, Russell’s Paradox, and the Halting Problem. From the paradise created for us by Cantor, no one will drive us out. – David Hilbert (1862 – 1943). Definition: Boolean Algebra.

kolya
Download Presentation

Discrete Structures

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. Discrete Structures Chapter 6: Set Theory 6.4 Boolean Algebras, Russell’s Paradox, and the Halting Problem From the paradise created for us by Cantor, no one will drive us out. – David Hilbert (1862 – 1943) 6.4 Boolean Algebras, Russell’s Paradox, and the Halting Problem

  2. Definition: Boolean Algebra • A Boolean algebra is a set B together with two operations, generally denoted + and , such that for all a and b in B both a + b and a  b are in B and the following properties hold: 1. Commutative Laws: For all a and b in B, a. a+ b = b+ a b. ab = ba 2. Associative Laws: For all a, b,and c in B, a. (a+ b) + c = A+ (b + c) b. (ab) c = a (bc) 6.4 Boolean Algebras, Russell’s Paradox, and the Halting Problem

  3. Definition: Boolean Algebra • A Boolean algebra is a set B together with two operations, generally denoted + and , such that for all a and b in B both a + b and a  b are in B and the following properties hold: 3. Distributive Laws: For all a, b,and c in B, a. a+ (bc)= (a+ b)  (a+ c) b. a (b + c)= (ab) + (ac) 4. Identity Laws:  distinct elements 0 and 1 in B, s.t.  a  B, a. a+ 0 = A b. a1 = A 6.4 Boolean Algebras, Russell’s Paradox, and the Halting Problem

  4. Definition: Boolean Algebra • A Boolean algebra is a set B together with two operations, generally denoted + and , such that for all a and b in B both a + b and a  b are in B and the following properties hold: 5. Complement Laws:  a  B, there exists an element in B denoted ā and called the complement or negation of a such that, a. a+ ā = 1 b. aā = 0 6.4 Boolean Algebras, Russell’s Paradox, and the Halting Problem

  5. Theorem 6.4.1 – Properties of Boolean Algebra Let B be any Boolean algebra. 1. Uniqueness of the Complement Law: For all a and x in B, if a + x =1 and a x = 0, then x = ā. 2. Uniqueness of 0 and 1: If there exists x in B such that for all a in B, then x = 0, and if there exists y in B such that a y = a for all a in B, then y = 1. 6.2 Properties of Sets

  6. Theorem 6.4.1 – Properties of Boolean Algebra Let B be any Boolean algebra. 3. Double Complement Law: For all aB, 6.2 Properties of Sets

  7. Theorem 6.4.1 – Properties of Boolean Algebra Let B be any Boolean algebra. 4. Idempotent Laws: For all a  B, a. A+ A = A b. AA = A 5. Universal Bound Laws: For all a  B,, a. a+ 1 = 1 b. a0 = 0 6.2 Properties of Sets

  8. Theorem 6.4.1 – Properties of Boolean Algebra Let B be any Boolean algebra. 6. De Morgan’s Laws: For all a and b B, a. b. 7. Absorption Laws: For all a and b B,, a. (a + b)  a = a b. (ab) + a = a 6.2 Properties of Sets

  9. Theorem 6.4.1 – Properties of Boolean Algebra Let B be any Boolean algebra. 6. Complements of 0 and 1: a. b. 6.2 Properties of Sets

  10. Examples – pg. 372 • For the examples below, assume that B is a Boolean algebra with operations + and . Prove each statements without using any parts of Theorem 6.4.1 unless they have already been proved. You may use any part of the definition of Boolean algebra and the results of previous exercises. 6.4 Boolean Algebras, Russell’s Paradox, and the Halting Problem

More Related