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ECE/CS 352: Digital System Fundamentals Lecture 5 – Basics of Boolean Algebra Based on slides by:Charles Kime & Thomas Kaminski © 2004 Pearson Education, Inc. Terms of Use (Hyperlinks are active in View Show mode) Outline Boolean Algebra Boolean Properties and Identities Boolean Algebraic Proofs Useful Theorems Algebraic Simplification Complementing Functions Function Evaluation Chapter 2 2 Boolean Algebra An algebraic structure defined on a set of at least two elements, B, together with three binary operators (denoted +, · and ) that satisfies the following basic identities: 1. 3. 5. 7. 9. X + 0 = X Define existence 2. X+1 =1 Of 0 and 1 X+X =X Idempotence Existence of X+X =1 Complement X = X Involution 10. X + Y = Y + X 12. (X + Y) + Z = X + (Y + Z) 14. X(Y + Z) = XY + XZ 16. X + Y = X . Y 4. 6. 8. X .1 =X X .0 =0 X .X = X Dual Functions: 1. Swap +/ · 2. Swap 0/1 X .X = 0 11. XY = YX Commutative Associative 13. (XY) Z = X(YZ) 15. X + YZ = (X + Y) (X + Z) Distributive DeMorgan’s 17. X . Y = X + Y Chapter 2 3 Some Properties of Identities & the Algebra If the meaning is unambiguous, we leave out the symbol “·” The identities above are organized into pairs. These pairs have names as follows: 1-4 Existence of 0 and 1 5-6 Idempotence 7-8 Existence of complement 9 Involution 10-11 Commutative Laws 12-13 Associative Laws 14-15 Distributive Laws 16-17 DeMorgan’s Laws The dual of an algebraic expression is obtained by interchanging + and · and interchanging 0’s and 1’s. The identities appear in dual pairs. When there is only one identity on a line the identity is self-dual, i. e., the dual expression = the original expression. Chapter 2 4 Some Properties of Identities & the Algebra (Continued) Unless it happens to be self-dual, the dual of an expression does not equal the expression itself. Example: F = (A + C) · B + 0 dual F = (A · C + B) · 1 = A · C + B Example: G = X · Y + (W + Z) dual G =((X+Y) · (W · Z)') = ((X+Y) ·(W' + Z') Example: H = A · B + A · C + B · C dual H = (A + B)(A + C)(B + C) Are any of these functions self-dual? (A + BC)(B + C) = AB + AC + BC Chapter 2 5 Boolean Operator Precedence The order of evaluation in a Boolean expression is: 1. Parentheses 2. NOT 3. AND 4. OR Consequence: Parentheses appear around OR expressions Example: F = A(B + C)(C + D) Chapter 2 6 Example 1: Boolean Algebraic Proof A + A·B = A (Absorption Theorem) Proof A + = A · = A · Justification Steps A·B 1 + A · B ( 1 + B) = A · 1 = A X = X · 1 X · Y + X · Z = X ·(Y + Z) (Distributive Law) 1 + X = 1 X · 1 = X Our primary reason for doing proofs is to learn: • • Careful and efficient use of the identities and theorems of Boolean algebra, and How to choose the appropriate identity or theorem to apply to make forward progress, irrespective of the application. Chapter 2 7 Example 2: Boolean Algebraic Proofs AB + A’C + BC = AB + A’C (Consensus Theorem) Proof Steps Justification AB + A’C + BC = AB + A’C + 1 · BC 1 . X = X = AB + A’C + (A + A’) · BC X + X’ = 1 = AB + A’C + ABC + A’BC X(Y + Z) = XY + XZ (Distributive Law) = AB + ABC + A’C + A’BC X + Y = Y + X (Commutative Law) = AB . 1 + ABC + A’C . 1 + A’C . B X . 1 = X, X . Y = Y (Commutative Law) = AB (1 + C) + A’C (1 + B) X(Y + Z) = XY +XZ (Distributive Law) = AB . 1 + A’C . 1 = AB + A’C X . 1 = X Chapter 2 . X 8 Useful Theorems x ×y + x ×y = y (x + y )(x + y )= y x (x + y ) = x x + xy = x x + x ×y = x + y x ×(x + y )= x ×y x ×y + x ×z + y ×z = x ×y + x ×z Minimization Absorption Simplification Consensus (x + y )×(x + z )×(y + z ) = (x + y )×(x + z ) x + y = x ×y x ×y = x + y DeMorgan' s Laws Chapter 2 9 Expression Simplification An application of Boolean algebra Simplify to contain the smallest number of literals (complemented and uncomplemented variables): Absorption: + A B AXC+D A BD + AC D + A BCD XY+ =X = AB + ABCD + A C D + A C D + A B D Simplification: + X’Y Y +ABD = AB + AB(CD) + AXC (D=+X +D) = AB + A C + A B D = B(A + AD) +AC = B (A + D) + A C 5 literals Chapter 2 10 Complementing Functions Use DeMorgan's Theorem to complement a function: 1. Interchange AND and OR operators 2. Complement each constant value and literal Example: Complement F = xy z + x y z F = (x + y + z)(x + y + z) Chapter 2 11 Boolean Function Evaluation F1 = xy z F2 = x + yz F3 = x y z + x y z + x y F4 = x y + x z x 0 0 0 0 1 1 1 1 y 0 0 1 1 0 0 1 1 z F1 0 0 1 0 0 0 1 0 0 0 1 0 0 1 1 0 F2 F3 F4 0 1 0 0 1 1 1 1 1 0 0 1 1 1 0 0 0 1 0 1 1 1 0 0 Chapter 2 12 Summary Boolean Algebra Boolean Properties and Identities Boolean Algebraic Proofs Useful Theorems Algebraic Simplification Complementing Functions Function Evaluation Chapter 2 13