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Chapter 2 - Part 1 - PPT - Mano & Kime - 2nd EdSlide 1 Lecture 5 Basics of Boolean AlgebraLecture 5 Basics of Boolean Algebra
Slide 2 OutlineOutline Boolean Algebra Boolean Properties and Identities Boolean Algebraic Proofs Useful Theorems Algebraic Simplification Complementing Functions Function Evaluation
Slide 3 Boolean Algebra1. 3. 5. 7. 9. 11. 13. 15. 17. Commutative Associative Distributive DeMorgan s 2. 4. 6. 8. X . 1 X = X . 0 0 = X . X X = 0 = X . X 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: 10. 12. 14. 16. X + Y Y + X = (X + Y) Z + X + (Y Z) + = X(Y + Z) XY XZ + = X + Y X . Y = XY YX = (XY) Z X(Y Z) = X + YZ (X + Y) (X + Z) = X . Y X + Y = X + 0 X = + X 1 1 = X + X X = 1 = X + X X = X Define existence Of 0 and 1 Idempotence Existence of Complement Involution Dual Functions: Swap +/ · Swap 0/1
Slide 4 Some Properties of Identities & the AlgebraThe 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 If the meaning is unambiguous, we leave out the symbol · Some Properties of Identities & the Algebra 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.
Slide 5 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 = Example: H = A · B + A · C + B · C dual H = Are any of these functions self-dual? Some Properties of Identities & the Algebra (Continued) ((X+Y) · (W · Z)') = ((X+Y) ·(W' + Z') (A + B)(A + C)(B + C) (A + BC)(B + C) = AB + AC + BC
Dual G = ((X+Y) · (W · Z)') = ((X+Y) ·(W' + Z') Dual H = (A + B)(A + C)(B + C). Using the Boolean identities, = (A +BC) (B+C) = AB + AC + BC. So H is self-dual.
Slide 6 Boolean Operator PrecedenceBoolean 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)
Slide 7 Example 1: Boolean Algebraic ProofExample 1: Boolean Algebraic Proof A + A·B = A (Absorption Theorem) Proof Steps Justification A + A·B = A · 1 + A · B X = X · 1 = A · ( 1 + B) X · Y + X · Z = X ·(Y + Z) (Distributive Law) = A · 1 1 + X = 1 = A 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.
Slide 8 Example 2: Boolean Algebraic ProofsAB + 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 . X (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 Example 2: Boolean Algebraic Proofs
Justification 1: 1 . X = X Justification 2: 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 . X (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
Slide 9 Useful Theoremsx y × y Useful Theorems ( ) ( ) n inimizatio M y y y x y y y x = + + = × × ( ) tion Simplifica y x y x y x y x × = + × + = × + Consensus z y x z y z y x × + × = × + × + × ( ) ( ) ( ) ( ) ( ) z y x z y z y x + × + = + × + × + Laws s DeMorgan' x x × = + + x x x x x x x x y x = + y
Slide 10 Expression SimplificationExpression Simplification An application of Boolean algebra Simplify to contain the smallest number of literals (complemented and uncomplemented variables): = AB + ABCD + A C D + A C D + A B D = AB + AB(CD) + A C (D + D) + A B D = AB + A C + A B D = B(A + AD) +AC = B (A + D) + A C 5 literals + + + + D C B A D C A D B A D C A B A Absorption: X + XY = X Simplification: X + X Y = X + Y
Slide 11 Complementing FunctionsComplementing 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 = F = (x + y + z)(x + y + z) x + z y z y x
Slide 12 Boolean Function EvaluationBoolean Function Evaluation z x y x F4 x z y x z y x F3 x F2 xy F1 + = + = = = z y z + y + 1 0 0 1 1 1 0 0 0 1 0 1 1 1 0 0 0 0 0 0 0 0 1 0 0 1 0 0 1 1 1 1
F3 is 1 for x y z , x yz, xy z and xy z => F3 = 1,0,0,1,1,1,0,0 F4 is 1 for xy z , xy z, x y z and x y z => F4 = 0,1,0,1,1,1,0,0
Slide 13 SummarySummary Boolean Algebra Boolean Properties and Identities Boolean Algebraic Proofs Useful Theorems Algebraic Simplification Complementing Functions Function Evaluation
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