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3.1 Multiplying Out and Factoring Expressions 3.2 Exclusive-OR and Equivalence Operations 3.3 The Consensus Theorem 3.4 Algebraic Simplification of Switching Expressions
Example 2
Also called, the equivalence operation, A B A B 0 1 1 0 Exclusive OR (XOR) and Equivalence A B 0 0 0 1 1 0 1 1
A xnor B 1 0 0 1 Also written, A XOR B It is called equivalence operation because it is equal to 1 when A=B A B = A’B + AB’ it’s equal to 1 at terms 01 (A’B) and 10 (AB’) A B = AB + A’B’ it’s equal to 1 at terms 00 (A’B’) and 11 (AB)
Exclusive OR (XOR)
Equivalence Theorems (0 0) = 1 (0 1) = 0 (1 0) = 0 (1 1) = 1 (X Y) = X’Y’ + XY
Show by Boolean algebra that XOR and equivalence are the complement of each other (X Y)’ = (X’Y + XY’)’ = (X+Y’).(X’+Y) = XY + X’Y’ = (X Y)
Apply DeMorgan’s (A+B)’ = A’B’, (AB)’ = A’+B’ Apply (A+B)(A’+C) = AC + A’B Here, A=X, B=Y’, C=Y
Example 3 Simplify this expression F = A’ B C F = [(A’)’B + A’B’] C = (AB + A’B’) C = (AB + A’B’)’C + (AB + A’B’)C’ = (A’B + AB’)C + (AB + A’B’)C’ = A’BC + AB’C + ABC’ + A’B’C’ X Y = X’Y + XY’ AB + A’B’ is the complement of A’B + AB’ The first one is ‘equivalence’ or XNOR, the second one is XOR
Consensus Theorem XY + X’Z + YZ = XY + X’Z (3-20) Proof: XY + X’Z + YZ = XY + X’Z + (X+X’)YZ = XY + X’Z + XYZ + X’YZ = XY(1+Z) + X’Z(1+Y) = XY + X’Z Dual Form: (X + Y)(X’ + Z)(Y + Z) = (X + Y)(X’ + Z) (3-21)
Last Part from Exercises
Simplifying Boolean Expressions
1. Combining terms. Use the theorem XY + XY′ = X to combine two terms. 2. Eliminating terms. Use the theorem X + XY = X to eliminate redundant terms if possible; then try to apply the consensus theorem ( XY + X′Z + YZ = XY + X′Z ) to eliminate any consensus terms. abc′d′ + abcd′ = abd′ [X = abd′, Y = c] (3-24) Section 3.4 (p. 68-69) a′b + a′bc = a′b [X = a′b] a′bc′ + bcd + a′bd = a′bc′ + bcd [X = c, Y = bd, Z = a′b] (3-24)
Simplifying Boolean Expressions
4. Adding redundant terms. Chose to combine or eliminate other terms , ex. Add YZ to XY+XZ or XY to X WX + XY + X′Z′ + WY′Z′ (add WZ′ by consensus theorem) = WX + XY + X′Z′ + WY′Z′ + WZ′ (eliminate WY′Z′) = WX + XY + X′Z′ + WZ′ (eliminate WZ′) = WX + XY + X′Z′ (3-27)
XY + X’Z + YZ = XY + X’Z
