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Unit 3 Solutions
23
3.10 (c)
Unit 3 Problem Solutions
(W + X’ + Z’) (W’ + Y’) (W’ + X + Z’) (W + X’) (W + Y + Z)
= (W + X’) (W’ + Y’) (W’ + X + Z’) (W + Y + Z)
3.6 (a)
3.7 (a) BCD + C’D’ + B’C’D + CD
= CD + C’(D’ + B’D) = (C’ + D) [C + (D’ + B’D)] {Using (X + Y) (X’ + Z) = X’Y + XZ with X = C}
= (C’ + D) [C + (D’ + B’) (D’ + D)] = (C’ + D) (C + D’ + B’)
3.9 A ⊕ BC = (A ⊕ B) (A ⊕ C) is not a valid distributive law. PROOF: Let A = 1, B = 1, C = 0.
LHS: A ⊕ BC = 1⊕ 1 · 0 = 1 ⊕ 0 = 1. RHS: (A⊕ B) (A ⊕ C) = (1 ⊕ 1) (1 ⊕ 0) = 0 · 1 = 0.
(A’ + C’ + D’) (A’ + B + C’) (A + B + D) (A + C + D)
= (A’ + C’ + D’) (B + C’ + D) (A’ + B + C’) (A + B + D) (A + C + D) Add consensus term
= (A’ + B + C’) (A + B + D)
= (A’ + C’ + D’) (B + C’ + D) (A + C + D) Removing consensus terms
Unit 3 Solutions
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3.13 (a) KLMN’ + K’L’MN + MN’ = K’L’MN + MN’ = M(K’L’N + N’) = M(N’ + K’L’) {Elimination with X = N’} = MN’ + K’L’M
3.11 (A + B’ + C + E’) (A + B’ + D’ + E) (B’ + C’ + D’ + E’) = [A + B’ + (C + E’) (D’ + E)] (B’ + C’ + D’ + E’)
= (A + B’ + D’E’ + CE) (B’ + C’ + D’ + E’) = B’ + (A + D’E’ + CE) (C’ + D’ + E’)
(K’ + L’ + M’) (K + M + N’) (K + L) (K’ + N) (K’ + M + N)
= [K’ + N (L’ + M’)] [K + L ( M + N’)] = KN (L’ + M’) + K’L (M + N’) = KNL’ + KNM’ + K’LM + K’LN’
3.15 (b)
KL’M’ + MN’ + LM’N’ = KL’M’ + N’(M + LM’) = KL’M’ + N’(M + L) = KL’M’ + MN’ + LN’
3.13 (b)
(K + L’) (K’ + L’ + N) (L’ + M + N’) = L’ + K (K’ + N) (M + N’) = L’ + KN (M + N’) = L’ + KMN
3.13 (c)
KL + K’L’ + L’M’N’ + LMN’ = L’ (K’ + M’N’) + L (K + MN’)
= (L + K’ + M’N’) (L’ + K + MN’) {Multiplying out and factoring with X = L}
= (L + K’ + M’) (L + K’ + N’) (L’ + K + M) (L’ + K + N’)
3.14 (b)
KL + K’L’M + L’M’N + LM’N’ = L’ [K’M + M’N] + L [K + M’N’] = L’ [(M + N) (M’ + K’)] + L [(K + M’) (K + N’)]
= [L + (M + N) (M’ + K’)][L’ + (K + M’) (K + N’)] = (L + M + N) (L + M’ + K’) (L’ + K + M’) (L’ + K + N’)
3.14 (c)
Unit 3 Solutions
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(B + C + D) (A + B + C) (A’ + C + D) (B’ + C’ + D’) = (A + B + C) (A’ + C + D) (B’ + C’ + D’)
3.21 (c)
W’XY + WXZ + WY’Z + W’Z’ = W’XY + WXZ + WY’Z +W’Z’ + XYZ = WY’Z + W’Z’ + XYZ
XYZ (add consensus term)
3.21 (d)
M’(K ⊕ N’) + MN + K’N = M’ [K’N’ + KN] + MN + K’N = K’M’N’ + KM’N + MN + K’N
= K’M’N’ + N(M + KM’ + K’)
= K’M’N’ + N(M + K’ + M’) = K’M’N’ + N = N + K’M’ = (K’ + N)(M’ + N)
3.16 (b)
(a) x ≡ 0 = x(0) + x’(0)‘ = x’
(b) x ≡ 1 = x(1) + x’(1)‘ = x
3.17
(a) x ⊕ 0 = x(0)‘ + x’(0) = x
(b) x ⊕ 1 = x(1)‘ + x’(1)‘ = x’
3.18
(a) x ⊕ y ⊕ xy = x ⊕ [y(xy)‘ + y’(xy)] = x ⊕ [yx’] = x(yx’)‘ + x’(yx’) = x(y’+x) +x’y = x + x’y = x + y
(b) x ≡ y ≡ xy = (xy + x’y’) ≡ xy = (xy + x’y’)xy + (xy + x’y’)‘(xy)‘ = xy + (xy’ +x’y)(x’ + y’) = xy + x’y + xy’
= x + y
3.19
(a) xy ⊕ xz = xy(x’ + z’) + (x’ + y’)xz = xyz’ + xyz = x(yz’ + yz’) = x(y ⊕ z)
3.20
(K + L + M) (K’ + L’ + N’) (K’ + L’ + M’) (K + L + N) = ( K + L + MN) (K’ + L’ + M’N’)
= K ( L’ + M’N’) + K’( L + MN) {Multiplying out and factoring with X = K} = KL’ + KM’N’ + K’L + K’MN
3.15 (d)
Unit 3 Solutions
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3.23
3.24 X’Y’Z’ + XYZ = (X + Y’Z’) (X’ + YZ) = (X + Y’) (X + Z’) (X’ + Y) (X’ + Z) (Y + Z’)
= (X + Y’) (X + Z’) (X’ + Y) (X’ + Z) (Y + Z’) = (X + Y’) (X + Z’) (X’ + Z) (Y + Z’)
= (X + Y’) (X’ + Z) (Y + Z’)
Alt.: (X’ + Y) (Y’ + Z) (X + Z’) by adding (Y’ + Z) as consensus in 3rd step
3.25 (c) xy’ + z + (x’ + y) z’
= x’y + (x’ + y) {Elimnation with Y = z}
= xy’ + x’ + y = x + x’ + y = 1 + y = 1
Alt.: xy’ + z + (x’ + y) z’ = (xy’ + z) + (xy’ + z)‘ = 1
3.25 (d)
a’d (b’ + c) + a’d’ (b + c’) +(b’ + c) (b + c’)
= a’b’d + a’cd + a’bd’ + a’c’d’ + b’c’ + bc
F = A’B + AC + BC’D’ + BEF + BDF = (A + B) (A’ + C) + B (C’D’ + EF + DF)
= [(A + B) (A’ + C) + B] [(A + B) (A’ + C) + C’D’ + EF + DF]
= (A + B) (A’ + C + B) (A + B + C’D’ + EF + DF) (A’ + C + C’D’ + EF + DF)
B + C C + D’
A’BC’ + BC’D’ + A’CD + B’CD + A’BD = BC’D’ + B’CD + A’BD
3.21 (e)
Unit 3 Solutions
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WXY’ + (W’Y’ ≡ X) + (Y ⊕ WZ)
= WXY’ + W’Y’X + (W’Y’)‘ X’ + Y (WZ)‘ + Y’WZ
= WXY’ + W’XY’ + (W + Y) X’ + Y (W’ + Z’) + Y’WZ
3.27 3.28 (a)
3.28 (b) NOT VALID. Counterexample: a = 0, b = 1, c = 0.
LHS = 0, RHS = 1. ∴ This equation is not always
valid.
3.28 (c) VALID. Starting with the right side, add consensus
terms
RHS = abc + ab’c’ + b’cd + bc’d + acd + ac’d
VALID: a’b + b’c + c’a
= a’b (c + c’) + (a + a’) b’c + (b + b’) ac’
= a’bc + a’bc’ + ab’c + a’b’c + abc’ + ab’c’
3.28 (d) VALID: LHS = xy’ + x’z + yz’
consensus terms: y’z, xz’, x’y
= xy’ + x’z + yz’ + y’z + xz’ + x’y
= y’z + xz’ + x’y = RHS
3.28 (e) NOT VALID. Counterexample: x = 0, y = 1, z = 0,
then LHS = 0, RHS = 1. ∴ This equation is not
always valid. In fact, the two sides of the equations
are complements.
LHS = (x + y) (y + z) (x + z)
= [(x + y)‘ + (y + z)‘ + (x + z)‘]‘
= (x’y’ + y’z’ + x’z’)‘ = [x’ (y’ + z’) + y’z’]‘
3.25 (f) 3.25 (g) [(a’ + d’ + b’c) (b + d + ac’)]‘ + b’c’d’ + a’c’d
= ad (b + c’) + b’d’ (a’ + c) +b’c’d’ + a’c’d
= abd + ac’d+ a’b’d’ + b’cd’ + b’c’d’ + a’c’d
c’d b’d’
= abd + a’b’d’ + b’d’ + c’d = abd + b’d’ + c’d
A’BCD + A’BC’D+ B’EF+ CDE’G+A’DEF+A’B’EF
= A’BD + B’EF + CDE’G + A’DEF (consensus)
= A’BD + B’EF + CDE’G
Unit 3 Solutions
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VALID. [A + B = C] ⇒ [D’ (A + B) = D’(C)]
[A + B = C] ⇒ [AD’ + BD’ = CD’]
3.32 (a) NOT VALID. Counterexample: A = 1, B = C = 0
and D = 1 then LHS = (0)(0) + (0)(0) = 0
RHS = (0)(1) = 0 = LHS
but B + C = 0 + 0 = 0; D = 1 ≠ B + C
∴ The statement is false.
3.32 (b)
3.31 (b) LHS = (W’ + X + Y’) (W + X’ + Y) (W + Y’ + Z) = (W’ + X + Y’) (W + (X’ + Y) (Y’ + Z))
= (W’ + X + Y’) (W + (X’Y’ + YZ)) = (W’ (X’Y’ + YZ) + W (X + Y ‘)) = W’X’Y’ + W’YZ + WX + WY’
consensus terms: X’Y’ XYZ
SUM = (X ⊕ Y) ⊕ Ci = (XY’ + X’Y) ⊕ Ci
3.29 X Y Ci SUM Co
1
1 0 0 1
1
1 1 1 1
VALID:
LHS = (X’ + Y’) (X ⊕ Z) + (X + Y) (X ⊕ Z)
3.31 (a)
A B C F
0
0 0 0
0
0 1 0
3.30
Unit 3 Solutions
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A’C’ + BC + AB’ + A’BD + B’C’D’ + ACD’
Consensus terms: (1) B’C’ using A’C’ + AB’
3.33 (a) A’C’D’ + BC’D + AB’C’ + A’BC
Consensus terms:
3.33 (b)
3.34 abd’f’ + b’cegh’ + abd’f + acd’e + b’ce
= (abd’f’ + abd’f) + (b’cegh’ + b’ce) + acd’e
= abd’ + b’ce + acd’e
= abd’ + b’ce (consensus)
= (b + ce)(b’ + ad’)
= (b + c)(b + e)(b’ + a)(b’ + d’)
3.35 (a + c)(b’+ d)(a + c’+ d’)(b’+ c’+ d’)
= (a + cd’)(b’ + c’d)
= ab’+ ac’d + b’cd’
3.36 abc’ + d’e + ace + b’c’d’
= (d’ + abc’ + ace + b’c’d’)(e + abc’ + ace + b’c’d’)
= (d’ + abc’ + ace)(e + abc’ + b’c’d’)
3.37 (a) (x ≡ y)‘ = (xy + x’y’)‘ = (x’ + y’)(x + y)
= x’y + xy’ = x ⊕ y
(b) a’b’c’ + a’bc + ab’c + abc’
(a) x(y + a′) = x(y + b′) implies that
0 = x(y + a′) ⊕ x(y + b′) = x(y + a′)(x′ + y′b) +
(x′ + y′a)x(y + b′) = xy′b(y + a′) + xy′a(y + b′) =
xy′a′b + xy′ab′ = xy′(a′b + ab′) = xy′(a ⊕ b)
which implies that a = b.
3.38 Let A = [(a + b) + c][a + (b + c)]
= a[(a + b) + c] + (b +c) [(a + b) + c]
= [(a + ab) + ac] + [(a + b)(b +c) + c(b +c)]
= a + [(a + b)(b + c) + c]
= a + [{(a + b)b + (a + b)c} + c]
3.39
Unit 3 Solutions
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