Unit 2 Solutions
17
(A + B + C + D) (A + B + C + E) (A + B + C + F)
= A + B + C + DEF
Apply second Distributive Law twice
See FLD p. 731 for solution.
2.1
Unit 2 Problem Solutions
Answer is in FLD p. 731
2.3
AB + C’D’ = (AB + C’) (AB + D’)
= (A + C’) (B + C’) (A + D’) (B + D’)
2.6 (a) WX + WY’X + ZYX = X(W + WY’ + ZY)
= X(W + ZY) {By Absorption}
= X(W +Z) (W + Y)
2.6 (b)
X
X Y
X
2.7 (a) WXYZ + VXYZ + UXYZ = XYZ (W + V + U)
By first Distributive Law
2.7 (b)
Y
X
Y Z
X
Z
X
Unit 2 Solutions
18
(A’ + B’ + C)(A’ + B’ + C)‘ = 0 By
Complementarity Law
AB + (C’ + D)(AB)‘ = AB + C’ + D
By Elimination Theorem
[AB’ + (C + D)‘ +E’F](C + D)
= AB’(C + D) + E’F(C + D) Distributive Law
F1 = A’A + B + (B + B) = 0 + B + B = B
F3 = [(AB + C)‘D][(AB + C) + D]
= (AB + C)‘D (AB + C) + (AB + C)‘ D
= (AB + C)‘ D By Absorption
AB(C’ + D) + B(C’ + D) = B(C’ + D) By
Absorption
(A’BF + CD’)(A’BF + CEG) = A’BF + CD’EG
By Distributive Law
A’(B + C)(D’E + F)‘ + (D’E + F)
= A’(B + C) + D’E + F By Elimination
2.11 (a)
2.13 (a) F2 = A’A’ + AB’ = A’ + AB’ = A’ + B’
Z = [(A + B)C]‘ + (A + B)CD = [(A + B)C]‘ + D
By Elimination with X = [(A + B) C]‘
= A’B’ + C’ + D’
2.11 (c)
2.11 (e)
2.11 (b)
2.11 (d)
2.11 (f)
2.13 (c)
2.13 (b)
2.13 (d)
X
Y
X
2.10 (a) 2.10 (b)
X
YX Y
Unit 2 Solutions
19
2.18 (b)
2.18 (d)
sum-of-products
sum term, sum-of-products, product-of-sums
2.18 (c)
2.18 (e)
none apply
product-of-sums
F = D[(A’ + B’ )C + AC’ ]
= D(A’ + B’ + AC’ )(C + AC’ )
2.20 (c)
A B C H F G
0 0 0 0 0 0
2.21
A’B’ + A’CD + A’DE’
= A’(B’ + CD + DE’)
= A’[B’ + D(C + E’)]
= A’(B’ + D)(B’ + C + E’)
2.22 (a)
A’B’ + (CD’ + E) = A’B’ + (C + E)(D’ + E)
= (A’B’ + C + E)(A’B’ + D’ + E)
= (A’ + C + E)(B’ + C + E)
(A’ + D’ + E)(B’ + D’ + E)
2.22 (d)
2.23 (a) 2.23 (b)
W + U’YV = (W + U’)(W + Y)(W + V)TW + UY’ + V
= (T+U+Z)(T+Y’+V)(W+U+V)(W+Y’+V)
Unit 2 Solutions
H = [W’X’(Y’ + Z’)]‘ = W + X + YZ
2.27
2.28 (a)
F = (V + X + W) (V + X + Y) (V + Z)
= (V + X + WY)(V + Z) = V + Z (X + WY)
By Distributive Law with X = V
F = ABC + A’BC + AB’C + ABC’
Beginning with the answer to (a):
2.28 (b)
2.26 (c)
+
+
W
YXZVF
[(XY’)‘ + (X’ + Y)‘Z] = X’ + Y + (X’ + Y)‘Z
= X’ + Y’+ Z By Elimination Theorem with X
= (X’ + Y)
2.24 (a) (X + (Y’(Z + W)‘)‘)‘ = X’Y’(Z + W)‘ = X’Y’Z’W‘
2.24 (b)
X Y Z X+Y X’+Z (X+Y)
(X’+Z)
XZ X’Y XZ+X’Y
0 0 0 0 1 0 0 0 0
0 0 1 0 1 0 0 0 0
2.29 (b)2.29 (a) X Y Z X+Y Y+Z X’+Z (X+Y)
(Y+Z)
(X’+Z)
(X+Y)
(X’+Z)
0 0 0 0 0 1 0 0
Unit 2 Solutions
21
2.29 (d)
2-29 (c)
W X Y Z W’XY WZ W’XY+WZ W’+Z W+XY (W’+Z)(W+XY)
0 0 0 0 0 0 0 1 0 0
0 0 0 1 0 0 0 1 0 0
0 0 1 0 0 0 0 1 0 0
0 0 1 1 0 0 0 1 0 0
2.29 (e)
X Y Z XY YZ X’Z XY+YZ+X’Z XY+X’Z
0 0 0 0 0 0 0 0
0 0 1 0 0 1 1 1
0 1 0 0 0 0 0 0
A B C A+C AB+C’ (A+C)
(AB+C’)
AB AC’ AB
+AC’
0 0 0 0 1 0 0 0 0
0 0 1 1 0 0 0 0 0
2.30 F = (X+Y’)Z + X’YZ’ (from the circuit)
= (X+Y’+ X’YZ’)( Z+X’YZ’) (Distributive Law)
= (X+Y’+X’)(X+Y’+Y)(X+Y’+Z’)(Z+X’)(Z+Y)(Z+Z’) (Distributive Law)