1

2-2.

a) X Y + XY + XY =X + Y

= (XY+ X Y ) + (X Y + XY)

= X(Y + Y) + Y(X + X) +

= X + Y

Verification of DeMorgan’s Theorem

X YZ XYZ XYZ X+Y+Z

0 0 0 0 1 1

001 0 1 1

0 1 0 0 1 1

0 1 1 0 1 1

100 0 1 1

101 0 1 1

110 0 1 1

111 1 0 0

The Second Distributive Law

X Y Z YZ X+YZ X+Y X+Z (X+Y)(X+Z)

0 000000 0

001 0 0 0 1 0

0 100010 0

0 111111 1

100 0 1 1 1 1

101 0 1 1 1 1

110 0 1 1 1 1

111 1 1 1 1 1

X YZX

YYZXZXY+YZ+XZ XY YZ XZXY+YZ+XZ

0 00000 0 000 0

001010 1 001 1

0 10100 1 010 1

0 11100 1 001 1

100001 1 100 1

101010 1 100 1

110001 1 010 1

111000 0 000 0

XYZ X Y Z++=

XYZ+XY+()XZ+()⋅=

XY YZ XZ++ XY YZ XZ++=

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2

Problem Solutions – Chapter 2

b) A B+ B C + AB + B C = 1

= (A B+ AB) + (B C + B C)

= B(A + A) + B(C + C)

= B + B

= 1

c) Y + X Z + X Y = X + Y + Z

= Y + X Y + X Z

= (Y + X)(Y + Y) + X Z

= Y + X + X Z

= Y + (X + X)(X + Z)

= X + Y + Z

d) X Y + Y Z + XZ + XY + Y Z = X Y + XZ + Y Z

= X Y + Y Z(X + X) + XZ + XY + Y Z

= X Y + X Y Z + X Y Z + XZ + XY + Y Z

= X Y (1 + Z) + X Y Z +XZ + XY + Y Z

= X Y + XZ(1 + Y) + XY + Y Z

= X Y + XZ + XY (Z + Z)+ Y Z

= X Y + XZ + XY Z +Y Z (1 + X)

= X Y + XZ(1 + Y) + Y Z

= X Y + XZ + Y Z

2-7.

a) X Y + XYZ + XY = X + XYZ = (X + XY)(X + Z)

= (X + X)(X + Y)(X + Z) = (X + Y)(X + Z) = X + YZ

b) X + Y(Z + X Z) = X + YZ + X Y Z = X + (YZ + X)(YZ + YZ) = X + Y(X + YZ)

= X + XY + YZ = (X + X)(X + Y) + YZ = X + Y + YZ = X + Y

c) WX(Z + YZ) + X(W + W YZ) = WXZ + WXYZ + WX + WXYZ

= WX + WXZ + WXZ = WX + WX = X

d)

=

=

= A + C + A(BCD)

= A + C + BCD

= A + C + C(BD)

= A + C + BD

2-9.

a)

b)

c)

d)

AB AB+()CD CD+()AC+

ABCD ABCD ABCD ABCD A C+++++

ACABCD++

FAB+()AB+()=

FVW+()XY+()Z=

FWX+YZ+()YZ+()+[]WX+YZ YZ++[]=

FABCAB+()CABC+()++=

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Digital Logic Design

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20/07/2012