Computer Science
Simplify the following expression and convert it into its canonical POS form:
(X.Y + Z)(Y + Z'.X)
Boolean Algebra
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Answer
Simplifying the expression:
(X.Y + Z)(Y + Z'.X)
= (X + Z).(Y + Z).(Y + Z').(X + Y)
= (XY + YZ + XZ + Z).(Y + Z').(X + Y)
= [XY + YZ + Z(X + 1)].(Y + Z').(X + Y)
= (XY + YZ + Z).(Y + Z').(X + Y)
= (XY + YZ + YZ + XYZ' + YZZ' + ZZ').(X + Y)
= [XY(1 + Z') + YZ + 0 + 0].(X + Y)*
= (XY + YZ).(X + Y)
= XY + XYZ + XY + YZ
= XY(1 + 1 + Z) + YZ
= XY + YZ
Converting to canonical POS form:
XY + YZ
= Y(X + Z)
= (Y + XX' + ZZ').(X + YY' + Z)
= (X + Y + Z).(X + Y + Z').(X' + Y + Z).(X' + Y + Z').(X + Y + Z).(X + Y' + Z)
= (X + Y + Z).(X + Y + Z').(X' + Y + Z).(X' + Y + Z').(X + Y' + Z)
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