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Boolean Algebra

Question
CBSEENCO12011601

Derive a Canonical POS expression for a Boolean function G, represented by the following truth table:

X Y X G(X,Y,Z) 
0 0 0 0
0 0 1 0
0 1 0 1
0 1 1 0
1 0 0 1
1 0 1 1
1 1 0 0
1 1 1 1

Solution

G(X,Y,Z)= (X+Y+Z).( X+Y+Z’).( X+Y’+Z’).(X’+Y’+Z)