Simplify the following functions using Boolean Algebra a. F(X, Y, Z) = XY+X'Y+XZ b. F(X, Y,...

 

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Simplify the following functions using Boolean Algebra a. F(X, Y, Z) = XY+X'Y+XZ b. F(X, Y, Z) = (X+Y) (X+Y+Z) c. F(X, Y, Z) = XY'Z+XYZ+Y'Z d. F(X, Y, Z) = XY+X'YZ e. F(X, Y, Z) = X'Y + XYZ' f. F(X, Y, Z) =(XY) + (X+Y+Z)³X + YZ g. F(X, Y, Z) = (XY)' +(X+Y+Z)' Simplify following functions using Boolean Algebra a. F(X,Y,Z) = YZ + (X+Y)' + (XYZ)' b. F(X,Y,Z) = (X+Y+Z)' (X+Y) Simplify the following functions using Boolean Algebra a. F(X, Y, Z) = XY+X'Y+XZ b. F(X, Y, Z) = (X+Y) (X+Y+Z) c. F(X, Y, Z) = XY'Z+XYZ+Y'Z d. F(X, Y, Z) = XY+X'YZ e. F(X, Y, Z) = X'Y + XYZ' f. F(X, Y, Z) =(XY) + (X+Y+Z)³X + YZ g. F(X, Y, Z) = (XY)' +(X+Y+Z)' Simplify following functions using Boolean Algebra a. F(X,Y,Z) = YZ + (X+Y)' + (XYZ)' b. F(X,Y,Z) = (X+Y+Z)' (X+Y)

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QUESTION a Solution Rewrite the equation XY XY XZ YXX XZ Use complement law XX1 with XX YXX XZY1XZ Apply the identity law X1 X with XY Y1XZYXZ FXYZXYXYXZ is simplified to YXZ which is the answer QUEST... View the full answer