Simplify Boolean Expressions Using Laws

Boolean expressions can be easily Simplified by using the laws of Boolean algebra. Let’s explain some examples to Simplify Boolean Expressions.

Example 01

F(X, Y, Z) = X.Z+X’.Z.Y

F(X, Y, Z) = X.Z+X’.Y.Z     {By Commutative and commutative law}

F(X, Y) = (X+X’.Y).Z         {By Distributive law}

F(X, Y) = (1.Y).Z              {by complement law}

F(X, Y) = (X.Y)                 {by definition of identity element}

Hence, the simplified Boolean function will be X.Y

Symbolically Simplification   

Logic diagram, before and simplification for the Boolean function F(X, Y, Z) = X.Z+X’.Z.Y is given below

Example 02

F(X, Y, Z) = X’.Y.Z+X.Y’+X’.Y’.Z

F(X, Y, Z) = X’.Y.Z + X’.Y’.Z + X.Y’ {By Commutative law}

F(X, Y, Z) = X’.Z(Y+Y’) + X.Y’  {By Distributive law}

F(X, Y, Z) = X’.Z (1) + X.Y’ {by complement law}

F(X, Y, Z) = X’.Z + X.Y’ {by definition of identity element}

Hence, the simplified Boolean function will be X’.Z + X.Y’

Symbolically Simplification   

Logic diagram, before and simplification for the Boolean function F(X, Y, Z) = X’.Y.Z+X.Y’+X’.Y’.Z is given below