Minimization of Boolean Functions

Simplification of Boolean Functions:

We discussed some of the laws, postulates and some theorems of boolean algebra.Now , let us simplify some boolean function 

Example1

Simplify the boolean function f(A,B,C)=ABC'+ABC+A'BC

Example 2

Simplify the boolean function f(A,B,C)=(A+B)(A+B')(A'+B)(A'+B')

Answer

Solution1 :

f(A,B,C)=ABC'+ABC+A'BC

= AB(C+C')+A'BC.........Using complement law(x+x'=1)

=  (A+A')(A+C)B..............Using distribution and complement law

=   B(A+C).... answer

 

Solution2:

f(A,B,C)=(A+B)(A+B')(A'+B)(A'+B')

=   (AA+AB'+BA+BB')(A'A'+A'B'+BA'+BB')...          (Use laws AA=A,BB'=0,A'A'=A')         

=    (A+A(B+B')+0)(A'+A'(B'+B)+0) (use laws b+b'=1)

=  (A+A)(A'+A')  ................... (use laws a+a=1)

=   A.A'................................(use  laws a.a'==0)                

=   0

OR

=   (A+BB')(A'+BB')

 

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