CODEWITHSUNDEEP
booleansimplification
Sergi
Sergi

Reputation: 501

How to simplify ( ~A * B) + C * (~B + A)

I have simplified a Boolean function up to a point but I got stuck on the last step, I can't see which rule (if any) I should apply to get to the simplified expression.

I want to simplify the following Boolean function:

( ~A * B) + C * (~B + A)

I know the simplified form is (B * ~A + C), but I can't identify which rules should I use, although I can see that ( ~A * B) and (~B + A) cancel each other, but I'm not sure if there is a rule for this or is this based on pure observation.

The rules I have tried to use are detailed here http://electronics-course.com/boolean-algebra

Anybody could give me a hand?

Thanks!

Upvotes: 0

Views: 374

Answers (1)

peer
peer

Reputation: 4719

                   ( ~A * B) + C * (~B + A)
~C * ( ~A * B) + C*( ~A * B) + C * (~B + A)    X = X*Y + X*~Y
~C * ( ~A * B) + C*( ~A * B) + C * ~(B * ~A)   De Morgan
~C * ( ~A * B) + C                             X*Y + X*~Y = X 
     ( ~A * B) + C                             Absorption

Not sure if X = X*Y + X*~Y has a name, you can derive it from:

x 
X * 1            Idempotence
X * (Y + ~Y)     Complement
X * Y + X * ~Y   Distributive Law

Edit: I found a simpler way:

                   ( ~A * B) + C * (~B + A) 
                   ( ~A * B) + C * ~(B * ~A)   De Morgan
                   ( ~A * B) + C * ~(~A * B)   Commutative Law
                   ( ~A * B) + C               Absorption

Upvotes: 2

Related Questions