AB+B((B+C')+B'C) can be simplified to

Question

The answer is B+C but When I tried to solve I got a different answer, which is B.

The way I solved

AB+B((B+C')+B'C)
=AB+B(B+C')+BB'C
=AB+BB+BC'+BB'C
=(BB'C=0,BB=B) then
AB+B+BC'
Taking B out frm all terms
B(A+1+C') =B
[A+1+C'=1]
What is the problem in my solution

@ganeshie8 actually this is not a text book qn. It was aksed in a competitive examination and B+C was there in the final answerkey :(. Thanks for you response — Abhijith, Aug 25 '19 at 02:33
@Abhijith ohk :) there's nothing wrong in your work. My guess is some mistake from the answer key or some typo in the given expression itself.. please see my answer... — AgentS, Aug 25 '19 at 02:36
I don't know what theorems you have available, but $(B+C')+B'C=True$, so then you have $AB+B=B$ — Ross Millikan, Aug 25 '19 at 02:40

AgentS · Answer 1 · 2019-08-25T10:06:25.937

1

Your solution looks all good. That means textbook answer $B+C$ must be incorrect, or there is a typo in the given expression.

Let's check this by plugging in $B=0$ and $C=1$:
$B+C = 0+1 = 1$
$AB+B((B+C')+B'C) = A0 + 0(...) = 0+0 = 0$

Thus $B+C$ cannot be the simplified form. However, you would get the textbook answer if the given expression had been: $$AB+B(B+C')+B'C $$

edited Aug 25 '19 at 10:06

answered Aug 25 '19 at 02:26

AgentS

12,195

1

Any Boolean Algebra can be represented as a family of subsets of a set $X$, with $AB=A\cap B$ and $A+B=A\cup B$, with $1=X$ and $0=\emptyset$. If we apply this to the formula of the Q it reduces to $B$..........+1 – DanielWainfleet Aug 25 '19 at 06:10

AB+B((B+C')+B'C) can be simplified to

The way I solved

1 Answers1