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The K-map method of simplifying the Boolean Expression A'B'C + A'BC + AB'C gives the answer to be A'C+B'C. But I am not able to solve this algebraically. Please help me out. What I have tried is

A'B'C + A'BC + AB'C
= A'C (B'+B) + AB'C
= A'C(1) + AB'C
= A'C + AB'C

This is the most simplified I get using Boolean algebra but this is not matching the result. Can someone please help me out.

mohd shoaib
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2 Answers2

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With the use of A+A=A we have

$$ A'B'C + A'BC + AB'C =\\ = \underbrace{\color{blue}{A'B'C} + A'BC} +\underbrace{\color{blue}{A'B'C}+ AB'C}=\\ =A'C (B'+B) + (A+A')B'C $$

user376343
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  • Thanks for the answer @user376343. This is a satisfying answer. Can you please also help me understand, how we get to know that we should add a new term A'B'C to reduce the expression? For example, suppose I did not know that the simplified form is A'C+B'C then how would I decide that this can be simplified further(using boolean algebra and without using k-map) – mohd shoaib Apr 19 '23 at 12:14
  • Karnaugh map is useful, I use it often to verify my results. If you want to simplify an expression without KM, you can look at each pair of terms separately and ask: can they merge? – user376343 Apr 19 '23 at 12:21
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    Thanks for all the clarifications and answer @user376343 – mohd shoaib Apr 19 '23 at 12:23
  • You're welcome. I am glad I could help :) – user376343 Apr 19 '23 at 12:26
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Since $C$ is everywhere, we can just forget about it. With what you have already done, you obtained $A' + AB'$. You want to prove that $A' + AB' = A' + B'$. Of course, this is OK from the truth table. I do not know what rules you used to define the $+$ operator, but maybe the following computation will be OK for you:

A'+B'
= A' + 1 B'
= A' + (A + A') B'
= A' (1 + B') + A B'
= A' + A B'
cs89
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  • thanks for the quick help @cs89 . I understand the idea you shared. But is there any other way by means of which we can reach the simplified form rather than proving it conversely? I mean to say what if I was not knowing that the final result should be A'C + B'C then how would I reach this simplified form? – mohd shoaib Apr 19 '23 at 12:11
  • This answer is far better than mine! – cs89 Apr 19 '23 at 12:14