How to simplify this Boolean expression?

C

chamarawix

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Mar 10, 2020
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5
(A.B.C)+(A'.B.C)+(A.B'.C)+(A.B.C')

I tried up to some extend in different ways. But got stuck in the middle.
(B.C) + A(B'.C + B.C')
...
...
...
Answer should be A.B+A.C+B.C
Please help.
 
Their result might be easier to obtain using Karnaugh maps (but only if you have already studied them in class).

Using algebra
(B.C) + A(B'.C + B.C')
(B.C) + A.B.C + A(B'.C + B.C') <--- see * below
(B.C) + A(B.C + B'.C + B.C')
...
can you take it from here?

* you can bring A.B.C back in because it does not affect the end result AND I spotted that it's a useful thing to do. I'm not sure I would have spotted this without seeing their end result!
 
Cubist said:
Their result might be easier to obtain using Karnaugh maps (but only if you have already studied them in class).

Using algebra
(B.C) + A(B'.C + B.C')
(B.C) + A.B.C + A(B'.C + B.C') <--- see * below
(B.C) + A(B.C + B'.C + B.C')
...
can you take it from here?

* you can bring A.B.C back in because it does not affect the end result AND I spotted that it's a useful thing to do. I'm not sure I would have spotted this without seeing their end result!
Click to expand...

Thanks for your solution. Am I right?

(B.C) + A(B.C + B'.C + B.C')
(B.C) + A(C(B + B') + B.C')
(B.C) + A(C + B.C')
(B.C) + A(C + B)
B.C + A.C + A.B

But, I'm not much clear about taking ABC in to this. Could you please explain.
 
chamarawix said:
Thanks for your solution. Am I right?

(B.C) + A(B.C + B'.C + B.C')
(B.C) + A(C(B + B') + B.C')
(B.C) + A(C + B.C')
(B.C) + A(C + B)
B.C + A.C + A.B
Click to expand...

With a bit more experience you might be able to spot that B.C + B'.C + B.C' = B+C which would lead to a quicker solution. Think about the truth table of the LHS. However your method is absolutely fine and you got the correct result, so WELL DONE.

chamarawix said:
But, I'm not much clear about taking ABC in to this. Could you please explain.
Click to expand...

This is hard to explain. You can probably see that A(B'.C + B.C') on its own can't be simplified further. But when this expression is ORd with another expression it can often lead to a simplification. An extreme example would be 1 + A(B'.C + B.C') which obviously simplifies to 1.

In your question I could see, from looking at the answer, that A(B'.C + B.C') had to become A(B+C). Look at the truth table for these...
Code: 
A  B  C  A(B'C+BC')  A(B+C)
1  0  0     0        0
1  0  1     1        1
1  1  0     1        1
1  1  1     0    --> 1 <--

So, in order to turn A(B'.C + B.C') into A(B+C) I just need to OR it with the highlighted "1" above, and luckily this "1" ( A.B.C ) could be "extracted" from the "BC+" on the left.

Can you see that +A.B.C can be added to B.C without changing the expression?
B.C + A.B.C = 1.B.C + A.B.C = (1+A).B.C = B.C
 
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