# Boolean simplification

Discussion in 'Homework Help' started by Yami, May 6, 2016.

1. ### Yami Thread Starter Member

Jan 18, 2016
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0
Hi I have to simply this eqn X=A'B'C'+A'BC'+A'BC+ABC using K-maps and I have to verify the simplified equation by Boolean Algebra. The answer I got for the K-maps is=A'C'+BC. So I tried with using Boolean Algebra I was able to get the same answer.
But if I take (A'B'C'+ABC) = 1, I would get a different answer. My question really is (A'B'C'+ABC) = 1 correct?

2. ### WBahn Moderator

Mar 31, 2012
18,093
4,920
I don't understand the point of your question. You have one equation for X and you reduced it to a second, equivalent equation. Fine so far.

Then you take a second, unrelated equation and ask if it is correct. Correct in what way? If (A'B'C' + ABC) = 1 then X=1, but if (A'B'C' + ABC) = 0 then X might still be 1 or it might be 0. So what is it you are asking?

3. ### Yami Thread Starter Member

Jan 18, 2016
54
0
Sorry that I was unclear earlier.
Basically the eqn is X=A'B'C'+A'BC'+A'BC+ABC
next step I did was
(A'B'C'+ABC)+A'BC'+A'BC
(1)+A'BC'+A'BC
......... so on

In the second step "cancelling" (A'B'C'+ABC) is it correct?

4. ### WBahn Moderator

Mar 31, 2012
18,093
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On what basis are you claiming that (A'B'C'+ABC) is always True? What if A is False and B is True? Do you even need to know what C happens to be to determine that the expression (A'B'C'+ABC) is False?

5. ### Yami Thread Starter Member

Jan 18, 2016
54
0
I've attached a picture of my working, which I get the same answer as the k-map simplification

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6. ### Yami Thread Starter Member

Jan 18, 2016
54
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Here is the other way which I'm talking about, which I dont get the right answer but I don't understand why it shouldn't work

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7. ### WBahn Moderator

Mar 31, 2012
18,093
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The work hear appears just fine.

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8. ### Yami Thread Starter Member

Jan 18, 2016
54
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Have a look at the second picture. I get a different answer

9. ### WBahn Moderator

Mar 31, 2012
18,093
4,920
Again, on what basis are you claiming that (ABC + A'B'C') is always True?

You seem to be basing this on the identity A + A' = 1, but that is a completely false equivalency.

Consider that A + A' represents two of the two possible combinations of one variable. Hence ALL of the possible combinations are covered.

But (ABC + A'B'C') represents just TWO of the EIGHT possible combinations of three variables. Hence 3/4 of the possible combinations yield a result of False for this expression.

10. ### Yami Thread Starter Member

Jan 18, 2016
54
0
I've got a another similar question.
Y=A'B'C'+A'B'C'+AB'C+ABC+ABC
Y=A'B'C'(1+1)+AB'C+ABC(1+1) -------->is this step correct?
Y=A'B'C'+AB'C+ABC--------- so on.

11. ### WBahn Moderator

Mar 31, 2012
18,093
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Yes.

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12. ### Yami Thread Starter Member

Jan 18, 2016
54
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thanks
and the rule would be A+A=A right?

appreciate the help so much

13. ### Yami Thread Starter Member

Jan 18, 2016
54
0
Here's is my working:-
Y=A'B'C'+A'B'C'+AB'C+ABC+ABC
Y=A'B'C'(1+1)+AB'C+ABC(1+1)
Y=A'B'C'+AB'C+ABC
Y=B'C(A'+A)+ABC
Y=B'C+ABC
Y=C(B'+AB)
for some reason i think its not done.

14. ### WBahn Moderator

Mar 31, 2012
18,093
4,920
Your answer says that in order for Y to be True, that C absolutely MUST be true.

But your original equation had the term A'B'C' which is True if all three input terms are False.

How do you go from

Y=A'B'C'+AB'C+ABC

to

Y=B'C(A'+A)+ABC

15. ### Yami Thread Starter Member

Jan 18, 2016
54
0
Oh sorry my bad. Here is the working I did, I also produced a truth table for it. The bit I don't understand is with the final equation A'B'C'+AC for the A'B'C its true but for the other variable AC does it work? and also what about the last ABC = 1 its not in the simplified equation. Sorry for the amateur questions but no way else to get the answers.

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16. ### WBahn Moderator

Mar 31, 2012
18,093
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To see if it works for AC, look at your truth table and identify all rows in which A and C are both 1. Is the output 1 for all of those rows? If not, then your answer does not match the original equation.

17. ### Yami Thread Starter Member

Jan 18, 2016
54
0
I've got a similar question which is quite a head scratcher for me. I've included my workings and the truth table. The question is X=AB'C+A'BC+A'B'C+A'B'C'+AB'C' . The simplified answer I get is X=A'C'+B' . I drew a K-map and got a whole different answer :S. Also with the truth table- I had a few questions too. For e.g. I get a true for all the B' except for 011 or A'BC ---->its also not C' so here is what I'm thinking correct me if I'm wrong since A'C' (in the simplified equation) is "AND" function A'BC term works?

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