Boolean simplification

Thread Starter

Yami

Joined Jan 18, 2016
290
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?
 

WBahn

Joined Mar 31, 2012
25,914
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?
 

Thread Starter

Yami

Joined Jan 18, 2016
290
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?
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?
 

WBahn

Joined Mar 31, 2012
25,914
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?
 

WBahn

Joined Mar 31, 2012
25,914
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
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.
 

Thread Starter

Yami

Joined Jan 18, 2016
290
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.
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.
 

Thread Starter

Yami

Joined Jan 18, 2016
290
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.
 

WBahn

Joined Mar 31, 2012
25,914
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.
RRitesh, you need to start looking at you work and asking if your answers make sense.
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
 

Thread Starter

Yami

Joined Jan 18, 2016
290
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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WBahn

Joined Mar 31, 2012
25,914
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.
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.
 

Thread Starter

Yami

Joined Jan 18, 2016
290
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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