# Thread: Simplifying circuit using Boolean Algebra

1. ## Simplifying circuit using Boolean Algebra

Hopefully I've uploaded the circuit properly

Using an apostrophe to denote the over-bar (Not), "+" is or, and "." is and, I wrote the initial circuit as:

r = (p'.(p.q).q)'
which I then simplified as follows:
r = p'' + (p.q)' + q'
= p + p' + q + q'
= 1

However, I'm not convinced this is correct. Help please.

2. The line

= p + p' + q + q' should be

= p + p' + q' + q',

which does not reduce to 1.

[EDIT]: See below for a correction.

3. Originally Posted by Ackbeet
The line

= p + p' + q + q' should be

= p + p' + q' + q',

which does not reduce to 1.
Hmm, I think it still does.

4. p + p' + q' + q'
= 1 + q' + q'
= q' + q'
= q', right?

[EDIT]: See below for a correction.

5. Thanks for your help.

When you think about it logically, it appears q' is the right answer.

To draw the simplified circuit, is it simply q, going into a "not" semi-circle thing and then to r, can p be ignored?

6. Originally Posted by Ackbeet
p + p' + q' + q'
= 1 + q' + q'
= q' + q'
= q', right?
I think the Earth is round or there is a mistake here.

7. Oh, now I'm with you. Clever pointing out of my mistake. So, 1 + x = 1, no matter what x is. Gotcha.

8. Originally Posted by Ackbeet
Oh, now I'm with you. Clever pointing out of my mistake. So, 1 + x = 1, no matter what x is. Gotcha.
Right so the answer is one then after all?

9. Yes. This could also be seen without applying the De Morgan's law (though it's not much of a change). In the original formula, (p'.(p.q).q)', conjunction is associative and p' . p is always false, so p' . p . q . q is false as well. Therefore, its negation is always true.

10. Originally Posted by emakarov
Yes. This could also be seen without applying the De Morgan's law (though it's not much of a change). In the original formula, (p'.(p.q).q)', conjunction is associative and p' . p is always false, so p' . p . q . q is false as well. Therefore, its negation is always true.
Cool, that way's simpler to be fair. Does no circuit exist then?

11. If you don't have a gate for 1, you can take p + p'.