# Simplifying circuit using Boolean Algebra

• Mar 29th 2011, 04:53 AM
MickHarford
Simplifying circuit using Boolean Algebra
Hopefully I've uploaded the circuit properly
Attachment 21291

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.
• Mar 29th 2011, 05:43 AM
Ackbeet
The line

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

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

which does not reduce to 1.

[EDIT]: See below for a correction.
• Mar 29th 2011, 06:08 AM
emakarov
Quote:

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.
• Mar 29th 2011, 06:14 AM
Ackbeet
p + p' + q' + q'
= 1 + q' + q'
= q' + q'
= q', right?

[EDIT]: See below for a correction.
• Mar 29th 2011, 06:42 AM
MickHarford

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?
• Mar 29th 2011, 06:48 AM
emakarov
Quote:

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. (Smile)
• Mar 29th 2011, 06:50 AM
Ackbeet
Oh, now I'm with you. Clever pointing out of my mistake. So, 1 + x = 1, no matter what x is. Gotcha.
• Mar 29th 2011, 07:02 AM
MickHarford
Quote:

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?
• Mar 29th 2011, 07:11 AM
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.
• Mar 29th 2011, 07:14 AM
MickHarford
Quote:

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?
• Mar 29th 2011, 07:29 AM
emakarov
If you don't have a gate for 1, you can take p + p'.