# Boolean Simplification

#### MWalden

Joined Apr 26, 2007
24
First I want to say whoever wrote the E-Book volumes on this website did a very good job. They make it very easy to understand. I am reading the Digital section. But it's missing one thing, maybe some tests or quizes written in javascript to help practice especially in the digital section. I'd be willing to help I can write a little in javascript and html. Just an idea.

Does anyone know where I can get some practice in Boolean simplification or any type of digital work, like tests/quizes/practice problems?

But anyway I have a question. To practice on the simplification I tried to simplify a simple adder circuit I got from HSW http://computer.howstuffworks.com/boolean2.htm They show a simple adder circuit but then say that it isn't the best way to do it that it can be done with fewer gates, so I have been trying to simplify it. Could someone tell me if I am doing this right or what:

First I converted it into an equation...
For Q:
(AB'+A'B)C' + (AB+A'B')C =
AB'C'+A'BC'+ABC+A'B'C
Since there are 3 variables grouped together I can't figure out how to simplify it any further than it already is? Am I doing something wrong?

The (AB'+A'B) is the XOR gate of course. The (AB+A'B') is the XNOR gate correct? (Which I got like this:
(AB'+A'B)' = (AB')'(A'B)' = (A'+B)(A+B') = AA'+A'B'+AB+BB' = A'B' + AB ? with DeMorgans theorem)

For CO (Carry out):
(AB)C'+(A+B)C =
ABC'+AC+BC =
A(C+C'B)+BC =
A(C+B)+BC OR AB+AC+BC

I got this one down a little, but can it go any further and am I doing it right?

Sorry if this is the wrong section of the forum.
Thanks for any help.

#### Dave

Joined Nov 17, 2003
6,969
First I want to say whoever wrote the E-Book volumes on this website did a very good job. They make it very easy to understand. I am reading the Digital section. But it's missing one thing, maybe some tests or quizes written in javascript to help practice especially in the digital section. I'd be willing to help I can write a little in javascript and html. Just an idea.

Does anyone know where I can get some practice in Boolean simplification or any type of digital work, like tests/quizes/practice problems?
Try our sister project the Socratic Project: http://www.ibiblio.org/kuphaldt/socratic/doc/topical.html

But anyway I have a question. To practice on the simplification I tried to simplify a simple adder circuit I got from HSW http://computer.howstuffworks.com/boolean2.htm They show a simple adder circuit but then say that it isn't the best way to do it that it can be done with fewer gates, so I have been trying to simplify it. Could someone tell me if I am doing this right or what:

First I converted it into an equation...
For Q:
(AB'+A'B)C' + (AB+A'B')C =
AB'C'+A'BC'+ABC+A'B'C
Since there are 3 variables grouped together I can't figure out how to simplify it any further than it already is? Am I doing something wrong?
No. Assuming you have the compliments available, this expression requires only 5 gates. I can't see a way to simply this more.

The (AB'+A'B) is the XOR gate of course. The (AB+A'B') is the XNOR gate correct? (Which I got like this:
(AB'+A'B)' = (AB')'(A'B)' = (A'+B)(A+B') = AA'+A'B'+AB+BB' = A'B' + AB ? with DeMorgans theorem)
Correct.

For CO (Carry out):
(AB)C'+(A+B)C =
ABC'+AC+BC =
A(C+C'B)+BC =
A(C+B)+BC OR AB+AC+BC

I got this one down a little, but can it go any further and am I doing it right?

Sorry if this is the wrong section of the forum.
Thanks for any help.
Your simplification looks fine here. You have options for your final expression, but would recommend you leave it in the form of A(C+B)+BC

Dave