# Reducing Boolean algebra

Discussion in 'General Electronics Discussion' started by cps13, Feb 25, 2013.

1. ### cps13

54
1
Feb 25, 2013
Hi,

I am new to the forum so hello to all.

I am studying a BTEC in EE and going onto a HND hopefully next year. I am struggling a bit with reducing boolean algebra. I seem to be able to look at the same problem 3 times and come up with three different answers.

Can someone give me some hints. If some is NOT for example NOT A I write it as A'

I have this problem.... ABC + A'B'C' + ABC' + A'C'

this is one way I have "solved" it.

ABC + A'B'C' + ABC' + A'C'
ABC + A'B'C' + ABC' + A' . 1 (last C' cancelled against first C, turns into 1 (A+A' = 1))
AB + A'B'C' + C' + A' . 1 (cancelled first AB against second AB in third group)
AB + A'B'C' + A' . 1 (cancelled second C' as same as C' in second group)
AB + B'C' + A' . 1 (A' in second group removed as same as last A')
AB + B'C' + A' (A' . 1 = A')

Any help would be greatly appreciated!

Thanks!

2. ### john monks

693
3
Mar 9, 2012
I cannot figure out how to answer your question other than to do your homework for you.
If worst comes to worst draw up a truth table and figure it out.
You probably should look at the Tutorial entitled "Converting Truth Tables into Boolean Expressions"

3. ### cps13

54
1
Feb 25, 2013
Are you able to explain to me how you would get the correct answer? Rather than just write the correct answer. Or if my answer is wrong where I have gone wrong?

I have read sections from two text books but I am still struggling. I cannot draw up a truth table to solve it because the exercise is to reduce a boolean expression using de-morgans theorum, boolean postulates etc...

4. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,501
2,841
Jan 21, 2010
OK, here's a hint that might get you started.

Look at ABC + ABC' -- Can you simplify that in your head?

You can't just cancel things willy nilly.

ABC + ABC' = AB(C + C') And C+C' = ?

693
3
Mar 9, 2012