# Boolean Algebra Question

#### DeMorgansDog

Joined Nov 10, 2015
2
Hi all,

Not new to electronics, but new to forum. Coming back after a few years out of the business and a bit rusty, I'm hoping someone can help me with the following question:

I have a textbook question that asks me to prove (A + !B).(!A + !B + C) = AC + !B using boolean algebra. I have verified this is correct quite easily using a truth table, but I'm struggling with the boolean algebra, and I can't seem to re-factor or minimise the L.H.S. satisfactorily, even after a couple of hours head scratching and doodling out smaller truth tables.

So far, using the distributive laws, I have come up with a broken down L.H.S.

A.!A + A.!B + AC + !A.!B + !B.!B + !B.C = AC + !B

I think A.!A resolves to 0, and !B.!B resolves to !B, so I think I am on the right track (as this gives me the AC + !B components of the R.H.S.), but I can't think where to go from here, particularly removing the !B.C component.

Alternatives I thought about were to convert the (!A + !B + C) to !(A .B) + C using De Morgans, or using inversion, but I think these would complicate the matter more than necessary. Am I using the distributive law correctly? Any hints on the right direction would be much appreciated.

Thanks and regards,

DMDog.

#### _zeb_

Joined Nov 10, 2015
1
Hi there,
I'm in the same boat! I'm trying to dust off my skills after a bunch of years. I found that DeMorgan way is more clear. Here are the two different methods I used:
------
Q=AA'+AB'+AC+A'B'+B'B'+B'C
Q=0+AB'+AC+A'B'+ B'+B'C (I moved B' to the first term in the next step)
Q=B' +AB'+AC+A'B'+B'C
Q=B' +AC+B'(A'+C)
Q=B'+B'(A'+C)+AC finally factor B' from first two terms B'(1+A'+C)= B' and we are left with
Q=B'+AC

-------
Q=(A+B').(A'+B'+C)
First let's complement the whole function: Q' =((A+B').(A'+B'+C))'
Using DeMorgan Q'=(A'B)+(ABC')
Factoring B out we get Q'=B (A'+AC') and the second term A'+AC' is equal to A'+C'. (remember X+X'Y= X+Y)
Now we have Q'=B (A'+C') all we have to do is complement Q' which would be Q= B'+AC.

#### MrAl

Joined Jun 17, 2014
7,656
Hi all,

Not new to electronics, but new to forum. Coming back after a few years out of the business and a bit rusty, I'm hoping someone can help me with the following question:

I have a textbook question that asks me to prove (A + !B).(!A + !B + C) = AC + !B using boolean algebra. I have verified this is correct quite easily using a truth table, but I'm struggling with the boolean algebra, and I can't seem to re-factor or minimise the L.H.S. satisfactorily, even after a couple of hours head scratching and doodling out smaller truth tables.

So far, using the distributive laws, I have come up with a broken down L.H.S.

A.!A + A.!B + AC + !A.!B + !B.!B + !B.C = AC + !B

I think A.!A resolves to 0, and !B.!B resolves to !B, so I think I am on the right track (as this gives me the AC + !B components of the R.H.S.), but I can't think where to go from here, particularly removing the !B.C component.

Alternatives I thought about were to convert the (!A + !B + C) to !(A .B) + C using De Morgans, or using inversion, but I think these would complicate the matter more than necessary. Am I using the distributive law correctly? Any hints on the right direction would be much appreciated.

Thanks and regards,

DMDog.
Hi,

This is simpler than most so you dont need inversion. You were on the right track, just didnt follow through.

And also, too many times when we get involved with Boolean Logic we forget it is Boolean Algebra, and when we get involved with Boolean Algebra we get stuck on the "Boolean" part and forget that this is still basically Algebra, even simpler than normal algebra.
So dont forget to use what you learned in your algebra classes because that comes into play here.
Simplify the expression so that everything you have to simplify is on one side and maybe "Y" is on the other side, so forget about the known solution for now:
Y=my_expession_to_simplify

and then try to apply the rules of regular algebra, and simplify when you recognize a pattern like A*A' as you saw already. Using factoring and stuff like that, you'll find this to be one of the easiest problems you will probably ever encounter #### WBahn

Joined Mar 31, 2012
25,899
Using factoring and stuff like that, you'll find this to be one of the easiest problems you will probably ever encounter I have to agree. Usually (or at least often enough) problems like this that people have problems with involve consensus terms. I was in a bit of a hurry and didn't look closely enough. That's not the case here, so I was steering you a bit off the simplest path (but you DO need to learn about consensus terms soon). Focus on factoring out common factors for this one.

Thanks for all your responses I am very grateful, I will work through this this afternoon. Just for a bit of perspective, I worked as an FPGA designer for several years (VHDL), and never really needed to use this level of boolean algebra. It's just I have been out of the game for a while and I thought it might be interesting to go back to first principles, a bit alarmed when I got stuck so early on! I guess 'use it or lose it' really applies here.. 