# Boolean Algebra help!

#### Codine

Joined Jan 29, 2007
2
Hi just a quick homework question Im having difficulty with

I need to simplify the following and display the laws used an applied etc.

A.B.C+ A.B.C + A.B.C + A.B.C + A.B.C

underlines are the inverts

#### thingmaker3

Joined May 16, 2005
5,084
What have you tried so far?

#### Codine

Joined Jan 29, 2007
2
Distributive law and negative absobtion, but I always end up with somthing stupid like A(B+C) which im pretty sure is wrong considering I have to build a circuit out of this

#### Dave

Joined Nov 17, 2003
6,970
Your working out would be a help so that we could guide you away from any errors you may be making. Note in the following, ' indicates negated function.

My first pointer would be to take A'.B' out as a common factor from the first two expressions, and take A'.B out as a common factor from the third and forth expressions. You can then say (C + C') = 1.

Then take out A' as a common factor, where (B + B') = 1.

This leaves you with: A' + A.B'.C'

If you follow my above explanation you should arrive at the above expression. If you wish to try the answer feel free to upload it here to let someone check it.

Dave

#### dragan733

Joined Dec 12, 2004
152
A'b'c'+ A'b'c + A'bc' + A'bc + Ab'c'=
A'b'(c'+c)+a'b(c'+c)+ab'c'=
A'b'+a'b+ab'c'=
A'(b'+b)+ab'c'=
A'+ab'c'=
A'+b'c'

#### Dave

Joined Nov 17, 2003
6,970
A'+ab'c'=
A'+b'c'
You are correct dragan733, I missed that final step. If it is NOT A' then it is A, hence the final simplification.

Dave

#### JOHNYBOY80

Joined Feb 7, 2007
5
Hello,
I am having so much trouble understanding how to simplify boolean expressions. I know how to use the k map but it is not helping me with this problem. i need to simplify an equation that i think is already reduced down it is
AB+A'B'C
i even tried the truth tables to see the output of the circuit and match the answers of the truth table to the other 5 possible answers that were given and i have no luck. I am lost. thanks in advance