# Math Help - Boolean expression simplification

1. ## Boolean expression simplification

I know this stuff should be way easy once I figure it out, but right now I am literally pulling out my hair trying to do so. The original expression is:

AD + ABC + A + AB + ABC + ABC

I see that A and A are common factors, so I pull them out:

A(D + BC + 1 + BC) + A(B + BC)

The fact that there is a "1" in the first parentheses makes everything in the parentheses 1, right?

A(1) + A(B + BC)

X and 1 = X

A + A(B + BC)

The only thing I can think to do is distribute the (B + BC) but I don't know if that is the right thing to do...

A + A(B+B)(B+C)

Now, (B+B) is equal to 1. A and 1 is equal A.

A + A(B+C)

Now I am stuck. I have no way to check my answer, but Wolfram says the simplified answer is:

A+B+C

Can you help me finish/correct the above work?

2. You're ok up until the moment you distribute. I don't think the way you did that was valid. If you multiply back out again your factorization, you end up with

A(BB + BC + BB + BC) = A(0 + BC + B + BC) = A((B + B)C + B) = A(C + B). This is not equal to A(B + BC). I don't think there's any way of simplifying any further. That is, A + A(B + BC) is as far as you get.

3. Originally Posted by Ackbeet
You're ok up until the moment you distribute. I don't think the way you did that was valid. If you multiply back out again your factorization, you end up with

A(BB + BC + BB + BC) = A(0 + BC + B + BC) = A((B + B)C + B) = A(C + B). This is not equal to A(B + BC). I don't think there's any way of simplifying any further. That is, A + A(B + BC) is as far as you get.
Actually B + C = B + BC.

To the original poster, use the identity A + AB = A + B on your last step and you will get "Wolfram"'s identity.

Additionally, whenever you wonder if two expression F and G are equal, I encourage you to write the truth tables of F and G and check if they are equal.
(A truth table is just a fancy name for testing the functions on all possible values in the domain. In Boolean cases, we only have to try 1 and 0 for each variable and thus testing equality in Boolean logic for small number of variables is quite simple).

4. Actually B' + C' = B' + BC'.
Hmm. You're right. Maybe it's been too long since I've done boolean simplifications!

Going off of Isomorphism's post, and your last step

A + A(B+C),

if it's true that x' + xy' = x' + y', then can you see that you only have one step from

A + A(B+C)

to

A + B+C`?