# Boolean Algebra Simplification

1. Aug 27, 2012

### bd411

1. The problem statement, all variables and given/known data

Simplify the following boolean expression:

A'C + A'C'D + ABC + ABC'D

I have a solution but when I checked against a boolean algebra simplifier online it gave me a different answer... is there a glaring mistake below ?

2. Relevant equations

None.

3. The attempt at a solution

A'C + A'C'D + ABC + ABC'D
= A'(C + C'D) + AB(C + C'D)

= A'(C + D) + AB(C + D) = A'C + A'D + ABC + ABD

= C(A' + AB) + D(A' + AB)
= C(A'B + 1) + D(A'B + 1)

= C(1) + D(1) = C + D.

Any help would be much appreciated !

2. Aug 27, 2012

### gabbagabbahey

This isn't correct. Consider that if $A$ is true and $B$ is false, then $A' + AB$ is false, while $A'B + 1$ is always true.

3. Aug 27, 2012

### rude man

You did so well with (C + C'D) = (C + D) at the start; use the same deal here.

4. Aug 28, 2012

### bd411

Hmm right so I could say that (A' + AB) = (A' + B)

In which case C(A' + B) + D(A'+B) = A'C + BC + A'D + BD.

I'm a little stumped as to how this simplifies further.

Any help would be much appreciated !

Sorry the answer should be BC + BD + A'C'D !

Last edited: Aug 28, 2012
5. Aug 28, 2012

### gabbagabbahey

I think the answer you have in front of you is wrong. I get the same thing as you, which can be rewritten as (A'+B)(C+D). Wolfram Alpha also agrees.

6. Aug 28, 2012

### bd411

Great, thanks so much !

7. Aug 28, 2012

### gabbagabbahey

You're welcome!

P.S. To see why BC + BD + A'C'D is incorrect, consider the case where A is false and C is true. To see why BC + BD + A'B' is incorrect, consider the case where both A and C are false.