# Simplify the boolean equation

I am trying to simplify this boolean expression:

F = A B ((A + C')') + B (A C + A' B) + (A + B)(A' + C D)


The resultant solution is supposed to be:

F = A C D + ('A B) + B C


But all I can get is

F = A C D + ('A B) + A B C + B C D


Help would be appreciated.

Checked on this page and it says minified form is what real solution is

From where you ended up:

$ACD + A'B + ABC + BCD =$ (Idempotence)

$ACD + A'B + A'B + ABC + BCD =$ (Adjacency)

$ACD + A'B + (A'BC + A'BC') + ABC + BCD =$ (Association and Commutation)

$ACD + (A'B + A'BC') + A'BC + ABC + BCD =$ (Absorption)

$ACD + A'B + (A'BC + ABC) + BCD =$ (Adjacency)

$ACD + A'B + (BC + BCD) =$ (Absorption)

$ACD + A'B + BC$

• How can one get this solution if the person doesn't know it in advance. For example step: (Association and Commutation) that is really confusing, why would I do it? Could you please care to explain. – ro ko Mar 23 '17 at 14:57
• @roko Excellent question!! Yes, obviously I was greatly guided by knowing what I had to obtain.. But in general there are a few principles you can always be on the lookout for: absorption, adjacency, and reduction being the main ones. Once you get a lot of experience with these, you start to see how you can rewrite terms in such a way that you can combine and simplify them, even if sometimes you first need to temporarily expand some terms as I did in my solution. And finally, the step where I did Association and Commutation was really just about reordering the terms, that's all. – Bram28 Mar 23 '17 at 15:23

The supposed solution is correct.

Check the Karnaugh map as confirmation:

Alex Kemper's answer is, I suppose, the de facto way to simplify the expression.
Another way to conclude that $$ABC+A'B+ACD+BCD = BC+A'B+ACD,$$ using Boolean Algebra axioms, is the following: \begin{align} A'B + ACD + BC &= A'B +ACD + (A+A')BC\\ &= A'B + ACD + ABC + A'BC\\ &= (A'B + A'BC) + ACD + ABC\\ &= A'B + ACD + ABC. \tag{absorption} \end{align} Now, \begin{align} ABC+A'B+ACD+BCD = A'B + ACD + ABC &\Leftrightarrow BCD \leq A'B + ACD + ABC\\ &\Leftrightarrow BCD (A'B + ACD + ABC) = BCD. \end{align} But $$BCD (A'B + ACD + ABC) = (A'BCD) + (ABCD) + (ABCD) = (A + A') BCD = BCD.$$