0
$\begingroup$

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

$\endgroup$
5
$\begingroup$

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)

$ACD + A'B + BC$

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – ro ko Mar 23 '17 at 14:57
  • $\begingroup$ @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. $\endgroup$ – Bram28 Mar 23 '17 at 15:23
2
$\begingroup$

The supposed solution is correct.

Check the Karnaugh map as confirmation:

enter image description here

$\endgroup$
1
$\begingroup$

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 $$BCD (A'B + ACD + ABC) = (A'BCD) + (ABCD) + (ABCD) = (A + A') BCD = BCD.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.