Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How do you simply the following equation? $$X = A'B'C'D' + A'B'CD' + A'BCD' + ABCD' + AB'CD'$$

Here is what I did:

$$\begin{eqnarray} X & = & A'B'C'D'+A'CD'(B'+B) + ACD'(B+B') \\ & =& A'B'C'D'+CD'(A'+A) \\ & = & D'(A'B'C'+C) \end{eqnarray}$$

Is this correct?

share|cite|improve this question
Relevant: Karnaugh maps. – user2468 Aug 23 '12 at 14:12
you have done everything fine,but you could simplify it more,please see my answer ,you need to know that $c'*c=0$ and $c'+c=1$ – dato datuashvili Aug 23 '12 at 14:22
@clueless you need to know that $C + C' Y = C + Y$ for any $Y$. – user2468 Aug 23 '12 at 14:27
up vote 4 down vote accepted

It looks great. The one improvement that could be made is that the $C'$ is redundant, owing to an identity:


You can deduce this using the absorbtion law $ZY+Y=Y$, and the complementary law $Y+Y'=1$.

Intuitively, when adding part of $Z$ outside of $C$ to $C$, you may as well add all of $Z$ to $C$, because the part already inside $C$ will be abosorbed anyway.

share|cite|improve this answer
I'm sorry. I did an edit. Could you please not use $X$ since it might confuse the OP (s/he already has an $X$ in the question). – user2468 Aug 23 '12 at 14:34
@JenniferDylan Oh thanks! Yeah I'll switch that. – rschwieb Aug 23 '12 at 14:35
does that means that (A′B′C′+C) = (A'B'+C) because of the absorbtion and complementary law? – clueless Aug 23 '12 at 14:45
@clueless Taking $Y=C$ and $Z=A'B'$, yes. – rschwieb Aug 23 '12 at 15:42

we can simplify it much $(A'B'C'+C)=(C+C')*(C+A'B')$ we can proof it if we open brackets,we get

so finally we get $D'*(C+A'B')=C*D'+A'B'D'$ because $C+C'=1$

share|cite|improve this answer
hmm i do not under this c+a′b′c+a′b′c′=c+a′b′c′ can u explain? – clueless Aug 23 '12 at 14:26
sure ,first what does equal to $c+a'b'c$?take c out of brackets,you get $c(1+a'b')$ right? but $1+k$ in boolean algebra is equal to $1$ regardles any value of $k$ – dato datuashvili Aug 23 '12 at 14:30
oh I see it now that you write it that way thanks alot – clueless Aug 23 '12 at 14:34
good lucks friend,you are welcome – dato datuashvili Aug 23 '12 at 14:35

As all above answer pointed towards Absorption Law which is


So here you can apply this law -

 = > D′(A′B′C′+C)
 = > D′(A′B′+C)
 = > D′A′B′+D′C

which is your answer

Also you can verify this by K-Map that, this is the simplest form possible of this Boolean expression.

enter image description here

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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