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

So far I've got:

$A'B'C'D' + A'BC'D' + A'BC'D + AB'C'D$

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

$= A'C'D'(1) + C'D(A \;\text{ XOR }\; B)$

$= C'[A'D' + D(A \;\text{ XOR }\; B)]$

Did I do this correctly? Is there a simpler solution?



share|cite|improve this question
Did you use the results from your previous question to help in this case? – abiessu Oct 11 '13 at 12:01
The result in the previous question was great, but when mapped, used the same amount of gates (including nots) as the solution I included with my question. I ended up keeping my solution. – KrispyK Oct 11 '13 at 12:11
Also, the solution in the previous question that someone provided used Karnaugh maps. We haven't gone over Karnaugh maps in class, so I'm not familiar with using them. – KrispyK Oct 11 '13 at 12:12
Then I think this solution is similarly correct. :-) – abiessu Oct 11 '13 at 12:16
up vote 4 down vote accepted

Yes, your work is correct.

Simplification of a Boolean expression depends on context, and what form you are seeking in your "simplification": for example, conjunctive normal form (product of sums) or disjunctive normal form (sum of products), etc. Typically, one does not introduce "xor" $\oplus$ unless expressing the entire function in terms of $\oplus$, $\land$, $'$.

See, for example, the following, without the use of $\oplus$ (xor):

$$\begin{align} A'C'D'(B' + B) + A'BC'D + AB'C'D &= A'C'D' + A'BC'D + AB'C'D \\ \\ &= C'[A'D' + D(A'B + AB')]\end{align}$$

But again, all your manipulations are indeed correct.

share|cite|improve this answer
This needs a TU! +! – Amzoti Oct 11 '13 at 12:56
You can't generally express Boolean functions in terms of xor and not alone -- perhaps you meant xor, not and and? – Henning Makholm Oct 11 '13 at 13:30
@Henning Thank you. Yes, indeed, that's what I meant! I've edited accordingly. – amWhy Oct 11 '13 at 13:33
Thank you for the help! You guys rock. – KrispyK Oct 11 '13 at 14:18

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.