1
$\begingroup$

I can't use K-maps. I've managed to bring it down to C + ABC', not sure if I did it right and I think I can simplify further, but don't know how to. I also feel like there's a simpler method I'm missing.

AB + A'C + B'C
= AB + A'BC + A'B'C + B'C
= AB + A'BC + B'C (1 + A')
= AB + A'BC + B'C
= ABC + ABC' + A'BC + B'C
= BC(A+A') + ABC' + B'C
= BC + B'C + ABC'
= C + ABC'

I have a second question, might as well add it here because it's also simplification of boolean algebra.

f = cx + ac'x + bc'x + a'b'c'x' (used a K-map to generate this, now I have to simplify further)
f = c'x(a+b) + cx + a'b'c'x' - no idea how to continue from here

$\endgroup$
2
$\begingroup$

$AB+A'C+B'C $

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

$=AB + C(AB)'$

$=AB+C$

$\endgroup$
  • $\begingroup$ I would remove the word hint, sorry!! $\endgroup$ – Satish Ramanathan Oct 2 '18 at 12:47
  • 1
    $\begingroup$ No harm done, feel reassured. $\endgroup$ – Yves Daoust Oct 2 '18 at 12:48
  • $\begingroup$ Wow, that was easier than I thought. I got to that second line on my first try but got stuck there, I forgot about De Morgan's Laws. Thanks for the help. Could you please help with my other question (edited into my main post) if possible? $\endgroup$ – Karan Bijani Oct 2 '18 at 13:27
  • $\begingroup$ Use the logic that I have used in step 3 of the first problem into your first two terms of the second problem. The last term cannot be reduced any further $\endgroup$ – Satish Ramanathan Oct 2 '18 at 13:35
2
$\begingroup$

$\begin{aligned}AB+A'C+B'C & =AB\left(C'+C\right)+A'\left(B+B'\right)C+\left(A+A'\right)B'C\\ & =ABC'+ABC+A'BC+A'B'C+AB'C+A'B'C\\ & =ABC'+ABC+A'BC+A'B'C+AB'C\\ & =ABC'+ABC+ABC+A'BC+A'B'C+AB'C\\ & =AB\left(C'+C\right)+\left(AB+A'B+A'B'+AB'\right)C\\ & =AB+\left(A+A'\right)\left(B+B'\right)C\\ & =AB+C \end{aligned} $

Constructive advice:

Make a Venn-diagram and observe that: $$(A\cap B)\cup(A^{\complement}\cap C)\cup(B^{\complement}\cap C)=(A\cap B)\cup C$$

$\endgroup$
  • $\begingroup$ Thanks for the help. That Venn diagram is pretty useful advice for checking the answer. Could you please help with the other question (edited into my main post) if possible? $\endgroup$ – Karan Bijani Oct 2 '18 at 13:28
  • $\begingroup$ $c+ac'+bc'=c+a+b$ leading to $\cdots=(a+b+c)x+a'b'c'x'=(a+b+c)x+(a+b+c)'x'$ $\endgroup$ – drhab Oct 2 '18 at 14:32

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.