I got some homework for Boolean Simplification, could someone please help me and solve this because I'm literally going to rip my head apart :D I've tried this multiple times but I keep ending up doing it wrong...

A'BC'D'+ AB'C'D' + AB'CD' + ABC'D + ABCD'

Any help would be greatly appreciated!


closed as off-topic by Namaste, C. Falcon, Arnaud D., астон вілла олоф мэллбэрг, John B Dec 4 '16 at 1:07

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  • $\begingroup$ How can you tell you're doing it wrong? Do you have an answer key? If so, please edit your post to include the answer you're aiming for. (It actually happens, though not very often, that a given solution might contain a typo). Why don't you show work from one of those multiple tries you made, so we can help steer you in the right direction. $\endgroup$ – Namaste Dec 3 '16 at 16:46
  • $\begingroup$ I destroyed my answers from stress and anger, I'm counting this to be my last resort. I'm sorry I can't update it with my last answer, but it was wrong. I started from this -> C'.D' (BC' + B'C') A (B'CD + BC'D) $\endgroup$ – Andrius Zapolskis Dec 3 '16 at 16:55
  • $\begingroup$ I'm aiming to simplify that expression as much as possible. There's no particular answer $\endgroup$ – Andrius Zapolskis Dec 3 '16 at 17:08
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    $\begingroup$ Then how can you tell your attempts ended "up doing it wrong", if there is not some answer/objective you aim to seek? Seems that your claim of "multiple" attempts, to the point that you're "literally going to rip my head apart" is an overly dramatic way of trying to make users here feel sorry for you. $\endgroup$ – Namaste Dec 3 '16 at 17:22
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    $\begingroup$ Perhaps you've encountered the expression "Be a man!" At the very least, Andrius, stop acting like a child. $\endgroup$ – Namaste Dec 3 '16 at 17:27

$A'BC'D'+ AB'C'D' + AB'CD' + ABC'D + ABCD'$

$=A'BC'D'+ AB'D'(C'+C) + ABC'D + ABCD'$

$=A'BC'D'+ AB'D' + ABC'D + ABCD'$

I don't think it can be simplified further unless we use other constructs like $\implies$ and XOR.

  • $\begingroup$ Is that the final answer? $\endgroup$ – Andrius Zapolskis Dec 3 '16 at 16:52
  • $\begingroup$ I have this question, and I have never done Computer Architecture before, this really has me baffled. And if ==> and XOR are part of Boolean Algebra, than we can try and simplify this further, can't we? $\endgroup$ – Andrius Zapolskis Dec 3 '16 at 17:02
  • $\begingroup$ This boils down to the meaning of simplification. Adding more constructs can surely make the expression smaller,but does it get simplified? Smaller is not always more simple. $\endgroup$ – Meet Taraviya Dec 3 '16 at 17:09
  • $\begingroup$ My question was "Using Boolean Algebra simplify the minterm as far as you can, showing each major step that you take. The rule you are applying at each step should also be made clear" $\endgroup$ – Andrius Zapolskis Dec 3 '16 at 17:13
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    $\begingroup$ Andrius, your not in any kind of position to demand Meet "tailor" the answer to your liking. I actually, in another answer field, just completed a proof using XOR. But given your last comment, and your lack of any sort of demonstrated effort on your part, I refuse to post it now. $\endgroup$ – Namaste Dec 3 '16 at 17:18

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