2
$\begingroup$

Original expression: $$D=A’BCD+AB’C’D+AB’CD’+AB’CD+ABC’D’+ABC’D+ABCD’+ABCD$$

This is the simplest form I could find: $$B\cdot C\cdot D+A\cdot(B'\cdot D+B \cdot C'+CD')$$ It is equivalent to the solution. This is the solution I got from an online calculator: $$A \cdot B + A \cdot C + A \cdot D + B \cdot C \cdot D$$

How can I do this? Do I need to take a step back and then simplify more?

$\endgroup$
  • $\begingroup$ Can you please post the original expression, not the one you wrote as "the simplest form I could fnd"? $\endgroup$ – amWhy Nov 3 '17 at 14:49
  • $\begingroup$ @amWhy Posted it $\endgroup$ – shurup Nov 3 '17 at 14:52
  • $\begingroup$ Thanks, Nick... $\endgroup$ – amWhy Nov 3 '17 at 17:21
2
$\begingroup$

$D=A'BCD+AB'C'D+AB'CD'+AB'CD+ABC'D'+ABC'D+ABCD'+ABCD$

$= A(B'C'D + B'CD' + B'CD + BC'D' + BC'D + BCD' + BCD) + A'BCD$

$= A(B(C'D'+C'D+CD'+CD) + B'(C'D + CD' + CD))+A'BCD$

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

$=A(B(C*1+C'*1)+B'(C*1+C'D))+A'BCD$

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

$= AB + AC + AC'D +A'BCD$

$=AB + AC + AD + A'BCD$

$= AB + AC + AD + BCD$

The trick is in the last 3 lines to think about what is covered in the previous cases. For example, if you have $AB + AB'C$, then if $A$ and $B$ are correct, you have 1 anyway. Or in other words you could write this as $A(BC + BC' + B'C)$. I hope my point gets clear here, when in doubt, a table will always help.

$\endgroup$
1
$\begingroup$

$$A’BCD+AB’C’D+AB’CD’+\color{red}{AB’CD}+ABC’D’+\color{green}{ABC’D}+\color{blue}{ABCD’}+\color{orange}{ABCD} = \text{(Idempotence x 6)}$$

$$A’BCD+AB’C’D+AB’CD’+\color{red}{AB’CD+AB’CD}+ABC’D’+\color{green}{ABC’D+ABC’D}+\color{blue}{ABCD’+ABCD’}+\color{orange}{ABCD +ABCD+ABCD+ABCD}= \text{(Commutation)}$$

$$\color{red}{ABC’D’+ABC’D+ABCD’+ABCD}+\color{green}{AB’CD’+AB’CD+ABCD’+ABCD} +\color{blue}{AB’C’D +ABC’D+AB’CD+ABCD}+\color{orange}{A’BCD+ABCD}= \text{(Adjacency)}$$

$$\color{red}{ABC’+ABC}+\color{green}{AB’C+ABC} +\color{blue}{AB’D+ABD}+\color{orange}{BCD}= \text{(Adjacency)}$$

$$\color{red}{AB}+\color{green}{AC} +\color{blue}{AD}+\color{orange}{BCD}$$

(note: the colors help connect lines 1 and 2, and also lines 3,4,5, but the colors do not reveal how line 2 goes to line 3: that step is really just a matter of reordering all the terms by Commutation)

$\endgroup$
  • 1
    $\begingroup$ Really, just being honest: The use of four colors (plus black) becomes very distracting. E.g., I have no inclination to make rhyme or reason of what you've written. Color, when used sparingly, is helpful. Too much color can overwhelm the entire substance of the answer. $\endgroup$ – amWhy Nov 3 '17 at 22:14
  • 1
    $\begingroup$ @amWhy Looking at it again, I see what you mean ... :/ $\endgroup$ – Bram28 Nov 3 '17 at 22:48

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.