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Basing on that problem. All I have in my solution is this:

mystep1:[AB(C +(B' + C')) + AB']CD'

mystep2:[AB(CB'+ CC') + AB']CD'

mystep3: [AB(CB') + AB']CD'

mystep4:[B(A+C+B') + AB']CD'

mystep5:[AB + AC + AB'] CD'

mystep6:[AC]CD'

mystep7: ACD'

F = ACD' (my simplified answer)

Please do tell me if I have followed all the necessary rules and I have the correct answer.. I am doubtful with this. For I have tried plotting it on livewire(software for logical designs) and then tried kmapping for much more easier and faster simplification. I got different answers. I believe I have wrong plots on my logical design. But this Step-by-step solution of mine is also hanging. Really not sure.

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Sha Lala
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3 Answers3

1

$$(AB(C + B' + C') + AB')CD'$$ but $C+C'=1$ and $1+B =1$

$$(AB + AB')CD'$$

$$A(B + B')CD'$$ $$ACD'$$

Suraj M S
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You switched operations in your second step. In general, $$C+(B'+C')\ne C(B'+C')=CB'+CC'.$$ Rather, $$C+(B+C')=C+(C'+B)=(C+C')+B=1+B=1.$$ Your third, fourth, fifth, and sixth steps are also faulty, but somehow, five wrongs made a right, and you came up with the correct answer.

Despite the correct answer, and despite the fact that you'll need to take a different path from the corrected second step, I strongly recommend that you take a hard look at your third, fourth, fifth, and sixth steps to see if you can find where you made your mistakes in reasoning (so you can avoid such mistakes in the future). Let me know if you have any difficulty spotting the mistakes. (Ignore the $CD'$ on the outside, as it never comes into play in those steps.)

Cameron Buie
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  • Let me see.. Here's my edited answer: (hoping I have followed the rules)

    step1 > [AB (C + (B' + C')) + AB'] CD' step2 > [AB ((C + C') + B') + AB'] CD' step3 > [AB ((1) + B') + AB'] CD' step4 > [AB (1 + B') + AB'] CD' step5 > [AB + AB'] CD' step6 > ABCD' + AB'CD' step7 > ACD' + ACD' step8 > ACD' ---> FINAL ANSWER

    First off, This time is this correct? Second question, What is this rule again? (1 + B) = 1 or (1 + B') = 1 Third, This will result to cancelling them when you're into solutions right?

     – Sha Lala Sep 18 '13 at 19:15
  • You can actually add an edited answer to your original post by clicking the edit button, or just following this link. Did you figure out what went wrong in any or all of the other four steps with errors? – Cameron Buie Sep 18 '13 at 19:20
  • Your edited answer is much better, though I don't follow step 7. Instead, I'd factor the $A$ out after step 5 and then use the fact that $B+B'=1$. The rule $B+1=1$ doesn't have a particular name that I'm aware of. It follows from complement, associativity, and idempotence rules: $$\begin{align}B+1 &= B+(B+B')\ &= (B+B)+B'\ &= B+B'\ &= 1.\end{align}$$ I'm not quite sure what you mean by your third question. – Cameron Buie Sep 18 '13 at 19:27
  • I see, I tend to overlook things.I'll answer my question using this answer button, using my final Answer. See if it's correct. Thank you Cameron.

    Note: I tried to show every process so that it's really clear for others also for me. I needed to see where are things going.

     – Sha Lala Sep 19 '13 at 14:33
  • Precisely right! – Cameron Buie Sep 19 '13 at 15:31
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Here's my in-depth solution - final answer:

Given F = [ AB ( C + (BC)' ) + AB' ] CD'

Solution:

  1. [AB (C + (B' + C')) + AB'] CD'
  2. [AB ((C + C') + B') + AB'] CD'
  3. [AB ((1) + B') + AB'] CD'
  4. [AB (1 + B') + AB'] CD'
  5. [AB + AB'] CD'
  6. [A(B + B')] CD'
  7. [A(1)] CD'
  8. [A]CD'

FINAL ANSWER: ACD'

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Sha Lala
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