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I'd like to simplify this equation: $ABC' + BC'D' + BC + C'D$ prove it to $B + C'D$

My attempt is : $$\begin{align} &= ABC' + BC'D'(A+A') + BC + C'D\\ &= ABC' + ABC'D' + A'BC'D' + BC + C'D\\ &= ABC'(1 + D') + A'BC'D' + BC + C'D\\ &= ABC' + A'BC'D' + BC + C'D\end{align}$$

and then i'm running out of idea.. can anyone help me what i suppose to do next step? Thank you

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  • $\begingroup$ can't you just use Karnaugh Map? if there is a restriction in problem that it has to be done algebraically and not with Karnaugh Map? $\endgroup$ – K.K.McDonald Mar 30 '16 at 12:00
  • $\begingroup$ the answer is $B+C'D$ with Karnaugh Map $\endgroup$ – K.K.McDonald Mar 30 '16 at 12:10
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ok here is solution algebraically( although i still insist Karnaugh Map is much more simple for the 4 variable case): $$ABC' + BC'D' + BC + C'D = ABC' + BC'D' + BC + C'D(1+B)=ABC' + BC'D' + BC + C'D+BC'D = B(AC'+C'D'+C+C'D)+C'D=B(AC'+C+C'(D'+D))+C'D=B(AC'+C+C')+C'D=B(AC'+1)+C'D=B+C'D$$

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  • $\begingroup$ Thank you so much. I were careless that we could put outside B. $\endgroup$ – Hermon Jay Apr 5 '16 at 14:06

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