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So I have this expression: $$ A'BD + B'C' + AC' + C'D $$ How can I simplify it using boole's algebra? I figure out the $C'D$ is included in the other terms through Karnaugh maps but I can't figure out the simplification.

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To show that $C'D \leq AC' + A'BD + B'C'$ is equivalent to show that their product is $C'D$.
Now, \begin{align} C'D ( AC' + A'BD + B'C' ) &= AC'D + A'BC'D + B'C'D \\ &= (A + A'B + B') C'D. \end{align} Now we prove that $A + A'B + B' = 1$, concluding the initial inequality. \begin{align} A + A'B + B' &= A + A'B + (A + A')B' \\ &= A + A'B + A'B' \\ &= A + A'( B + B') \\ &= A + A' \\ &= 1. \end{align}

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  • $\begingroup$ Perfect, solved it;) $\endgroup$ – Carlos Sá Nov 3 '16 at 19:44
  • $\begingroup$ Just a request, if you did not know from karnaugh maps that $C'D$ was included how would you simplify this expression? I was thinking about expanding all terms and cut equal ones but there ought to be a more beautiful solution. Right? @amrsa $\endgroup$ – Carlos Sá Nov 3 '16 at 20:16
  • $\begingroup$ @CarlosCosta That's a tricky one :) If I didn't know, I really can't tell how I would come to this final result. The original expression was already in the form of a disjunction of conjunctions, although it was not the DNF because it could be further simplified. Your clue using Karnaugh maps was crucial. Then it was easy to prove that it was so, but otherwise, I'm not sure I could reach this solution. But of course you can use the above proof to give a more direct deduction of the equality... $\endgroup$ – amrsa Nov 3 '16 at 20:22

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