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I've got a problem here that I could use help solving. I have simplified it to this point. Using Wolfram Alpha, I know it is still possible. My lecturer did it but I didn't catch all of it. It is frustrating me like mad and I didn't want to come here for just one question but I feel that it is my last resort.

I am stuck on;

(B&&C)||(¬B&&C&&D)||(¬A&&C&&D)||(A&&D)||(A&&B)

For a more eye friendly version;

BC + !BCD + !ACD + AD + AB

I need to extract CD from that but I don't know a rule that can do it for me. Wolfram Alpha is telling me that I can still work out the final answer from here (Which is (A + C)(B + D)).

Please Stack Exchange, you're my only hope.

Pejman Poh
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1 Answers1

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First note that $$B+B\,'D=(B+BD)+B\,'D=B+(BD+B\,'D)=B+(B+B\,')D=B+D\;.$$ Similarly, $A'C+A=A+C$. Thus, $A'CD+AD=(A'C+A)D=(A+C)D$, and $BC+B\,'CD=(B+B\,'D)C=(B+D)C$, so

$$\begin{align*} BC+B\,'CD+A'CD+AD+AB&=(B+D)C+(A+C)D+AB\\ &=BC+CD+AD+CD+AB\\ &=AB+AD+CB+CD\\ &=A(B+D)+C(B+D)\\ &=(A+C)(B+D)\;. \end{align*}$$

Brian M. Scott
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  • I'm not sure I understand why B+B'D = (B+BD)+B'D? How did you get to that in the first place? Otherwise I think I understand the rest of the proof. – Pejman Poh Oct 21 '13 at 23:40
  • @Pejman: That’s one of the absorption laws: $X+XY=X$ for any $X$ and $Y$. The other is $X(X+Y)=X$. – Brian M. Scott Oct 21 '13 at 23:47
  • I'm googling around and I understand that I can use that rule. But no where does it explain why that can happen? Sorry that it's taking me a long time to get this. – Pejman Poh Oct 22 '13 at 10:15
  • @Pejman: Think of it in terms of a Boolean algebra of sets. It says that $X\cup(X\cap Y)=X$, which is clear: anything in $X\cap Y$ is already in $X$. Or in terms of truth values of statements: then it says that $X\text{ of }(X\text{ and }Y)$ is equivalent to $X$. This is also clear, since the alternative $X\text{ and }Y$ adds nothing to the alternative $X$. – Brian M. Scott Oct 22 '13 at 20:27
  • I solved it! Thanks a lot Brian! – Pejman Poh Oct 27 '13 at 14:51
  • @Pejman: You’re welcome! – Brian M. Scott Oct 28 '13 at 04:02