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I have solved the equation like this:

 (A + C)(AD + AD) + AC + C
=(A + C)(AD) + (A + C)(AD) + AC + C
=AAD + ACD + AAD + ACD + AC + C
=AAD + AAD + ACD + ACD + AC + C
=AAD + ACD + AC + C 
=AAD + ACD + C(A+1)
=AAD + ACD + C
=AD + ACD + C
=AD(1+C) + C
=AD + C

Am I right?

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  • $\begingroup$ Definitely boolean algebra you want? $\endgroup$ Oct 3, 2013 at 14:50
  • $\begingroup$ @stevemarvell I want to solve the equation using laws of Boolean algebra. Have I solved it correctly? $\endgroup$
    – user221458
    Oct 3, 2013 at 14:51
  • $\begingroup$ You might want to consider the truth of AA $\endgroup$ Oct 3, 2013 at 14:53
  • $\begingroup$ @stevemarvell Just edited it. Is it perfect now? $\endgroup$
    – user221458
    Oct 3, 2013 at 14:55
  • 1
    $\begingroup$ consider the simplification of (AD + ACD) $\endgroup$ Oct 3, 2013 at 15:00

2 Answers 2

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It looks just fine! Well done!

A bit more simply, $$\begin{align}(A+C)(AD+AD)+AC+C &= (A+C)AD+AC+C\\ &= AAD+CAD+AC+C\\ &= AD+(AD+A+1)C\\ &= AD+C\end{align}$$

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  • $\begingroup$ It can be simplified a great deal more than that. $\endgroup$ Oct 3, 2013 at 14:59
  • $\begingroup$ Really? How can $AD+C$ be simplified? $\endgroup$ Oct 3, 2013 at 15:00
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    $\begingroup$ The question has been edited since my comment. $\endgroup$ Oct 3, 2013 at 15:01
  • $\begingroup$ You probably just needed to refresh your page. I had the same thing happen to me. Initially, I mentioned that two pairs of terms had common factors, which allowed further simplification, but after posting the answer, the page automatically refreshed and showed that the OP had noticed that and acted accordingly, so I edited my answer. $\endgroup$ Oct 3, 2013 at 15:03
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I would have taken the following steps: $$ \begin{align} (A + C)(AD + AD) + AC + C &= (A + C)(AD) + (AC + C)\\ &= AD + ADC + C\\ &= AD + C\\ \end{align} $$

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