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$\bar{A}.\bar{C}+\bar{A}.B+A.\bar{B}.C+B.C$
$=>\bar{A}.\bar{C}+\bar{A}.B+A.\bar{B}.C+\color{Orchid }{(A+\bar{A}) }.B.C$
$=>\bar{A}.\bar{C}+\color{blue}{\bar{A}.B}+\color{green}{A}.\bar{B}.\color{green}{C}+\color{green}{A}.B.\color{green}{C}+\color{blue}{\bar{A}.B}.C$
$=>\bar{A}.\bar{C}+\bar{A}.B+A.C$

From here onwards I tried using $(B+\bar{B})$, $(A+\bar{A})$, and $(C+\bar{C})$ with the terms, but no matter what it all deduce to the last expression only. But the answer has only two terms $\bar{A}.B+C$. How to minimize this further?

2 Answers2

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Karnaugh map for AB along the top and C on the left:

$\begin {array} {c | c c c c} & 00 & 01 & 11 & 10 \\ \hline 0 & 1 & 1 & 0 & 0 \\ 1 & 0 & 1 & 1 & 1 \\ \end {array}$

The zeros indicate the locations where the equation is not satisfied. So your equation is: $$\lnot (\bar A \bar B C + A \bar C)$$ $$(A + B + \bar C) (\bar A + C)$$

Work backwards now and get your original problem.

DanielV
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  • I am sorry it has to be in the SOP form and the answer is $\bar{A}.B+C$ –  Oct 25 '13 at 11:51
  • $\bar AB + C$ is wrong. It is satisfied by $\bar A, \bar B, C$, whereas the original equation is not. It does not satisfy $\bar A, \bar B, \bar C$, whereas the original equation does. – DanielV Oct 25 '13 at 16:58
  • Hmmm you are right, I also get the same deduction. This must be it. Thank you very much. –  Oct 25 '13 at 17:44
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$ \bar{A}.\bar{C}+\bar{A}.B +A.C $ is the minimized expression as

$\bar{A}.\bar{C}+\bar{A}.B+A.C $ = {0,2,3,5,7},

where as $\bar{A}.B + C $ = {1,2,3,5,7}

Please verify yourself by expanding the both to the canonical SOP form.......

sai kiran grandhi
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