1) I just want to know that my steps are correct or not? what are the missing steps. please help me.
2) Actually I can not simplify this is. So what are the missing steps at my trying path?
please help me.....
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1$\begingroup$ You can use for example \$ \overline{ab} \$ to write ab with a bar and to clear your question. $\endgroup$– BIS HDNov 27, 2013 at 12:12
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2 Answers
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A few hints:
1) Your steps don't seem to be correct. As far as I can see, you can write
$$\bar b ( a(c + \bar c) + \overline{ac}) = \bar b (a + \bar{ac}) = \bar b a + \overline{abc}$$ How to proceed from here? How did you come to your answer?
2) Consider that $a + \bar a = $ true, for every $a$. So you can simplify a lot in your last expression.
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$\begingroup$ yes but i can not understand this....now i read the Boolean rules. but not a good result. You are how to get that "a"? "
-b(a(c+ -c) + -a-c)" $\endgroup$ Nov 27, 2013 at 12:52
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- $\overline{abc + ab\overline{c} + \overline{a}b\overline{c}}$ = $\overline{b(ac + a\overline{c} + \overline{a}\overline{c})}$
=$\overline{b([ac + a\overline{c}]+ [a\overline{c}+ \overline{a}\overline{c}])}$ as $a + a =a$
=$\overline{b(a + \overline{c})}$
=$\overline{b} + \overline{a}c$ Using the De-Morgans Law
2.From 1, 3 terms you can eliminate a, 2&3 b, 3&4 c resulting in
$\overline{\overline{b}c + ac + ab} = (b + \overline{c})(\overline{a} +\overline{c})(\overline{a} +\overline{b})$
This can be multiplied and simplified further to the result
