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I have a function, $$ f = (A+B\cdot \overline C) $$

I am trying to simplify it this form using the inverse function $\overline f$ from the truth table (by anding the rows which form a '0' result).

$$ \overline f = (\overline A + \overline B + \overline C) \cdot (\overline A + \overline B + C) \cdot (\overline A + B + C) $$

My steps are as follows, $$ \overline f = ((\overline A + \overline B) + (C \cdot \overline C)) \cdot (\overline A + B + C) $$ $$= (\overline A + \overline B) \cdot (\overline A + B + C)$$ $$= \overline A + (\overline B \cdot (B + C))$$ $$= \overline A + \overline B \cdot C$$ now, $$ \overline{\overline f} = f = \overline {\overline A + \overline B \cdot C} = A \cdot (B + \overline C) $$

But from directly inverting the function, f $$ \overline f = \overline {(A+B\cdot \overline C)}$$ $$= \overline A \cdot (\overline B +C)$$ and this checks out in the truth table.
I wanted to get this as the answer to f in POS form.
Using Karnaugh map, the answer becomes $$ f = (A+B) \cdot (A + \overline C) = A + B \cdot C $$

So did I do something wrong in the simplification ?

Prasanth Ravi
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  • Sorry about the confusion, your use of the distributive law checks out on that first step. I usually use the (A + B)(A + B') = A theorem, but that is a perfectly valid approach. – Derek W Aug 02 '14 at 04:19

1 Answers1

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Given : $$ f = (A+B\cdot \overline C) $$

When you do it the following way, you get this what you already got :

$$ \overline f = (\overline A + \overline B + \overline C) \cdot (\overline A + \overline B + C) \cdot (\overline A + B + C) $$

$$ \ = (\overline A +\overline B) \cdot (\overline A + B + C) $$ $[Since (P+Q).(P+\overline Q)=P].$ $ Here$ $P = \overline A + \overline B,$ so $$ (\overline A + \overline B)\cdot (\overline A + B + C) $$ $$= \overline A + \overline B \cdot C$$

But when you are directly inverting the function, f, you get $ $ $\overline f = \overline {(A+B\cdot \overline C)}$ $= \overline A \cdot ( \overline B +C)$

creative
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