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I am trying to simplify a logic expression but I think I simplified it too much. The expression is as follows:

$$\overline{\overline{(A \cdot B)} \cdot C \cdot (\overline{A}+\overline{(B+C)})}$$

This is what I got after I did the simplification: But Logic Friday says that the answer should be:

$$A \cdot B+A \cdot C+\overline{C}$$

Which one is correct? Am I allowed to do the manipulations I did in this case?

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2  
I'd suggest filling two karnaugh tables to confirm that the two are equivalent (truth tables would work for that, Karnaugh tables would help fix the potential issue) –  AProgrammer Aug 18 at 8:46
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@user2357111317192329 I think this is an homework. You reach a better answer than what they (who assigned you the homework) expected to reach. So, great job! (Sorry for my bad english) –  Danny Aug 18 at 8:58
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@user2357111317192329: funny that you minimized the function better than Logic Fridays. –  CommuSoft Aug 18 at 15:02
    
@CommuSoft thanks to Vladimir Cravero, he taught me this method of simplification! –  user51414 Aug 19 at 4:01

2 Answers 2

up vote 10 down vote accepted

Both answers are correct.

Let: $$ f_1(A,B,C) = AB+AC+\overline{C}\\ f_2(A,B,C) = A+\overline{C} $$ Let's build the thruth table: $$\begin{array}{|c|c|c|c|c|} \hline A & B & C & f_1 & f_2 \\ \hline 0 & 0 & 0 & 1 & 1\\ \hline 0 & 0 & 1 & 0 & 0\\ \hline 0 & 1 & 0 & 1 & 1\\ \hline 0 & 1 & 1 & 0 & 0\\ \hline 1 & 0 & 0 & 1 & 1\\ \hline 1 & 0 & 1 & 1 & 1\\ \hline 1 & 1 & 0 & 1 & 1\\ \hline 1 & 1 & 1 & 1 & 1\\ \hline \end{array}$$

As you can see the two functions correspond.

Please note that this tabular method of proving that two functions are the same is perfectly valid and is called Proof by exhaustion.

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Thank you again Vladimir Cravero. –  user51414 Aug 18 at 8:49
    
You're welcome of course. Each question gives you a new weapon to defeat logic expressions, go for it! –  Vladimir Cravero Aug 18 at 8:49

Simply use De Morgan's Theorem twice. Using it once we get

$$A \cdot B + \overline{C} + \overline{(\overline{A} + \overline{(B+C)})}$$

Using it on the final term then gives us

$$A \cdot B + \overline{C} + A \cdot (B+C)$$

which simplifies to the expression given

$$A \cdot B + A \cdot C + \overline{C}$$

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2  
The question is about the possible further simplification. –  AProgrammer Aug 18 at 8:47
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Can't you simplify that last expression you got further like I did? –  user51414 Aug 18 at 8:48

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