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