Boolean Algebra simplify minterms I have this equation 
$$\bar{A}\cdot\bar{B}\cdot\bar{C} + A\cdot\bar{B}\cdot C + A\cdot B\cdot \bar{C} + A \cdot B\cdot C$$
and need to simplify it. I have got as far as I can and spent a good 2 hours at it. I've realized I now need to use De Morgan's law to continue however I am baffled as to which rule to use. If someone could send me in the right direction that would be great!
$$\bar{A}\cdot\bar{B}\cdot\bar{C} + A\cdot\bar{B}\cdot C + A\cdot B\cdot \bar{C} + A \cdot B\cdot C$$
$$\bar{A}\cdot\bar{B}\cdot\bar{C} + A(\bar{B}\cdot\bar{C} + B\cdot\bar{C} + B\cdot C$$
$$\bar{A}\cdot\bar{B}\cdot\bar{C} + A(\bar{B}\cdot C + B(\bar{C} + C)) $$
$$\bar{A}\cdot\bar{B}\cdot\bar{C} + A(\bar{B}\cdot C + B) $$ 
$$\bar{A}\cdot\bar{B}\cdot\bar{C} + A(C + B) $$ 
Thanks

If you use a karnaugh map:
$$
\begin{array}{c|c|c|c|c}
C, AB  & 00 & 01 & 11 & 10 \\ \hline
0 & 1  &   & 1 &  \\ \hline
1 &    &   & 1 & 1\\ \hline
\end{array}
\equiv
\bar{A}\bar{B}\bar{C} + A\bar{B} C + AB\bar{C} + ABC
$$
Which suggests Xor of the 2 groups:
$$
\begin{array}{c|c|c|c|c}
C, AB  & 00 & 01 & 11 & 10 \\ \hline
0 &    &   & 1 & 1\\ \hline
1 &    &   & 1 & 1\\ \hline
\end{array}
\equiv A
$$
$$
\begin{array}{c|c|c|c|c}
C, AB  & 00 & 01 & 11 & 10 \\ \hline
0 & 1  &   &   & 1\\ \hline
1 &    &   &   &  \\ \hline
\end{array}
\equiv \overline B \cdot \overline C
$$
Which gives:
$$A \oplus (\overline B \cdot \overline C)$$
