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
 A: Picking up where you (initially) left off (your work thus far is correct, save for the last line were you remove $\overline B$)...
$$\begin{align} &\overline{A}\cdot\overline{B}\cdot\overline{C} + A\cdot\overline{B}\cdot C + A\cdot B\cdot \overline{C} + A \cdot B\cdot C \\ \\
&\vdots \\ \\
&=\overline A \cdot \overline B \cdot \overline C + A(\overline B\cdot C+B) \\ \\
& = \overline A \cdot \overline B \cdot \overline C + A\cdot \overline B \cdot C + AB\tag{as good as it gets!}\\ \\ 
& = \overline B(\overline A \cdot \overline C + A\cdot C) + AB\tag{doesn't help any}\\ \\
\end{align}$$
A: The $\mathrm{ExOR}$ function can be denoted by: $\oplus$ ; $X \oplus Y=\overline {X}\cdot Y +X \cdot \overline{Y} $. 
Also $\overline{X \oplus Y}=\overline {X}\cdot \overline{Y} +X \cdot {Y} $
Hence $$\begin{align}
\bar{A}\cdot\bar{B}\cdot\bar{C} + A\cdot\bar{B}\cdot C + A\cdot B\cdot \bar{C} + A \cdot B\cdot C &=\left(\bar{A}\cdot\bar{B}\cdot\bar{C} + A\cdot\bar{B}\cdot C \right)+ \left(A\cdot B\cdot \bar{C} + A \cdot B\cdot C \right) \\
&=\bar{B}\left(\bar{A}\cdot\bar{C} + A\cdot C \right)+A\cdot B\left(\bar{C} + C \right)\\
&=\bar{B}\left(\overline{A \oplus C}\right)+A\cdot B\\
\end{align}$$
A: 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:
$$A \oplus (\overline B \cdot \overline C)$$
A: $\bar{A}\bar{B}\bar{C} + A\bar{B}C + AB\bar{C} + ABC $
$= \bar{A}\bar{B}C + A\bar{B}C + AB(\bar{C} + C) $
$= \bar{A}\bar{B}C + A\bar{B}C + AB $
$= C(\bar{A}\bar{B} + A\bar{B}) + AB $
$= C(\bar{B}(\bar{A} + A)) + AB $
$= C\bar{B} + AB$
$\bar{A} = \text{not}\space A$
