Boolean Algebra simplify minterms

Question

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

Answer 1

Picking up where you (initially) left off (your work thus far is correct, save for the last line were you remove $\overline B$)...

$$\begin{align} &\overline{A}\cdot\overline{B}\cdot\overline{C} + A\cdot\overline{B}\cdot C + A\cdot B\cdot \overline{C} + A \cdot B\cdot C \\ \\ &\vdots \\ \\ &=\overline A \cdot \overline B \cdot \overline C + A(\overline B\cdot C+B) \\ \\ & = \overline A \cdot \overline B \cdot \overline C + A\cdot \overline B \cdot C + AB\tag{as good as it gets!}\\ \\ & = \overline B(\overline A \cdot \overline C + A\cdot C) + AB\tag{doesn't help any}\\ \\ \end{align}$$

Answer 2

$\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 $ $= C\bar{B} + AB$

$\bar{A} = \text{not}\space A$

Answer 3

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 $$\begin{align} \bar{A}\cdot\bar{B}\cdot\bar{C} + A\cdot\bar{B}\cdot C + A\cdot B\cdot \bar{C} + A \cdot B\cdot C &=\left(\bar{A}\cdot\bar{B}\cdot\bar{C} + A\cdot\bar{B}\cdot C \right)+ \left(A\cdot B\cdot \bar{C} + A \cdot B\cdot C \right) \\ &=\bar{B}\left(\bar{A}\cdot\bar{C} + A\cdot C \right)+A\cdot B\left(\bar{C} + C \right)\\ &=\bar{B}\left(\overline{A \oplus C}\right)+A\cdot B\\ \end{align}$$

Answer 4

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: $$A \oplus (\overline B \cdot \overline C)$$