I have been working on a boolean algebra problem. I have to simplify the following:$$\overline{A(\overline{C+D})}+BE$$
The only part I don't understand is how to deal with the double bar, and I can't find any examples of this from my lecture. Thank you.
The bar is the complementation operator. (logical not).
When asked to "simplify", you are expected to use de Morgan's laws and such until the only negations are of literals.
$$\begin{align}\overline {~X~Y~} ~&=~ \overline X+\overline Y\\[1ex] \overline {~X + Y~} ~&=~ \overline X~\overline Y\end{align}$$
Recall also that in Boolean logic, double negation of an expression equals the expression itself.
$$\begin{align}\overline {~\overline X~} ~&=~ X\end{align}$$
So substituting $Z=C+D$, can you simplify $\overline{A\, \overline Z}+B\,E$ ?