I have the following Boolean expression that I would like to simplify. I know that I could use a Karnaugh map to simplify, but I need to use Boolean Algebra and simplify step by step.
$'$ is NOT
$+$ is OR $$AB'C+A'BC+ABD+CD'+D'$$ I was able to only get his far. $$AB'C+A'BC+ABD+D'$$
Looking on Wolfram Alpha, I see that this can be more simplified to the following, I'm unsure of what theorems to use to reduce this anymore.
$$AB+AC+BC+D'$$
To start with, I suppose you know that in Boolean algebras, $XY \leq X$, so that $X + XY = X$.
(This is what you used in that little bit you advanced.)
Now,
\begin{align}
ABC (ABD + D')
&= ABCD + ABCD'\\
&= ABC(D+D') = ABC,
\end{align}
so that $ABC \leq ABD + D'$, whence $ABC + ABD + D' = ABD + D'$.
We also use the fact that $ABD' + D' = D'$.
Using this, we get
\begin{align}
AB'C + A'BC + ABD + CD' + D'
&= AB'C + A'BC + ABD + D' \\
&= AB'C + A'BC + ABC + ABD + D' \\
&= AB'C + A'BC + ABD + ABC + ABD' + D'\\
&= ABD + ABD' + ABC + AB'C + ABC + A'BC + D'\\
&= AB(D+D') + AC(B+B') + BC(A+A') + D'\\
&= AB + AC + BC + D'.
\end{align}
