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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.

$$AB+AC+BC+D'$$

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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}

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