How would you simplify the following sum of products expression using algebraic manipulations in boolean algebra?
$$A'B'C' + A'B'C + ABC'$$
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$\begingroup$Essentially, all that's involved here is using the distributive law (DL), once.
Distributive Law, multiplication over addition: $$PQ + PR = P(Q + R)\tag{DL}$$
In your expression, in the first two terms, put $P = A'B'$:
We also use the identity $$\;P + P' = 1\tag{+ID}$$
$$\begin{align} A'B'C' + A'B'C + ABC' & = A'B'(C' + C) + ABC' \tag{DL}\\ \\ &= A'B'(1) + ABC' \tag{+ ID}\\ \\ & = A'B' + ABC'\end{align}$$
$\endgroup$ $\begingroup$Hint: the first two terms are the same except for the $C'$ or $C$. Put those two terms together.
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