Complex Boolean simplification 
Simplify the following Boolean expression:
  $$F=A'B'C'+A'B'C+A'BC'+AB'C'+AB'C.$$  

I have tried and am getting stuck at $A'+B'+C'+A'BC'$.
According to the K-Map it should get to $B'+A'C'$.
 A: $F$ is NOT equal to 
$$A'+B'+C'+A'BC'=A'+B'+C'+A'BC'=A′+B′+C′(1+A′B)=A′+B′+C′.$$ 
If $A$ and $B$ are true and $C$ is false then $F$ is false whereas  $A′+B′+C′$ is true.
In order to simplify the given Boolean expression, note that  $A'B'C'=A'B'C'+A'B'C'$ and
\begin{align*}
F&=(A'B'C'+A'B'C+AB'C'+AB'C)+(A'BC'+A'B'C')\\
&=B'(A'(C'+C)+A(C'+C))+A'C'(B+B')\\
&=B'(A'+A)+A'C'=B'+A'C'\\
\end{align*}
where we used the fact that $A+A'=B+B'=C+C'=1$.
A: There is a handy equivalence rule called Adjacency:
Adjacency
$PQ + PQ' = P$
'Adjacency' is of course named after the fact that it reflects exactly what you do in a K-map: you group together adjacent cells in order to get a simpler expression.
Another very useful principle is called Reduction:
Reduction
$P + P'Q = P + Q$ (that is, in the context of $P$, the term $P'Q$ reduces to just $Q$)
Applied to your problem:
$$A'B'C'+A'B'C+A'BC'+AB'C'+AB'C= \text{ (Adjacency) }$$
$$A'B'+A'BC'+AB'= \text{ (Adjacency) }$$
$$B'+A'BC'\text{ (Reduction) }$$
$$B'+A'C'$$
If you didn't know these two equivalences yet ... put them in your toolbox immediately!
And here is one more:
Absorption
$P+PQ=P$
$P(P+Q)=P$
