Trying to simplify $A'B'C'D + A'B'CD + A'BC'D + AB'CD + ABCD$ My solution so far:
$A'(B'C'D + B'CD + BC'D) + A(B'CD + BCD)$
$= A'(C'D(B' + B) + B'CD) + A(CD(B'+B))$
$= A'(C'D(1) + B'CD) + A(CD(1))$
$= A'C'D + A'B'CD + ACD$
$= D(A'C' + AC) + A'B'CD$
$= D(A\text{ xnor }C) + A'B'CD$
$= D[(A\text{ xnor }C) + A'B'C]\text{ <<< Can't seem to get any further than this.}$
Did I perhaps screw up somewhere? Or not do it as well as I could have? Maybe missed using a rule?
Thanks!
K
 A: I'd start by plotting the Karnaugh map.
$A′B′C′D+A′B′CD+A′BC′D+AB′CD+ABCD$ is an sum of five minterms, thus the Karnaugh map will show five ones (I've left out zeroes for greater visual contrast):
$$
\begin{matrix}
     & A'B' & A'B & AB & AB' \\
C'D' &                       \\
C'D  &  1   &  1             \\
CD   &  1   &     &  1 &  1  \\
CD'  &
\end{matrix}
$$
This can be covered as a sum of three products, of which two are essential, in two ways. 
$$
F = A'C'D + ACD + B'CD = ((A\ xnor\ C) + B'C)D \\
F = A'C'D + ACD + A'B'D = ((A\ xnor\ C) + A'B')D
$$
Which are both one term shorter than your original: $((A\ xnor\ C) + A'B'C)D$. Note how you used disjoint terms as a covering for the function. This often happens when you simplify as expression without looking at the Karnaugh map.

Starting from the same Karnaugh map, you can also choose to start with a product of sums. As a product of three sums, all essential:
$$
F = D(A'+C)(A+B'+C')
$$
Or if you prefer, an inverted sum of products:
$$ F = not(D'+AC'+A'BC) $$
A: It simplifies to $[(A+B)'C+A' + B'D +BD]'$.
