F = AB'C + A'BC + ABD + CD' + D'
I started by doing D'(C+1) and reducing (C+1) into 1, resulting with D'1 which is just D'. I now have AB'C + A'BC + ABD + D', and cannot seem to find another way to reduce this further, though I know this cannot be the case. What am I missing? I tried to factor out a C from the first 2 terms, but did not know what to do after.
Well $ABD+D'=AB+D'$, after which you can set the $D'$ off to the side and focus on the rest.
Next, distributing out the common term is a good plan. However, recall $AB=ABC+AB$, so do that first and don't forget that $AB=AB+AB$.
Okay, that sounds like it is making things less simple, but look what happens.
$\begin{split}F ~=~& AB'C + A'BC + ABD + CD' + D'&\qquad\qquad&\text{as provided}\\=~ &AB'C+A'BC + ABD+D' &&\text{by absorption} \\= ~&AB'C+A'BC+AB&+D' &\text{by redundancy} \\=~&AB'C+A'BC+ABC+AB&+D' &\text{by absorption} \\= ~&(AB'+AB+A'B)C + AB~&+D'&\text{by distribution}\\=~&(AB'+AB~+~AB+A'B)C + AB~&+D'&\text{by idempotence} \\ =~&\text{well, you can take it from here.}\end{split}$
It's by the redundancy rule. $$\begin{split}ABD+D' &= (AB+D')(D+D')\qquad & \text{distribution}\\ &= (AB+D')1 & \text{inversion}\\ & = AB+D' & \text{identity}\end{split}$$
Remember your Boolean Algebra $$\begin{array}{|r:l:l:l|}\hline \text{Law} & \text{AND} &\text{OR}\\\hline 1 & \text{Identity} & {1A=A} & 0+A=A\\\hdashline 2 & \text{Annihilation} & 0A=0 & 1+A=1\\\hdashline 3 & \text{Inversion} & {AA'=0} & {A+A'=1} \\\hdashline 4 & \text{Double Inversion} & (A')'=A & (A')'=A \\\hdashline 5 & \text{Idempotence} & AA=A & A+A = A\\\hdashline 6 & \text{Commutation} & AB=BA & A+B=Ba\\\hdashline 7 & \text{Association} & (AB)C {=A(BC)\\=ABC} & (A+B)+C{=A+(B+C)\\=A+B+C}\\\hdashline 8 & \text{Distribution} & A(B+C)=AB+AC & A+BC = (A+B)(A+C)\\\hdashline 9 & \text{Duality} & (AB)'= A'+B' & (A+B)'=A'B'\\\hline 10 &\text{Absorption} & A(A+B)=A & A+AB=A\\\hdashline 11 & \text{Redundancy} & A(A'+B)=AB & A+A'B=A+B\\\hline \end{array}$$