I started with a big problem and through various simplifications I've arrived at a point where I don't quite know what else to do. I've tried to further simplify but I keep running into issues.
Original
(A+B+!C+!D)(A+!B+C+D)(A+!B+!C+D)(A+!B+!C+!D)(!A+B+!C+!D)(!A+!B+C+D)(!A+!B+C+!D)(!A+!B+!C+D)(!A+!B+!C+!D)
Simplified
= (A+B+!C+!D)(!A+B+!C+!D)(A+!B+C+D)(A+!B+!C+D)(A+!B+!C+!D)(!A+!B+C+D)(!A+!B+C+!D)(!A+!B+!C+D)(!A+!B+!C+!D)
= (B+!C+!D)+{(A)(!A)}(A+!B+D)+{(C)(!C)}(A+!B+!C+!D)(!A+!B+C)+{(D)(!D)}(!A+!B+!C)+{(D)(!D)}
= (B+!C+!D)+{0}(A+!B+D)+{0}(A+!B+!C+!D)(!A+!B+C)+{0}(!A+!B+!C)+{0}
= (B+!C+!D)(A+!B+D)(A+!B+!C+!D)(!A+!B+C)(!A+!B+!C)
= (B+!C+!D)(A+!B+D)(A+!B+!C+!D)(!A+!B)+[(C)(!C)]
= (B+!C+!D)(A+!B+D)(A+!B+!C+!D)(!A+!B)+[0]
= (B+!C+!D)(A+!B+D)(A+!B+!C+!D)(!A+!B)
This is where my issues arise. I don't know how to further simplify it. I use KMaps and I got to my answer but I don't know how to do it through Boolean algebra
$\begin{align}&\sf= (B+!C+!D)~(A+!B+D)(A+!B+!C+!D)(!A+!B) &&\text{you are here} \\&\sf= (B+!C+!D)~\big(!B+(A+D)(A+!C+!D)(!A)\big)&&\text{distribute out common disjunct}\sf!B \\&\sf= (B+!C+!D)~\big(!B+\big(A+D(!C+!D)\big)(!A)\big)&&\text{distribute out common disjunct }\sf A \\&\sf= (B+!C+!D)~\big(!B+(A+!CD)(!A)\big)&&\text{absorption of }\sf !D \\&\sf= (B+!C+!D)~(!B+!A!CD)&&\text{absorption of }\sf A \\&\sf=&&\text{keep on going on} \end{align}$