Well, here's my process:
$$\begin{align*}F&=\overline{W}Y + W\overline{Y}Z + \overline{W}\overline{X}\overline{Z}\\\\&=\overline{\overline{\overline{W}Y + W\overline{Y}Z + \overline{W}\overline{X}\overline{Z}}}\\\\&=\overline{W Y + W \overline{Z} + \overline{W}\overline{Y}Z+X\overline{Y}\overline{Z}}\\\\&=\overline{\overline{\overline{W Y + W \overline{Z}}} + \overline{\overline{\overline{W}\overline{Y}Z+X\overline{Y}\overline{Z}}}}\\\\&=\overline{\overline{\overline{W}+\overline{Y}Z} + \overline{Y+W Z + \overline{X}\overline{Z}}}\\\\&=\overline{\overline{\overline{W}+\overline{Y+\overline{Z}}} + \overline{Y+\overline{\overline{W Z + \overline{X}\overline{Z}}}}}\\\\&=\overline{\overline{\overline{W}+\overline{Y+\overline{Z}}} + \overline{Y+\overline{X \overline{Z}+\overline{W}Z}}}\\\\&=\overline{\overline{\overline{W}+\overline{Y+\overline{Z}}} + \overline{Y+\overline{\overline{\overline{X}+ Z}+\overline{W+\overline{Z}}}}}\end{align*}$$
I've no promise that it's optimized in any way. And it's only for 2-in NORs. But it does follow a process that should work for you.
You seem already familiar with the basic idea. But just getting stuck here and there. Hopefully, the above clears up that log-jam.
Added
I hacked out the following behavioral Python code recently:
def fsimplify(e): f = distribute(e) # distribute products of sums: (a+b)*(a'+c)=aa'+ac+a'b+bc f = removefalse(e) # remove always-false terms: ac+a'b+bc f = expand(f) # expand: abc+ab'c+a'bc+a'bc'+abc+a'bc f = removeduplicates(f) # remove duplicates: abc+ab'c+a'bc+a'bc' return reduce(f) # combine: ac+a'bdef fgenerateNOR(e): f = fsimplify(e) if f is true: return 1 if f is false: return 0 if termcount(f) = 1: # termcount counts the number of terms if onevar(f): # onevar determines if a term only has one variable if inverted(f): # inverted checks to see if the variable is not'd return NORINV(f) # if so, just use a NOR as an inverter return f # otherwise just pass the variable out return fgenerateNOR(fnegate(f)) # compound term is a product -- recurse to resolve fL, fR = split(f) # split expression into two sums-of-terms return NOR(fgenerateNOR(fnegate(fL)),fgenerate(fnegate(fR)))The above is a recursive process that should produce the right results. I may have missed a detail. But it may capture the basic idea.
