 Solution

Yes, under the finite total assumption in part (d). The pair result retains enough information to reconstruct the tail; the ~ pattern keeps its two components lazy:Take $f=\mathrm{spanP}p$, $g=(:)$ and $a=[]$. The base image is $([],[])$. If $p(x)$ is true, the recursive equation gives the first branch. If it is false, part (d), applied to the tail, gives $ys++zs=xs$, so the second branch becomes $([],x:xs)$. These are precisely the fusion compatibility equations. Finite structural induction establishes the desired equality; it does not require extending the restricted-domain hypothesis to undefined predicate applications.

The qualification matters. If $p(v)$ is false but $p(u)$ is undefined, ordinary spanP p [v,u] can return the untouched list suffix immediately, whereas the reconstruction through spanFold p [u] cannot recover that list suffix. The weaker condition needed only for the single reconstruction identity in (d) is therefore insufficient for this particular list fold on all list suffixes. The stronger totality condition stated there supplies exactly the missing premise.