Every composite bull-free graph has a nontrivial module

By [F1] and [L2], after passing to the complement if needed we may assume that $A$ is an odd hole with vertices $h_1,\dots ,h_k$ in cyclic order, together with a vertex $c$ complete to $V(A)$ and a vertex $a$ anticomplete to $V(A)$. To apply [L1], it is enough to show that $V(A)$ is split.

Let $x\in V(G)\setminus V(A)$ be neither complete nor anticomplete to $V(A)$. By cyclic symmetry, assume $x$ is adjacent to $h_1$ and nonadjacent to $h_2$. If $x$ is adjacent to $h_k$, then the path $h_k$-$h_1$-$h_2$ gives the first split alternative from [F2]. So assume $x$ is nonadjacent to $h_k$. If $x$ is also nonadjacent to $h_{k-1}$, then $h_1{h}_2\in E(G)$ while $h_1{h}_{k-1},h_{k-1}{h}_2\mathrm{\notin }E(G)$, so the triple $(h_1,h_{k-1},h_2)$ gives the second split alternative. Otherwise $x$ is adjacent to $h_{k-1}$, and then $h_{k-1}{h}_k\in E(G)$ while $h_{k-1}{h}_2,h_2{h}_k\mathrm{\notin }E(G)$, so the triple $(h_{k-1},h_2,h_k)$ gives the second split alternative. Hence every mixed outside vertex satisfies [F2], so $V(A)$ is split.

Step 1.2 shows that the odd hole $A$ is a split set, and step 1.1 supplies a complete and an anticomplete outside vertex for it. Therefore [L1] gives a nontrivial module in $G$.