In a basic bull-free graph, an odd hole with a complete outside vertex has tightly constrained neighbors

Because $G$ is basic, $u$ cannot be anticomplete to $V(H)$: otherwise the odd hole $H$ would have the complete outside vertex $c$ and the anticomplete outside vertex $u$, making $G$ composite and contradicting [F1]. Assume neither outcome of the Statement holds. By cyclic symmetry choose $h_1$ adjacent to $u$. Suppose that $u$ is adjacent to $h_2$. Since $\{h_k,h_1,u,h_2,h_3\}$ is not a bull, $u$ is adjacent to at least one of $h_k,h_3$; after reversing and shifting the cyclic labels in the second case, we may assume that $u$ is adjacent to $h_k$. If $k=5$, the three distinct neighbors $h_1,h_2,h_k$ give outcome 2, contrary to our assumption. Thus $k>5$, and outcome 1 failing means $u$ is not complete to $H$. Since $\{u,h_2,h_3,c,h_{k-1}\}$ is not a bull, $u$ is adjacent to at least one of $h_3,h_{k-1}$; reflecting the cyclic labels through $h_1$ if necessary, we may assume that $u$ is adjacent to $h_3$. Let $i>3$ be minimal with $u$ nonadjacent to $h_i$. Minimality makes $u$ adjacent to $h_{i-2},h_{i-1}$, while $u$ is adjacent to $h_k$. If $i\mathrm{\ne }k-1$, then $\{h_i,h_{i-1},h_{i-2},u,h_k\}$ is a bull, so $i=k-1\ge 5$. But then $\{h_i,h_{i-1},h_{i-2},u,h_1\}$ is a bull, a contradiction. Thus $u$ is not adjacent to $h_2$; applying the same argument to any putative consecutive pair shows that $u$ has no two consecutive neighbors on $H$.

Since $u$ is adjacent to $h_1$ but to no consecutive pair on $H$, the vertices $\{u,h_1,h_2,c,h_i\}$ with $4\le i\le k-1$ would induce a bull unless $u$ were adjacent to every such $h_i$. Reflecting the cycle through $h_1$ gives the same conclusion for $h_3,\dots ,h_{k-2}$. In particular $u$ is adjacent to both $h_3$ and $h_4$, a consecutive pair, contradicting step 1.1. Therefore our assumption that neither outcome holds was impossible.

One of the two stated outcomes must hold.