In a basic bull-free graph, an odd hole with an anticomplete outside vertex forbids consecutive neighbors

Because $G$ is basic, $u$ cannot be complete to $V(H)$: together with the anticomplete outside vertex $a$, that would make the odd hole $H$ a composite witness, contrary to [F1]. So $u$ has a nonneighbor on $H$. Suppose $u$ had two consecutive neighbors, say $h_1$ and $h_2$. Let $i$ be minimal with $u$ nonadjacent to $h_i$; then $i\ge 3$, and minimality gives $u$ adjacent to $h_{i-1}$ and $h_{i-2}$. Since $a$ is anticomplete to $H$, the five vertices $\{a,u,h_{i-2},h_{i-1},h_i\}$ induce a bull, contradicting bull-freeness. Therefore $u$ has no two consecutive neighbors on $H$.

On a cycle of length $k$, any vertex subset with no two consecutive vertices has size at most $\lfloor k\mathrm{/}2\rfloor$. Step 1.1 therefore bounds the number of neighbors of $u$ on $H$ by $\lfloor k\mathrm{/}2\rfloor$, so the number of nonneighbors is at least $k-\lfloor k\mathrm{/}2\rfloor =\lceil k\mathrm{/}2\rceil$.