A split set in a bull-free graph

Let $G$ be a finite graph and let $S\subseteq V(G)$ with $\mathrm{\mid }S\mathrm{\mid }>1$. We say that $S$ is split when, for every vertex $x\in V(G)\setminus S$ that is neither complete nor anticomplete to $S$, there exist distinct vertices $u,v,w\in S$ such that one of the following holds:

1. $u$-$v$-$w$ is an induced path in $G[S]$ (so $uv,vw$ are edges and $uw$ is a nonedge), with $x$ adjacent to $u$ and $v$ and nonadjacent to $w$; or
2. $u$ and $w$ are adjacent in $G[S]$, while $u$ is nonadjacent to $v$ and $v$ is nonadjacent to $w$, and $x$ is adjacent to $u$ and nonadjacent to $v$ and $w$.

The definition is complement-invariant: $S$ is split in $G$ if and only if it is split in $\bar{G}$, because clause $1$ in $G$ is clause $2$ in $\bar{G}$, clause $2$ in $G$ is clause $1$ in $\bar{G}$, and completeness swaps with anticompleteness.