The left six-vertex prime $\mathcal{H}$-graph: primeness and leaf/co-leaf deletions

By [F1], exactly $c,f$ have degree one and exactly $b$ has degree four. Deleting $f$ leaves edges $ab,bc,ad,bd,be,de$: on $a,b,d,e$ these form a complete graph minus $ae$, with $c$ attached only to $b$. Deleting $b$ leaves exactly $ad,de,ef$ and isolated $c$. Deleting either leaf removes one of the seven edges, leaving six, so neither resulting graph is the five-edge bull.

If a module $M$ contains a set $S$ and a vertex outside $S$ has both a neighbour and a nonneighbour in $S$, that vertex must also belong to $M$. Otherwise it would violate the defining uniform outside adjacency of a module. Thus such vertices can be forced into $M$ successively.

The following table covers every unordered pair of vertices. Starting from the pair in the first column, append the vertices in the second column from left to right. At each append, [F1] shows that the new vertex has a neighbour and a nonneighbour in the set already obtained. Consequently step 1.2 forces all six vertices into every module containing the initial pair.

Every set with at least two vertices contains one of those pairs, so step 2.1 excludes every proper module of size at least two. By [L1], $L$ is prime. Together with step 1.1 this proves all the stated claims.