The left six-vertex prime $\mathcal{H}$-graph is prime, and deleting any pendant leaf gives the bull

Deleting any leaf gives the bull. For instance, after deleting ${\mathrm{\ell }}_1$ the triangle $t_1{t}_2{t}_3$ remains, with leaves ${\mathrm{\ell }}_2$ at $t_2$ and ${\mathrm{\ell }}_3$ at $t_3$; by [L2] this is the bull. The same argument works for deleting ${\mathrm{\ell }}_2$ or ${\mathrm{\ell }}_3$.

To check primeness, let $M$ be a nontrivial module. First, $M$ cannot contain two leaves: if it contains ${\mathrm{\ell }}_i,{\mathrm{\ell }}_j$ and omits one support, that support sees its own leaf but not the other; if it contains both supports as well, then either the remaining triangle vertex or the remaining leaf splits the set. Hence $M$ contains at most one leaf.

Now $M$ cannot contain one leaf together with another vertex. If ${\mathrm{\ell }}_i,t_i\in M$, then another triangle vertex is adjacent to $t_i$ but not to ${\mathrm{\ell }}_i$. If ${\mathrm{\ell }}_i\in M$ and $t_i\mathrm{\notin }M$, then any other vertex of $M$ is either another leaf, excluded by step 1.2, or some $t_j$ with $j\mathrm{\ne }i$, and then ${\mathrm{\ell }}_j$ is outside $M$ and adjacent to $t_j$ but not to ${\mathrm{\ell }}_i$. Therefore a module containing a leaf must be the singleton $\{{\mathrm{\ell }}_i\}$.

Consequently a nontrivial module contains no leaves, so it is a subset of $\{t_1,t_2,t_3\}$ with at least two vertices. But if $t_i,t_j\in M$, then the outside leaf ${\mathrm{\ell }}_i$ is adjacent to $t_i$ and not to $t_j$, so $M$ is not a module. This contradiction shows that no nontrivial module exists. By [L1], the graph is prime.