In a special-vertex comb of a co-$E$-free graph, vertices in other comb blocks remain pure to every $H_5$-overlap quotient block

Write ${\mathcal{L}}^r$ for iterate $r\ge 1$. For any external comb-block vertex $u$, the comb and special-vertex hypotheses give $u\in N(v)\setminus N(a_i)$, with $v,a_i$ nonadjacent and complete to $B_i$. Thus [F1] makes $u$ pure to every induced $H_5$ in $B_i$, and [F2] makes it pure to each block of ${\mathcal{L}}^1$.

All blocks of every ${\mathcal{L}}^r$ are connected: start with the initial classes and repeatedly apply connectedness preservation in [F6].

Fix $s\ge 2$ and assume the assertion for $s-1$. Suppose an external vertex $u$ is mixed on a block $L$ of ${\mathcal{L}}^s$. By the induction hypothesis, each member block of ${\mathcal{L}}^{s-1}$ inside $L$ is complete or anticomplete to $u$, and both labels occur. By [F7], a mixed chain inside $L$ joins blocks of opposite labels; at a change of label, consecutive mixed blocks $D_1,D_2$ have $u$ complete to $D_1$ and anticomplete to $D_2$.

Consider any mixed blocks $D_1,D_2$ at level $r\ge 1$ with an outside triple $x,y,z$ satisfying: $xy$ is a nonedge, $x,y$ are complete to $D_1\cup D_2$, and $z\in N(x)\setminus N(y)$ is complete to $D_1$ and anticomplete to $D_2$. No vertex $b\in D_1$ is mixed on $D_2$. Indeed, if one were, connectedness and [F4] give an edge $c{c}^{\mathrm{\prime }}$ in $D_2$ with $bc$ an edge and $b{c}^{\mathrm{\prime }}$ a nonedge. Then $b-c-c^{\mathrm{\prime }}$ is induced, $x,y$ are outside and complete to it, and $z$ is mixed on it with consecutive nonneighbours $c,c^{\mathrm{\prime }}$, contrary to [F5].

The pair in step 1.3 has the required triple $(x,y,z)=(v,a_i,u)$ at level $s-1$. Whenever its current level $r$ exceeds one, apply [F3] to ${\mathcal{L}}^{r-1}$: step 1.2 supplies connected member blocks and step 2.1 supplies the directional no-mixed-vertex hypothesis. The resulting mixed blocks at level $r-1$ have an outside triple with all the same adjacency conditions. Repeating this finite descent reaches mixed initial classes $A_1,A_2$ and an outside triple $x^{\mathrm{\prime }},y^{\mathrm{\prime }},u^{\mathrm{\prime }}$ with $x^{\mathrm{\prime }}{y}^{\mathrm{\prime }}$ a nonedge, $x^{\mathrm{\prime }},y^{\mathrm{\prime }}$ complete to $A_1\cup A_2$, and $u^{\mathrm{\prime }}\in N(x^{\mathrm{\prime }})\setminus N(y^{\mathrm{\prime }})$ complete to $A_1$ and anticomplete to $A_2$. If $s=2$, the initial pair already has these properties.

Apply step 2.1 to $A_1,A_2,x^{\mathrm{\prime }},y^{\mathrm{\prime }},u^{\mathrm{\prime }}$. Every vertex of $A_1$ is pure to $A_2$. Since the pair $(A_1,A_2)$ is mixed, some vertex $p\in A_1$ is complete to $A_2$ and some vertex $q\in A_1$ is anticomplete to $A_2$; otherwise all vertices have the same label and the pair is pure. Therefore any $u_2\in A_2$ is adjacent to $p$ and nonadjacent to $q$, and is mixed on $A_1$. Such a vertex exists because a blockade block is nonempty.

The vertices $y^{\mathrm{\prime }},u^{\mathrm{\prime }}$ are outside $A_1$, nonadjacent, and both complete to $A_1$. Also $u_2\in N(y^{\mathrm{\prime }})\setminus N(u^{\mathrm{\prime }})$. Apply [F1] with $(x,y,u)=(y^{\mathrm{\prime }},u^{\mathrm{\prime }},u_2)$ to every induced $H_5$ contained in $A_1$, then [F2] to the overlap class $A_1$. It follows that $u_2$ is pure to $A_1$, contradicting step 4.1. Thus no external vertex is mixed on any block at level $s$. Together with the base case this proves the assertion for every iterate.