Relative to a complete nonedge pair in a co-$E$-free graph, every one-sided vertex is pure to an induced $H_5$

Suppose $u$ is mixed on the $H_5$. If $u$ is complete to the rim and some $u{v}_j^{\mathrm{\prime }}$ is a nonedge, then $u$ is mixed on the induced path $v_j^{\mathrm{\prime }}-v_j-v_{j+1}-v_{j+2}$ and has three consecutive neighbours there, contrary to [F2]. Thus $u$ is adjacent to every $v_j^{\mathrm{\prime }}$. If $uw$ failed, $\{v_1^{\mathrm{\prime }},u,v_1,v_2,v_3,w\}$ induces co-$E$; hence $uw$ holds and $u$ is complete to $H_5$, a contradiction.

If $u$ is anticomplete to the rim and some $u{v}_j^{\mathrm{\prime }}$ is an edge, then $u$ is mixed on $v_j^{\mathrm{\prime }}-v_j-v_{j+1}$ with two consecutive nonneighbours, contrary to [F2]. Thus every $u{v}_j^{\mathrm{\prime }}$ is a nonedge. If $uw$ were an edge, then $u$ would be mixed on $v_1^{\mathrm{\prime }}-v_1-w$ with two consecutive nonneighbours, again contrary to [F2]. Hence $uw$ is a nonedge, so $u$ is anticomplete to $H_5$, also a contradiction.

It remains that $u$ is mixed on the rim. Any cyclic run of two rim nonneighbours or three rim neighbours, together with a vertex of the opposite adjacency supplied by mixedness, lies in an induced rim subpath to which [F2] applies. Thus the two run restrictions force, up to cyclic relabeling, $N(u)\cap V(C)=\{v_1,v_3,v_4\}$. If $uw$ were a nonedge, then $u$ would be mixed on $v_4-w-v_2$ with two consecutive nonneighbours; if $u{v}_1^{\mathrm{\prime }}$ were a nonedge, then $u$ would be mixed on $v_1^{\mathrm{\prime }}-v_1-w-v_3$ with three consecutive neighbours. Hence [F2] gives $uw,u{v}_1^{\mathrm{\prime }}\in E(G)$; then $\{u,w,v_1^{\mathrm{\prime }},v_1,v_2,v_3\}$ induces co-$E$, impossible.

Every possible rim relation contradicts mixedness, so $u$ is pure to the induced $H_5$.