The $E$-graph and the Bird graph are wonderful

For the $E$-graph, take the $1$-subdivision of $K_{1,3}$ with center $c$, subdivision vertices $s_1,s_2,s_3$, and leaves $t_1,t_2,t_3$. On the six-vertex subset $\{t_1,s_1,c,s_2,t_2,s_3\}$, the induced edges are $t_1{s}_1$, $s_{1}c$, $c{s}_2$, $s_2{t}_2$, and $c{s}_3$, which is exactly the edge set of the $E$-graph from The $E$-graph and co-$E$. Thus $E$ is an induced subgraph of the $1$-subdivision of $K_{1,3}$, so [L1] makes $\{E\}$ wonderful.

In the graph $H$ from [A1], the vertices $v_2$ and $v_5$ are adjacent to each other and both have the same neighbourhood outside $\{v_2,v_5\}$, namely $\{v_1,v_3,v_4\}$. Hence $\{v_2,v_5\}$ is a homogeneous clique. Let $Q$ be the five-vertex graph on $\{x,v_1,v_3,v_4,v_6\}$ with edge set $\{x{v}_1,x{v}_3,x{v}_4,v_1{v}_3,v_1{v}_6,v_4{v}_6\}$. Then $H$ is obtained from $Q$ by substituting $K_2$ for the vertex $x$. By [L2], both $Q$ and $K_2$ have the Erdős-Hajnal property, so [L3] gives the Erdős-Hajnal property for $H$.

Form $H^{+}$ from [A1] by adjoining a new vertex $v$ adjacent to $v_1$ and $v_2$, and delete $v_3$. On the remaining six vertices $\{v,v_1,v_2,v_4,v_5,v_6\}$ the edge set is $\{v{v}_1,v{v}_2,v_1{v}_2,v_1{v}_5,v_1{v}_6,v_2{v}_4,v_2{v}_5,v_4{v}_5,v_4{v}_6\}$. Under the relabelling $x_1=v$, $x_2=v_6$, $x_3=v_5$, $y=v_4$, $z=v_2$, and $w=v_1$, the six missing edges are exactly $x_1{x}_2$, $x_1{x}_3$, $x_2{x}_3$, $x_{1}y$, $x_{2}z$, and $yw$, which are precisely the Bird edges. Therefore $H^{+}-v_3$ is co-Bird, so $H^{+}$ is not co-Bird-free.

Step 1.2 shows that $\{H\}\cup \{\mathrm{c}\mathrm{o}\text{-}\mathrm{B}\mathrm{i}\mathrm{r}\mathrm{d}\}$ has the Erdős-Hajnal property, and step 1.3 shows that $H^{+}$ is not co-Bird-free. Therefore [L1] applies to the singleton family $\{\mathrm{B}\mathrm{i}\mathrm{r}\mathrm{d}\}$ and proves that Bird is wonderful. Together with step 1.1, this proves the statement.