The leaf/co-leaf corollary recovers the $P_5$ case from the $P_4$ case

The graph $P_5$ has leaf $1$. By [L5], the complement $\overline{P_5}$ has edges $13$, $14$, $15$, $24$, $25$, and $35$, so vertex $1$ has degree $3=5-2$. By [L4], vertex $1$ is therefore a co-leaf of $\overline{P_5}$. Delete that leaf of $P_5$ and that co-leaf of $\overline{P_5}$. The resulting smaller families are ${\mathcal{F}}_1=\{P_4,\overline{P_5}\}\text{and}{\mathcal{F}}_2=\{P_5,\overline{P_4}\}\mathrm{.}$ From the path edges $12$, $23$, and $34$, [L5] gives complement edges $13$, $14$, and $24$, which form the path $3$-$1$-$4$-$2$. Thus $\overline{P_4}\cong P_4$.

The path $P_4$ is a forest, and step 1.1 shows that $\overline{P_4}\cong P_4$. Therefore [L3] applied with $H=P_4$ gives the Erdős-Hajnal property for the hereditary class of $P_4$-free graphs. By [L6], the classes defined by forbidding $\{P_4,\overline{P_5}\}$ and $\{P_5,\overline{P_4}\}$ are hereditary. Each is a subclass of the $P_4$-free class, using step 1.1 for the second family. Hence [L2] gives the Erdős-Hajnal property for both ${\mathcal{F}}_1$ and ${\mathcal{F}}_2$.

Applying [L1] to the family $\mathcal{F}=\{P_5,\overline{P_5}\}$ and the two smaller families from steps 1.1 and 2.1 yields the Erdős-Hajnal property for graphs with no induced $P_5$ and no induced $\overline{P_5}$.