Leaf-reducible families yield a large anticomplete pair or a deeper restricted induced subgraph

By [L1], fix $H$ and $v$ so that the modified family ${\mathcal{F}}^{\mathrm{\prime }}:=\{H\setminus \{v\}\}\cup (\mathcal{F}\setminus \{H\})$ has the Erdős-Hajnal property. By the implication from assertion 1 to assertion 3 in [L2], the family ${\mathcal{F}}^{\mathrm{\prime }}$ is viral.

Apply [L3] with both leaf-deletion slots equal to the same graph $H$ and with the same leaf $v$. The two modified families are both ${\mathcal{F}}^{\mathrm{\prime }}$, so step 1.1 makes them viral. Therefore $\mathcal{F}$ itself is viral. Using the implication from assertion 3 to assertion 2 in [L2], choose $d>0$ such that every $\mathcal{F}$-free graph has an $ϵ$-restricted induced subgraph on at least ${ϵ}^d$ times its number of vertices for every $ϵ\in (0,\frac{1}{2})$. Set $h:=1$.

Since $G$ is $\mathcal{F}$-free by [L4], step 2.1 applies to $G$ with $ϵ:=y^b\in (0,\frac{1}{2})$. We obtain a $y^b$-restricted induced subgraph of $G$ with at least $(y^b{)}^d\mathrm{\mid }V(G)\mathrm{\mid }=y^{bd}\mathrm{\mid }V(G)\mathrm{\mid }$ vertices. Since $y\in (0,\frac{1}{2})$, one has $y^{bd}\ge y^{bd+1}$, so this induced subgraph also has at least $y^{bd+1}\mathrm{\mid }V(G)\mathrm{\mid }$ vertices. Hence outcome 2 holds.

Because outcome 2 always holds, the displayed dichotomy is satisfied.