Property (*) and leaf reducibility yield five comb outcomes in a restricted graph

Let $c_1,c_2,c_3$ be the constants from [L1]. Let $d>0$ and $h\ge 1$ be the constants from [L2], and set $c_4:=\max \{12d+3,4\},c_5:=\max \{h,4\}\mathrm{.}$

[assume-case dense-side] Suppose first that $\bar{G}$ is $y^3$-sparse. Because $G$ is $\bar{\mathcal{F}}$-free, the complement graph $\bar{G}$ is $\mathcal{F}$-free. Applying [L2] to $\bar{G}$ with the parameter $y^3$ and $b=4$, we obtain either:

[assume-case sparse-side] We may therefore assume that $G$ itself is $y^3$-sparse. If $\mathrm{\mid }G\mathrm{\mid }\le x^{-9}$, then $(x^9\mathrm{\mid }G\mathrm{\mid }{)}^{c_1}\le 1$, so any vertex of $G$ already gives outcome 3. Hence we may further assume that $\mathrm{\mid }G\mathrm{\mid }\ge x^{-9}\ge y^{-4}$.

Apply [L3] to the sparse graph $G$. If its first branch holds, then outcome 1 holds immediately. If its second branch holds, then outcome 2 holds immediately. So only the comb branch remains.

In that comb branch, [L3] gives an integer ${\mathrm{\ell }}_0\in [y^{-1},\lceil x^{-2}\rceil ]$, a width $w:=y^4\mathrm{\mid }G\mathrm{\mid }\mathrm{/}{\mathrm{\ell }}_0^2$, an $({\mathrm{\ell }}_0,w)$-comb $((a_i,B_i):i\in [{\mathrm{\ell }}_0])$, and a vertex $v$ complete to ${\bigcup }_i{B}_i$ and anticomplete to the teeth. Since $x\le y\le 2^{-4c_5}$ and $c_5\ge 4$, one has $x\le 2^{-16}$ and hence ${\mathrm{\ell }}_0\ge y^{-1}\ge 4$. Also $\lceil x^{-2}\rceil \le x^{-2}+1\le 2x^{-2}$, so $w=\frac{y^4\mathrm{\mid }G\mathrm{\mid }}{{\mathrm{\ell }}_0^2}\ge \frac{y^4\mathrm{\mid }G\mathrm{\mid }}{(2x^{-2}{)}^2}\ge \frac{x^8\mathrm{\mid }G\mathrm{\mid }}{4}\ge x^9\mathrm{\mid }G\mathrm{\mid },$ because $y\ge x$ and $x\le \frac{1}{4}$. Using $\mathrm{\mid }G\mathrm{\mid }\ge x^{-9}$ from step 2.1 and $x\le 2^{-16}$ again, this also gives $w\ge x^{-1}\mathrm{/}4\ge 4$.

Apply [L1] to this special-vertex comb. If it yields a clique or stable set of size at least $w^{c_1}$, then step 4.1 gives $w^{c_1}\ge (x^9\mathrm{\mid }G\mathrm{\mid }{)}^{c_1}$, so outcome 3 holds.

If [L1] yields a complete or anticomplete $(k,w\mathrm{/}{k}^{c_2})$-blockade with $k\ge {\mathrm{\ell }}_0^{c_3}$, then $\frac{w}{k^{c_2}}\ge \frac{\mathrm{\mid }G\mathrm{\mid }}{{\mathrm{\ell }}_0^6{k}^{c_2}}\ge \frac{\mathrm{\mid }G\mathrm{\mid }}{k^{c_2+6\mathrm{/}{c}_3}},$ and also $k\ge {\mathrm{\ell }}_0^{c_3}\ge y^{-c_3}$. So outcome 4 holds.

If [L1] yields a pure $({\mathrm{\ell }}_0,w\mathrm{/}{\mathrm{\ell }}_0^2)$-blockade, set $\mathrm{\ell }:=\sqrt{{\mathrm{\ell }}_0\mathrm{/}y}\mathrm{.}$ Because ${\mathrm{\ell }}_0\ge y^{-1}$, one has $\mathrm{\ell }\ge y^{-1}$ and also ${\mathrm{\ell }}_0\ge \mathrm{\ell }$, so the same blockade has length at least $\mathrm{\ell }$. Since ${\mathrm{\ell }}_0\le 2x^{-2}$ and $y\ge x$, ${\mathrm{\ell }}^2={\mathrm{\ell }}_0\mathrm{/}y\le 2x^{-2}\mathrm{/}x=2x^{-3}\le x^{-4},$ because $x\le \frac{1}{2}$, and therefore $\mathrm{\ell }\le x^{-2}$. Finally, $\frac{w}{{\mathrm{\ell }}_0^2}=\frac{y^4\mathrm{\mid }G\mathrm{\mid }}{{\mathrm{\ell }}_0^4}=\frac{\mathrm{\mid }G\mathrm{\mid }}{{\mathrm{\ell }}^8}\mathrm{.}$ Hence outcome 5 holds.

Steps 1.2, 2.1, 3.1, 5.1, 5.2, and 5.3 exhaust all cases, so one of the five stated outcomes always holds.