The structural comb-partition criterion implies property (*)

Set $c_1=c_3=c\mathrm{/}4$ and $c_2=10\mathrm{/}c$. We verify the three alternatives required by [F1] for an arbitrary given comb.

Suppose outcome one and outcome three both fail. By [F3], choose the resulting partition of some $X_i$ and relabel its finitely many blocks in nonincreasing order of size. Relabelling preserves the partition, its block bounds, purity, and the isomorphism type of its pattern graph, as well as the cross-block condition in the structural hypothesis. The relabelled partition is therefore decreasing and still structural; form its integral layers.

Otherwise every preterminal layer has a block below its threshold; [F5] then gives $\mathrm{\mid }{X}_i\mathrm{\mid }<w\mathrm{/}2$, contradicting [F3].

If some $Y_i$ has size at least $w\mathrm{/}2$, [F2] gives a clique or stable set of size at least $w^{c\mathrm{/}2}\ge w^{c\mathrm{/}4}=w^{c_1}$; this is outcome one.

If a preterminal layer $C_r$ is wide at the threshold $w\mathrm{/}{\mathrm{\ell }}^{5r\mathrm{/}2}$, [F4] gives a complete or anticomplete blockade of width at least $w\mathrm{/}{k}^{c_2}$. Since $r\ge 1$, its length parameter satisfies $k\ge {\mathrm{\ell }}^{cr\mathrm{/}4}\ge {\mathrm{\ell }}^{c\mathrm{/}4}={\mathrm{\ell }}^{c_3}$, so outcome two holds.

Thus failure of outcomes one and three forces outcome two, while step 2.1 handles the remaining case. The three outcomes in [F1] therefore always hold, proving property $(\ast )$.