Large almost-pure pair hypotheses yield a complete or anticomplete blockade

Let $n$ be maximal such that $G$ has a blockade $(B_1,\dots ,B_n)$ with $\mathrm{\mid }{B}_i\mathrm{\mid }\ge y^{a+2}\mathrm{\mid }G\mathrm{\mid }$ for all $i\in [n]$, with $\mathrm{\mid }{B}_n\mathrm{\mid }\ge (1-by{)}^n\mathrm{\mid }G\mathrm{\mid }$, and with the property that for each $i\in [n]$, either every later block is complete to $B_i$, or every later block is anticomplete to $B_i$. This is possible because $0<y^{a+2}\le 1$, so $B_1:=V(G)$ already satisfies the required lower bounds.

Suppose $n<2y^{-1}$. The bound $y\le c=2^{-4b}$ implies $by<\frac{1}{2}$, and the elementary inequality $1-t\ge 2^{-2t}$ for $t\in [0,\frac{1}{2}]$ gives $(1-by{)}^n\ge (1-by{)}^{2y^{-1}}\ge 2^{-4b}=c$. Hence $\mathrm{\mid }{B}_n\mathrm{\mid }\ge c\mathrm{\mid }G\mathrm{\mid }$, so the hypothesis applies to $G[B_n]$. Choose disjoint $X,Y\subseteq B_n$ with $\mathrm{\mid }X\mathrm{\mid }\ge y^a\mathrm{\mid }{B}_n\mathrm{\mid }\ge y^{a}c\mathrm{\mid }G\mathrm{\mid }\ge y^{a+2}\mathrm{\mid }G\mathrm{\mid },$ where the last inequality uses $c\ge y\ge y^2$, and with $\mathrm{\mid }Y\mathrm{\mid }\ge (1-by)\mathrm{\mid }{B}_n\mathrm{\mid }\ge (1-by{)}^{n+1}\mathrm{\mid }G\mathrm{\mid }$, and $Y$ complete or anticomplete to $X$. Moreover, $\mathrm{\mid }Y\mathrm{\mid }\ge (1-by)c\mathrm{\mid }G\mathrm{\mid }>\frac{1}{2}c\mathrm{\mid }G\mathrm{\mid }\ge \frac{1}{2}y\mathrm{\mid }G\mathrm{\mid }\ge y^{a+2}\mathrm{\mid }G\mathrm{\mid },$ where the last inequality follows from $a>1$ and $y\le \frac{1}{2}$. Because $X\cup Y\subseteq B_n$, every earlier block has the same pure relation to both $X$ and $Y$ that it had to $B_n$. Thus $(B_1,\dots ,B_{n-1},X,Y)$ is a longer blockade of the same type, contradicting the maximality of $n$. So $n\ge 2y^{-1}$.

Let $Q$ be the set of indices $i$ such that every later block is complete to $B_i$, and let $R$ be the set of indices $i$ such that every later block is anticomplete to $B_i$. By construction every index lies in $Q\cup R$, so one of $Q$ or $R$ has size at least $n\mathrm{/}2\ge y^{-1}$.

If $\mathrm{\mid }Q\mathrm{\mid }\ge y^{-1}$, choose $\lceil y^{-1}\rceil$ indices from $Q$ in their inherited order. The corresponding blocks form a complete blockade, and every block has size at least $y^{a+2}\mathrm{\mid }G\mathrm{\mid }$ by step 1.1. If instead $\mathrm{\mid }R\mathrm{\mid }\ge y^{-1}$, the same construction with $R$ gives an anticomplete blockade. In either case we obtain a complete or anticomplete $(\lceil y^{-1}\rceil ,y^{a+2}\mathrm{\mid }G\mathrm{\mid })$-blockade.

Therefore the stated blockade exists.