Large sparse-pair hypotheses yield an $x$-sparse or complete 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 $i$, with $\mathrm{\mid }{B}_n\mathrm{\mid }\ge (1-by{)}^n\mathrm{\mid }G\mathrm{\mid }$, and such that for every $i\in [n]$, either every later block is $x$-sparse to $B_i$ or every later block is complete to $B_i$. This is possible because $n=1$ and $B_1=V(G)$ already satisfy the conditions.

Suppose that $n<2y^{-1}$. Since $y\le c=2^{-4b}$, one has $by\le b{2}^{-4b}<1\mathrm{/}2$. For $t\in [0,1\mathrm{/}2]$ the elementary inequality $1-t\ge 2^{-2t}$ holds, so with $t=by$ we get $(1-by{)}^{2y^{-1}}\ge (2^{-2by}{)}^{2y^{-1}}=2^{-4b}=c$. Therefore $\mathrm{\mid }{B}_n\mathrm{\mid }\ge (1-by{)}^n\mathrm{\mid }G\mathrm{\mid }\ge c\mathrm{\mid }G\mathrm{\mid }$. Applying the hypothesis to the induced subgraph $G[B_n]$, choose disjoint $X,Y\subseteq B_n$ with $\mathrm{\mid }X\mathrm{\mid }\ge y^a\mathrm{\mid }{B}_n\mathrm{\mid }\ge y^{a+2}\mathrm{\mid }G\mathrm{\mid }$ and $\mathrm{\mid }Y\mathrm{\mid }\ge (1-by)\mathrm{\mid }{B}_n\mathrm{\mid }\ge (1-by{)}^{n+1}\mathrm{\mid }G\mathrm{\mid }$, and with $Y$ $x$-sparse or complete to $X$. Because $X\cup Y\subseteq B_n$, the relation of every earlier block $B_i$ to $B_n$ restricts to the same relation to both $X$ and $Y$. Hence $(B_1,\dots ,B_{n-1},X,Y)$ is a larger 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 $x$-sparse to $B_i$, and let $R$ be the set of indices $i$ such that every later block is complete to $B_i$. By construction every index lies in $Q\cup R$, so one of $Q$ or $R$ has cardinality at least $n\mathrm{/}2\ge y^{-1}$.

Since one of $\mathrm{\mid }Q\mathrm{\mid },\mathrm{\mid }R\mathrm{\mid }$ is an integer at least $y^{-1}$, step 3.1 makes that cardinality at least $\lceil y^{-1}\rceil$. [step 3.1, given] If it is $\mathrm{\mid }Q\mathrm{\mid }$, choose $\lceil y^{-1}\rceil$ indices from $Q$ in their inherited order; the corresponding blocks form an $x$-sparse blockade. If it is $\mathrm{\mid }R\mathrm{\mid }$, the same choice from $R$ gives a complete blockade. Every selected block has size at least $y^{a+2}\mathrm{\mid }G\mathrm{\mid }$ by step 1.1. Thus one of the two required blockades exists.

Therefore $G$ contains an $x$-sparse or complete $(\lceil y^{-1}\rceil ,y^{a+2}\mathrm{\mid }G\mathrm{\mid })$-blockade.