A polynomial homogeneous set in the auxiliary pattern yields a $y^4$-restricted union

Let $m$ be the common block size. Since $\mathrm{\mid }I\mathrm{\mid }\ge y\mathrm{\ell }$, $\mathrm{\ell }\ge y^{-a}$, and $r\ge \mathrm{\mid }I{\mathrm{\mid }}^c$, we have $r^{-1}\le \mathrm{\mid }I{\mathrm{\mid }}^{-c}\le (y\mathrm{\ell }{)}^{-c}\le y^{c(a-1)}\le y^5$.

Now suppose that $R$ is a stable set in $J$. For any $x\in B_i\subseteq S$, the neighbours of $x$ inside its own block contribute fewer than $m=r^{-1}\mathrm{\mid }S\mathrm{\mid }$ vertices. If $j\in R\setminus \{i\}$, then $ij\mathrm{\notin }E(J)$, so the pairs $(B_i,B_j)$ are mutually $y^a$-sparse and [L2] gives at most $y^a\mathrm{\mid }{B}_j\mathrm{\mid }$ neighbours of $x$ in $B_j$. Summing over all other selected blocks, $x$ has at most $r^{-1}\mathrm{\mid }S\mathrm{\mid }+y^a{\sum }_{j\in R\setminus \{i\}}\mathrm{\mid }{B}_j\mathrm{\mid }\le (r^{-1}+y^a)\mathrm{\mid }S\mathrm{\mid }$ neighbours in $S$.

First suppose that $R$ is a clique in $J$. Then [L3] makes every two distinct selected blocks complete. For any $x\in B_i\subseteq S$, the only possible nonneighbours of $x$ inside $S$ lie in $B_i$, so $x$ has fewer than $m=r^{-1}\mathrm{\mid }S\mathrm{\mid }$ nonneighbours in $S$. Step 1.1 gives $r^{-1}\mathrm{\mid }S\mathrm{\mid }\le y^5\mathrm{\mid }S\mathrm{\mid }\le y^4\mathrm{\mid }S\mathrm{\mid }$, so $S$ is $y^4$-dense and hence $y^4$-restricted by [L1].

Since $a\ge 5\mathrm{/}c+1$ and $y<\frac{1}{2}$, step 1.1 yields $r^{-1}+y^a\le y^5+y^a\le y^5+y^5\le y^4$. Hence every vertex of $S$ has at most $y^4\mathrm{\mid }S\mathrm{\mid }$ neighbours inside $S$, so $S$ is $y^4$-sparse and therefore $y^4$-restricted by [L1].

Steps 2.1 and 2.2 show that whether $R$ is a clique or a stable set, the union $S$ is $y^4$-restricted. Also $\mathrm{\mid }S\mathrm{\mid }=rm\ge m$, so $S$ has at least the common block size.