of size at most $k \cdot g(k)$. To obtain a contradiction, it suffices to derive that then $G-F$ also has no long cycles. (With slight abuse of notation, we refer to $F$ as a subset of both $E(G)$ and $E(G^*)$.) Suppose for a contradiction that $G-F$ has a long cycle $C$. Then $C$ must use an edge of $T_2$. Since $T_2$ has no long cycle, we know that $C$ consists of two internally disjoint paths $P_1, P_2$ that each join $v$ and $t_2$, where $P_1$ is a path in the graph $G-\left( V(T_2)\backslash\left\{ v, t_2 \right\}\right)$, while $P_2$ is a path in $T_2$. Moreover, because $B_r(x)\subseteq H(x)$ has no long cycle, $P_1$ must pick up a vertex outside of $B_r(x)$. Together with the fact that $V(T_2)$ is contained in $B_{r- l/2}(x)$, this implies that $P_1$ has length at least $\ell$. It follows that the cycle obtained from $C$ by contracting $P_2$ is a long cycle in $G^*$. Hence $F$ intersects $P_1$. But then $C$ cannot be a subgraph of $G-F$; contradiction.

At last, we are ready to prove Claims \[clm:0\], \[clm:1\], \[clm:2\] and \[clm:3\].

Each pre-leaf is the ancestor-block of some leaf-block of $H(x)$, for otherwise $G$ would have a 