a counterexample, contradicting the minimality of $G$.

\[clm:consecutiveblocks\] Let $x\in X$ and let $T_1$ and $T_2$ be two blocks of $H(x)$ that both intersect the ball $B_{r-\ell}(x)$. Suppose furthermore that $V(T_1)\cap V(T_2)= \left\{v\right\}$, for some cut-vertex $v$ of $H(x)$. Then at least one of $T_1, T_2$ contains at least three cut-vertices of $H(x)$.

Suppose for a contradiction that $T_1, T_2$ both contain at most two cut-vertices of $H(x)$. First we deal with the case that one of them, say $T_1$, contains at most one cut-vertex of $H(x)$. Since $T_1$ has diameter at most $(\ell-1)/2$ and intersects $B_{r-\ell}(x)$, it follows that $T_1$ has no neighbours outside $H(x)$ and is thus a leaf-block of $G$. This contradicts Claim \[clm:noleafblock\].

We may thus assume that both $T_1$ and $T_2$ in fact contain exactly two cut-vertices of $H(x)$, one of which must be their common vertex $v$. Let $t_i \in V(T_i)$ denote the cut-vertex of $H(x)$ which is not $v$, for each $i\in \left\{1,2\right\}$. Observe that $t_1\neq t_2$, for else $T_1$ and $T_2$ would not be blocks.

For $i\in \left\{1,2\right\}$, we write $p_i$ for the maximum number of edge-disjoint paths in $T_i$ between $t_i$ and $v$. Without loss of generality, $p_1\leq p_2$. Next, let 