from $x$. So $B_m(x)$ and $T(x)$ have at least $d(d-1)^{m-1}$ leaves. Hence $T(x)$ sends at least $d(d-1)^m$ edges to vertices outside $T(x)$. Because $G$ has girth at least $8m+3$, two distinct trees $T(x), T(y)$ are joined by at most one edge. Thus the graph obtained from $G$ by contracting the trees $(T(x))_{x \in X}$ has minimum degree at least $d(d-1)^m $, as desired.

We will also need the following lemma.

\[lem:mindegreepacking\] Let $k$ be a nonnegative integer and let $G$ be a graph of minimum degree at least $3k$. Then $G$ contains $k$ vertex-disjoint cycles.

We proceed by induction on $k$. The statement is clearly true for $k=0$. Suppose $k\geq 1$. Since $G$ has minimum degree at least $3$, it has a cycle. Let $C$ be a shortest cycle in $G$. Every vertex in $V(G)\backslash V(C)$ has at most three neighbours in $V(C)$, since otherwise there would be a shorter cycle. Thus the minimum degree of $G-V(C)$ is at least $3(k-1)$ and so, by induction, there exists a collection of at least $k-1$ vertex-disjoint cycles in $G-V(C)$. Adding $C$ to this collection, we deduce that $G$ contains $k$ vertex-disjoint cycles.

Now everything is ready for the proof. We provide it 