${\ensuremath{\mathfrak{U}}}_{k-1}^+$ is the union of ${\ensuremath{\mathfrak{U}}}_{k-1}$ and the set of unpaired neighbours of the roots. Now, let $\{{\ensuremath{\mathcal{G}}}_k\}_{k\geq 0}$ be the natural filtration associated with the exploration stage. Note that $\tau$ is a stopping time, that ${\ensuremath{\mathbf{w}}}(z_k)$ is ${\ensuremath{\mathcal{G}}}_{k-1}-$measurable, and that the conditional law of $z_k'$ given ${\ensuremath{\mathcal{G}}}_{k-1}$ is uniform on ${\ensuremath{\mathcal{X}}}\setminus\{z_1,\ldots,z_k,z_1'\ldots,z_{k-1}'\}$. Thus, $$\begin{aligned}
{\ensuremath{\mathbb{E}}}[W_{k}-W_{k-1}|{\ensuremath{\mathcal{G}}}_{k-1}] & = & -{\bf 1}_{(k\leq\tau)}\frac{{\ensuremath{\mathbf{w}}}(z_k)(|{\ensuremath{\mathfrak{U}}}_{k-1}^+|-2)+W_{k-1}}{N-2k+1}.\\
{\ensuremath{\mathbb{E}}}\left[(W_{k}-W_{k-1})^2|{\ensuremath{\mathcal{G}}}_{k-1}\right] & = & {\bf 1}_{(k\leq\tau)}\frac{{\ensuremath{\mathbf{w}}}(z_k)^2(|{\ensuremath{\mathfrak{U}}}_{k-1}^+|-4)+2{\ensuremath{\mathbf{w}}}(z_k)W_{k-1}+\sum_{z\in{\ensuremath{\mathfrak{U}}}_{k-1}}{\ensuremath{\mathbf{w}}}(z)^2}{N-2k+1}.\end{aligned}$$ To bound those quantities, observe that each half-edge in ${\ensuremath{\mathfrak{U}}}_{k-1}$ has weight at least $\frac{{\ensuremath{\mathbf{w}}}(z_k)}{\Delta}$ because its parent has been selected at an earlier iteration and our selection rule ensures that the quantity ${\ensuremath{\mathbf{w}}}(z_k)$ is non-increasing with $k$. Consequently, $$\begin{aligned}
 |{\ensuremath{\mathfrak{U}}}_{k-1}|\frac{{\ensuremath{\mathbf{w}}}(z_k)}{\Delta}\leq \sum_{z\in{\ensuremath{\mathfrak{U}}}_{k-1}}{\ensuremath{\mathbf{w}}}(z)\leq 2.\end{aligned}$$ Combining this with the bound $|{\ensuremath{\mathfrak{U}}}_{k-1}^+|\leq |{\ensuremath{\mathfrak{U}}}_{k-1}|+2\Delta$, we arrive at $$\begin{aligned}
{\ensuremath{\mathbb{E}}}[W_{k}-W_{k-1}|{\ensuremath{\mathcal{G}}}_{k-1}] & \geq & -{\bf 1}_{(k\leq \tau)}\frac{4\Delta}{N-2k+1}\\
{\ensuremath{\mathbb{E}}}\left[(W_{k}-W_{k-1})^2|{\ensuremath{\mathcal{G}}}_{k-1}\right]
& \leq & {\bf 1}_{(k\leq \tau)}\frac{4\Delta{\ensuremath{\mathbf{w}}}(z_k)+2}{N-2k+1}.\end{aligned}$$ Now recall that ${\ensuremath{\mathbf{w}}}(z_k)\geq {{w_{\textsc{min}}}}$ and ${\ensuremath{\mathbf{h}}}(z_k)<\frac t2$ as per our selection rule, implying $$\begin{aligned}
{{w_{\textsc{min}}}}\tau & \leq & \sum_{k\geq 1}{{\ensuremath{\mathbf{w}}}(z_k)}{\bf 1}_{(\tau\geq k)}\ \leq \ 
\sum_{z\in{\ensuremath{\mathfrak{F}}}}{\ensuremath{\mathbf{w}}}(z){\bf 1}_{\left({\ensuremath{\mathbf{h}}}(z)<\frac t2\right)}\ \leq t.\end{aligned}$$ The right-most inequality follows from the fact that the total weight at a given height in ${\ensuremath{\mathfrak{F}}}$ is at most $2$ (the total weight being preserved from a parent to its children, if any). We conclude 