n,\nu}$, let $\mathcal{A} = \{v_{j_1},v_{j_2},\dots,v_{j_m}\}$, ordered so that $j_1 < j_2 < \dots < j_m$, be the subset of prunable nodes of $T$ which are maximal among prunable nodes under the partial order $\leq_T$. By definition, $\mathcal{A}$ is an antichain, and all of its elements are in the attic of $T$. Recall the notation defined in Definition~\ref{def: Tree Decomp}.

    Let $\nu^{(h)}$ be the unique labeled lattice path whose steps are $w_{\nu^{(h)}} = w^{(h)}$ and whose inorder reading word is $\inorder(\nu^{(h)}) = v_{k_{h}}^\ell v_{k_h+1}^\ell \cdots v_{i_{h+1}}^\ell$. Define
   \[
     \Phi_{n,\nu}(T) \coloneq (\nu^{(0)},T_1,\nu^{(1)},\dots,\nu^{(m-1)},T_{m},\nu^{(m)}),
   \]
   which lies in $\mathscr{P\!T}^\ell_{n,\nu}$ by Lemma~\ref{lem: partition}.
 \end{definition}

 Figure~\ref{fig: Infected Tree} shows a tree $T$ with its prunable nodes highlighted in green. In this labeled tree, we have $\mathcal{A}=\{v_{2}, v_{4}, v_{12}\}$. Figure~\ref{fig: Phi Decomp} shows the image of the tree $T$ under the map $\Phi_{n,\nu}$.
 The reader can check that for $T$ in Figure~\ref{fig: Infected Tree} and $\Phi_{n,\nu}(T)$ in Figure~\ref{fig: Phi Decomp},
 \begin{align*}
 \wt{T} &= \bar\lambda^3\lambda^6\bar\rho^3\rho^3= \bar\lambda\lambda^3\bar\rho\rho(\bar\lambda\bar\rho\,\wt{T_1})(\bar\lambda\bar\rho\,\wt{T_2})(\lambda\rho\,\wt{T_3})= \wt{\Phi_{n,\nu}(T)},\\ \inorder(T) &= 3235461765768131 = \inorder(\Phi_{n,\nu}(T)).
 \end{align*}

 \begin{figure}[H]
   \includegraphics[scale=0.5]{Example-Infected-Tree-v2.pdf}
   \caption{A tree $T\in \pbtl_{16}$ whose prunable nodes are highlighted in green with $\can{T} = DUDUUUDUUDUDUUD$.}
   \label{fig: Infected Tree}
 \end{figure}

   \begin{figure}[H]
     \includegraphics[scale=0.8]{Example-Phi-Decomp.pdf}
     \caption{The image under $\Phi_{n,\nu}$ of the tree in Figure~\ref{fig: Infected Tree}.}
     \label{fig: Phi Decomp}
   \end{figure}
     %



 %
 \begin{lemma}\label{lem: Weight-Preserving}
   For all $T\in\pbtl_{n,\nu}$, we have that $\wt{T} = \wt{\Phi_{n,\nu}(T)}$ and $\inorder(T)=\inorder(\Phi_{n,\nu}(T))$.
 \end{lemma}
 %
 \begin{proof}
   Definition~\ref{def: Phi} states that $\inorder(\nu^{(h)}) = v_{k_h}^\ell v_{k_h+1}^\ell\cdots v_{i_{h+1}}^\ell$. Therefore, the fact that $\inorder(T) = \inorder(\Phi_{n,\nu}(T))$ follows from the partition of $[n]$ in \eqref{eq: partition of nodes}.

  Recall from \eqref{eq: S weight} that
   \begin{equation}\label{eq: Phi weight}
     \wt{\Phi_{n,\nu}(T)} = \wt{\nu^{(0)}} \prod_{h=1}^m \gamma_h\,\wt{T_h}\,\wt{\nu^{(h)}},
   \end{equation}
   where
   \begin{align}
     \gamma_h &= \begin{cases}
     \bar\lambda \bar\rho & \text{ if }\mathrm{end}(\nu^{(h-1)})^\ell \leq \mathrm{start}(\nu^{(h)})^\ell,\\
     \lambda\rho & \text{ if }\mathrm{end}(\nu^{(h-1)})^\ell > \mathrm{start}(\nu^{(h)})^\ell.
   \end{cases}
 \end{align}
 Observe that $v_{i_h}^\ell = \mathrm{end}(\nu^{(h-1)})^\ell$ and $v_{k_h}^\ell = \mathrm{start}(\nu^{(h)})^\ell$. Since $v_{j_h}$ is maximal among prunable nodes in $T$, then all ancestors of $v_{j_h}$ are life-sustaining. Applying Lemma~\ref{lem: infection lemma} with $i = i_h$, $j=j_h$, $k=k_h$, $e=e_h$ and $f=f_h$, we have
 \begin{align}
 \wt{e_h}\wt{f_h} &= \begin{cases}
 \bar\lambda\bar\rho &\text{if }v_{i_h}^\ell \leq v_{k_h}^\ell,\\
 \lambda\rho &\text{if }v_{i_h}^\ell > v_{k_h}^\ell,
 \end{cases}\\
 &= \begin{cases}
     \bar\lambda \bar\rho & \text{ if }\mathrm{end}(\nu^{(h-1)})^\ell \leq \mathrm{start}(\nu^{(h)})^\ell,\\
     \lambda\rho & \text{ if }\mathrm{end}(\nu^{(h-1)})^\ell > \mathrm{start}(\nu^{(h)})^\ell,
   \end{cases}\\
   &= \gamma_h.
 \end{align}

  Let us define a map on the disjoint union of the sets of edges of the paths $\nu^{(h)}$,
  \begin{align}
  \Lambda : \bigsqcup_{h=0}^m \edge{\nu^{(h)}}\to \edge{T}\setminus\left(\bigcup_{h=1}^{m}\{e_h,f_h\}\cup \edge{T_h})\right)
  \end{align}
 with the property that $\wt{e} = \wt{\Lambda(e)}$.
   Given $e\in\edge{\nu^{(h)}}$, let its endpoints be the $j$th and $(j+1)$th nodes of $S$ in inorder. First, suppose that $e$ is a $U$ step. Since $\nu^{(h)}$ is defined so that $w_{\nu^{(h)}} = w^{(h)}$, then the $j$th letter of $\can{T}$ is a $U$. It follows from Lemma~\ref{lem: steps}(a) that $v_{j+1}$ has a left edge, which we then define to be $\Lambda(e)$.
   Similarly, if $e$ is a $D$ step, then the $j$th letter of $\can{T}$ must be a $D$. It follows from Lemma~\ref{lem: steps}(b) that $v_j$ must have a right edge, which we then define to be $\Lambda(e)$.
   It can be checked that $\Lambda$ is a bijection by construction.

    In the case that $e$ is a $U$ step, then by Lemma~\ref{lem: path edge weight}(a) we have $\wt{e} = \wt{\Lambda(e)}$.  In the case that $e$ is a $D$ step, then by Lemma~\ref{lem: path edge weight}(b) we have $\wt{e} = \wt{\Lambda(e)}$. Therefore, $\Lambda$ is a weight-preserving bijection.
  Hence,
  \begin{align}
     \wt{\Phi_{n,\nu}(T)} &= \prod_{h=0}^m\wt{\nu^{(h)}} \prod_{h=1}^m \gamma_h\wt{T_h}\\
     &= \prod_{h=0}^m \wt{\nu^{(h)}} \prod_{h=1}^m \wt{e_h}\wt{f_h} \wt{T_h}\\
     &= \wt{T}.
   \end{align}
  \end{proof}
 %
 Next, we prove that the map $\Phi_{n,\nu}$ is a bijection by constructing an inverse map. Our inverse map to $\Phi_{n,\nu}$ involves an algorithm which is inspired by the Push-Gliding algorithm of Pr\'eville-Ratelle and Vien