$W'$ is sensible, well-behaved, shallow and outer-terminal having the same potential as $W$, and $\sum_{\hat{S} \in \text{Seq}(W')} \text{Volume}(\hat{S}) < \sum_{\hat{S} \in \text{Seq}(W)} \text{Volume}(\hat{S})$.
Proof. We first argue that the cycle move/pull operation is applicable to $(W, C)$. Let $P_1, P_2$ and $P_3$ be the partition of $C$ in Definition 8.17. Note that because $S$ is innermost, its projecting cycle does not contain any edge in $E(W)$ that is not parallel to some edge of $R$, and hence so does $C$ as it is drawn in the interior (including the boundary) of the projecting cycle. Furthermore, the only edges parallel to $R$ that $C$ can contain are those parallel to the edge $e_i$ of $P_3$ whose subscripts have absolute value larger than $|i|$. However, none of these edges belong to $W$ because $i \in {-n+\ell-1, n-h+1}$ where $\ell$ and $h$ are as in Definition 8.16, and $W$ is shallow. Lastly, note that $P_1$ is either $S$ (which might be a cycle) or a subpath of $S$, and hence it is a subwalk of $W$. Thus, the cycle move or pull (depending on whether $P_1$ is a cycle) is applicable to $(W, C)$. Furthermore, the new walk $W'$ that results from the application is the modification of $W$ that replaces $P_1$ by the path consisting of $P_2$ and $P_3$.
Because $\mathcal{W}$ is sensible and the endpoints of no walk in $\mathcal{W}$ are changed in $\mathcal{W}'$, we have that $\mathcal{W}'$ is sensible as well. Moreover, the vertices of $P_2$ are not used by any walk in $\mathcal{W}$ apart from $\mathcal{W}'$ and only in its subwalk that traverses $P_2$, and therefore, as $\mathcal{W}$ is well-behaved, so is $\mathcal{W}'$. Additionally, note that $\mathcal{W}$ is shallow and that each edge belongs to at most as many projecting cycles of sequences in $\mathcal{W}'$ as it does in $\mathcal{W}$. Thus, if $P_3$ does not contain an edge (the only edge parallel to an edge of $R$ that might be used by $\mathcal{W}'$ but not by $\mathcal{W}$ is the edge $e_i$ of $P_3$, if it exists), it is immediate that $\mathcal{W}'$ is shallow. Now, suppose that $e_i$ exists. Let $b \in {-1, 1}$ have the same sign as $i$. Recall that $i \in {-n+\ell-1, n-h+1}$ where $\ell$ and $h$ are as in Definition 8.16, thus to conclude that $\mathcal{W}'$ is shallow, we only need to argue that $e_b$ belongs to the interior of fewer projecting cycles of sequences in $\mathcal{W}'$ as it does in $\mathcal{W}$. However, this holds since the only difference between the sequences of $\mathcal{W}$ and $\mathcal{W}'$ is that the sequence $S$ occurs in $\mathcal{W}$ (and contains $e_b$ in the interior of its projecting cycle), but is transformed into (one or two) other sequences in $\mathcal{W}'$, and these new sequences, by the definition of $\mathcal{W}'$, no longer contain $e_b$ in their projecting cycles. In this context, also note that the projecting cycles of the (one or two) new sequences enclose disjoint areas contained in the area enclosed by the projecting cycle of $S$, and the projecting cycles of the new sequences do not enclose the faces enclosed by $C$, but the projecting cycle of $S$ does enclose them. Thus, $\sum_{\hat{S} \in \text{Seq}(\mathcal{W})} \text{Volume}(\hat{S}) < \sum_{\hat{S} \in \text{Seq}(\mathcal{W})} \text{Volume}(\hat{S})$.
It remains to show that $\mathcal{W}'$ is outer-terminal and that has the same potential as $\mathcal{W}$. The second claim is immediate since $\mathcal{W}$ and $\mathcal{W}'$ have precisely the same crossings with $R$. For the first claim, note that since $\mathcal{W}$ is outer-terminal, it uses exactly one edge incident to $t^*$. The only vertex of $R$ that can possibly be incident to more edges in $E(\mathcal{W}')$ that in $E(\mathcal{W})$ is the other endpoint, say, $w$, of the edge of $P_3$ in the case where $P_3$ contains an edge. So, suppose that $P_3$ does contain an edge and that $w = t^*$, else we are done. Since $t^*$ is a leaf or $R$ that belongs to the boundary of the outer-face of $H$, it cannot be enclosed in the strict interior of the projecting cycle of $S$ and therefore it must be a vertex of $S$. However, this together with the maximality of the number of cycles enclosed by the shrinking cycle $C$ implies that the $C$ is equal to the projecting cycle of $S$. Thus, by the definition of $\mathcal{W}'$, the only difference between the edges incident to $t^*$ in $\mathcal{W}$ compared to $\mathcal{W}'$ is that in $\mathcal{W}$ it is incident to an edge of $S$, while in $\mathcal{W}'$ it is incident to the edge of $P_3$. In particular, this means that $\mathcal{W}'$ has exactly one edge incident to $t^*$ and therefore it is outer-terminal. □
Having Lemmas 8.2, 8.4, 8.5 and 8.6 at hand, we are ready to push a solution onto $R$. Since this part is only required to be existential rather than algorithmic, we give a simpler proof by contradiction rather than an explicit process to push the solution. Notice that once the solution has already been pushed, rather than using the notion of shallowness, we only demand to have multiplicity at most 2$n$.
Lemma 8.7. Let $(G, S, T, g, k)$ be a good Yes-instance of Planar Disjoint Paths, and $R$ be a backbone Steiner tree. Then, there exists a sensible outer-terminal weak linkage $\mathcal{W}$ in $H$ that