Figure 23: U-Turn with $e_i$ and $e_j$ on the same side
the first nor the last edge of $\tilde{W}$. Thus, when we traverse $\tilde{W}$ so that when we visit $e_\ell$, we first visit $u$ and then $v$, we next visit an edge $e'$. Because $\mathcal{W}$ is a weak linkage, this edge must belong to the strict interior of $C'$ (because otherwise we obtain that $(v, e_x, e_y, e_\ell)$ is a crossing or an edge is used more than once). However, this implies that $e'$ is parallel to the edges $e_\ell, e_x, e_y$, and $\tilde{U} = {e_\ell, e'}$ is a U-turn whose edges lie in the interior (including the boundary) of the cycle $C$ and which forms a cycle $\tilde{C}$ that contains fewer edge of $H$ than $C'$ in its interior. This is a contradiction to the choice of $U'$. $\square$
We now prove that an innermost U-turn corresponds to a cycle on which we can perform the cycle pull operation, and consider the result of its application.
Lemma 8.11. Let $(G, S, T, g, k)$ be an instance of Planar Disjoint Paths, and $R$ be a Steiner tree. Let $\mathcal{W}$ be a sensible, outer-terminal, extremal weak linkage that is pushed onto $R$, and let $U = {e_i, e_j}$ be an innermost U-turn. Let $\mathcal{W}$ be the walk in $\mathcal{W}$ that uses $e_i$ and $e_j$, and $C$ be the cycle in $H$ that consists of $e_i$ and $e_j$. Then, the cycle pull operation is applicable to $(W, C)$. Furthermore, the resulting weak linkage $\mathcal{W}'$ is sensible, outer-terminal, extremal, pushed onto $R$, having fewer edges than $\mathcal{W}$, $|\text{Seg}(\mathcal{W}')| \le |\text{Seg}(\mathcal{W})|$, and if $U$ is special, then also its potential is upper bounded by the potential of $\mathcal{W}$.
Proof. Because $U$ is innermost, there does not exist an edge in the strict interior of $C$ that belongs to $E(\mathcal{W})$. Therefore, the cycle pull operation is applicable to $(W, C)$. The only difference between $\mathcal{W}'$ and $\mathcal{W}$ is that $\mathcal{W}'$ does not use the edge $e_i$ and $e_j$ and hence the vertex, say, $v$, that $\mathcal{W}$ visits between them. Therefore, because $\mathcal{W}$ be a sensible, outer-terminal, extremal and pushed onto $R$, so is $\mathcal{W}'$. Moreover, the walks in $\mathcal{W}'$ have the same endpoints as their corresponding walks in $\mathcal{W}$, and thus because $\mathcal{W}$ is sensible, so is $\mathcal{W}'$. Let $u$ be the other endpoints of the edges $e_i$ and $e_j$, and let $\mathcal{W}'$ be the walk in $\mathcal{W}'$ that resulted from $\mathcal{W}$. Observe that $\mathcal{W}'$ has at most as many crossings with $R$ as $\mathcal{W}$ has—indeed, if the elimination of $e_i$ and $e_j$ created a new crossing at $u$ (this is the only new crossing that may be created), then $\mathcal{W}$ crosses $R$ between $e_i$ and $e_j$ and this crossing does not occur in $\mathcal{W}'$. Thus, $|\text{Seg}(\mathcal{W}')| \le |\text{Seg}(\mathcal{W})|$.
Now, suppose that $U$ is special. Then, at least one among the following conditions holds: (i) $i$ and $j$ have the same sign; (ii) $u, v \notin V_{=1}(R) \cup V_{\ge 3}(R)$. We first consider the case where $i$ and $j$ have the same sign, say positive (without loss of generality; see Fig. 23). Let $e'_x, \hat{e}_y \in E(\mathcal{W})$ be such that $e'_x$ and $e_i$ are consecutive in $\mathcal{W}$ (if such an edge $e'_x$ exists) and denote the segment that contains $e'_x$ by $S_x$, and $\hat{e}_y$ and $e_j$ are consecutive in $\mathcal{W}$ (if such an edge $\hat{e}_y$ exists) and denote the segment that contains $\hat{e}_y$ by $S_y$. Possibly some edges among $e'_x, \hat{e}y$ and $e_i$ are parallel. In case $e'x$ does not exist (see Fig. 23(a)), then $u \in V{=1}(R)$ and therefore $e_i$ and $e_j$ belong to a segment that is in a singleton segment group. When we remove $e_i$ and $e_j$, either this segment shrinks (and remains in a singleton group) or it is removed completely together with its segment group. If the segment shrinks, the potential clearly remains unchanged, and otherwise the reduction of segment groups makes the potential decrease by 1 (the potential of the consecutive segment group remains unchanged as it is a singleton segment group because it has an endpoint in $V{=1}(R)$). The case where $\hat{e}_y$ does not exist is symmetric, thus we now assume that both $e'_x$ and $\hat{e}_y$ exist. In case both $x$ and $y$ are on the same side as $e_i$ and $e_j$ (see Fig. 23(b)), then the removal of $e_i$ and $e_j$ only shrinks the segment $S_x = S_y$ where all of the four edges $e_i, e_j, e'_x$ and