Simplify d=0 case in outerplanar lemma proof

Co-Authored-By: Claude Sonnet 4.6 <noreply@anthropic.com>

papers/plane_depth_sequencing/paper.tex

@@ -109,7 +109,7 @@ Let $G$ be a graph with a plane embedding and outer cycle $C$. For each $d \geq
 \begin{proof}
 Let $H = G[V_d]$ with the plane embedding inherited from $G$. It suffices to show every vertex of $H$ lies on the outer face of $H$.
 
-For $d = 0$, we have $V_0 = V(C)$ and $H$ is a subgraph of the cycle $C$, hence a disjoint union of paths, which is outerplanar.
+For $d = 0$, we have $V_0 = C$, so $H$ is outerplanar.
 
 For $d \geq 1$, let $U$ be the open subset of the plane obtained by removing all vertices and edges of $H$. We show every $v \in V_d$ lies on the boundary of the component $U_{\mathrm{out}}$ of $U$ containing the outer face of $G$.