$a(G'-v) \geq k$. Thus $a(G'-v)=k$, and $G'-v$ is a spanning subtree of a $k$-tree $H$. As $v$ is $k$-simplicial in $G'$, it follows $G'$ (and thus $G$) is a spanning subgraph of a $k$-tree, which contradicts our assumption. Finally, show ${\textsf{\textup{tw}}}(G) \leq b(G)$. $G$ is a spanning subgraph of a $H$, chordal graph with no $(b(G)+2)$-clique. There is a tree decomposition of $H$ where every bag is a clique; this means it has width at most $b(G)$. This tree decomposition is also a tree decomposition for $G$, so ${\textsf{\textup{tw}}}(G) \leq b(G)$.

Hence, it follows that $b(G) \leq a(G) \leq {\textsf{\textup{tw}}}(G) \leq b(G)$, which is sufficient to prove our desired result.

Separators {#section:separators}
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For a graph $G$, a set $S \subseteq V(G)$, and some $c \in [\frac{1}{2},1)$, a $(k,S,c)$-*separator* is a set $X \subseteq V(G)$ with $|X| \leq k$, such that no component of $G-X$ contains more than $c|S-X|$ vertices of $S$. Note that a $(k,S,c)$-separator is also a $(k,S,c')$-separator for all $c' \geq c$. Define the *separation number* ${\textsf{\textup{sep}}}_c(G)$ to be the minimum integer $k$ such that there is a $(k,S,c)$-separator for all $S \subseteq V(G)$. We also consider the following variant: a $(k,S,c)^*$-*separator* is a set $X \subseteq 