in the ordering, and note at most $a(G)$ other vertices are in the clique.) Thus $b(G) \leq a(G)$.

Second, show $a(G) \leq {\textsf{\textup{tw}}}(G)$. Assume for the sake of a contradiction that $G$ is a vertex-minimal counterexample, and say $G$ has treewidth $k$. It is easy to see $a(G) \leq {\textsf{\textup{tw}}}(G)$ when $G$ is complete, so assume otherwise. Let $T$ be a tree decomposition of $G$ with minimum width. By Lemma \[lemma:normal\], assume $T$ is normalised. Note since $G$ is not complete, $T$ contains more than one bag. Let $G'$ be the graph created by taking $G$ and adding all edges $vw$, where $v$ and $w$ share some bag of $T$. So $G$ is a spanning subgraph of $G'$ and $T$ is a tree decomposition of $G'$ as well as $G$. By the normalisation, there is a vertex $v \in V(G')$ such that $v$ appears in a leaf bag $B$ of $T$ and nowhere else. Hence $v$ has exactly $k$ neighbours in $G'$, which form a clique as they are all in $B$. Since it is smaller than the minimal counterexample, $a(G'-v) \leq {\textsf{\textup{tw}}}(G'-v) \leq k$. Since $G'-v$ contains a $(k+1)$-clique (consider a bag of $T$ other than $B$), it follows 