minimizing the number of vertices in $\operatorname{ct}(G)$. Note that $e(G) > 5$ since ${\chi_{s}'(G)}> 5$, and let $n$ be the number of vertices in $\operatorname{ct}(G)$. By using the discharging method, we will show that $\operatorname{mad}(\operatorname{ct}(G)) \geq 2+\frac{1}{7}$, which is a contradiction, so no such minimal counterexample exists.

Observe that $G$ does not contain any of the reducible configurations addressed in Claim \[clm:Reducible\]. We also have the following additional structure on $\operatorname{ct}(G)$.

\[lma:cutedge2\] $\operatorname{ct}(G)$ is 2-connected.

Suppose that $\operatorname{ct}(G)$ contains a cut-edge $uv$. In $G$, the vertices $u$ and $v$ have degree at least two. There are exactly two components, $G_1$ and $G_2$, in $G - uv$, with $u \in V(G_1)$ and $v \in V(G_2)$. Let $u_1,u_2$ be neighbors of $u$ in $G_1$ and $v_1,v_2$ be neighbors of $v$ in $G_2$; let $u_1 = u_2$ only when $u$ has a unique neighbor in $G_1$, and $v_1 = v_2$ only when $v$ has a unique neighbor in $G_2$. Let $G_1' = G_1 + \{ uv, vv_1, vv_2\}$ and $G_2' = G_2 + \{ uv, uu_1, uu_2\}$.

If $G_1' = G$, then consider $G' = G - v_1 - v_2$. Since $n(G') < n(G)$ and $\operatorname{mad}(G') \leq \operatorname{mad}(G)$, there is a strong 5-edge-coloring 