figure \[fig:easy-cycle\] is a rainbow cycle of length $t-1$.

(-5.2,-2)(5.2,2) (-5,0)[.08]{}[x1]{} (-1,0)[.08]{}[xpi]{} (0,0)[.08]{}[xi]{} (1,0)[.08]{}[xsi]{} (5,0)[.08]{}[xt]{}

(-3.25,0)[a]{} (-2.75,0)[b]{}

(2.75,0)[a1]{} (3.25,0)[b1]{}

By assumption, $P$ could not have been nice, so there is a vertex $v \in V(P)^c$ with $$\abs{{\Gamma_{\mathrm{new}}(v,P)}\cap V(P)} \geq \abs{{\Gamma_{\mathrm{new}}(v,P)}} - a\text.$$ Some thought shows that $$\abs{{\Gamma_{\mathrm{new}}(v,C)} \cap V(C)}\geq \abs{{\Gamma_{\mathrm{new}}(v,C)}}-a-5\text,$$ because at most three vertices of ${\Gamma_{\mathrm{new}}(v,P)}\cap V(P)$ are in not in ${\Gamma_{\mathrm{new}}(v,C)}\cap V(C)$, and at most two vertices of ${\Gamma_{\mathrm{new}}(v,C)}$ are not in ${\Gamma_{\mathrm{new}}(v,P)}$.

But no two vertices $u,u'\in {\Gamma_{\mathrm{new}}(v,C)}\cap V(C)$ can be adjacent in $C$, because otherwise, the cycle $(C - \{u,u'\}) \cup (u,v, u')$ would be a rainbow cycle of length $t$, contradicting Proposition \[cons:no-cycles\]. In other words, no edge $e\in E(C)$ is incident to two vertices in ${\Gamma_{\mathrm{new}}(v,C)} \cap V(C)$. Hence, if we write $X = \{e \in E(C) : e\cap {\Gamma_{\mathrm{new}}(v,C)} \neq \emptyset\}$, then $$\begin{aligned}
    \abs{X} &= 2\abs{{\Gamma_{\mathrm{new}}(v,C)}\cap V(C)}\\
    &\geq 2\abs{{\Gamma_{\mathrm{new}}(v,C)}}-2a-10\\
    &\geq 2n-2t-2a-12\text.
  \end{aligned}$$

Now let $Y = V(C)^c \cap {\Gamma_{\mathrm{new}}(v,C)}^c$. We have $$\begin{aligned}
    \abs{Y} &= \abs{V(C)^c \cap {\Gamma_{\mathrm{new}}(v,C)}^c}\\
    &= \abs{V(C)^c} - \abs{{\Gamma_{\mathrm{new}}(v,C)} \cap V(C)^c}\\
    &= n-t+1 - \abs{{\Gamma_{\mathrm{new}}(v,C)}} + \abs{{\Gamma_{\mathrm{new}}(v,C)}\cap V(C)}\\
    &\geq n-t+1 - \abs{{\Gamma_{\mathrm{new}}(v,C)}} + \abs{{\Gamma_{\mathrm{new}}(v,C)}}-a-5\\
    &= n-t-a-4\text.
  \end{aligned}$$ Clearly, $c(X) \subseteq c(C)$ and $c(v,Y\setminus \{v\})\subseteq c(C)$. Then $c(X)$ and $c(v,Y\setminus \{v\})$ have 