C'.$$ By construction any two successive chains in this sequence have a non-trivial intersection. This shows that $\Delta((1,\lcm(u,v)))$ is connected.

Conversely, assume that $\Delta((1,\lcm(u,v)))$ is connected for all $u,v \in G(I)$ with $\deg(\lcm(u,v))>d+1$. By induction on $\deg (\lcm(u,v))$ we prove that $G_I^{(u,v)}$ is connected for all $u,v \in G(I)$ with $\deg(\lcm(u,v))>d+1$.

In order to prove this, let $u,v \in G(I)$ with $\deg(\lcm(u,v))>d+1$, and let $w,w'\in G^{(u,v)}_I$ with $w\neq w'$. There exist maximal chains $C$ and $D$ in $(1,\lcm(u,v))$ with $\min(C)=w$ and $\min(D)=w'$. Since $\Delta((1,\lcm(u,v)))$ is connected, there exist maximal chains $C_1,\ldots,C_r$ in $\Delta((1,\lcm(u,v)))$ with $C=C_1$ and $C_r=D$ and such that $C_j\sect C_{j+1}\neq \emptyset$ for $j=1,\ldots,r-1$. Let $w_{j}=\min(C_j)$ for $j=1,\ldots,r$. Let $j$ be such that $w_{j}\neq w_{j+1}$. Then $d+1 \leq \deg (\lcm(w_{j}, w_{j+1}))<\deg (\lcm(u,v))$ because $\lcm(w_{j}, w_{j+1})$ divides $\lcm(u,v)$ and $\lcm(w_{j}, w_{j+1})\in (1,\lcm(u,v))$ because $C_j\sect C_{j+1}\neq \emptyset$. If $\deg(\lcm(u,v))=d+2$, it follows that $\deg (\lcm(w_{j}, w_{j+1}))=d+1$ for all $j$ with $w_{j}\neq w_{j+1}$. This shows that $G_I^{(u,v)}$ is connected whenever $\deg(\lcm(u,v))=d+2$ and establishes the proof of the induction begin.

Suppose now that $\deg (\lcm(u,v))>d+2$. Since $\Delta((1,\lcm(w_{j},w_{j+1})))$ is connected and $\deg (\lcm(w_{j}, w_{{j+1}}))<\deg (\lcm(u,v))$ we may apply our induction hypothesis and deduce that $w_{j}$ and $w_{j+1}$ are connected in $G_I^{(w_{j},w_{j+1})}$. Since $G_I^{(w_{j},w_{j+1})}$ is an 