We will prove this by induction on $k$. If $k=1$, consider $G^{(u,v)}_{I}$ for $u,v \in G(I)$. Hence $u=x_ix_j$ and $v=x_{i'}x_{j'}$ with $\{i, j\}, \{i', j'\}$ edges in $G$. Since $\deg(\lcm(u,v))>3$ we have $\deg (\lcm(u,v))=4$. Our assumption implies that at least one of the edges $\{i, i'\}, \{i, j'\}, \{i', j\}, \{i', j'\}$ is in $E(G)$. Without loss of generality we may assume that $\{i, i'\}\in E(G)$. Set $w=x_ix_{i'}$. So $w\in V(G^{(u,v)}_{I})$ and $w$ connects $u$ to $v$.

Now let $k>1$, and suppose that $G^{(u,v)}_{I^{k-1}}$ is connected for all $u, v\in G(I^{k-1})$ with $\deg(\lcm(u,v))>2(k-1)+1$.

Assume that $u/\tilde{w}, v/\tilde{w}\in G(I^{k-1})$ for some $\tilde{w}\in G(I)$. Since $\deg (\lcm(u/\tilde{w}, v/\tilde{w}))> 2(k-1)+1$, our induction hypothesis implies that there is a path $w_0, w_1, \ldots, w_r$ in $G^{({u}/{\tilde{w}},{v}/{\tilde{w}})}_{I^{k-1}}$ with $w_0={u}/{\tilde{w}}$ and $w_r={v}/{\tilde{w}}$. Since $\deg (\lcm(w_i,w_{i+1})) =2(k-1)+1$ it follows that $\deg (\lcm(\tilde{w}w_i, \tilde{w}w_{i+1}))=2k+1$ for all $0\leq i\leq r-1$, and since $\tilde{w}w_j\in V(G^{(u,v)}_{I^{k}})$ for all $j$, the sequence $u=\tilde{w}w_0, \tilde{w}w_1, \ldots, \tilde{w}w_r=v $ is a path in $G^{(u,v)}_{I^{k}}$ connecting $u$ and $v$.

We may now suppose that $u/\tilde{w}, v/\tilde{w}\notin G(I^{k-1})$ for all $\tilde{w}\in G(I)$. Since $u\neq v$ and $\deg u=\deg v$, there is an index $ i$ with $\deg_{x_i}v>\deg_{x_i}u$. In particular, there exists $\tilde{v}\in G(I)$ such that $v/\tilde{v}\in 