a contradiction. It follows that $i_1\neq i$ and $i_2\neq j$.

Let $w=(u/\tilde{u})\tilde{v}$. Then $w\in G(I^k)$.

In case (i), $\deg_{x_i}w= \deg_{x_i}u+1$. Since $\deg_{x_i}v>\deg_{x_i}u$, it follows that $\deg_{x_i}w\leq \deg_{x_i}v$. We also note that $\deg_{x_j}w=\deg_{x_j}u$ and $\deg_{x_{i_1}}w=\deg_{x_{i_1}}u-1\leq \deg_{x_{i_1}}u$.

In case (ii), $\deg_{x_i}w= \deg_{x_i}u$. Moreover, $\deg_{x_j}w=\deg_{x_j}u+1=1$ because $x_j$ does not divide $u$. However, since $x_j$ divides $v$, it follows that $\deg_{x_j}w\leq \deg _{x_j}v$. Finally $\deg_{x_{i_2}}w=\deg_{x_{i_2}}u-1\leq \deg_{x_{i_2}}u$.

In case (iii), $\deg_{x_i}w= \deg_{x_i}u+1=1$ because $x_i$ does not divide $u$. However, since $x_i$ divides $v$, it follows that $\deg_{x_i}w\leq \deg _{x_i}v$. Moreover, $\deg_{x_j}w=\deg_{x_j}u$ and $\deg_{x_{i_1}}w=\deg_{x_{i_1}}u-1\leq \deg_{x_{i_1}}u$.

Thus in all the three cases $\deg_{x_t}w \leq\deg_{x_t}(\lcm (u, v))$ for all variables $x_t$. Therefore $w$ divides $\lcm(u,v)$, and so by definition $w\in V(G^{(u,v)}_{I^{k}})$.

Note that $\deg (\lcm(u, w))=2k+1$, and $w/\tilde{v}=u/\tilde{u}$ which implies that $w/\tilde{v}\in G({I^{k-1}})$. Therefore the assertion follows in these three cases.

Now we consider case (iv). Let $\tilde{u}\in G(I)$ with $u/\tilde{u}\in G(I^{k-1})$, and let $w=(u/\tilde{u})\tilde{v}$ with $\tilde{v}$ as above. Then $w\in G(I^k)$. Since neither $x_i$ nor $x_j$ divides $u$, we have $\deg_{x_i}w=1=\deg_{x_j}w$. Thus $\deg_{x_t}w \leq\deg_{x_t}(\lcm (u, v))$ for all variables $x_t$. It follows that $w$ divides $\lcm(u,v)$ and so $w\in V(G^{(u,v)}_{I^{k}})$.

Moreover, $w\neq v$. Indeed, suppose that $w= v$. Then $v/\tilde{v}=u/\tilde{u}$. Let $\tilde{w}\in G(I)$ with $(v/\tilde{v})/\tilde{w}\in G(I^{k-2})$. Then 