$1\notin Y$ or $n\notin Y$, and $\ker \eta \in \mathcal{Q}_{m}$ otherwise. Thus, there is $\delta \in \left\langle \O(X,Y), B \right\rangle $ with $\ker \delta
=\ker \eta$. Then we obtain as above that $\eta =\delta^{l}\in \left\langle \O(X,Y),B \right\rangle$, for a suitable natural number $l$. Consequently, $\left\langle
\O(X,Y),\mathcal{A}\right\rangle \subseteq \left\langle \O(X,Y),B \right\rangle$. By Proposition \[le2\], we obtain $\OP(X,Y)=\left\langle \O(X,Y),B \right\rangle$. The generating set $B$ is minimal by properties $(i)$ and $(ii)$ together with Lemma \[le3\] and by the property $(iii)$ of $\widetilde{B}$.

Conversely, let $B$ be a minimal relative generating set of $\OP(X,Y)$ modulo $\O(X,Y)$. By Lemma \[le3\], there is a set $\overline{B}\subseteq B$ such that $\mathcal{P}_{m}=\{\ker \beta :\beta \in \overline{B}\}$ and $\left\vert
\overline{B}\right\vert =\left\vert \mathcal{P}_{m}\right\vert$. Since $\OP(X,Y) = \left\langle \O(X,Y),B \right\rangle$, there are $\beta _{1},\ldots ,\beta _{k}\in \O(X,Y)\cup B$ such that $\eta =\beta
_{1}\cdots \beta _{k}$. Without loss of generality, we can assume that there is not $\gamma \in \{\beta _{i}:1\leq i\leq k$, $\ker \beta
_{i}\in \mathcal{Q}_{m}\}=: \widehat{B}$ such that $\eta$ is a product of transformations in $\overline{B}\cup (\widehat{B}\setminus \{\gamma \})$. In the first part of the proof, we have shown that $\overline{B}\cup \widehat{B}$ is a relative generating set of $\OP(X,Y)$ modulo $\O(X,Y)$. Because of the minimality of $B$, we have 