Lemma 2

Let $x \in A[P]$ and let $(x_p e^p)_{p \in X}$ be the family of maximal terms of x. Let $q \in P$ and let $y \in A[P]$ be such that $e^q$ is the unique maximal term of y. Then, the family of maximal terms of $xy$ is $(x_p e^{p+q})_{p \in X}$.

Put $x = \sum_p x_p e^p, y = \sum_r y_r e^r$ and $xy = \sum_t z_t e^t$. Then $r \leq q$ for all $r \in P$ such that $y_r \neq 0$ and $z_t = \sum_{p+r=t} x_p y_r$.

If $t = p + q = p' + r$ with $p \in X$ and $x_{p'} y_r \neq 0$, then $r \leq q$, hence $p' \geq p$ and consequently $p' = p$. Thus $z_{p+q} = x_p y_q = x_p \neq 0$. This shows that $X + q$ is contained in the support of the product $xy$.

On the other hand, if $t = p' + r$ with $x_{p'} y_r \neq 0$, there exists $p \in X$ such that $p' \leq p$ and we have $t \leq p + q$. The maximal support of $xy$ is therefore contained in $X + q$. Since no two elements of $X + q$ are comparable, it follows that $X + q$ is exactly the maximal support of $xy$ and we have seen above that $z_{p+q} = x_p$ for $p \in X$, which completes the proof of the lemma.