Lemma 1

Let $i_1, \ldots, i_n$ be in $I$ and $p_\alpha$ in $P_{i_\alpha}$ for $1 \leq \alpha \leq n$. For all $a \in A$ there exists $a' \in A$ and a sequence $(p'_\alpha)_{1 \leq \alpha \leq n}$ with $p'_\alpha \in P_{i_\alpha}$ such that

$h(a) = \phi_{i_n}(h_{i_n}(a))$ and there exists $a_n \in A$ and $p'_n \in P_{i_n}$ with It follows that $\phi_{i_n}(p_n)h(a) = h(a_n)\phi_{i_n}(p'_n)$, whence the lemma follows from this by induction on $n$.

(C) End of the proof:

Let $x \in M$ and let $n$ be the minimum of the lengths of decompositions of $x$. We shall prove that every decomposition $\sigma$ of $x$ of length $n$ is reduced. This will establish the existence of a reduced decomposition of $x$; the uniqueness of the reduced decomposition then implies $l(\sigma') > l(\sigma)$ for every decomposition $\sigma' \neq \sigma$ of $x$.

The case $n = 0$ being trivial, suppose $n > 0$. Let

be a decomposition of $x$ of length $n$. If there existed an integer $\alpha$ with $1 \leq \alpha \leq n$ and $p_\alpha = e_{i_\alpha}$, the sequence

would be a decomposition of $x$ of length $n - 1$, which is excluded. Suppose that there exists an integer $\alpha$ with $1 \leq \alpha < n$ and $i_{\alpha} = i_{\alpha + 1}$ and let

with $a' \in \mathbf{A}$ and $p'_{\alpha} \in P_{i_{\alpha}}$; by Lemma 1 there exists elements $a'' \in \mathbf{A}$, such that and the sequence is a decomposition of $x$ of length $n - 1$, which is a contradiction.

We have thus proved that $\sigma$ is reduced.