Lemma 1

Let $s, s' \in S$, with $s \neq s'$, and let $v \in W_{s,s'}$. Then $v(A_s \cap A_{s'})$ is contained in either $A_s$ or in $s(A_s)$, and in the second case $l(sv) = l(v) - 1$.

The proof will be given in no. 5.

We apply the lemma to the element $v = s'u$. There are two possibilities: either

or Moreover, in the second case, $l(ss'u) = l(s'u) - 1$. Hence and we know that this implies that $l(sw) = l(w) - 1$.

Proof that ((P_{n+1}) and (Q_n)) \Longrightarrow (Q_{n+1}).

Let w \in W, with l(w) = n + 1, and s, s' \in S, s \neq s'. If w(C) is contained in A_s \cap A_{s'}, condition (Q_{n+1}) is satisfied with u = 1. For if not, suppose for example that w(C) is not contained in A_s. By (P_{n+1}), w(C) \subset s(A_s) and l(sw) = n. By (Q_n), applied to sw, there exists v \in W_{s,s'} such that

sw(C) \subset v(A_s \cap A_{s'}) \quad \text{and} \quad l(sw) = l(v) + l(v^{-1}sw).

Then

w(C) \subset sv(A_s \cap A_{s'})

and

l(w) = 1 + l(sw) = 1 + l(v) + l(v^{-1}sw) \geq l(sv) + l((sv)^{-1}w) \geq l(w),

so the inequalities above must be equalities. It follows that (Q_{n+1}) is satisfied with u = sv.

Proof of the theorem.

Let w \in W, with w \neq 1. We can write w in the form sw' with s \in S and l(w') = l(w) - 1. By (P_n), applied to w' and n = l(w'), we have w'(C) \subset A_s, since the case w'(C) \subset s(A_s) is excluded because l(sw') = l(w) = l(w') + 1. So w(C) = sw'(C) \subset s(A_s), and since A_s and s(A_s) are disjoint, we have C \cap w(C) = \varnothing. Q.E.D.