Proving the Deletion Condition using the Strong Exchange Condition

Question

I wish to prove the following statement.

Let $w=s_{r_1}\cdots \; s_{r_k}$ with $\ell (w) < k$. Then there exist $i < j$ such that $w=s_{r_1}\cdots \; \hat{s}_{r_i}\cdots \; \hat{s}_{r_j}\cdots \; s_{r_k}.$

Here is my proof.

Let $w=s_{r_1}\cdots \; s_{r_k}$ with $\ell (w) < k$.

Since $s_{r_1}\cdots \; s_{r_k}$ is not a reduced expression for $w$, it follows that multiplying on the left by generators $s_{r_i}$ will produce a reduced expression for $w$. And since $\ell(w) < k$, this means, in particular, that there exists $j$ with $1 \leqslant j \leqslant k$ such that $$\ell(s_{r_j} \cdots \; s_{r_k}) < \ell(s_{r_{j+1}}\cdots \; s_{r_k}).$$ Let $i$ be a maximal such number. Then by the Strong Exchange Condition, $s_{r_i}(s_{r_{i+1}}\cdots \; s_{r_k})=s_{r_i}s_{r_{i+1}}\cdots \;\hat{s}_{r_j}\cdots \; s_{r_k}$ with $j > i$. How do I complete the proof? The proofs I have seen just state that it follows then that $$w=s_{r_1}\mathbin{\cdots}\;\hat{s}_{r_i}\cdots \;\hat{s}_{r_j}\cdots \;s_{r_k},$$ but how does this follow?

Answer 1

The result of utilizing the Strong Exchange Condition should be $s_{r_i}(s_{r_{i+1}}\cdots \; s_{r_k})=s_{r_{i+1}}\cdots \;\hat{s}_{r_j}\cdots \; s_{r_k}$, and the equation wanted follows by times $s_1\cdots s_{r_{i-1}}$ to both sides of the former equation.