Finite Weyl strong exchange and deletion

Let $w=s_{i_1}\cdots s_{i_m}$ be any word, and suppose $\beta >0$ but $w^{-1}\beta <0$. Starting with $\beta$, apply $s_{i_1},s_{i_2},\dots ,s_{i_m}$ successively; the resulting final vector is $w^{-1}\beta$. At the first positive-to-negative change, say index $j$, F2 implies $s_{i_{j-1}}\cdots s_{i_1}\beta ={\alpha }_{i_j}$. With $u=s_{i_1}\cdots s_{i_{j-1}}$ this says $\beta =u{\alpha }_{i_j}$. The reflection identity $s_{\beta }=u{s}_{i_j}{u}^{-1}$ therefore cancels precisely the $j$th letter in $s_{\beta }w$. This proves the claimed deletion for every word, including an initially nonreduced one.

If $w^{-1}\beta <0$, apply step 1.1 to a reduced word to get $\mathrm{\ell }(s_{\beta }w)\le \mathrm{\ell }(w)-1$. If $w^{-1}\beta >0$, put $v=s_{\beta }w$; then $v^{-1}\beta =-w^{-1}\beta <0$, so the same argument gives $\mathrm{\ell }(w)\le \mathrm{\ell }(v)-1$. These two cases prove the reflection-descent criterion and strong exchange. Reversing a word proves $\mathrm{\ell }(w^{-1})=\mathrm{\ell }(w)$. Applying the criterion to $w^{-1}$ and $\beta ={\alpha }_i$ gives the sign of $\mathrm{\ell }(w{s}_i)-\mathrm{\ell }(w)$. Appending or removing a single $s_i$ bounds its absolute value by one, so the difference is exactly $\pm 1$.

Write $n(w)=\mathrm{\mid }\mathrm{Inv}(w)\mathrm{\mid }$. Since $s_i$ permutes ${\mathrm{\Phi }}^{+}\setminus \{{\alpha }_i\}$, counting those roots first gives $n(w{s}_i)=n(w)+1$ if $w{\alpha }_i>0$, and $n(w{s}_i)=n(w)-1$ if $w{\alpha }_i<0$. To prove $n(w)=\mathrm{\ell }(w)$, induct on reduced word length. The empty word has both numbers zero. If $w=u{s}_i$ is reduced, its prefix $u$ is reduced, since a shorter prefix would shorten $w$. Step 2.1 gives $u{\alpha }_i>0$, as the negative case would make $\mathrm{\ell }(u{s}_i)=\mathrm{\ell }(u)-1$. Thus $n(w)=n(u)+1=\mathrm{\ell }(u)+1=\mathrm{\ell }(w)$. This also proves that an element preserving all positive roots is the identity.

Consider a nonreduced word and its first nonreduced prefix $u{s}_i$, where the given word for $u$ is reduced. Step 2.1 forces $\mathrm{\ell }(u{s}_i)=\mathrm{\ell }(u)-1$, so $u{\alpha }_i<0$. Apply step 1.1 to the reversed word for $u^{-1}$ and the positive root ${\alpha }_i$. It deletes one letter to express $s_i{u}^{-1}$. Reverse this equality to express $u{s}_i$ by the word for $u$ with that letter deleted. In the original prefix this removes that letter and its terminal $s_i$, hence two letters without changing its value. Reattach the remaining suffix of the original word. All arguments are finite, include rank zero and the identity, and use no AC.