Deletion-labeled Bruhat intervals are lexicographically shellable, with the explicit earlier/later chain comparison

Conditions (N) and (L). For $n\le 1$ the interval has a single maximal chain by [F9], whose empty or one-entry word is increasing, lexicographically first, and has no repeated entry. For $n\ge 2$, by [F1] the labels of any maximal chain of a rooted interval are pairwise distinct, which is (N). By [F5] every rooted interval of $[u,v]$ has exactly one increasing maximal chain and its label word is lexicographically first among the maximal chains of that rooted interval, which is (L). This proves (i); the rooted intervals of $[u,v]$ with their induced labeling are exactly the rooted intervals to which [F2] attaches the deleted-position labels.

The label word determines the chain. Let $m$ be a maximal chain of a rooted interval with retained expression $t_1\cdots t_r$. By the recursion of [F1], each element $m_i$ is the product of $t_1\cdots t_r$ with the positions ${\lambda }_1(m),\dots ,{\lambda }_i(m)$ deleted, so the label word determines every element of $m$ and hence the chain; consequently distinct maximal chains have distinct label words, and the lexicographic order of label words is a linear order on the maximal chains.

Small ranks. If $u=v$ there is no element $x$ with $u<x<u$, and if $n=1$ there is no $x$ with $u<x<v$, because such an $x$ would satisfy $\mathrm{\ell }(u)<\mathrm{\ell }(x)<\mathrm{\ell }(v)$ [F10] while $\mathrm{\ell }(v)=\mathrm{\ell }(u)+1$; so $(u,v)$ is empty, its order complex has the single facet $\mathrm{\varnothing }$ [F4], and the shelling condition of [F3] is vacuous for a one-facet complex. If $n=2$, then $[u,v]$ has exactly four elements [F6], the open interval consists of the two middle elements, which have the same length $\mathrm{\ell }(u)+1$ and are therefore incomparable [F10], and $\mathrm{\Delta }((u,v))$ has the two facets $\{a\},\{b\}$, the empty set and the two singletons being the only chains of a two-element antichain; listing the facets in either order, say $F_1=\{a\}$, $F_2=\{b\}$, the criterion of [F3] holds for $i=1$, $k=2$ with $j=1$ and the vertex $b$ of $F_2$, because $F_1\cap F_2=\mathrm{\varnothing }=F_2\setminus \{b\}$.

Earlier/later chain comparison. By step 1.1 the deleted-position labeling of $[u,v]$ satisfies (N) and (L) on every rooted interval, and by [F9] the poset $[u,v]$ is finite and graded with the covering relation of [F9]; these are exactly the hypotheses of the abstract comparison lemma [F7], which therefore yields, for all maximal chains $m^{\mathrm{\prime }},m$ of $[u,v]$ with $\lambda (m^{\mathrm{\prime }})\prec \lambda (m)$, a maximal chain $k$ with $\lambda (k)\prec \lambda (m)$, $m^{\mathrm{\prime }}\cap m\subseteq k\cap m$ and $\mathrm{\mid }k\cap m\mathrm{\mid }=\mathrm{\mid }m\mathrm{\mid }-1$. In that argument $k$ is obtained by replacing a two-step segment at a descent position by the increasing chain of the corresponding rooted rank-two interval, which is the local descent replacement of At most one increasing chain, rank-two diamonds, the lexicographically first chain, and the local descent replacement (iv).

Shelling of the open interval. For $n\le 1$ the conclusion is step 1.3. Suppose $n\ge 2$ and put $E:=\{u,v\}$ and $F_m:=m\setminus E$ for each maximal chain $m$ of $[u,v]$. Every maximal chain of $(u,v)$ becomes maximal in $[u,v]$ upon adjoining the endpoints, and conversely: any missing intermediate element would enlarge either chain. Thus $m\mapsto F_m$ is a bijection onto the facets of $\mathrm{\Delta }((u,v))$ by [F4]. Give $F_m$ the label word of its endpoint extension $m$; this orders the facets linearly by step 1.2. For $m^{\mathrm{\prime }}\prec m$, step 2.1 supplies $k\prec m$ with $m^{\mathrm{\prime }}\cap m\subseteq k\cap m=m\setminus \{z\}$ for one vertex $z$ of $m$; the cardinality equality there gives the last equality, and $z\mathrm{\notin }E$ since both endpoints lie in every chain. Removing $E$ yields $F_{m^{\mathrm{\prime }}}\cap F_m\subseteq F_k\cap F_m=F_m\setminus \{z\}$, and explicitly $\mathrm{\mid }{F}_k\cap F_m\mathrm{\mid }=\mathrm{\mid }k\cap m\mathrm{\mid }-2=\mathrm{\mid }m\mathrm{\mid }-3=\mathrm{\mid }{F}_m\mathrm{\mid }-1$. This is exactly [F3], proving (iii); (ii) is step 2.1 and (iv) is step 1.3.