Skip roots form a basis, negative skips are cover roots, and the cover decomposition of sortable elements

Fix a word for $c$. All inductions below are on (rank, length); the empty sorting word has skips $r$, roots $e_r$, no negative roots and no covers. It supplies the base cases. Decreasing blocks mean that for each letter the selected occurrences form an initial segment of all its occurrences. If initial $s$ is selected first, deleting it gives the $scs$-sorting word of $sv$; otherwise sortable $v$ lies in $W_{S\setminus \{s\}}$ and its word is the $sc$-sorting word. The skip positions then give $C_c(v)=\rho (s)C_{scs}(sv)$ in the first case and $C_c(v)=\{e_s\}\cup C_{sc}(v)$ in the second. This proves agreement with the recursive description by induction.

Signs and inversions of a skip. Write $p=a_1\cdots a_i$, $v=pz$, and $\gamma =\rho (p)e_r$. The root-length test says $\gamma$ is negative exactly when $pr$ is not reduced. At the skipped occurrence the greedy remainder $z$ has no left descent $r$, so $\rho (z^{-1})e_r$ is positive. Hence $\rho (v^{-1})\gamma =\rho (z^{-1})e_r$ is positive. Thus if $\gamma ={\beta }_t$ then $t\mathrm{\notin }I(v)$, and if $\gamma =-{\beta }_t$ then $t\in I(v)$. The dictionary gives $\gamma =\pm {\beta }_t$ for $t=pr{p}^{-1}$. This proves all signs in (1) and the stated signed-root sets.

Endpoint inequalities. For initial $s$ and ${\beta }_t=\sum a_r{e}_r\in {\mathrm{\Phi }}_{+}$, the triangular Euler formula gives ${\omega }_c(e_s,{\beta }_t)=-{\sum }_{r\mathrm{\ne }s}2a_{r}B(e_r,e_s)\ge 0$. Equality means the root is supported on generators commuting with $s$ (including $s$), so [F8] puts its reflection in the subgroup generated by them; it commutes with $s$. The final-letter formula reverses the sign and has the same equality implication. Thus both inequalities are strict when the reflections do not commute.

Cover transport. For $v=sq$ with $\mathrm{\ell }(q)=\mathrm{\ell }(v)-1$, a right descent $r$ of $v$ has negative root $\rho (v)e_r$. Applying $\rho (s)$ preserves its negative sign except when it is $-e_s$, equivalently when its cover reflection is $s$. Conversely a positive $\rho (v)e_r$ could change to negative only if it were $e_s$, which would imply $\rho (v^{-1})e_s=e_r>0$ and contradict $s{\le }_{R}v$. Hence the right descents of $q$ correspond exactly to the cover reflections of $v$ other than $s$, and $\mathrm{C}\mathrm{o}\mathrm{v}(q)=\{sts:t\in \mathrm{C}\mathrm{o}\mathrm{v}(v)\setminus \{s\}\}$. If $s$ is a cover, $I(q)=I(v)\setminus \{s\}$ by [F7]; the general inversion transport also gives $I(q)=s(I(v)\setminus \{s\})s$. In particular $I(v)\setminus \{s\}$ is conjugation-invariant when $s$ is a cover.

The recursion of step 1.1 gives a basis: it either applies the invertible map $\rho (s)$ to a smaller-length basis, or adjoins $e_s$ to a smaller-rank basis of $V_{S\setminus \{s\}}$. Independence under a commuting swap in the word for $c$ follows from the scan comparison [F2]: prefix products after the two positions agree; if an omitted $r$ exchanges position with a selected commuting $q$, the prefix changes by $q$ but $\rho (q)e_r=e_r$, so its skip root agrees; if both are omitted no prefix changes. Iterating these swaps proves word independence, hence independence of any initial-letter recursion choices. This proves (2).

Euler orthogonality. Order the skips by their actual first omitted positions. In the descent branch their order is unchanged by removing the first selected $s$, and the transport identity for $E$ reduces every pair to smaller length. In the non-descent branch $e_s$ is the first root and all other roots lie in $V_{S\setminus \{s\}}$, so $E_c(e_s,\beta )=0$; the remaining pairs reduce by restriction and smaller rank. The empty word has $E_c(e_{s_i},e_{s_j})=0$ for $i<j$. This proves (4).

Unforced-skip preparation. If the first omitted $r$ follows prefix $p$, and $pr$ is reduced, then its selected positions together with this $r$ are the sorting positions of $pr$ and have decreasing blocks. Indeed all earlier $r$-occurrences were selected, so adding this occurrence preserves the initial-segment property. To verify the greedy assertion, suppose an earlier omitted $q$ would be selected for $pr$. At its current prefix $h$, a forced omission cannot be selected for $pr$: its negative root is the negative of an inversion of $h$, which is already below $pr$. Thus this differing omission is unforced, and its conjugate reflection $u=hq{h}^{-1}$ is an inversion of $pr$ but not of $v$ by step 1.2. Since $I(pr)=I(p)\cup \{t\}$ and $I(p)\subseteq I(v)$, necessarily $u=t=pr{p}^{-1}$. Write $p=hz$. Strong exchange, or direct cancellation of the unique last prefix reflection of $pr$, gives $qzr=z$. But $q$ never occurs in $z$, since sortable $v$ has no selected $q$ after this omitted occurrence. Support invariance in the reduced equality $zr=qz$ therefore forces $q=r$; this contradicts that the given $r$ is its first omitted occurrence. No earlier omission is selected, and the selected prefix positions remain greedy since $p{\le }_{R}pr$. Thus $pr$ is sortable with the asserted sorting word. Also no later selected letter is $r$.

For an unforced skip with reflection $t$ after $i$ selections, ${\omega }_c({\beta }_t,{\beta }_{t_j})\ge 0$ for every $j>i$, strictly when $t,t_j$ do not commute. Induct along step 1.1. If $v\mathrm{̸}{\ge }_{R}s$ and $r=s$, this is the initial-root inequality in step 1.3; otherwise restrict to the smaller parabolic. If $v{\ge }_{R}s$, both the skip and every later selected root transport by $\rho (s)$ from the shorter sorting word; the skip remains unforced because its positive root is not $e_s$ (it is not an inversion of $v$ by step 1.2). The form identity transfers the inductive inequality and commutation data.

If a group element commutes with all prefix reflections of a reduced word $b_1\cdots b_m$, it commutes with that word: from the first prefix reflection obtain commutation with $b_1$; successively conjugating the next reflection by the already commuting prefix obtains commutation with each $b_j$. If $s$ is final in $c$ and $s\in I(v)$, let $t_i=s$ in the sorting reflection sequence. Uniform positivity and the final-root inequality of step 1.3 force $s$ to commute with every $t_j$ for $j>i$. Conjugating by $a_1\cdots a_i$ and applying the preceding observation shows $a_i$ commutes with the suffix $a_{i+1}\cdots a_k$. Therefore $sv$ is the reduced word with $a_i$ removed, and $v=(sv)a_i$; thus $s$ is a cover reflection. This proves (5)(i).

Negative skips are covers, descent case with $s$ a cover. Put $q=sv$. First $v$ is also $scs$-sortable. Indeed step 1.4 gives $I(v)\setminus \{s\}=s(I(v)\setminus \{s\})s$, hence $I(v)=sI(v)s$. For a rank-two subsystem not containing $s$, conjugation by $s$ preserves its positive roots and extreme rays; form transport and this invariance transfer $c$-alignment of $v$ to $scs$-alignment on the conjugate subsystem. In a noncommutative subsystem containing $s$, its positive simple-coordinate cone makes $e_s$ an extreme ray. Let its other endpoint be $p$. Since $s$ is initial, step 1.3 orders its angular list as $u_1=s,u_2=sps,\dots ,u_m=p$ with positive skew value. Alignment and $s\in I(v)$ make the intersection an initial segment. If it contains $u_2$, conjugation invariance puts $p=s{u}_{2}s$ in it, so it is the full list; otherwise it is $\{s\}$. For $scs$, where $s$ is final, the positive orientation is reversed and both the full list and the singleton final endpoint $\{s\}$ are allowed. Thus $v$ is $scs$-aligned in every noncommutative subsystem and is $scs$-sortable by [F3]. Now scan the fixed $scs$ word for $v$ and $q$ until their first different selection. Since $I(v)=I(q)\cup \{s\}$, the difference is the reflection $s$, selected for $v$ and omitted for $q$ after a common prefix $p$, with letter $r$ and $pr{p}^{-1}=s$. It is unforced for $q$ since $pr$ is a reduced prefix for $v$. No earlier omitted $r$ was common to the two scans: its later selection for sortable $v$ would violate decreasing blocks. Consequently this position is the first omitted $r$ for $q$, and $C_{scs}^r(q)=\rho (p)e_r=e_s$. Length induction identifies the negative skip roots of $q$ with $\mathrm{C}\mathrm{o}\mathrm{v}(q)$; transport now adds $-e_s$ and takes the other negative roots to the covers of $v$ by step 1.4.

An insertion criterion. Suppose the reflection sequence of a word commutation-equivalent to a sorting word for sortable $v$ is $t_1,\dots ,t_k$. Insert a distinct $t$ after $j$ entries, assume $t_1,\dots ,t_j,t$ is a reduced-word reflection sequence, and assume all earlier roots have nonnegative omega with ${\beta }_t$ and all later roots have nonnegative omega after ${\beta }_t$, strictly for noncommuting pairs. Then $t$ is an unforced skip of $v$. For initial $s$ with $v\mathrm{̸}{\ge }_{R}s$, uniform positivity makes the element with inversions $\{t_1,\dots ,t_j,t\}$ sortable; if $t$ is outside $W_{S\setminus \{s\}}$, its only inversion outside that parabolic must be $s$ by the sortable recursion, so $t=s$, the first unforced skip. Otherwise rank induction applies. If $v{\ge }_{R}s$, move the first $s$ to the front of the original commutation class; letters crossed commute with $s$ and have zero skew value by the initial-root formula, as in the uniform proof. If the inserted $t$ lies before this $s$, the two inequalities with the initial root from step 1.3 force ${\omega }_c({\beta }_t,e_s)=0$ and commutation; it too can be moved across $s$, preserving the prefix-reduced condition. Delete the first $s$ and conjugate all remaining reflections by $s$; positivity is preserved because none is $s$, and length induction applies to $sv$ and $scs$. Its unforced skip transports back to the asserted skip of $v$. This proves the criterion by rank/length induction.

Descent case with $s$ not a cover. If $e_s$ were an unforced skip root of $q=sv$, step 2.4 for the final letter $s$ in $scs$ would force $s$ to commute with all selected reflections after that skip. Conjugating by its prefix and using step 2.5 shows its skipped letter $r$ commutes with the remaining suffix. If $q=pz$ and $s=pr{p}^{-1}$, then $v=sq=prz=pzr=qr$ is reduced and covers $q=sv$, contrary to the assumption. Thus $e_s$ is absent. Also $-e_s$ cannot be a skip root of $q$: step 1.2 would put $s$ in $I(q)$, although $q\mathrm{̸}{\ge }_{R}s$. All other root signs are preserved by $\rho (s)$, so step 1.4 and length induction identify the negative skips with the covers of $v$. The non-descent branch restricts to the parabolic and adjoins the positive root $e_s$; its covers are those of that parabolic by support and intrinsic length. Together with step 2.6 this proves (3).

Confinement preparation. Suppose $s$ is initial or final and a cover of $v$. By (3), $-e_s$ is a skip root. Euler orthogonality says for every other skip root $\gamma =\pm {\beta }_t$ that either $E_c(e_s,\gamma )=0$ or $E_c(\gamma ,e_s)=0$. The coordinate formulas and initial-letter conjugation then imply either ${\beta }_t\in V_{S\setminus \{s\}}$ or $\rho (s){\beta }_t\in V_{S\setminus \{s\}}$, hence either $t$ or $sts$ lies in that parabolic. For example, for initial $s$, the first equality reads off the coordinate of $\gamma$, while the second becomes $E_{scs}(\rho (s)\gamma ,-e_s)=0$ with $s$ final; For final $s$, apply the same initial-letter identity in $scs$ and then its inverse; the coordinate equalities give the same alternatives.

If $s,t$ do not commute, take the rank-two plane spanned by their roots, with the generic perpendicular point supplied by [F9]. One extreme root is $e_s$, since nonnegative simple coordinates make its ray extreme. The other canonical reflection $p$ lies in $W_{S\setminus \{s\}}$: one of $t,sts$ lies there by step 4.1, and its root has zero $s$-coordinate; this is the other extreme ray. Thus $\{t,sts\}=\{p,sps\}$. If $t$ is a cover, step 1.4 shows both $t,sts$ are inversions; together with $s$ this forces the full rank-two inversion set by [F9]. Deleting an internal angular root would leave a set which is neither initial nor final, whereas deleting a cover leaves an inversion set. Hence the cover $t$ must be the endpoint $p$, and lies in the parabolic.

If $t$ is unforced, it is not an inversion by step 1.2. Conjugation invariance from step 1.4 shows neither $p$ nor $sps$ is an inversion of $v$; its rank-two inversion set is therefore just $\{s\}$. The sortable element $v^{\mathrm{\prime }}=a_1\cdots a_{i}r$ from step 2.3 has inversions equal to the prefix inversions together with $t$, so its rank-two inversion set is either $\{p\}$ with $t=p$, or $\{s,sps\}$ with $t=sps$, by recognition. If $s$ is initial, its first sorting position is selected, so $s$ is in that prefix and only the second possibility holds; thus $sts=p$ lies in the parabolic. If $s$ is final, the endpoint order with positive omega is $(p,s)$, so alignment of $v^{\mathrm{\prime }}$ excludes $\{s,sps\}$ and forces $t=p$ in the parabolic. The commuting case has $t=sts$ and is already covered by step 4.1. These are the full confinement assertions needed below.

For final $s$ with $v{\ge }_{R}s$, step 2.5 makes $s$ a cover; for initial $s$ assume it is a cover. In either case step 5.1 puts every other cover in $W_{S\setminus \{s\}}$, so the local cover-join formulas [F7] give $v=s\vee v_{S\setminus \{s\}}$ and $\mathrm{C}\mathrm{o}\mathrm{v}(v)=\{s\}\cup \mathrm{C}\mathrm{o}\mathrm{v}(v_{S\setminus \{s\}})$.

For final $s$, step 6.1 puts every unforced skip $t$ in that parabolic. Insert $t$ in the sorting reflection sequence as in steps 2.3-2.4 and restrict the sequence to parabolic reflections. The intrinsic parabolic reflection-sequence restriction in the proof of Omega-positive words are commutation-equivalent to sortable sorting words; sortable equals aligned; parabolic restriction (3), and the prefix inversion formula identify the restricted sequence with a reduced word for $v_{S\setminus \{s\}}$; uniform positivity makes it commutation-equivalent to its $cs$-sorting word. The same restriction of the prefix-plus-$t$ sequence is reduced by recognition, and the omega inequalities restrict with the form. Thus the insertion criterion in step 2.7 makes $t$ an unforced skip of that prefix. Both unforced sets have the same cardinality: the bases have respectively $n$ and $n-1$ roots and the cover sets differ by the one reflection $s$. The inclusion is therefore equality.

For initial $s$, transport each unforced skip $t$ to $sts$ for $sv$; it is in the parabolic by step 6.1, and the restriction argument of step 7.1 makes it an unforced skip of $(sv{)}_{S\setminus \{s\}}$. Since $s$ is a cover, $I(sv)=I(v)\setminus \{s\}$ and the two parabolic prefixes have equal inversion sets, hence $(sv{)}_{S\setminus \{s\}}=v_{S\setminus \{s\}}$. Equal cardinalities as in step 7.1 give $uf{s}_c(v)=\{sts:t\in uf{s}_{sc}(v_{S\setminus \{s\}})\}$. This proves all of (5).

Steps 1.1-2.2 prove (1),(2),(4); steps 1.4, 2.6 and 3.1 prove (3); steps 2.5 and 2.7-8.1 prove (5). All selections are individual witnesses from finite sets or fixed words, and no Choice is used.