Every trivial braid word combs as W_1W_2

Prefix insertion. By [F3] and $\phi (W)=1$, there are positions $j_0=j_m=n$ with $j_k=s_{i_k}(j_{k-1})$ and $W\equiv {\prod }_{k=1}^m{F}_k$, $F_k:={\alpha }_{j_{k-1}}^{-1}{\sigma }_{i_k}^{{\epsilon }_k}{\alpha }_{j_k}$.

Reducing the factors. Fix $k$ and apply [F4] to $F_k$, whose letter has index $i_k$ and whose connector position is $j_{k-1}$ with $j_k=s_{i_k}(j_{k-1})$; the factor is equivalent to the empty word, or to a single letter $x_{i_k}$ or $x_{i_k}^{-1}$, or to a single letter ${\sigma }_{i_k}^{{\epsilon }_k}$ with $i_k\le n-2$, or to a single letter ${\sigma }_{i_k-1}^{{\epsilon }_k}$ with $i_k-1\le n-2$. Deleting the factors that reduce to the empty word and choosing one such reduced word in each remaining factor, we obtain a word $U$ in the mixed alphabet ${\sigma }_1^{\pm 1},\dots ,{\sigma }_{n-2}^{\pm 1},x_1^{\pm 1},\dots ,x_{n-1}^{\pm 1}$ with $W\equiv U$, because $\equiv$ is compatible with concatenation by [F1].

Collection of the lower-rank letters. We show: for every word $U$ in the mixed alphabet ${\sigma }_1^{\pm 1},\dots ,{\sigma }_{n-2}^{\pm 1},x_1^{\pm 1},\dots ,x_{n-1}^{\pm 1}$ there are a word $u$ in the $x$-letters and a word $v$ in ${\sigma }_1^{\pm 1},\dots ,{\sigma }_{n-2}^{\pm 1}$ with $U\equiv uv$. Proceed by induction on the number $r$ of $\sigma$-letters occurring in $U$. If $r=0$, take $u:=U$ and $v$ empty. If $r\ge 1$, let $s$ be the last (rightmost) $\sigma$-letter of $U$ and write $U=PsA$, where $A$ is the (possibly empty) $x$-word following $s$, so that no $\sigma$-letter occurs in $A$. While $A$ is nonempty, let $x$ be its first letter and replace the adjacent pair $sx$ by $ws$, where $s={\sigma }_i^{\epsilon }$ with $i\le n-2$, $x=x_j^{\delta }$ and ${\sigma }_i^{\epsilon }{x}_j^{\delta }=w{\sigma }_i^{\epsilon }$ is the identity of [F5]; this is a permitted rewrite, and the new word again has $s$ as its rightmost $\sigma$-letter, now followed by $A$ with its first letter deleted, because the $x$-word $w$ stands immediately to the left of $s$. Hence the number of letters strictly to the right of $s$ decreases by exactly one at each rewrite, so after finitely many steps we obtain a word $U^{\mathrm{\prime }}\equiv U$ whose letters strictly to the right of the rightmost $\sigma$-letter $s$ are none, that is, $U^{\mathrm{\prime }}=P^{\mathrm{\prime }}s$ where $P^{\mathrm{\prime }}$ is a word in the mixed alphabet with exactly $r-1$ $\sigma$-letters. By the induction hypothesis applied to $P^{\mathrm{\prime }}$, there are an $x$-word $u^{\mathrm{\prime }}$ and a $\sigma$-word $v^{\mathrm{\prime }}$ with $P^{\mathrm{\prime }}\equiv u^{\mathrm{\prime }}{v}^{\mathrm{\prime }}$; then $U\equiv U^{\mathrm{\prime }}\equiv u^{\mathrm{\prime }}(v^{\mathrm{\prime }}s)$ by [F1], where $u^{\mathrm{\prime }}$ is an $x$-word and $v^{\mathrm{\prime }}s$ is a word in the $\sigma$-letters of rank at most $n-2$, so the induction is complete.

Assembly. By step 1.2 there is a mixed word $U$ with $W\equiv U$, and by step 1.3 there are an $x$-word $W_1$ and a lower-rank $\sigma$-word $W_2$ with $U\equiv W_1{W}_2$; since $\equiv$ is transitive by [F1], $W\equiv W_1{W}_2$, which is the required product.

Conclusion. Given $W$ with $\phi (W)=1$, steps 1.1-1.3 rewrite it, using only the two Artin relations and free insertions and deletions of adjacent inverse pairs, first into the product of its combing factors, then into a word $U$ in the mixed alphabet, and finally into a product $W_1{W}_2$ with $W_1$ a word in $x_1^{\pm 1},\dots ,x_{n-1}^{\pm 1}$ and $W_2$ a word in ${\sigma }_1^{\pm 1},\dots ,{\sigma }_{n-2}^{\pm 1}$; the name $n-2$ in the statement is justified because [F4] bounds every surviving $\sigma$-index by $n-2$. ∎