Each combing factor reduces to a lower-rank letter or an x-letter

The four cases in which the letter moves the tracked point. Assume $j=k$ or $j=k+1$; by [F2] we have the word identities ${\alpha }_k={\sigma }_k{\alpha }_{k+1}$, ${\alpha }_k^{-1}={\alpha }_{k+1}^{-1}{\sigma }_k^{-1}$, $x_k={\alpha }_{k+1}^{-1}{\sigma }_k^2{\alpha }_{k+1}$ and $x_k^{-1}={\alpha }_{k+1}^{-1}{\sigma }_k^{-2}{\alpha }_{k+1}$. Substituting ${\alpha }_k$ or ${\alpha }_{k+1}$ for the two connectors and cancelling the adjacent inverse pair ${\sigma }_k^{-1}{\sigma }_k$, or its inverse pair, by [F1]: for $j=k$ and $\epsilon =1$, $F={\alpha }_k^{-1}{\sigma }_k{\alpha }_{k+1}={\alpha }_{k+1}^{-1}{\sigma }_k^{-1}{\sigma }_k{\alpha }_{k+1}\equiv {\alpha }_{k+1}^{-1}{\alpha }_{k+1}\equiv 1,$ the empty word; for $j=k$ and $\epsilon =-1$, $F={\alpha }_k^{-1}{\sigma }_k^{-1}{\alpha }_{k+1}={\alpha }_{k+1}^{-1}{\sigma }_k^{-2}{\alpha }_{k+1}=x_k^{-1};$ for $j=k+1$ and $\epsilon =1$, $F={\alpha }_{k+1}^{-1}{\sigma }_k{\alpha }_k={\alpha }_{k+1}^{-1}{\sigma }_k{\sigma }_k{\alpha }_{k+1}={\alpha }_{k+1}^{-1}{\sigma }_k^2{\alpha }_{k+1}=x_k;$ and for $j=k+1$ and $\epsilon =-1$, $F={\alpha }_{k+1}^{-1}{\sigma }_k^{-1}{\alpha }_k={\alpha }_{k+1}^{-1}{\sigma }_k^{-1}{\sigma }_k{\alpha }_{k+1}\equiv {\alpha }_{k+1}^{-1}{\alpha }_{k+1}\equiv 1.$ This gives (a), (b), (c) and (d).

The case $k<j-1$: far commutation. Here $j\mathrm{\notin }\{k,k+1\}$, so $j^{\mathrm{\prime }}=j$ and $F={\alpha }_j^{-1}{\sigma }_k^{\epsilon }{\alpha }_j$. Every letter ${\sigma }_r^{\pm 1}$ occurring in the word ${\alpha }_j$ of [F2] has index $r\ge j\ge k+2$, hence $\mathrm{\mid }r-k\mathrm{\mid }\ge 2$ and ${\sigma }_k$ commutes with that letter by the far-commutation relation of [F1]; iterating over the letters of ${\alpha }_j$ (whose length is $n-j$, possibly $0$ when $j=n$), we get ${\sigma }_k^{\epsilon }{\alpha }_j\equiv {\alpha }_j{\sigma }_k^{\epsilon }$, whence $F={\alpha }_j^{-1}{\sigma }_k^{\epsilon }{\alpha }_j\equiv {\alpha }_j^{-1}{\alpha }_j{\sigma }_k^{\epsilon }\equiv {\sigma }_k^{\epsilon }$ by free cancellation. This is (e); note $k\le j-2$ and $j\le n$, so $k\le n-2$.

The case $k>j$: sliding the letter to the right. Here again $j\mathrm{\notin }\{k,k+1\}$, so $j^{\mathrm{\prime }}=j$ and $F={\alpha }_j^{-1}{\sigma }_k^{\epsilon }{\alpha }_j$. First take $\epsilon =1$. Since $j<k\le n-1$, the word ${\alpha }_j={\sigma }_j\cdots {\sigma }_{n-1}$ splits, with the three groups possibly empty, as the word ${\alpha }_j=({\sigma }_j\cdots {\sigma }_{k-2}){\sigma }_{k-1}{\sigma }_k({\sigma }_{k+1}\cdots {\sigma }_{n-1}),$ where the first group contains exactly the letters with indices $r\le k-2$ and the last exactly those with indices $r\ge k+1$. Every letter of the first group has $k-r\ge 2$, so ${\sigma }_k$ commutes with each of them and moves right past them; every letter of the last group has $r-(k-1)\ge 2$, so ${\sigma }_{k-1}$ commutes with each of them and moves right past them; and the three middle letters satisfy the braid relation ${\sigma }_k{\sigma }_{k-1}{\sigma }_k\equiv {\sigma }_{k-1}{\sigma }_k{\sigma }_{k-1}$ by [F1]. Combining the three moves gives the chain ${\sigma }_k{\alpha }_j\equiv ({\sigma }_j\cdots {\sigma }_{k-2}){\sigma }_k{\sigma }_{k-1}{\sigma }_k({\sigma }_{k+1}\cdots {\sigma }_{n-1})\equiv ({\sigma }_j\cdots {\sigma }_{k-2}){\sigma }_{k-1}{\sigma }_k{\sigma }_{k-1}({\sigma }_{k+1}\cdots {\sigma }_{n-1})\equiv ({\sigma }_j\cdots {\sigma }_{k-2}){\sigma }_{k-1}{\sigma }_k({\sigma }_{k+1}\cdots {\sigma }_{n-1}){\sigma }_{k-1}={\alpha }_j{\sigma }_{k-1}\mathrm{.}$ For $\epsilon =-1$, left-multiply this identity in the group $B_n$ by ${\sigma }_k^{-1}$: it becomes ${\alpha }_j={\sigma }_k^{-1}{\alpha }_j{\sigma }_{k-1}$, hence ${\sigma }_k^{-1}{\alpha }_j={\alpha }_j{\sigma }_{k-1}^{-1}$. In both signs, therefore, ${\sigma }_k^{\epsilon }{\alpha }_j\equiv {\alpha }_j{\sigma }_{k-1}^{\epsilon }$ and $F={\alpha }_j^{-1}{\sigma }_k^{\epsilon }{\alpha }_j\equiv {\alpha }_j^{-1}{\alpha }_j{\sigma }_{k-1}^{\epsilon }\equiv {\sigma }_{k-1}^{\epsilon }$ by free cancellation. This is (f); here $k>j\ge 1$ gives $k-1\ge 1$ and $k\le n-1$ gives $k-1\le n-2$.

Transfer to the geometric braid group. Since $\phi$ is a homomorphism with $\phi ({\sigma }_r)=[{\sigma }_r]$ by [F4], applying $\phi$ to each of the reductions of steps 1.1, 1.2 and 1.3 turns it into the corresponding identity in $G_n$: the free cancellations become $[\gamma ][\gamma {]}^{-1}=1$, and the two Artin relations used are the geometric relations supplied by the published The geometric three strand braid relation and Far commutativity of elementary geometric half twists.

Conclusion. The conditions of the six cases are exactly: $j=k$ (with either sign), $j=k+1$ (with either sign), $j\mathrm{\notin }\{k,k+1\}$ with $k\le j-2$, and $j\mathrm{\notin }\{k,k+1\}$ with $k\ge j+1$; if $j\mathrm{\notin }\{k,k+1\}$ then either $j>k+1$, that is $k<j-1$, or $j<k$, that is $k>j$, so the list is exhaustive, and the conditions are visibly mutually exclusive. Steps 1.1, 1.2 and 1.3 establish the reductions (a)-(f), and the reduced forms are the empty word, $x_k^{\pm 1}$, or ${\sigma }_k^{\epsilon }$ with $k\le n-2$, or ${\sigma }_{k-1}^{\epsilon }$ with $k-1\le n-2$, so all of them are words in ${\sigma }_1^{\pm 1},\dots ,{\sigma }_{n-2}^{\pm 1},x_1^{\pm 1},\dots ,x_{n-1}^{\pm 1}$; in particular no copy of ${\sigma }_{n-1}$ survives the reduction outside an $x$-letter. Step 2.1 transfers each reduction to $G_n$. ∎