Boone commutator extracts an auxiliary history

Set $g={\mathrm{\Sigma }}^{-1}t\mathrm{\Sigma }\in G_3$. It has one $t$, hence is nonidentity by [F2]. In the identity word $kg{k}^{-1}{g}^{-1}$ the only two $k$ letters must form a pinch. For the identity edge map of $D$ this says precisely $g\in D$, with membership in the embedded base $G_3$.

Write $g^{-1}=R_0(q^{-1}{t}^{e_1}q)R_1\cdots (q^{-1}{t}^{e_n}q)R_n,$ where $R_j\in C$ are auxiliary words and $e_j\in \{1,-1\}$, and choose the least possible $n$. Such finite expressions exist by $g\in D$ and the definition of generated subgroup. If $n=0$, the equality ${\mathrm{\Sigma }}^{-1}t\mathrm{\Sigma }{R}_0=1$ has exactly one $t$ and violates [F2]. Thus $n\ge 1$.

Apply [F2] to the displayed word for $g{g}^{-1}$ in $G_3$. If a pinch uses its first $t$, it pairs that $t$ with $t^{e_1}$, so $e_1=-1$ and $\mathrm{\Sigma }{R}_0{q}^{-1}=P\in C$ in $G_2$. Multiplying gives $P^{-1}\mathrm{\Sigma }{R}_0=q$. This is the desired auxiliary equation.

Any other pinch pairs consecutive $t^{e_j},t^{e_{j+1}}$ with $e_{j+1}=-e_j$ and $q{R}_j{q}^{-1}\in C$. Since $t$ commutes with $C$, for either sign $e_j$ the corresponding subexpression satisfies $(q^{-1}{t}^{e_j}q)R_j(q^{-1}{t}^{-e_j}q)=q^{-1}(q{R}_j{q}^{-1})q=R_j\mathrm{.}$ Replacing it combines the neighboring auxiliary factors and gives an expression for $g^{-1}$ with $n-2$ such occurrences, contrary to minimality. Thus this kind of pinch cannot occur.

A pinch exists, so step 3.1 must apply. Freely reducing $P^{-1}$ and $R_0$ changes neither represented element nor alphabet, and yields the claimed $L,R$. All coefficient equalities were obtained in $G_2$, by the embedded-base clause in [F1]. Least finite length and finite free reduction need no additional choice.