Boone reduced auxiliary words have no rule pinches

Here is also the intersection check directly. Undoing the state-twisting automorphism in [F1] fixes $H$. A reduced word with a state syllable then cannot lie in $H$, so an element of $A_i\cap ⟨x⟩$ must lie in $T_1=⟨sx⟩$. The tape retraction sends a nonempty reduced basis word there to a nonempty reduced tape word, whereas it sends $x^m$ to identity. Therefore the basis word is empty and $x^m=1$, whence $m=0$. For $B_i$ the same computation uses $T_{-1}=⟨s{x}^{-1}⟩$, with the identical retraction image. This verifies the intersections used in [F2].

An internal pinch in a reduced auxiliary word must pair $r_i^{ϵ},r_i^{-ϵ}$ across a pure power $x^m$. By step 1.1 membership in the requisite edge subgroup forces $m=0$. But then the two letters freely cancel, contrary to reducedness. There is thus no internal pinch.

Rewrite the given equality as $L{X}^{\mathrm{\#}}{q}_j=q{R}^{-1}{Y}^{-1}$. Both sides are rule-reduced by step 2.1, since all their rule letters occur in their single auxiliary block. The multiple-letter comparison of [F1], derived by successively pairing letters across the seam, shows their signed rule sequences agree. In particular their lengths agree. This is consistent with the single-letter normal-form invariant [F3], with choices licensed by [A1]; the finite comparison here uses the proved multiple-letter version.

If the common length is positive, multiple-letter Britton applied to $L{X}^{\mathrm{\#}}{q}_{j}YR{q}^{-1}=1$ supplies a pinch. It cannot be wholly within either auxiliary block by step 2.1. The only other consecutive pair of rule letters is the last one of $L$ and the first one of $R$, with intervening coefficient $x^m{X}^{\mathrm{\#}}{q}_{j}Y{x}^n$ for their adjacent terminal/initial powers. Thus it has exactly the central form in the statement. If the common length is zero, the whole equality is in $G_0$ by base embedding.