Boone positive history pushing

For every integer $a$, $x^{a}s=s{x}^{2a}$ follows by taking powers in $s^{-1}xs=x^2$. The rule relation also gives $r_i^{-1}s=s{x}^{-1}{r}_i^{-1}{x}^{-1}$: from $r_{i}s=sx{r}_{i}x$, obtain $r_i^{-1}sx=s{x}^{-1}{r}_i^{-1}$ and multiply on the right by $x^{-1}$. Thus for either $ϵ=1$ or $-1$, $r_i^{ϵ}s=s{x}^{ϵ}{r}_i^{ϵ}{x}^{ϵ}$.

For a positive word $V$ of length $m$, set $d_m=2^m-1$. At $m=0$, $r_i^{ϵ}V=V{x}^{ϵd_m}{r}_i^{ϵ}{x}^{ϵd_m}$. If this holds for $V$ and $a=ϵd_m$, then $r_i^{ϵ}Vs=V{x}^a{r}_i^{ϵ}{x}^{a}s=V{x}^a{r}_i^{ϵ}s{x}^{2a}=Vs{x}^{2a+ϵ}{r}_i^{ϵ}{x}^{2a+ϵ}\mathrm{.}$ Since $2a+ϵ=ϵd_{m+1}$, this proves the formula for all $m$, for both signs.

Let $U$ be positive of length $n$, let $Z$ be its reversal and put $e=2^n-1$. Then $U^{\mathrm{\#}}=Z^{-1}$. Applying step 2.1 to $r_i^{ϵ}Z=Z{x}^{ϵe}{r}_i^{ϵ}{x}^{ϵe}$ and multiplying by $Z^{-1}$ gives $U^{\mathrm{\#}}{r}_i^{ϵ}=x^{ϵe}{r}_i^{ϵ}{x}^{ϵe}{U}^{\mathrm{\#}}\mathrm{.}$ Together with step 2.1 these are all four signed pushing identities, including $n=0$ and $m=0$.

Every word in a history beginning at ${\mathrm{\Sigma }}^{\ast }$ has exactly one positive state letter, since every relation preserves that count. Hence a contextual forward replacement has old word $U{F}_i{q}_{a(i)}{G}_{i}V$ and new word $U{H}_i{q}_{b(i)}{K}_{i}V$ with positive tape contexts $U,V$. With $e=2^{\mathrm{\mid }U\mathrm{\mid }}-1$, $d=2^{\mathrm{\mid }V\mathrm{\mid }}-1$, the corresponding special spellings $Z_{\mathrm{o}\mathrm{l}\mathrm{d}},Z_{\mathrm{n}\mathrm{e}\mathrm{w}}$ satisfy $Z_{\mathrm{n}\mathrm{e}\mathrm{w}}=U^{\mathrm{\#}}{r}_i^{-1}{a}_i{r}_{i}V=(x^{-e}{r}_i^{-1}{x}^{-e})Z_{\mathrm{o}\mathrm{l}\mathrm{d}}(x^d{r}_i{x}^d)\mathrm{.}$ The reverse replacement uses $a_i=r_i{b}_i{r}_i^{-1}$ and gives $Z_{\mathrm{o}\mathrm{l}\mathrm{d}}=(x^e{r}_i{x}^e)Z_{\mathrm{n}\mathrm{e}\mathrm{w}}(x^{-d}{r}_i^{-1}{x}^{-d})\mathrm{.}$ These equalities use only relations of $G_2$.

Along the finite symmetric path choose, at each edge, the applicable equality expressing the earlier spelling as an auxiliary left factor times the later spelling times an auxiliary right factor. Substitution multiplies the left factors in path order and the right factors in reverse path order. Since the last spelling is $q$, it yields $\mathrm{\Sigma }=LqR$. For a path of length zero both factors are empty. This is a finite product of explicitly given factors, with no choice of infinite histories.

Put $g=q^{-1}tq$. Since $t$ commutes with $L$, ${\mathrm{\Sigma }}^{-1}t\mathrm{\Sigma }=R^{-1}gR$. Since $k$ commutes with $R$ and $g$, $W(\mathrm{\Sigma })=k{R}^{-1}gR{k}^{-1}{R}^{-1}{g}^{-1}R=R^{-1}(kg{k}^{-1}{g}^{-1})R=1.$ These equalities hold in $\mathcal{B}$ by the embedded tower, under its stated AC assumption.