Ascending HNN extensions admit the one-sided normal form

Choose a right-coset transversal $S$ for $\varphi (A)\mathrm{\backslash }A$ containing $e$, and apply [L2] with the transversal for the negative associated subgroup equal to $\{e\}$. Every coefficient following a letter $t$ is then $e$, so a sign pattern $+,-$ cannot occur: it would create a pin. Thus every transversal normal form has all negative stable letters before all positive stable letters.

Such a normal form has the shape $a_0{t}^{-1}{s}_1\cdots t^{-1}{s}_m{t}^n$, where $s_i\in S$ and the positive-letter coefficients are identities. Repeatedly use $b{t}^{-1}=t^{-1}\varphi (b)$ to move $a_0$ and then the intervening coefficients to the right. This gives one expression $t^{-m}a{t}^n$. If $m,n>0$, normality gives $s_m\mathrm{\ne }e$; the resulting middle coefficient has the form $\varphi (b)s_m$, so it lies outside $\varphi (A)$.

Conversely, start with $t^{-m}a{t}^n$ and recursively decompose the coefficient immediately following the last $t^{-1}$ into its unique form $ds$ with $d\in \varphi (A)$ and $s\in S$, moving $d$ left by $t^{-1}d={\varphi }^{-1}(d)t^{-1}$. This recovers a unique transversal normal form. When $m,n>0$, the condition $a\mathrm{\notin }\varphi (A)$ makes the last representative nonidentity, so no cancellation occurs at the sign change. This reverse construction and the construction of step 2.1 are inverse, and uniqueness in [L2] therefore gives uniqueness of $m,n,a$.