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The <HNN extension> in the convention used here is
$$
G_{*\phi}=\langle G,t\mid tat^{-1}=\phi(a)\text{ for every }a\in A\rangle.
$$
The new generator $t$ is its <stable letter>, and $A,B$ are its <associated subgroups of an HNN extension>. We prove the <normal form theorem for an HNN extension>, which also proves that the natural map $G\to G_{*\phi}$ is injective.
Set $C_+=A$, $C_-=B$, $\psi_+=\phi$, and $\psi_-=\phi^{-1}$. For each sign $\epsilon$, choose a <right coset transversal> $R_\epsilon$ for $C_\epsilon\backslash G$, containing $1$ as the representative of $C_\epsilon$. Thus every $g\in G$ has a unique decomposition $g=cr$, with $c\in C_\epsilon$ and $r\in R_\epsilon$. A normal form is
$$
\boxed{g_0t^{\epsilon_1}r_1\cdots t^{\epsilon_k}r_k,\quad
r_i\in R_{\epsilon_i},\quad
r_i\ne1\text{ if }i<k\text{ and }\epsilon_{i+1}=-\epsilon_i,}
$$
where $g_0\in G$ is unrestricted and the case $k=0$ is just $g_0$.
Here is a constructive proof. Let $\mathcal N$ be the set of these formal sequences. Left multiplication by $h\in G$ changes only $g_0$ to $hg_0$. To prepend $t^\epsilon$, uniquely decompose $g_0=cr$ using $R_\epsilon$ and use
$$
t^\epsilon cr=\psi_\epsilon(c)t^\epsilon r.
$$
If the old first stable letter is $t^{-\epsilon}$ and $r=1$, cancel this pair and multiply $\psi_\epsilon(c)$ into the now-leading coefficient $r_1$. Otherwise retain the new first stable letter. Both cases give a sequence in $\mathcal N$; after a cancellation the leading coefficient is unrestricted, so no further normalization is needed at that end. Prepending the letters of a word from right to left proves existence.
These operations define <permutations> of $\mathcal N$. The base-group operations satisfy $L_hL_{h'}=L_{hh'}$. The operations $L_{t^+}$ and $L_{t^-}$ are inverse: if prepending did not cancel, the reverse operation decomposes the new leading coefficient $\psi_\epsilon(c)$ with representative $1$ and cancels the newly inserted letter; if it did cancel, the reverse operation decomposes $\psi_\epsilon(c)r_1$ with representative $r_1$, restoring the original prefix. The normal-form restriction rules out an unwanted second cancellation in the latter case.
For $a\in A$, decomposing $ag_0$ uses the same representative as decomposing $g_0$, and changes its subgroup coefficient from $c$ to $ac$. Both the cancelling and noncancelling cases therefore give
$$
L_tL_a=L_{\phi(a)}L_t.
$$
Thus all defining relations act identically on $\mathcal N$, and we obtain a <group action> of the presented <HNN extension>. Each written normal form sends the empty form, whose leading coefficient is $1$, to that normal form itself. Equal group elements must have the same image of the empty form, proving uniqueness and the embedding of $G$.
A <reduced sequence in an HNN extension> is a word $g_0t^{\epsilon_1}g_1\cdots t^{\epsilon_k}g_k$ containing no <pinch in an HNN extension>: no $tat^{-1}$ with $a\in A$ and no $t^{-1}bt$ with $b\in B$. Equivalently, whenever $\epsilon_{i+1}=-\epsilon_i$, one requires $g_i\notin C_{\epsilon_i}$.
<Britton's lemma> states that
$$
\boxed{\text{a reduced sequence with }k>0\text{ is not the identity}.}
$$
More strongly it cannot represent an element of $G$. To prove this from the <normal form theorem for an HNN extension>, normalize the coefficients from right to left. Splitting $g_i=cr$ moves $\psi_{\epsilon_i}(c)$ to the coefficient immediately on its left. If the preceding stable letter has the opposite sign, this transported factor belongs to the subgroup relevant to that inverse pair; multiplying by it cannot turn a coefficient outside that subgroup into one inside it. If the signs agree, cancellation is impossible anyway. Hence no stable letter disappears during normalization of a reduced sequence. Its unique normal form has $k>0$, whereas every element of $G$ has stable-letter length zero. This proves <Britton's lemma>. The same proof works with finitely many stable letters, with pinches requiring the same letter and its inverse.
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