The proof of the Adian-Rabin theorem involves constructing a class of groups $W_w$ for a word $w$ in some group $U$ with undecidable word problem, and showing that this group either has some Markov property $M$ if $w\neq1$, or is trivial if $w=1$.
$W_w$ is constructed by taking HNN extensions and amalgamated free products of smaller groups, but in order to understand the proof I tried to write down the presentation of $W_w$ and got this:
Generators:
$$y_0,y_1,y_2...$$ $$t_0,t_1,t_2...$$ $$z,j,k,l$$
Relations:
$$t_iy_i{t_i}^{-1}={y_i}^2$$ $$zt_i{z}^{-1}={t_i}^2$$ $$kjk^{-1}=j^2$$ $$lkl^{-1}=k^2$$ $$[w,y_0] = l$$ $$j=z$$
Where $y_0,y_1,y_2...$ generate the group $U\ast G_{-} \ast \langle y_0 \rangle$, for a group $G_{-}$ not embedding into any group satisfying $M$.
It seems like the point of the proof is that this group is designed to embed $G_{-}$ unless $w=1$, but I have another candidate that I think embeds $G_{-}$ unless $w=1$ for similar reasons.
Generators: $$y_0,y_1,y_2...$$ $$t_0,t_1,t_2...$$ $$z$$ Relations: $$t_iy_i{t_i}^{-1}={y_i}^2$$ $$zt_i{z}^{-1}={t_i}^2$$ $$[w,y_0] = z$$
Without the relation $[w,y_0] = z$, the proof of the theorem argues that this group embeds $G_{-}$ because it's an (iterated) HNN extension of $U\ast G_{-}\ast\langle y_0\rangle$. It's not clear to me how enforcing $[w,y_0] = z$ at might make this group no longer embed $G_{-}$, and intuitively, it seems like it shouldn't since $z$ and $y_0$ seem to have nothing to do with elements of $G_{-}$ just going by the relations (more of a hunch than a formal argument here.)
Can anyone make it clear why I should worry that $[w,y_0] = z$ might somehow make the group not embed $G_{-}$?