FALSE: recursively enumerable trivial words already give a decision algorithm

Enumerate binary strings first by length, then lexicographically, starting with the empty string as index zero. Both the indexing and its inverse are computable by finite scans. Let $S\subseteq \mathbb{N}$ be the indices of the halting language from [L1]. Dovetail its recognizer on all strings to enumerate $S$. A decider for $S$ would decide the halting language by computing an index. Hence $S$ is enumerable and undecidable.

In $F(a,b)$ put $u_i=a^{-i}b{a}^i$ for $i\in \mathbb{N}$. A nonempty reduced word in the formal symbols $u_i^{\pm 1}$ can be collected as $u_{i_1}^{e_1}\cdots u_{i_r}^{e_r}$ with $e_j\mathrm{\ne }0$ and adjacent indices distinct. Its expansion is $a^{-i_1}{b}^{e_1}{a}^{i_1-i_2}{b}^{e_2}\cdots a^{i_{r-1}-i_r}{b}^{e_r}{a}^{i_r}\mathrm{.}$ Every interior power of $a$ is nonzero and every displayed power of $b$ is nonzero, so this is a nonempty reduced word after omitting any zero end power. Thus the $u_i$ freely generate their subgroup. In particular $u_j$ lies in $U=⟨u_i:i\in S⟩$ exactly when $j\in S$: otherwise equality with a word on the selected basis elements would give a nonempty reduced relation.

In the disjoint free group $F(c,d)$ put $v_i=c^{-i}d{c}^i$ and $V=⟨v_i:i\in S⟩$. The same proof applies, and $u_i\mapsto v_i$ for $i\in S$ extends to an isomorphism $U\to V$. By [L4], amalgamating this isomorphism gives the recursive presentation ${\mathcal{P}}_S=⟨a,b,c,d\mid a^{-i}b{a}^i{c}^{-i}{d}^{-1}{c}^i,i\in S⟩\mathrm{.}$ Its relators can be listed by the enumeration in step 1.1.

Set $w_j=u_j{v}_j^{-1}$. If $j\in S$, then $w_j=1$ by its defining relation. If $j\mathrm{\notin }S$, step 2.1 and its counterpart for $V$ put the two factors outside the amalgamated subgroup, so [L4] makes $w_j\mathrm{\ne }1$. The required coset transversals exist without a choice axiom: order the finite-alphabet reduced words by length and then lexicographically, and take the unique least word in each coset. This selection need not be computable; it is only used for the normal-form proof. Rewriting either non-subgroup factor in these transversals gives a nonidentity representative, and absorbing the intervening subgroup element into the other factor leaves that factor outside its subgroup, yielding a positive-length normal word. Thus $w_j{=}_{{\mathcal{P}}_S}1⟺j\in S\mathrm{.}$

The map $j\mapsto w_j$ is computable by writing the four indicated powers. A word-problem decider for this one presentation would therefore decide $S$, contrary to step 1.1. Yet its trivial words are enumerable by [L2] and step 3.1. This presentation is the required counterexample. Its relator list is allowed to be infinite, as the Statement imposes no finite-presentation hypothesis.