 Solution

Take a recursively enumerable set $S\subseteq {\mathbb{N}}_{>0}$ that is not recursive, for example the diagonal halting set from Question 2 shifted by one. To obtain a finitely generated group with recursive presentation but undecidable word problem for a group, first define the countable Right-angled Artin groupTranslation of the indices gives an automorphism $a_i\mapsto a_{i+1}$. The semidirect product with its infinite cyclic shift is generated by $a=a_0,t$ and has presentationTo verify this presentation, the semidirect product satisfies its relators, while the reverse map sends $a_i$ to $t^{i}a{t}^{-i}$; all translated commutation relations follow by conjugation. These maps are inverse on their generators. Thus $B_S$ embeds in $G_S$.

For $n\in S$, the word $[a_0,a_n]$ is trivial. For $n\notin S$, map $a_0,a_n$ to the free generators $x,y$ of $F(x,y)$ and every other $a_i$ to one. Every defining commutator of $B_S$ maps to one, since the only possible relation involving both surviving generators would have separation $n$, which is absent. The resulting group homomorphism sends $[a_0,a_n]$ to the nontrivial reduced word $[x,y]$. HenceThis proves the shift-invariant graph groups encode arbitrary sets construction rather than supposing that a missing relator automatically means a nontrivial word. A word algorithm for $G_S$ would decide $S$, so none exists. Since $S$ is enumerable, the displayed two-generator presentation is a recursive presentation of a group. By the Higman embedding theorem, $G_S$ embeds in a finitely presented group $G_0$. A word algorithm for $G_0$, composed with substitution of the two fixed generator-image words, would solve the word problem for $G_S$. Therefore $\boxed{G_0\text{ is finitely presented with insoluble word problem}}$.