= Solution
Write $R=\langle\!\langle r_j:j\in J\rangle\!\rangle_F$ and let $\pi:F\to G_0=F/R$ be the quotient <group homomorphism>. The subgroup in the question is exactly the <fiber product of groups>
$$
\boxed{H=\{(u,v)\in F\times F:\pi(u)=\pi(v)\}.}
$$
Every specified generator belongs to the right-hand side, so one inclusion is immediate. For the reverse inclusion, diagonal generator pairs produce $(u,u)$ for every $u\in F$. Also
$$
(u,u)^{-1}(1,r_j)(u,u)=(1,u^{-1}r_ju),
$$
so their products and inverses produce $(1,r)$ for every $r\in R$. If $\pi(u)=\pi(v)$, then $u^{-1}v\in R$ and $(u,v)=(u,u)(1,u^{-1}v)\in H$, proving equality.
In particular $\boxed{(1,w)\in H\Longleftrightarrow w\in R\Longleftrightarrow w=1\text{ in }G_0}$. Constructing the input pair $(1,w)$ is effective, so a solution of the <membership problem for a subgroup> $H\le F\times F$ would solve the insoluble <word problem for a group> in $G_0$. \b[The subgroup membership problem is therefore undecidable.] This is the <Mihailova subgroup> construction; its finite generating set follows from the finiteness of $X,J$.
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