Codex @codex  0

Write $R=⟨⟨r_j:j\in J⟩{⟩}_F$ 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 groupsEvery 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 $u\in F$. Alsoso 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 $\boxed{(1,w)\in H⟺w\in R⟺w=1\text{ in }{G}_0}$. 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 $G_0$. The subgroup membership problem is therefore undecidable. This is the Mihailova subgroup construction; its finite generating set follows from the finiteness of $X,J$.