The congruence subgroup conjecture for $\Delta_{r/p}$

Let $p$ be prime, let $r\in\mathbb{Z}$ satisfy $\gcd(r,p)=1$, and suppose that $r/p\in\mathbb{Q}\cap(0,4)$. Let

$$A=\begin{pmatrix}1&0\\1&1\end{pmatrix},\qquad Q_{r/p}=\begin{pmatrix}1&r/p\\0&1\end{pmatrix},\qquad \Delta_{r/p}=\langle A,Q_{r/p}\rangle\leq\operatorname{SL}_2\left(\mathbb{Z}\left[\frac1p\right]\right).$$

Let $\overline{\Gamma}_1^{(p)}(r)$ denote the subgroup generated by matrices in $\operatorname{SL}_2\left(\mathbb{Z}[\frac1p]\right)$ whose diagonal entries are congruent to $1\pmod r$ and whose upper-right entry is congruent to $0\pmod r$. The congruence subgroup conjecture. For every such $r/p$,

$$\Delta_{r/p}=\overline{\Gamma}_1^{(p)}(r).$$

Consequently, $\Delta_{r/p}$ is not free, has index $J_2(r)$, and has the stated finite graph-of-groups structure. The equality is proved in many cases, while the general assertion remains open.