For a group $G$ we denote by $d(G)$ the cardinality of a smallest set of generators.

A finitely generated group $G$ is said to satisfy Schreier's formula if for every subgroup $H \subseteq G$ of index $k \in \mathbb{N}$ we have: $d(H) - 1 = k(d(G) - 1)$. If this equality does not hold, we say that $H$ violates Schreier's formula in $G$. For example, a finitely generated free group satisfies Schreier's formula.

Let $F$ be a free group on a finite number of generators, and $r \in \mathbb{N}$. Denote by $F^{(r)}$ the $r$-th term in the derived series of $F$. Set $Q_{r} = F/F^{(r)}$.

It has already been shown in: Schreier's index formula that $Q_r$ doesn't satisfy Schreier's formula, and here I ask for a generalization.

Is it true that every descending chain of subgroups of $Q_r$ eventually violates Schreier's formula?

More formally let, $H_1 \supseteq H_2 \supseteq \dots$ be an infinite countable properly descending chain of subgroups of $Q_r$. Must there be some $n \in \mathbb{N}$ such that $H_n$ violates Schreier's formula in $Q_r$?

Note that the case $r=1$ is trivial because every subgroup of finite index violates Schreier's formula, but the case $r=2$ is already nontrivial.