Schreier rewriting is invariant under free reduction

It is enough to treat one elementary cancellation. By symmetry it suffices to consider $u=px{x}^{-1}q$ and $v=pq$ with $x\in X$. Let $t=\bar{p}$ and $u_1=\overline{px}$. In the rewrite of $u$, the letter $x$ contributes $s(t,x)$ and the following letter $x^{-1}$ contributes $s(t,x{)}^{-1}$, so these two adjacent letters freely cancel.

After those two letters are read, the current coset is again $Hp$, so the successive representatives used for the remaining suffix $q$ are exactly the same whether one starts from $u$ or from $v$. Thus one elementary free cancellation turns $\tau (u)$ into $\tau (v)$. Repeating this argument along a finite chain of elementary cancellations and reverse insertions proves that the two rewrites are freely equivalent, and hence represent the same subgroup element.