Reidemeister-Schreier relators are independent of word representatives

If $r^{\mathrm{\prime }}$ is obtained from $r$ by one elementary cancellation or reverse insertion, then for every transversal element $t$ the word $t{r}^{\mathrm{\prime }}{t}^{-1}$ is obtained from $tr{t}^{-1}$ by the same local free reduction inside the middle block. Therefore [L2] gives $\tau (tr{t}^{-1})=\tau (t{r}^{\mathrm{\prime }}{t}^{-1})$.

Any freely equivalent pair $r,r^{\mathrm{\prime }}$ is connected by finitely many such moves, so the equality from step 1.1 persists through the whole chain. Thus every rewritten relator produced from $r$ is identical to the one produced from $r^{\mathrm{\prime }}$.

By [L1], replacing a generator of a normal closure by the same group element does not change that normal closure. Hence the presentation described in [L3] is independent of which freely equivalent word is chosen to represent each ambient relator.