Magnus commutator degree and finite truncation

In any filtered associative ring, write $U=1+a$ and $V=1+b$ with $a\in {\mathfrak{m}}^p$ and $b\in {\mathfrak{m}}^q$. Since $U,V$ are units, $[U,V]-1=(UV-VU)U^{-1}{V}^{-1}=(ab-ba)U^{-1}{V}^{-1}\mathrm{.}$ By [F1], $ab-ba\in {\mathfrak{m}}^{p+q}$, and multiplication by units preserves this two-sided ideal. Therefore $[U,V]\in 1+{\mathfrak{m}}^{p+q}$. This is a ring calculation conditional on the two degree bounds; no lower-central inclusion is assumed here.

By [F1], $E({\mathrm{\Gamma }}_1{F}_r)=E(F_r)\subseteq 1+\mathfrak{m}$. Assume $E({\mathrm{\Gamma }}_q{F}_r)\subseteq 1+{\mathfrak{m}}^q$. For a generator $[u,v]$ of ${\mathrm{\Gamma }}_{q+1}{F}_r=[F_r,{\mathrm{\Gamma }}_q{F}_r]$, [F1] gives $E(u)-1\in \mathfrak{m}$ and the induction hypothesis gives $E(v)-1\in {\mathfrak{m}}^q$. The ring calculation in step 1.1 therefore puts its image in $1+{\mathfrak{m}}^{q+1}$. Products and inverses remain in that subgroup: their images modulo the two-sided ideal ${\mathfrak{m}}^{q+1}$ are all $1$. Thus $E({\mathrm{\Gamma }}_{q+1}{F}_r)\subseteq 1+{\mathfrak{m}}^{q+1}$, and induction proves the inclusion for every $q$. Now for arbitrary $u\in {\mathrm{\Gamma }}_p{F}_r$ and $v\in {\mathrm{\Gamma }}_q{F}_r$, the established inclusions give both hypotheses of step 1.1, which yields $E([u,v])\in 1+{\mathfrak{m}}^{p+q}$.

If two words represent the same element of $F_r\mathrm{/}{\mathrm{\Gamma }}_{d+1}{F}_r$, their quotient lies in ${\mathrm{\Gamma }}_{d+1}{F}_r$. By step 2.1 its expansion is $1$ modulo ${\mathfrak{m}}^{d+1}$. Multiplicativity of $E$ then makes the two truncated expansions equal, proving representative-independence of every coefficient of degree at most $d$.