I have a question. I found the following statement in some paper.
Let $F$ be free group on two generator $x$ and $y$ and $F_i$ be $i$-th component of lower central series of $F$. If $F$ is generated mod $F_1=[F,F]$ by $b_1,...,b_n$ then $F_r$ is generated mod $F_{r+1}$ by elements of the form $[b_i,w]$ with $w \in F_{r-1}$. So for an element $c$ in $F_r$, Using standard commutator identities, we can express $c$ in the form $$c=[b_1,w_1]...[b_n,w_n] d$$ where $d\in F_{r+1}$ and $w_i\in F_{r-1}$. I need to find algorithmically elements $w_1,...,w_n\in F_{r-1}$ and $d\in F_{r+1}$. For example in the free group $F<{x,y}>$, let $b_1=x^{-2}y^{-3}$, $b_2=x^{-2}(xy)^5$. I know that $c=b_2^{-1}b_1^{-1}xy^2\in F_1=[F,F]$. How could I express $c=[b_1,w_1][b_1,w_2]d$ where $w_1,w_2\in F$ and $d\in F_2$?