Let $L(x,y)$ be the free Lie algebra generated by the Hall basis. We assume $\deg(x) =2$ and $\deg(y) =2$ and for any $\alpha, \beta \in L(x,y)$ we have $\deg([\alpha, \beta])= \deg(\alpha)+\deg(\beta)-1$.
We know that for any $\alpha, \beta$, and $\gamma \in L(x,y)$ the Lie algebra satisfies the following properties:
$[\alpha, \beta] = (-1)^{\deg(\alpha)\deg(\beta)}[\beta, \alpha]$;
$(-1)^{\deg(\alpha)\deg(\gamma)}[[\alpha, \beta],\gamma]+(-1)^{\deg(\beta)\deg(\alpha)}[[\beta, \gamma],\alpha] +(-1)^{\deg(\gamma)\deg(\beta)}[[\gamma,\alpha],\beta]=0$;
$[\alpha,[y,y]]= (-1)^{(\deg(\alpha)+1)}2[[\alpha, y],y]$.
Is there any algorithm available such that any element $\alpha \in L(x,y)$ can be written as $\sum_{\alpha} x_{\alpha} h_{\alpha}$, where $h_{\alpha}$ is the Hall basis of $L(x,y)?$
For example, consider the element $\alpha= [[[[y,x],x],[y,x]],[y,y]] \in L(x,y)$. Then using properties (2) and (1) we can decompose $\alpha$ as
$$-2[[[[y,x],y],y],[[y,x],x]]-2[[[[[y,x],x],y],y],[y,x]]$$
Note that here the terms $[[[[y,x],y],y],[[y,x],x]], [[[[[y,x],x],y],y],[y,x]]$ are members of the Hall basis of $L(x,y)$. (I consider $x<y$.)
You can find such an algorithm in the book [1], starting from Chapter 4. It would be difficult to summarise the content of this book in a post.
[1] Reutenauer, Christophe. Free Lie algebras, London Mathematical Society Monographs. New Series, 7. Oxford Science Publications. The Clarendon Press, Oxford University Press, New York, 1993. xviii+269 pp. ISBN: 0-19-853679-8