The standard twists commute and fix the complementary basic arcs

Commutation. Choose representatives $T_i,T_j$ of ${\tau }_i,{\tau }_j$ supported in closed annular neighbourhoods of $l_i,l_j$; since $l_i\cap l_j=\mathrm{\varnothing }$ and the annuli can be chosen disjoint and contained in $D^{\circ }\setminus \mathrm{\Delta }$ [L1], the composites $T_i{T}_j$ and $T_j{T}_i$ agree: on the support of $T_i$ the map $T_j$ is the identity, and conversely. Hence $[{\tau }_i,{\tau }_j]=1$ in $G$ for all $i,j$, including $i=j$.

The twists fix the complementary arcs. If $k\mathrm{\ne }j$, the fixed basic arc $b_k$ is disjoint from the chosen supporting annulus of ${\tau }_j$ by [L1]: for $k<j$ it lies outside the enclosed suffix disk, and for $k>j$ it lies inside that disk away from its boundary. Thus the chosen representative is the identity on $b_k$, and ${\tau }_j(b_k)\simeq b_k$.

The composite clause. Let $e_0,\dots ,e_{m-1}\in \mathbb{Z}$ and fix $j$. For $i\mathrm{\ne }j$ the twist ${\tau }_i$ fixes $b_j$ up to isotopy by step 1.2; by induction on the number of factors and the commutation of step 1.1, $\left({\prod }_{i\mathrm{\ne }j}{\tau }_i^{e_i}\right)(b_j)\simeq b_j$ and the factors can be moved past ${\tau }_j^{e_j}$ with the identities ${\tau }_i{\tau }_j={\tau }_j{\tau }_i$; hence $\left({\prod }_i{\tau }_i^{e_i}\right)(b_j)\simeq {\tau }_j^{e_j}(b_j)$.

Conclusion. The nested twists commute, fix the complementary basic arcs, and their composites act on $b_j$ through ${\tau }_j^{e_j}$ alone. No choice principle is used; all incidences are finite checks in the fixed standard picture.