Ko atomic braid relations are exactly the finite BH rank-two relations

By [F5], the words in [F2] are the two sides of Ko's relation for $(T,a,\bar{b})$. Their common minimum is the same finite-relative-longest element that [F3] uses for $v_{a,b,I}$. By [F1] they give standard expressions, with distinct initial labels $v_{b,I}$ and $v_{a,I}$. Since [F3] supplies exactly two standard expressions, these are that pair, up to interchanging the two sides. Their equality is therefore precisely the BH atomic braid relation.

Conversely assume the BH relation has finitary $I$, and put $T=I\bigsqcup \{a,b\}$. The group $W_I$ is finite. Its distinct simple generators indexed by $I$ lie in $W_I$, so $I$ is finite, and hence $T$ is finite. Apply [F4] to $I$ to obtain finite ${\mathrm{\Phi }}_I$. The BH hypothesis in [F3] gives finite ${\mathrm{\Phi }}_T\setminus {\mathrm{\Phi }}_I$, so ${\mathrm{\Phi }}_T={\mathrm{\Phi }}_I\cup ({\mathrm{\Phi }}_T\setminus {\mathrm{\Phi }}_I)$ is finite. Applying [F4] to the finite vertex set $T$ proves that $W_T$ is finite. The rotation construction in [F2] now applies with $s=a,t=\bar{b}$, and step 1.1 recovers the given BH relation. Thus no finitary BH relation is omitted.