, a single reduced OU tangle $T$, and a chain of generators of $\vcalPB_n$ where each divides the quotient of $T$ by all the previous ones. Finally, a diamond in $\calX_n$ is determined by a single $\beta\in\vcalPB_n$, a single $T\in\vcalROU_n$, two chains Alternatively, for $X=(\beta,T)\in\calX$ and $\gamma$ a virtual braid, set $X^\gamma=(\beta\gamma,\gamma^{-1}T)$, and then all the $\to$-related pairs in $\calX_n$ are of the form $(\beta,T)\to(\beta,T)^{\sigma_{ij}^{\pm 1}} \quad\text{provided}\quad \sigma_{ij}^{\pm 1}\mid T.$ ===== recycled on Mon Jul 13 20:46:01 EDT 2020 by drorbn on Ubuntu-on-X2 ====== Let $I$ denote both the the 0-crossing pure virtual braid on $n$ strands (the identity element of $\vcalPB_n$) and the 0-crossing OU tangle on $n$ strands. By (1) of the Division Lemma (\ref{lem:divquo}), if $\beta',\beta''\in\vcalPB_n$ are virtual braids and $g=\sigma_{ij}^{\pm 1}$ is a generator of $\vcalPB_n$ then either $(\beta'g,\Ch(\beta'')) \to (\beta',g\Ch(\beta'')) = (\beta',\Ch(g\beta''))$ (if $g^{-1}\mid\Ch(\beta'')$) or $(\beta',\Ch(g\beta'')) \to (\beta'g,\Ch(\beta''))$ (if $g\mid g\Ch(\beta'')$), and so by induction on the length of a presentation of $\beta\in\vcalPB_n$, $(\beta,I)$ and $(I,\Ch(\beta))$ are in the same connected component of $\calX_n$. Hence $f(I,\Ch(\beta)) = f(\beta,I) = (\beta,I)$, by the Diamond Lemma and as $\xi(I)=0$ implies that $(\beta,I)$ is final. ===== recycled on Mon Jul 13 20:52:40 EDT 2020 by drorbn on Ubuntu-on-X2 ====== Let $I$ denote both the the 0-crossing pure virtual braid on $n$ strands (the identity element of $\vcalPB_n$) and the 0-crossing OU tangle on $n$ strands. By (1) of the Division Lemma (\ref{lem:divquo}), if $\beta',\beta''\in\vcalPB_n$ are virtual braids and $g=\sigma_{ij}^{\pm 1}$ is a generator of $\vcalPB_n$ then either $(\beta'g,\Ch(\beta'')) \to (\beta',g\Ch(\beta'')) = (\beta',\Ch(g\beta'')) \qquad\text{if }g^{-1}\mid\Ch(\beta'')$ or $(\beta',\Ch(g\beta'')) \to (\beta'g,\Ch(\beta'')) \qquad\text{if }g\mid g\Ch(\beta''),$ and so by induction on the length of a presentation of $\beta\in\vcalPB_n$, $(\beta,I)$ and $(I,\Ch(\beta))$ are in the same connected component of $\calX_n$. Hence $f(I,\Ch(\beta)) = f(\beta,I) = (\beta,I)$, by the Diamond Lemma and as $\xi(I)=0$ implies that $(\beta,I)$ is final.