The classical Markov theorem tells us that the closures of two braids are isotopic links if and only if the braids are related by a sequence of Markov moves
(MI) $b \sim aba^{-1} $
(MII) $b \sim b \sigma_n^{\pm 1}$
where $a,b \in B_n$ and $\sigma_n^{\pm 1} \in B_{n+1}$ is the usual Artin generator.
I am interested in studying not links but rather tangles that are produced by some partial closure of a braid, and I would like to decide when two braids give rise to the same tangle in this fashion.
Question: Is this known?
More particularly, given $b \in B_{k+n}$, let $\mathrm{cl}_n(b)$ be the tangle with $k$ open components obtained by closing up the rightmost $n$ strands of $b$. Now, given braids $b \in B_{k+n}$, $b' \in B_{k+m}$, when are the tangles $\mathrm{cl}_n(b)$ and $\mathrm{cl}_m(b')$ isotopic?
Surely the following versions of (MI) and (MII) must belong to the "partial closure Markov moves" (here I'm leaving the number of open components $k$ fixed):
(MI) $b \sim (1_k \otimes a)b(1_k\otimes a^{-1}) $
(MII) $b \sim b \sigma_n^{\pm 1}$
where $a \in B_n$, $1_k \in B_k$ is the unit and $\otimes$ denotes the concatenation of braids.
Real question: Are these two moves enough to generate the desired equivalence relation?
I feel that there are some moves that I am missing...