Band exchanges decompose into ordinary Markov moves

Empty bands and conventions. If $p=0$ or $q=0$, both interchanges are $1$ by [F3], and both displayed endpoints are literally $PQ$ in the same group; no padding of a nonexistent band is attempted. Assume henceforth $p,q>0$. All products below are braid-group products with the action convention in [F4]. Core strands are fixed by the local words, whose indices are shifted by $c$. Descending and ascending products of length zero are $1$.

The equal-width braid relation. First suppose $p=q=m$. In $B_{c+3m}$ put $t=Q_{m,m}$ on blocks 1,2 and $s=Q_{m,m}$ on blocks 2,3. Let their free generators be $x_j,y_j,z_j$, $1\le j\le m$, and write $a=x_1\cdots x_m$, $b=y_1\cdots y_m$. The rectangular row induction from the packet proof gives $\rho (t)(x_j)=a{y}_j{a}^{-1}$, $\rho (t)(y_j)=x_j$ and fixes $z_j$; for $s$ it gives $\rho (s)(y_j)=b{z}_j{b}^{-1}$, $\rho (s)(z_j)=y_j$ and fixes $x_j$. To recall that induction explicitly, for $Q_{m,k}$ on $m+k$ strands put $a=x_1\cdots x_m$; its images are $a{x}_{m+i}{a}^{-1}$ for $1\le i\le k$ and $x_j$ for the input generator $x_{k+j}$, $1\le j\le m$. At $k=0$ this is identity. Adjoin a fixed generator and multiply by the next descending row: that row sends $x_{k+1}$ to $(x_{k+1}\cdots x_{k+m})x_{k+m+1}(x_{k+1}\cdots x_{k+m}{)}^{-1}$ and $x_{k+1+j}$ to $x_{k+j}$ for $1\le j\le m$; the previous substitution sends the parenthesized product to $a$. This proves the induction and the displayed equal-block formulas at $k=m$. Now both $tst$ and $sts$ send $x_j$ to $ab{z}_j{b}^{-1}{a}^{-1}$, $y_j$ to $a{y}_j{a}^{-1}$ and $z_j$ to $x_j$; they fix the core generators. For example $t(a)=ab{a}^{-1}$ and $t(b)=a$, giving the first image in $tst$; the other substitutions follow directly. Faithfulness in [F4] proves $tst=sts$ for every $m\ge 1$, including $m=1$.

Padding one strand is an ordinary stabilization. Suppose $p<q$, and allow either crossing sign: set $v_{+}=Q_{p,q}$ or $v_{-}=Q_{q,p}^{-1}$, shifted by $c$. At the cyclic cut after the crossing, $P{v}_{ϵ}Q{v}_{ϵ}^{-1}$ is conjugate to $Q{v}_{ϵ}^{-1}P{v}_{ϵ}$. Add one strand after the $p$-band, set $P^{+}=P{\sigma }_{c+p}$ and replace the crossing by $v_{+}^{+}=Q_{p+1,q}$ or $v_{-}^{+}=Q_{q,p+1}^{-1}$. Then $v_{ϵ}^{+}=R_{ϵ}{v}_{ϵ}$, where $R_{+}={\sigma }_{c+p+1}\cdots {\sigma }_{c+p+q}$ and $R_{-}={\sigma }_{c+p+1}^{-1}\cdots {\sigma }_{c+p+q}^{-1}$. For the positive word, interleave the $j$th letter of $R_{+}$ before the $j$th row of $Q_{p,q}$: it commutes past every earlier old-row letter, whose maximum index is $c+p+j-2$, at distance at least two. This makes precisely the rows of $Q_{p+1,q}$. For the negative word, split the last descending row of $Q_{q,p+1}$ and invert; its inverse is exactly $R_{-}$. The largest index in $P$ is $c+p-1$, so $P$ commutes with either $R_{ϵ}$. Whole-block naturality [F3] gives ${\sigma }_{c+p}{v}_{ϵ}^{+}=v_{ϵ}^{+}{\sigma }_n$, because this generator acts within the enlarged first block of width $p+1$. Consequently $Q(v_{ϵ}^{+}{)}^{-1}{P}^{+}{v}_{ϵ}^{+}=Q{v}_{ϵ}^{-1}P{v}_{ϵ}{\sigma }_n$. The left is the padded braid at its cyclic cut, and the right is exactly one positive ordinary stabilization of the original braid at its cyclic cut. The box $Q$ remains on the first $c+q$ old strands. Thus padding changes no component by a tensor identity; it is an actual Markov stabilization with explicit word equality.

A compensated equal-width exchange. For $m=1$, [F5] supplies the sequence directly, with the trivial full twist $T_1=1$. The following calculation covers $m\ge 2$. Write $z_i$ for the positive full twist $T_m$ on block $i$, $i=1,2,3$. The boxes $P,Q\in B_{c+m}$ commute with $s$, since their largest generator is $c+m-1$ and the smallest generator of $s$ is $c+m+1$. They also commute with $z_2,z_3$. By [F3], $s{z}_3=z_{2}s$ and $t^2{z}_2=z_2{t}^2$; these transport twists on entire blocks and do not pass a partial twist through a box. Put $A=P{t}^2$, $B=t^{-1}Q{t}^{-1}$ and $X=t^2{\beta }^{\mathrm{\prime }}{t}^{-2}$. Then $BA=A^{-1}\beta A$. Rearranging $sts=tst$ from step 1.2 gives $t^{-1}st=st{s}^{-1}$ and $s{t}^{-1}{s}^{-1}=t^{-1}{s}^{-1}t$; consequently $t{s}^{-1}{t}^{-1}s=t^2{s}^{-1}{t}^{-1}$ and $s^2{t}^{-1}{s}^{-1}=t^{-1}{s}^{-1}{t}^2$. These identities and the box commutations give $E_1=A{s}^{-1}B=Pt{s}^{-1}{t}^{-1}sQ{t}^{-1}$ and $t^{2}s{E}_1{s}^{-1}{t}^{-2}=X{s}^{-1}$; explicitly $s{E}_1{s}^{-1}=P{t}^{-1}Q{t}^{-1}{s}^{-1}{t}^2$ using $t{s}^{-1}{t}^{-1}s=t^2{s}^{-1}{t}^{-1}$ and $s^2{t}^{-1}{s}^{-1}=t^{-1}{s}^{-1}{t}^2$. Conjugate $\beta$ to $BA$ and apply the reverse mirrored packet of [F2], producing $BA{s}^{-1}{z}_2=BA{z}_3{s}^{-1}$ by $m$ negative ordinary stabilizations. Conjugate by $B^{-1}$ to $A{z}_3{s}^{-1}B$. Since $A$ is an old-strand braid, it commutes with $z_3$. Conjugation by $t^{2}s$ therefore yields $z_{2}X{s}^{-1}$: move $z_3$ left past $A$, use $s{z}_3=z_{2}s$ and $t^2{z}_2=z_2{t}^2$, and use the untwisted equality above. Conjugate next by $z_2^{-1}$ to $X{s}^{-1}{z}_2$, exactly a mirrored compensated packet with arbitrary surrounding braid $X$. Its $m$ negative ordinary destabilizations give $X$, and conjugation by $t^{-2}$ gives ${\beta }^{\mathrm{\prime }}$. This completes the equal-width exchange with every box retained. Reversing this sequence gives the reverse exchange with the same numbers and signs.

Unequal widths and move counts. For $0<p<q$, repeat step 1.3 exactly $\delta =q-p$ times. At stage $j$, the enlarged box lies in $B_{c+p+j}$, the other box stays in $B_{c+q}$, and the ordinary added generator is ${\sigma }_{c+p+j+q}$; hence every use satisfies the same support inequalities. The two padded endpoints have equal band width $m=q$. Step 2.1 relates them with $m$ negative stabilizations and $m$ negative destabilizations; apply step 1.3 backwards for the negative-crossing endpoint to remove the $\delta$ positive kinks. For $p>q>0$, cyclically start with $Q$ instead: the first endpoint becomes $Q{t}^{-1}Pt$ on block order $(q,p)$, and the second becomes $QuP{u}^{-1}$. These are respectively the negative and positive endpoints with the smaller first band $q$; pad that first band by step 1.3 and use the reverse of step 2.1 before removing the padding. For $p=q$ no padding is needed. Thus for every positive pair of widths the stated $\delta$ positive and $m$ negative stabilizations and matching destabilizations suffice, with conjugations accounted separately. Zero widths were settled in step 1.1, so all permitted cases are covered. AC is used only through the faithful-action and packet suppliers; no Markov closure-equivalence theorem is invoked.