Ordinary exchange moves are Markov sequences

Supports and mixed relations. Append strand $n+1$ and write $s={\sigma }_n$. Every letter of $P,Q$ has index at most $n-2$, so both boxes commute with $s$ and $s^{-1}$, including when $n=2$ and the boxes are trivial in $B_1$. From $sts=tst$ obtain $t^{-1}st=st{s}^{-1}$, hence $t{s}^{-1}{t}^{-1}s=t^2{s}^{-1}{t}^{-1}$. Also $s{t}^{-1}{s}^{-1}=t^{-1}{s}^{-1}t$ by inversion and rearrangement of the same relation, and therefore $s^2{t}^{-1}{s}^{-1}=t^{-1}{s}^{-1}{t}^2$: indeed $s^2{t}^{-1}{s}^{-1}=s{t}^{-1}{s}^{-1}t=t^{-1}{s}^{-1}{t}^2$, where the first equality follows from $s{t}^{-1}{s}^{-1}=t^{-1}{s}^{-1}t$. These identities hold without moving either box across $t$.

Stabilization and the first weaving. Put $A=P{t}^2$ and $B=t^{-1}Q{t}^{-1}$ in $B_n$. Then $BA=A^{-1}\beta A$. Conjugate $\beta$ to $BA$, negatively stabilize to $BA{s}^{-1}$, and conjugate by $B^{-1}$ to $A{s}^{-1}B$. The latter equals $E_1=Pt{s}^{-1}{t}^{-1}sQ{t}^{-1}$ by step 1.1. Thus $E_1$ is reached by exactly one negative ordinary stabilization and conjugations. With Figure 11's ports numbered from outermost to innermost, the old innermost strand is $n$ and the added inner strand is $n+1$; the two boxes occupy the first $n-1$ old ports. The successive crossing letters of the right-hand weaving are $t,s^{-1},t^{-1},s$, followed after $Q$ by $t^{-1}$, giving precisely the chronological record $E_1$. For these ordinary strands the indicated full twist on a single added strand is $1$.

The exchange calculation and the last weaving. Define $E_5=P{t}^{-1}Qs{t}^{-1}{s}^{-1}t$. Its successive left-hand weaving letters read chronologically, with the same ports and cut, are $s,t^{-1},s^{-1},t$, after the initial $t^{-1}$ and $Q$. Since $P,Q$ commute with $s$, use step 1.1 to compute $s{E}_1{s}^{-1}=P{t}^{-1}Q{s}^2{t}^{-1}{s}^{-1}=P{t}^{-1}Q{t}^{-1}{s}^{-1}{t}^2=E_5$. This gives an explicit conjugation from the first weaving to the last; it proves the comparison without presuming any unverified intermediate diagram arrow. Conjugating $E_5$ by $t^2$ now gives $(t^{2}P{t}^{-1}Q{t}^{-1})s^{-1}=(t^2{\beta }^{\mathrm{\prime }}{t}^{-2})s^{-1}$. Its parenthesis belongs to $B_n$, so an ordinary negative right destabilization deletes $s^{-1}$; conjugating by $t^{-2}$ yields ${\beta }^{\mathrm{\prime }}$. No later Garside or Markov theorem is used.

Signs and endpoints. Steps 2.1-3.1 use exactly one negative stabilization and one negative destabilization. The map ${\sigma }_i\mapsto {\sigma }_i^{-1}$ preserves both Artin relations and all strand placements, so it is an involutive automorphism. Apply the negative sequence to the mirrored boxes and then mirror every word and move: it gives a positive sequence from ${\beta }^{\mathrm{\prime }}$ to $\beta$. Reversing that sequence gives the asserted positive sequence from $\beta$ to ${\beta }^{\mathrm{\prime }}$. All groups and conjugators are explicit finite words, so the proof is choice-free. At $n=2$ the same formulas hold with $P=Q=1$; no ${\sigma }_0$ is used.