The second four-band comparison is a compensated band destabilization

Placements and zero widths. If $d=0$, $Y_1=V=1$, $F=Q_{b,a}$, $U=F^{-1}$, $K=1$, and $\alpha =\beta$. Assume $d\ge 1$. The boxes $A_1,B_1$ fit in $B_n$, while $C_{\mathrm{i}\mathrm{n}},D_{\mathrm{i}\mathrm{n}}$ use the first $a+d,b+d$ strands. Their largest generator is at most $a+d-1,b+d-1$, respectively, whereas $Y_1$ starts at generator $a+b+d+1$; hence both commute with $Y_1$. The word $F$ uses the first $a+b+d$ strands, and $U$ the first $a+b+2d$ strands of $B_{n+d}$. Both factors of $K$ fit in $B_n$, and $V$ occupies the last $d$ old and $d$ new strands. If $c=0$, $Y_1$ and the second factor of $K$ are identity; all $c$-rows below are absent.

Complete substitution rules. On ordered input blocks $(f_1,\dots ,f_q,g_1,\dots ,g_p)$, the action of $Q_{p,q}$ has output blocks $(x_1,\dots ,x_p,y_1,\dots ,y_q)$ and images $f_i\mapsto X{y}_i{X}^{-1}$, $g_j\mapsto x_j$, where $X=x_1\cdots x_p$. Its inverse sends $x_j\mapsto g_j$, $y_i\mapsto G^{-1}{f}_{i}G$, $G=g_1\cdots g_p$. To verify these rules, adjoin the descending row for the next input $f_i$: that row conjugates its last generator by the intervening consecutive product and shifts the other generators one place; induction on $q$ sends that product to $X$. This checks each individual input generator, with empty blocks giving identity. The action of $T_r$ conjugates all $r$ generators by their ordered product. For $\delta ={\sigma }_1\cdots {\sigma }_{r-1}$ and $P_h=x_1\cdots x_h$, direct substitution gives $\delta (P_h)=P_{h+1}{x}_1^{-1}$; induction gives ${\delta }^h(x_j)=P_h{x}_{1+((j+h-1)<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>r)}{P}_h^{-1}$ for $0\le h\le r$. At $h=r$ this proves the assertion. Each local word fixes the product of all its generators.

The routing identity on every generator. We prove $J:=F{Y}_{1}U{Y}_1^{-1}=KV{K}^{-1}$. Within this calculation capitals denote products of free generators, not braid boxes. Use the consecutive numeric basis $(b_i,d_j,a_j,c_h,e_j)$ of sizes $(b,d,a,c,d)$, and write its products $B,D,A,C,E$. Thus $b_i=x_i$, $d_j=x_{b+j}$, $a_j=x_{b+d+j}$, $c_h=x_{b+d+a+h}$ and $e_j=x_{n+j}$. Put $N=C{E}^{-1}{C}^{-1}D$. Both sides fix $b_i,c_h$ and have the following remaining individual images: $d_j⟼C{e}_j{C}^{-1},a_j⟼N{a}_j{N}^{-1},e_j⟼E^{-1}{C}^{-1}D{d}_j{D}^{-1}CE\mathrm{.}$ Here is the full left-side substitution. Apply the factors from right to left by step 1.2. After $Y_1^{-1}$ the ordered intermediate blocks are $(B,D,A,G,C)$; the $c$-row is unchanged and each new $e_j$ has image $C^{-1}{g}_{j}C$. The next $U$ gives order $(A,G,B,D,C)$: $b_i,d_j$ move without conjugation, while $a_j$ maps to $(BD{)}^{-1}{a}_j(BD)$ and the new row to $C^{-1}(BD{)}^{-1}{g}_j(BD)C$. The factor $Y_1$ interchanges the last $(D,C)$ blocks, giving order $(A,G,B,C,D)$, sends $d_j$ to $C{d}_j{C}^{-1}$ and fixes $c_h$, so the intervening product $BD$ becomes $BCD{C}^{-1}$. Finally $F=H{\iota }_a(T_d)$ maps the first blocks $(A,G,B)$ to $(B,D,A)$: first $T_d$ conjugates each $g_j$ by $G$, and $H$ sends $a_j$ to $BG{a}_j{G}^{-1}{B}^{-1}$, $g_j$ to $B{g}_j{B}^{-1}$ and $b_i$ to $b_i$. The final output names are $(B,D,A,C,E)$, so the intermediate last $D$ is now numeric $E$, and the intermediate $G$ is numeric $D$. Substitute into all five rows. The old $d_j$ becomes $C{e}_j{C}^{-1}$; the $a_j$ conjugator reduces to $C{E}^{-1}{C}^{-1}D=N$; the new row reduces to $E^{-1}{C}^{-1}D{d}_j{D}^{-1}CE$; $b_i,c_h$ stay fixed.

The other side on the same indexed basis. Temporarily name the old input slots of $K$ by $(B,A,C,G)$, with final $d$-block product $G$, and keep the new block $E$. Step 1.2 gives $K(a_j)=D{a}_j{D}^{-1}$, $K(c_h)=c_h$, $K(g_j)=C^{-1}{d}_{j}C$, with the first $b$ and new $e$ rows fixed. Its inverse therefore sends numeric $d_j$ to $C{g}_j{C}^{-1}$, numeric $a_j$ to $C{G}^{-1}{C}^{-1}{a}_{j}CG{C}^{-1}$, and numeric $c_h$ to $c_h$. The packet $V$ sends $g_j$ to $e_j$, sends $e_j$ to $E^{-1}G{g}_j{G}^{-1}E$, and fixes $a_j,c_h,b_i$. Applying $K$ last gives $d_j\mapsto C{e}_j{C}^{-1}$, $a_j\mapsto (C{E}^{-1}{C}^{-1}D)a_j(C{E}^{-1}{C}^{-1}D{)}^{-1}$, and $e_j\mapsto E^{-1}{C}^{-1}D{d}_j{D}^{-1}CE$, with the other two rows fixed. These are all images of the same numeric basis used in step 2.1, not merely block products or permutations. Faithfulness now proves $J=KV{K}^{-1}$.

Arbitrary boxes and the ordinary sequence. By step 1.1, $Y_1$ commutes with $D_{\mathrm{i}\mathrm{n}},C_{\mathrm{i}\mathrm{n}}$. Direct cancellation gives $Y_1\beta Y_1^{-1}=D_{\mathrm{i}\mathrm{n}}{A}_{1}F{B}_1{C}_{\mathrm{i}\mathrm{n}}{Y}_{1}U{Y}_1^{-1}=\alpha (F{Y}_{1}U{Y}_1^{-1})=\alpha KV{K}^{-1}\mathrm{.}$ This uses no commutation with $A_1,B_1$ and no partial-band assumption about any box. Conjugate $\beta$ by $Y_1$ and then by the old-strand word $K^{-1}$; the result is $(K^{-1}\alpha K)V$. Apply [F4] to this arbitrary surrounding braid, removing exactly $d$ last new strands by negative ordinary destabilizations and conjugations, and conjugate by $K$ to obtain $\alpha$. The strand number decreases from $n+d$ to $n$. For $d=1$ only $T_1=1$ is used, with $K$ retained. For $c=0$ the free-product name $C$ is the empty product, although the arbitrary box $C\in B_{a+d}$ is retained throughout the displayed braid equality. At $d=0$ use step 1.1. AC enters solely through [F3], [F4]; no Markov equivalence theorem is a supplier.