The first four-band comparison is a compensated band stabilization

Zero widths and complete embeddings. If $d=0$, $Y=Y_1=1$, $R=V=1$, $H=X$, $U=X^{-1}$, and all primed boxes are their unprimed counterparts; therefore $\beta =\alpha$ and the asserted equality and count hold. Now assume $d\ge 1$. All words are defined: $A_0,C_0$ fit inside the last $\mathrm{\ell }$ old strands after the first $b$, $B_0,D_0$ fit after the first $a$, their primed versions are shifted by $d$, $H$ uses the first $a+b+d$ strands, and $U$ uses the first $a+b+2d$ strands. The unshifted part of $K$ uses only the $\mathrm{\ell }$ old strands and $Q_{d,c}$ uses the last $c+d$ old strands, so $K\in B_n$. When $c=0$, $Y=Y_1=1$ and $Q_{d,c}=1$, with no padding or generator indexed zero; all subsequent products on the $c$-block are empty and all corresponding individual-generator rows are absent.

Substitution rules used below. For $Q_{p,q}$ on $p+q$ generators write $P=x_1\cdots x_p$. Its images are $P{x}_{p+i}{P}^{-1}$ for input $x_i$, $1\le i\le q$, and $x_j$ for input $x_{q+j}$, $1\le j\le p$. Induct on the number of rows $q$, adjoining a fixed generator each time: the next descending row sends $x_{q+1}$ to $(x_{q+1}\cdots x_{q+p})x_{q+p+1}(x_{q+1}\cdots x_{q+p}{)}^{-1}$ and $x_{q+1+j}$ to $x_{q+j}$; the preceding substitution sends that parenthesized product to $P$. This proves every individual-generator formula, including $p=0$ or $q=0$. Equivalently, in the ordered input basis $(f_1,\dots ,f_q,g_1,\dots ,g_p)$ and output names $(x_1,\dots ,x_p,y_1,\dots ,y_q)$, the positive interchange sends $f_i$ to $P{y}_i{P}^{-1}$ and $g_j$ to $x_j$; its inverse sends each $x_j$ to $g_j$ and each $y_i$ to $G^{-1}{f}_{i}G$, $G=g_1\cdots g_p$. The names on each side designate the corresponding consecutive indices of the same free basis, rather than a change of generator sign. For $r\ge 1$, $T_r$ conjugates each of its $r$ free generators by their product. Indeed for $\delta ={\sigma }_1\cdots {\sigma }_{r-1}$, $\delta (x_j)=x_1{x}_{1+(j<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>r)}{x}_1^{-1}$; setting $P_h=x_1\cdots x_h$, 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 $0\le h\le r$, since $\delta (P_h)=P_{h+1}{x}_1^{-1}$. At $h=r$ this is conjugation by the full product. Every Artin generator fixes the full boundary product, so all local words do too. These rules follow solely from [F3], and apply after any displayed shift.

The upper routing identity on all generators. Put $R_B={\iota }_a(Q_{b+c+d,d})$ and $Z=Q_{a,d}{Q}_{d,a}{\iota }_a(T_d)$. We prove $RH{\iota }_a(T_d)=X{R}_{B}Z$. Within this substitution calculation, capitals denote products of free generators, not the braid boxes. For this comparison label the input consecutive blocks by sizes $(a,d,b,c,d)$ and label the output free generators by $(b,a,c,d,d)$ as $b_i,a_j,c_h,d_j,e_j$, with products $B,A,C,D,E$ and $P=ACD$. The domain rows consist respectively of $x_j$, $x_{a+j}$, $x_{a+d+i}$, $x_{a+d+b+h}$ and $x_{n+j}$ in their specified ranges. These are explicit output index assignments: for example output $a_j=x_{b+j}$, $c_h=x_{b+a+h}$, $d_j=x_{b+a+c+j}$, and $e_j=x_{n+j}$. Both sides send an input generator in the first $a$-block to $BPE{P}^{-1}{a}_{j}P{E}^{-1}{P}^{-1}{B}^{-1}$; a generator in the next $d$-block to $BPE{e}_j{E}^{-1}{P}^{-1}{B}^{-1}$; a generator in the following $b$-block to $b_i$; and the final $c,d$ blocks to $c_h,d_j$ respectively. Here is the full substitution verification. By step 1.2, $H$ sends the first $a$ generators to $BF{a}_j{F}^{-1}{B}^{-1}$, the next $d$ to $B{f}_j{B}^{-1}$, and the next $b$ to $b_i$, in the intermediate order $(b,d,a,c,d)$ with middle product $F$; it leaves the last two blocks in place. The rightmost $T_d$ first conjugates that input $d$-block by its own product, and $R$ then sends $f_j$ to $P{e}_j{P}^{-1}$ and the remaining $\mathrm{\ell }$ intermediate generators to the corresponding $a,c,d$ generators, giving the stated five rows. On the other side the monodromy $Q_{a,d}{Q}_{d,a}$ sends $a_j$ to $AF{A}^{-1}{a}_{j}A{F}^{-1}{A}^{-1}$ and $f_j$ to $A{f}_j{A}^{-1}$; consequently $Z$ sends $f_j$ to $AF{f}_j{F}^{-1}{A}^{-1}$ while its $a$-row is unchanged. The word $R_B$ sends $f_j$ to $S{e}_j{S}^{-1}$, $S=BCD$, and the remaining $(b,c,d)$ generators to those same output blocks. Finally $X$ sends each current $a_j$ to $B{a}_j{B}^{-1}$ and each current $b_i$ to $b_i$ in the output order $(b,a)$, fixing $c,d,e$; in particular $X(AS)=BACD=BP$. Substituting gives exactly the same five rows above, including the full internal $d$-block conjugation. Faithfulness gives the asserted routing identity.

The compensated lower routing identity on all generators. We prove $RH{\iota }_a(T_d)U{R}^{-1}=KV{K}^{-1}$. Capitals again denote free-generator products in this calculation. Use this time the ordered free basis $(b_i,a_j,d_j,c_h,e_j)$ of sizes $(b,a,d,c,d)$, with products $B,A,D,C,E$, and set $P=ADC$, $M=D^{-1}PE$, $N=M{P}^{-1}$. Both sides fix every $b_i,c_h$, send $a_j$ to $N{a}_j{N}^{-1}$, send $d_j$ to $M{e}_j{M}^{-1}$ and send $e_j$ to $M^{-1}{d}_{j}M$. For the left side, apply the words from right to left using step 1.2. After $R^{-1}$ the intermediate order is $(b,d,a,d,c)$: each old $a,d,c$ generator is the corresponding generator in the shifted last $\mathrm{\ell }$ slots, and a new $e_j$ is $L^{-1}{f}_{j}L$, $L$ the product of those last $\mathrm{\ell }$ slots. The inverse interchange $U$ moves the $(b,d)$ group past $(a,d)$; its rule conjugates the latter by the inverse product of the relocated $(b,d)$ group. The next $T_d$ conjugates the middle $d$-block by its product. After applying $H$, in the intermediate order $(b,d,a,d,c)$ with products $(B,G,A,F,C)$, the old $a_j$ image is $F^{-1}G{a}_j{G}^{-1}F$, the old $d_j$ image is $F^{-1}G{g}_j{G}^{-1}F$, the old $c_h$ is $c_h$, and the new image is $L_1^{-1}{f}_j{L}_1$, $L_1=F^{-1}GAFC$; the $b$-row is fixed. These follow by cancellation of the conjugating $(B,F)$ product against the leading $B$ in the $H$ substitutions of step 2.1. The final $R$ sends $g_j$ to $P{e}_j{P}^{-1}$, each $f_j$ to $d_j$, and the remaining $a,c$ generators to themselves. It sends $L_1$ to $D^{-1}(PE{P}^{-1})ADC=D^{-1}PE=M$. Substitution gives precisely the five rows asserted, since $R(F^{-1}G)=D^{-1}PE{P}^{-1}=N$. For the right side, split the old slots temporarily as $(b,a,c,d)$ with last two products $F,G$. By step 1.2, $K={\iota }_b(T_{\mathrm{\ell }}^{-1}){\iota }_{a+b}(Q_{d,c})$ sends $a_j$ to $P^{-1}{a}_{j}P$, each generator of the $c$-block to $P^{-1}D{c}_h{D}^{-1}P$, and each last-$d$ generator to $P^{-1}{d}_{j}P$; it fixes the new $e_j$ and preserves $P$. Thus $K^{-1}(a_j)=P{a}_j{P}^{-1}$, $K^{-1}(d_j)=P{g}_j{P}^{-1}$, and $K^{-1}(c_h)=P{G}^{-1}{f}_{h}G{P}^{-1}$. The packet $V=T_d{Q}_{d,d}^{-1}$ fixes $a,f$, sends $g_j$ to $e_j$, and sends $e_j$ to $E^{-1}G{g}_j{G}^{-1}E$. In particular $K(V(P))=K(AFE)=D^{-1}PE=M$. Applying $K$ to these substituted images gives $N{a}_j{N}^{-1}$, $M{e}_j{M}^{-1}$, $c_h$ and $E^{-1}{P}^{-1}D{d}_j{D}^{-1}PE=M^{-1}{d}_{j}M$, with the $b$-row fixed. All individual generators have now been checked, so faithfulness proves the identity.

Strip arbitrary boxes by whole-block naturality. The word $R$ interchanges the whole old $\mathrm{\ell }$-block with the new $d$-block. Hence [F2] gives $R{C}_1{Y}_1^{-1}{A}_1=C_0{Y}^{-1}{A}_{0}R$, because the three old words lie wholly in that $\mathrm{\ell }$-block and their primed embeddings are shifted by $d$. Insert the upper routing identity of step 2.1 into $R\beta R^{-1}$. The factor $Z$ acts only on the first $a+d$ strands, whereas $B_1,D_1$ start after that block and $Y_1$ is farther out; their smallest possible generator is at least $a+d+1$, while $Z$ has largest at most $a+d-1$. Thus every letter of $Z$ commutes with every letter, including inverse letters, of $B_1{Y}_1{D}_1$. Next $R_B$ interchanges the entire old $(b+c+d)$ block after the first $a$ with the new $d$ block, so [F2] gives $R_B{B}_1{Y}_1{D}_1=B_{0}Y{D}_0{R}_B$. The calculation is therefore $R\beta R^{-1}=C_0{Y}^{-1}{A}_{0}X{B}_{0}Y{D}_0[R_{B}ZU{R}^{-1}]$. By steps 2.1 and 3.1, the bracket equals $X^{-1}KV{K}^{-1}$. This yields exactly $\alpha KV{K}^{-1}$. Every box is arbitrary on its stated whole block; no full twist on only part of a box is commuted through it.

The ordinary move sequence and all boundaries. Conjugate $\alpha$ in $B_n$ by $K^{-1}$. By the reverse of [F4], append $V$ to this arbitrary surrounding braid by exactly $d$ negative ordinary stabilizations: $K^{-1}\alpha K\mapsto (K^{-1}\alpha K)V$. Conjugate by $K$ and then by $R^{-1}$ to obtain $\beta$ using step 4.1. The strand number rises from $n$ to $n+d$; $K$ remains in the original $B_n$ and every intermediate use of [F4] has its selected last $d$ old strands and $d$ new strands. At $d=1$ all full twists on the single copied band are $1$, but the necessary old full twist in $K$ on $\mathrm{\ell }$ strands is retained. At $c=0$, the $c$ rows and cap words disappear, and every substitution and naturality argument above still holds with $C=1$ as a free-product name; this does not set the arbitrary braid box $C$ to identity. At $d=0$ use step 1.1, with no move and no nonexistent generator. These distinctions cover all stated widths and trivial or nontrivial boxes. AC is inherited only from the faithful-action and packet suppliers; no closure-equivalence or Markov theorem is used.