The four-band case is a Markov sequence

The two multiple reductions. The specified peak is $X\leftarrow Y\to Z$, with the two first ordinary reducing moves fixed as part of the given data. In Traczyk's Figure 7, one uniform multiple reduction begins with the specified move along ${\alpha }_r$ to $X$ and is completed by the uniform reduction along ${\alpha }_p$; the other begins with the specified move along ${\alpha }_s$ to $Z$ and is completed along ${\alpha }_u$. These are distinct arc pairs, rather than merely the same two moves in reversed order. They give the two height-zero diagrams $X^{\mathrm{\prime }},Z^{\mathrm{\prime }}$ shown in the bottom of Traczyk's Figure 7 and in Birman--Brendle's Figure 12, printed p. 26. We must compare their braid words, preserving the source's uniform over/under convention and arbitrary internal boxes.

Chronological records and oriented frames. The geometric product runs its rightmost factor first. In the source picture we may list crossings and boxes chronologically along the oriented braid; call this list its chronological record. The actual braid-group element is its word reversal $\mathrm{rev}$, with each individual generator's sign unchanged. Reversal preserves the Artin relations, which are palindromic or far commutations, so it defines an involutive anti-automorphism. It takes a conjugation by $g$ to one by $\mathrm{rev}(g{)}^{-1}$, and takes a right stabilization to a left stabilization; cyclic conjugation converts the latter to a right stabilization of the same sign. Hence any ordinary Markov sequence for chronological records gives one for the actual elements with the same strand-changing counts. Moreover $\mathrm{rev}(Q_{p,q})=Q_{q,p}$: label a cell of $Q_{p,q}$ by row $j=1,\dots ,q$ and entry $i=1,\dots ,p$, with index $p+j-i$. After reversal put $I=p-i+1$, $J=q-j+1$; the entry is ${\sigma }_{q+I-J}$, ordered first by increasing $J$ then $I$. The rows of $Q_{q,p}$ order this same grid first by $I$ then $J$. The pairs that change order have $I>I^{\mathrm{\prime }}$ and $J<J^{\mathrm{\prime }}$, with index difference $(I-I^{\mathrm{\prime }})+(J^{\mathrm{\prime }}-J)\ge 2$, so only far commutations are needed. Empty widths are included. At a Figure 8 cut we use increasing radial coordinate and positive depth as the transverse frame $(e_r,e_Z)$, so ports are numbered inner to outer. The clockwise braid tangent is $-e_{\theta }$, and $(e_r,e_Z,-e_{\theta })$ preserves ambient orientation. At a Figure 11 cut we instead use $(-e_r,-e_Z)$, so ports are numbered outer to inner; the two transverse frames differ by an orientation-preserving half rotation, not a reflection. Positive anticlockwise half twists therefore retain the frozen generator convention: the first inner port passes through negative physical depth in the former frame, while the first outer port passes through positive physical depth in the latter. All chronological formulas below are read in their stated frame and are explicitly reversed to interpret them as actual braid-group words.

The ordinary exchange comparison. If the ambiguity has the ordinary exchange form of [F3], put $t={\sigma }_{n-1}$ and $s={\sigma }_n$. Its first and last weaving words are $E_1=Pt{s}^{-1}{t}^{-1}sQ{t}^{-1}$ and $E_5=P{t}^{-1}Qs{t}^{-1}{s}^{-1}t$. The supplier proves $s{E}_1{s}^{-1}=E_5$ using only the adjacent Artin relation and the commutation of $P,Q$ with $s$. It also writes $E_1=A{s}^{-1}B$ with $A=P{t}^2$, $B=t^{-1}Q{t}^{-1}$ and $BA=A^{-1}(PtQ{t}^{-1})A$, and $t^2{E}_5{t}^{-2}=(t^2(P{t}^{-1}Qt)t^{-2})s^{-1}$. These expose an ordinary stabilization and destabilization with both arbitrary boxes retained. This verifies the ordinary Figure 11 exchange. The multiple-band case is supplied by [F8], with the source endpoint trace below.

The right-column arbitrary-box slide. Retain the first comparison's placements and put $F=H{\iota }_a(T_d)$, $C_{\mathrm{o}\mathrm{u}\mathrm{t}}=C_1$, $D_{\mathrm{o}\mathrm{u}\mathrm{t}}=D_1$, $C_{\mathrm{i}\mathrm{n}}=C$ on the first $a+d$ strands, $D_{\mathrm{i}\mathrm{n}}=D$ on the first $b+d$ strands, and $G=Y_1^{-1}{A}_{1}F{B}_1{Y}_1$. The second panel is ${\beta }_{R2}=C_{\mathrm{o}\mathrm{u}\mathrm{t}}G{D}_{\mathrm{o}\mathrm{u}\mathrm{t}}U$ and the third panel is ${\beta }_{R3}=D_{\mathrm{i}\mathrm{n}}G{C}_{\mathrm{i}\mathrm{n}}U$. The inverse whole-block naturality [F5] gives $U{C}_{\mathrm{o}\mathrm{u}\mathrm{t}}=C_{\mathrm{i}\mathrm{n}}U$ and $D_{\mathrm{o}\mathrm{u}\mathrm{t}}U=U{D}_{\mathrm{i}\mathrm{n}}$. Also $D_{\mathrm{o}\mathrm{u}\mathrm{t}}$ and $C_{\mathrm{i}\mathrm{n}}$ are on disjoint blocks, so they commute by [F7]. Thus conjugating by $C_{\mathrm{o}\mathrm{u}\mathrm{t}}^{-1}$ gives $G{D}_{\mathrm{o}\mathrm{u}\mathrm{t}}U{C}_{\mathrm{o}\mathrm{u}\mathrm{t}}=G{C}_{\mathrm{i}\mathrm{n}}U{D}_{\mathrm{i}\mathrm{n}}$, and conjugating this by $D_{\mathrm{i}\mathrm{n}}$ gives ${\beta }_{R3}$. These are two explicit conjugations with arbitrary internal boxes; there is no strand change and no twist is commuted through part of a box.

Inversion transfers signed comparisons. Word inversion reverses product order and inverts every letter. It takes conjugate braids to conjugate braids. If a chronological right stabilization is $\gamma \mapsto \gamma {\sigma }_n^{ϵ}$, its inverse is ${\gamma }^{-1}\mapsto {\sigma }_n^{-ϵ}{\gamma }^{-1}$, and cyclic conjugation turns the latter into ${\gamma }^{-1}{\sigma }_n^{-ϵ}$. Thus inversion transfers any ordinary Markov sequence, reversing its crossing signs and keeping its strand-changing counts; reversal of the sequence additionally interchanges stabilization and destabilization. This is an assertion about words and move sequences, not an identification of a link with the closure of its inverse braid. Universal identities with arbitrary boxes may be applied to inverse boxes first and then inverted.

The Figure 11 band-exchange endpoints. Cut immediately before the upper box $P$ in the initial and final panels of Traczyk's Figure 11, and number the ports from outermost to innermost. Write their widths as $(c,p,q)$: the outer core has width $c$, the middle exchanged band width $p$, and the inner exchanged band width $q$. The upper box acts on the first $c+p$ ports. Read the right-hand crossing in the braid direction: it uniformly interchanges $(p,q)$ with $(q,p)$, so the lower box $Q$ acts on the first $c+q$ ports; the left-hand crossing restores $(p,q)$. With the first crossing positive in this outer-first frame, the initial chronological record is $P{Q}_{p,q}Q{Q}_{p,q}^{-1}$. In the final panel the two crossing signs are reversed, while the two boxes remain in the same positions with the same oriented endpoint frames. The negative first interchange with input $(p,q)$ is $Q_{q,p}^{-1}$, so the final chronological record is $P{Q}_{q,p}^{-1}Q{Q}_{q,p}$. By step 1.2, their actual words, after a cyclic cut, are respectively $\mathrm{rev}(P)Q_{q,p}^{-1}\mathrm{rev}(Q)Q_{q,p}$ and $\mathrm{rev}(P)Q_{p,q}\mathrm{rev}(Q)Q_{p,q}^{-1}$; [F8] applies in reverse direction, with the same core-plus-band supports. If the projection convention reads the first crossing negative, reverse the same comparison. By [F8] these endpoints have a full ordinary Markov sequence for arbitrary boxes and unequal widths: the after-crossing cyclic-cut identity $Q(v_{ϵ}^{+}{)}^{-1}(P{\sigma }_{c+p})v_{ϵ}^{+}=(Q{v}_{ϵ}^{-1}P{v}_{ϵ}){\sigma }_{c+p+q}$ makes padding an actual stabilization, and the equal-width computation retains the compensation through its weaving conjugation and restores it before destabilizing. Zero widths give identical endpoints by [F8]. This proves the source's complete Figure 11 exchange by an independently reproducible computation, without requiring acceptance of its intermediate arrows. The relevant reducing-choice endpoint factorization is given below.

The first right-column strand-changing arrow. Cut the initial Figure 8 right-column diagram immediately before the striped box $C$, with inner-to-outer ports $(b,a,d,c)$ in the frame of step 1.2. Put $n=a+b+c+d$, $X=Q_{b,a}$ and $Y={\iota }_{a+b}(Q_{c,d})$. The chronological record is ${\alpha }_C=C_0{Y}^{-1}{A}_{0}X{B}_{0}Y{D}_0{X}^{-1}$, exactly $\alpha$ of [F12]; it is the cyclic record before $C$ of the initial before-$A$ record. At the corresponding cut in the second right-column panel the ports are $(b,d_{\mathrm{n}\mathrm{e}\mathrm{w}},a,d_{\mathrm{o}\mathrm{l}\mathrm{d}},c)$, with total $n+d$. The striped and black boxes now begin after the first $b+d$ ports, and the grid and textured boxes after the first $a+d$ ports. The left cap remains the negative interchange $Y_1^{-1}$ of the last $(d,c)$ blocks, the upper weave is the positive reversal of $(b,d_{\mathrm{n}\mathrm{e}\mathrm{w}},a)$ given by $H=Q_{b,d+a}{Q}_{d,a}$, its $+$ square is the full $T_d$ on the middle copied band after $H$, the right cap is $Y_1$, and the lower weave is the negative uniform interchange of the two whole blocks $(a+d,b+d)$ given by $U=Q_{b+d,a+d}^{-1}$. Thus the complete chronological record is ${\beta }_{R2}=C_1{Y}_1^{-1}{A}_{1}H{\iota }_a(T_d)B_1{Y}_1{D}_{1}U$, with every shifted embedding specified in [F12]. The old and copied $d$ port labels interchange at the end of this record; they are not asserted to remain the same physical strands inside arbitrary boxes. By the full general identity of [F12], $R{\beta }_{R2}{R}^{-1}={\alpha }_{C}KV{K}^{-1}$, where $\mathrm{\ell }=a+d+c$, $R={\iota }_b(Q_{\mathrm{\ell },d})$, $K={\iota }_b(T_{\mathrm{\ell }}^{-1}){\iota }_{a+b}(Q_{d,c})$ and $V={\iota }_{n-d}(T_d{Q}_{d,d}^{-1})$. In particular the old full twist in $K$ and the reverse compensation order in $V$ are retained. Conjugating ${\alpha }_C$ by $K^{-1}$, adding this packet by [F6], and conjugating by $K$ and $R^{-1}$ gives the second panel with exactly $d$ negative ordinary stabilizations. Step 1.2 reverses this complete sequence to interpret the actual geometric words, preserving those counts and signs. For $d=0$ the words coincide; for $c=0$ the caps disappear but the same general identity applies. This verifies the first strand-changing source arrow with all boxes and frame compensation present.

The second right-column strand-changing arrow. At the third panel cut before $D_{\mathrm{i}\mathrm{n}}$, the inner-to-outer ports are $(b,d_{\mathrm{o}\mathrm{l}\mathrm{d}},a,d_{\mathrm{n}\mathrm{e}\mathrm{w}},c)$. The first inner box is $D_{\mathrm{i}\mathrm{n}}$; the left cap $Y_1^{-1}$ moves $(d_{\mathrm{n}\mathrm{e}\mathrm{w}},c)$ to $(c,d_{\mathrm{n}\mathrm{e}\mathrm{w}})$, and the outer black box $A_1$ acts on the now consecutive $(a,c)$ block. After the upper routing and the grid box, $Y_1$ returns $(c,d_{\mathrm{n}\mathrm{e}\mathrm{w}})$ to $(d_{\mathrm{n}\mathrm{e}\mathrm{w}},c)$, so $C_{\mathrm{i}\mathrm{n}}$ acts on $(a,d_{\mathrm{o}\mathrm{l}\mathrm{d}})$ and $U$ closes the record with the two $d$ blocks interchanged. The cap is the inverse of a $(c,d)$ interchange and therefore requires input $(d,c)$, also when the widths differ. The upper positive routing $H$ takes $(b,d_{\mathrm{o}\mathrm{l}\mathrm{d}},a)$ to $(a,d_{\mathrm{o}\mathrm{l}\mathrm{d}},b)$ and its positive square is ${\iota }_a(T_d)$. The grid box is $B_1$, the right cap is $Y_1$, the inner striped box is $C_{\mathrm{i}\mathrm{n}}$, and the lower uniform negative routing is $U$. Hence its complete chronological record is ${\beta }_{R3}=D_{\mathrm{i}\mathrm{n}}{Y}_1^{-1}{A}_{1}F{B}_1{Y}_1{C}_{\mathrm{i}\mathrm{n}}U$. In the bottom-right panel the same cut has only $(b,d,a,c)$: the cap crossings have disappeared and the lower three-band routing with its negative square is precisely $F^{-1}$. Its record is ${\alpha }_{R4}=D_{\mathrm{i}\mathrm{n}}{A}_{1}F{B}_1{C}_{\mathrm{i}\mathrm{n}}{F}^{-1}$. These are exactly [F13], whose old-strand conjugator is $K_2={\iota }_b(Q_{d,a}){\iota }_{a+b}(Q_{c,d}^{-1})$. Therefore the third to fourth right-panel arrow is $d$ negative ordinary destabilizations and conjugations. The source's copied and old $d$ ports interchange when this closed record returns to its cut; the boxes have the specified supports irrespective of internal permutations. Step 1.2 interprets the actual group words without changing signs or counts. At $c=0$ there are no caps; at $d=0$ the panels coincide, as [F13] proves.

Both left-column strand-changing arrows. Apply the universal right-column computations to $A^{-1},B^{-1},C^{-1},D^{-1}$ and invert their complete words, then use a cyclic cut. The old bottom-left record before $A_1$ is ${\alpha }_{L_4}=A_1{D}_{\mathrm{i}\mathrm{n}}F{C}_{\mathrm{i}\mathrm{n}}{B}_1{F}^{-1}$; the third left-panel record is ${\beta }_{L_3}=A_1{Y}_1{D}_{\mathrm{i}\mathrm{n}}{U}^{-1}{C}_{\mathrm{i}\mathrm{n}}{Y}_1^{-1}{B}_1{F}^{-1}$. These are the cyclic inverses of ${\alpha }_{R4},{\beta }_{R3}$ with inverse box inputs. Their source ports at the cut before $A_1$ are $(b,d_{\mathrm{o}\mathrm{l}\mathrm{d}},a,c,d_{\mathrm{n}\mathrm{e}\mathrm{w}})$. The left positive cap gives $(b,d_{\mathrm{o}\mathrm{l}\mathrm{d}},a,d_{\mathrm{n}\mathrm{e}\mathrm{w}},c)$; the upper positive routing $U^{-1}$ then gives $(a,d_{\mathrm{n}\mathrm{e}\mathrm{w}},b,d_{\mathrm{o}\mathrm{l}\mathrm{d}},c)$; the right negative cap gives $(a,d_{\mathrm{n}\mathrm{e}\mathrm{w}},b,c,d_{\mathrm{o}\mathrm{l}\mathrm{d}})$. The lower negative routing and negative square are $F^{-1}$ and return $(b,d_{\mathrm{n}\mathrm{e}\mathrm{w}},a,c,d_{\mathrm{o}\mathrm{l}\mathrm{d}})$. The boxes encountered are respectively outer $A_1$, inner $D_{\mathrm{i}\mathrm{n}}$, inner $C_{\mathrm{i}\mathrm{n}}$, outer $B_1$, as the displayed record specifies. By steps 1.5 and 2.3, the source arrow $L_4\to L_3$ is $d$ positive ordinary stabilizations and conjugations. Next the second left-panel record before $C_1$ is ${\beta }_{L_2}=C_1{U}^{-1}{D}_1{Y}_1^{-1}{B}_1{F}^{-1}{A}_1{Y}_1$, while the initial left-panel record before $C_0$ is ${\alpha }_{L_1}=C_{0}X{D}_0{Y}^{-1}{B}_0{X}^{-1}{A}_{0}Y$. These are the cyclic inverses of ${\beta }_{R2},{\alpha }_C$ with inverse boxes. For ${\beta }_{L_2}$ start with ports $(b,d_{\mathrm{o}\mathrm{l}\mathrm{d}},a,d_{\mathrm{n}\mathrm{e}\mathrm{w}},c)$; the upper $U^{-1}$ gives $(a,d_{\mathrm{n}\mathrm{e}\mathrm{w}},b,d_{\mathrm{o}\mathrm{l}\mathrm{d}},c)$, the right $Y_1^{-1}$ gives $(a,d_{\mathrm{n}\mathrm{e}\mathrm{w}},b,c,d_{\mathrm{o}\mathrm{l}\mathrm{d}})$, the lower $F^{-1}$ gives $(b,d_{\mathrm{n}\mathrm{e}\mathrm{w}},a,c,d_{\mathrm{o}\mathrm{l}\mathrm{d}})$, and the left $Y_1$ gives $(b,d_{\mathrm{n}\mathrm{e}\mathrm{w}},a,d_{\mathrm{o}\mathrm{l}\mathrm{d}},c)$. Thus the outer striped and textured boxes have exactly $C_1,D_1$ placements and the lower black/grid boxes exactly $A_1,B_1$ placements; all crossing signs and the negative square are those of the inverse records. By steps 1.5 and 2.2, $L_2\to L_1$ is $d$ positive ordinary destabilizations and conjugations. The $L_3\to L_2$ slide is the inverse-box/inversion transfer of step 1.4, so consists only of conjugations. The zero-width cases transfer as well. Finally step 1.2 turns every chronological comparison into the actual word comparison.

Classify the commissioned reducing choices. Use the source's band convention in [F1]: every multiple reduction has all its individual slides over, or all under, the other band. The given peak specifies the first ordinary reducing move to $X$ or $Z$, including its crossing choice. Therefore the sign of the initial uniform multiple reduction $r$ or $s$ is fixed by that first move; the remaining copies in that uniform slide cannot independently change signs. The free choices are the final uniform reductions $p$ and $u$. Write the four original band widths as $a,b,c,d$, with the left and right bands $a,b>0$, and top/bottom $c,d\ge 0$, and retain $A,B,C,D$ for the black, grid, striped and textured boxes. For the right-column Figure 8 endpoint arising from $p$, cut before $A$ with inner-to-outer ports $(b,a,c,d)$. The upper central crossing interchanges $b,a$, leaving $A$ on $(a,c)$ and $B$ on $(b,c)$; thus $P_p=A{Q}_{b,a}B$ acts on the first $k+c$ strands, $k=a+b$. After the outer right cap interchanges $c,d$, the lower portion has $D$ on $(b,d)$, its central inverse interchange returns the core order, and $C$ acts on $(a,d)$; hence $Q_p=D{Q}_{b,a}^{-1}C$ acts on the first $k+d$ strands. In the inner-first frame of step 1.2, the outer band passes over the inner at the right cap, so that cap is positive $Q_{c,d}$; the left cap is its negative inverse. The endpoint's chronological record is therefore $P_p{Q}_{c,d}{Q}_p{Q}_{c,d}^{-1}$, with the cap words shifted by $k$. Reversing the uniform over/under choice of $p$ reverses precisely these two cap crossing signs; the four boxes and the central crossings from the fixed first reduction are unchanged in their oriented frames. The other chronological record is $P_p{Q}_{d,c}^{-1}{Q}_p{Q}_{d,c}$, exactly the typed exchange of [F8] with core $k$ and widths $(c,d)$. For the left-column endpoint arising from $u$, cut before $C$ with ports $(b,a,d,c)$. The upper portion is $P_u=C{Q}_{b,a}D\in B_{k+d}$; the lower portion after the right cap is $Q_u=B{Q}_{b,a}^{-1}A\in B_{k+c}$. Its displayed cap signs in that same inner-first frame give the chronological record $P_u{Q}_{c,d}^{-1}{Q}_u{Q}_{c,d}$, and changing the uniform $u$ choice gives $P_u{Q}_{d,c}{Q}_u{Q}_{d,c}^{-1}$, the typed exchange with widths $(d,c)$. All these chronological products use the consecutive strand placements fixed at their respective cuts; word reversal from step 1.2 converts them to actual elements and reverses the exchange direction, preserving each box's support. Thus [F8] resolves every free choice under the commissioned uniform convention for both final reductions, with arbitrary internal boxes and no arbitrary-suffix or mixed-choice assertion. If $c=0$ or $d=0$, the corresponding last multiple reduction has no individual operations and the two cap words are $1$, so there is no such ambiguity.

The initial uniform choices and the old frame rotation. The actual first reducing moves $r,s$ in the given peak may have the opposite crossing choices to the conveniently drawn source reference. We compare the completed braids, without changing those first moves. First justify changing the old $n$-strand frame. Let $R_v(x)=e^{\pi iv}x$ for $0\le v\le 1$. Because the fixed real base configuration is symmetric, the distinct paths $R_v(q_j)$ give a geometric $n$-braid $\delta$ returning to the same unordered configuration. For any braid path $\gamma (t)$, the square $(v,t)\mapsto R_v(\gamma (t))$ is a configuration homotopy whose two endpoint edges are this same path $\delta$. Its boundary identity, with the rightmost-first stacking convention [F11], is $R_1(\gamma )=\delta \gamma {\delta }^{-1}$. Rotation preserves transverse orientation and carries the fixed positive half twist on adjacent points $i,i+1$ to the positive half twist on points $n-i,n-i+1$. By [F14] choose an Artin word for $\delta$; completeness transfers the square identity to the Artin group, giving $\delta {\sigma }_i{\delta }^{-1}={\sigma }_{n-i}$. The assertion extends to inverse letters and arbitrary words. Thus index reversal $J_n({\sigma }_i)={\sigma }_{n-i}$ is an old-strand conjugation, not a mirror or a Markov strand change. It carries a box on the last $r$ strands to one on the first $r$. On a rectangular interchange it gives $J_{p+q}(Q_{p,q})=Q_{q,p}$: reflecting generator indices turns cell $(j,i)$ into index $q-j+i$; exchange the rectangular row and column order, moving only incomparable cells with index difference at least two, just as in step 1.2. Now cut the right completed diagram before the grid box $B$. Its two central uniform interchanges are the ones created by the initial reduction $r$, and its remaining record factors as $P{X}^{-1}QX$, with $P=BYD$ acting on the last $b+c+d$ ports, $Q=C{Y}^{-1}A$ on the last $a+c+d$ ports, and $X=Q_{b,a}$ on the inner $a+b$ ports. At this cut the fixed final reduction is contained wholly in the two outside cap routings $Y,Y^{-1}$. Change to the outer-first frame using $J_n$. With core width $c+d$, the boxes $J_n(P),J_n(Q)$ act on the first $c+d+b,c+d+a$ ports, and the central inverse interchange is ${\iota }_{c+d}(Q_{a,b}^{-1})$. The endpoint is therefore the negative endpoint of the typed exchange [F8] with active widths $(b,a)$. Reversing the initial uniform choice gives its positive endpoint with ${\iota }_{c+d}(Q_{b,a})$ and inverse return; the box braids, outside cap choices and their oriented frames are retained. For the left completed diagram the cut before $D$ gives the same factorization with $P=D{Y}^{-1}B$ and $Q=AYC$, with the same last-block supports and the same outside core. Hence [F8] also compares its actual initial $s$ choice with the reference choice. Conjugation by the old $\delta$ and step 1.2 transfer these comparisons back to the actual braid words. Together with step 3.2 this compares every commissioned uniform initial and final choice to the reference, without modifying the specified peak, selecting independent signs inside a uniform band slide, or assuming partial-band box commutation. Empty outside bands merely reduce the core width; $a,b\ge 1$ ensure both active initial reductions are present.

The middle comparison. In the bottom-right and bottom-left panels, all four boxes act on old $n$ strands. The black and textured boxes $A_1,D_{\mathrm{i}\mathrm{n}}$ are disjoint; likewise the grid and striped boxes $B_1,C_{\mathrm{i}\mathrm{n}}$ are disjoint, as their stated placements show. At the cut before $A_1$, the right record is $A_{1}F{B}_1{C}_{\mathrm{i}\mathrm{n}}{F}^{-1}{D}_{\mathrm{i}\mathrm{n}}$ and the left record is $A_1{D}_{\mathrm{i}\mathrm{n}}F{C}_{\mathrm{i}\mathrm{n}}{B}_1{F}^{-1}$. The routing $F$ is the full positive three-band weave including its middle-band positive full twist; the lower routing reverses that entire weave and twist and is $F^{-1}$. Moving the routed twist on its own whole band uses [F5]. By [F7], conjugation by $D_{\mathrm{i}\mathrm{n}}$ changes the right record into $D_{\mathrm{i}\mathrm{n}}{A}_{1}F{B}_1{C}_{\mathrm{i}\mathrm{n}}{F}^{-1}=A_1{D}_{\mathrm{i}\mathrm{n}}F{C}_{\mathrm{i}\mathrm{n}}{B}_1{F}^{-1}$, the left record. This retains every arbitrary box, makes no partial-band commutation, and has no strand change. Step 1.2 transfers the conjugation to the actual words.

Assembly of the source comparison. The right comparison block comprises the first arrow of step 2.2, the box slide of step 1.4, and the second arrow of step 2.3: one multiple negative stabilization and one multiple negative destabilization, each expanded into $d$ ordinary strand changes. Step 4.2 joins it to the left block, whose arrows and box slide are verified in step 3.1: one multiple positive stabilization and one multiple positive destabilization, each likewise expanded into $d$ ordinary strand changes. These are finite sequences even when a band's width is greater than one; at $d=0$ the strand-changing parts are empty. The reducing-choice exchanges of steps 3.2 and 4.1 contribute additional finite ordinary Markov sequences, so no fixed bound of two individual stabilizations and two individual destabilizations is asserted. All diagrams therefore have the claimed finite comparison, and the specified peak can be cancelled by Markov moves. AC is inherited from the peak-lowering, faithful-action and compensated-packet suppliers.