*Proof of Equations (346) to (352):* The particular cases of these equations that are on one or two qubits follow directly from Lemmas 7.1 and 7.6,<fnref n="22" /> therefore we only need to address the cases where there are at least three qubits.

By Equations (333) and (334) (together – if there are white controls – with Equation (81) and maybe with Equation (315) and/or (316)), any multi-controlled ZX or XZ gate with arbitrary combination of black and white controls can be written as a product of multi-controlled (−I) and H gates with a hole. Thus, we can deduce Equations (346) to (352) from Equations (335) to (345) whenever there is a black control in front of a white control, that is, in all cases except maybe in the particular case of Equations (346) and (351) where $x = y$. For Equation (351), this particular case is trivial if $a = b$, and follows directly from Equation (317) if $a \neq b$. Therefore it only remains to address it for Equation (346). By Equation (81), we can assume without loss of generality that $x$ (and therefore also $y$) is the vector consisting only of 1's; and by Lemma B.5 (used together with Equation (317)), we can assume without loss of generality that $a = 0$. Then we have

<figure><img src="image_1138.png" /></figure>

<figure><img src="image_1139.png" /></figure>

## D.8 Properties of Multi-Controlled ZX and XZ Gates

**Lemma D.14.** The following equations are consequences of the equations of Figure 4:

<figure><img src="image_1140.png" /></figure>

<footnote n="22">²²One can notice that those cases are relevant only for Equations (346) and (351).</footnote>
(312)
$=$

<figure><img src="image_1141.png" /></figure>

(311)
$=$

<figure><img src="image_1142.png" /></figure>

(310)(307)(339)
$=$

<figure><img src="image_1143.png" /></figure>

(309)
$=$

<figure><img src="image_1144.png" /></figure>

(274)(313)
$=$

<figure><img src="image_1145.png" /></figure>

same as
previous steps
$=$

<figure><img src="image_1146.png" /></figure>

*Proof of Equation (354):* By Lemma B.5 (used together with Equation (317)), we can assume without loss of generality that $a = 0$, and by Equation (353), we can assume without loss of generality that $x = \epsilon$. If $y$ is of length 0, then the result follows from Lemma 7.6. If $y$ is of length 1, then the result follows from Equations (136) and (137) together with Equation (81). If $y$ is of length at least 2, then by Equation (81) we can assume without loss of generality that is the vector consisting only of 1's. Then we have

<figure><img src="image_1147.png" /></figure>

(333)
$=$

<figure><img src="image_1148.png" /></figure>
<figure><img src="image_1149.png" /></figure>

□

## D.9 Additional Useful Auxiliary Equations on $n$ Qubits

**Lemma D.15.** The following equations are consequences of the equations of Figure 4:

$$ \begin{tikzpicture}[baseline=(current bounding box.center), scale=0.35] \node (A) at (0,0) {$\begin{tikzpicture}[baseline=-0.5ex, scale=0.7] \node (top) at (0,0) {$\Lambda^x$}; \node (bottom) at (0,1) {$\Lambda^x$}; \draw (-0.5,0) -- (0.5,0); \draw (-0.5,1) -- (0.5,1); \node[above] at (0.25,0.5) {$\oplus XZ$}; \node[below] at (0.25,0.5) {$\oplus XZ$}; \end{tikzpicture}$}; \node (B) at (2,0) {$\begin{tikzpicture}[baseline=-0.5ex, scale=0.7] \node (top) at (0,0) {$\Lambda^x$}; \node (bottom) at (0,1) {$\Lambda^x$}; \draw (-0.5,0) -- (0.5,0); \draw (-0.5,1) -- (0.5,1); \node[above] at (0.25,0.5) {$\oplus ZX$}; \node[below] at (0.25,0.5) {$\oplus ZX$}; \end{tikzpicture}$}; \draw (A) -- (B); \node[above] at (1,0.5) {$=$}; \node[below] at (1,0.5) {$=$}; \end{tikzpicture} \tag{355} $$

$$ \begin{tikzpicture}[baseline=(current bounding box.center), scale=0.35] \node (A) at (0,0) {$\begin{tikzpicture}[baseline=-0.5ex, scale=0.7] \node (top) at (0,0) {$\Lambda^x$}; \node (bottom) at (0,1) {$\Lambda^x$}; \draw (-0.5,0) -- (0.5,0); \draw (-0.5,1) -- (0.5,1); \node[above] at (0.25,0.5) {$\oplus ZX$}; \node[below] at (0.25,0.5) {$\oplus ZX$}; \end{tikzpicture}$}; \node (B) at (2,0) {$\begin{tikzpicture}[baseline=-0.5ex, scale=0.7] \node (top) at (0,0) {$\Lambda^x$}; \node (bottom) at (0,1) {$\Lambda^x$}; \draw (-0.5,0) -- (0.5,0); \draw (-0.5,1) -- (0.5,1); \node[above] at (0.25,0.5) {$\oplus ZX$}; \node[below] at (0.25,0.5) {$\oplus ZX$}; \end{tikzpicture}$}; \draw (A) -- (B); \node[above] at (1,0.5) {$=$}; \node[below] at (1,0.5) {$=$}; \end{tikzpicture} \tag{356} $$

$$ \begin{tikzpicture}[baseline=(current bounding box.center), scale=0.35] \node (A) at (0,0) {$\begin{tikzpicture}[baseline=-0.5ex, scale=0.7] \node (top) at (0,0) {$\Lambda^x$}; \node (bottom) at (0,1) {$\Lambda^x$}; \draw (-0.5,0) -- (0.5,0); \draw (-0.5,1) -- (0.5,1); \node[above] at (0.25,0.5) {$\oplus ZX$}; \node[below] at (0.25,0.5) {$\oplus ZX$}; \end{tikzpicture}$}; \node (B) at (2,0) {$\begin{tikzpicture}[baseline=-0.5ex, scale=0.7] \node (top) at (0,0) {$\Lambda^x$}; \node (bottom) at (0,1) {$\Lambda^x$}; \draw (-0.5,0) -- (0.5,0); \draw (-0.5,1) -- (0.5,1); \node[above] at (0.25,0.5) {$\oplus ZX$}; \node[below] at (0.25,0.5) {$\oplus ZX$}; \end{tikzpicture}$}; \draw (A) -- (B); \node[above] at (1,0.5) {$=$}; \node[below] at (1,0.5) {$=$}; \end{tikzpicture} \tag{357} $$

$$ \begin{tikzpicture}[baseline=(current bounding box.center), scale=0.35] \node (A) at (0,0) {$\begin{tikzpicture}[baseline=-0.5ex, scale=0.7] \node (top) at (0,0) {$\Lambda^x$}; \node (bottom) at (0,1) {$\Lambda^x$}; \draw (-0.5,0) -- (0.5,0); \draw (-0.5,1) -- (0.5,1); \node[above] at (0.25,0.5) {$\oplus ZX$}; \node[below] at (0.25,0.5) {$\oplus ZX$}; \end{tikzpicture}$}; \node (B) at (2,0) {$\begin{tikzpicture}[baseline=-0.5ex, scale=0.7] \node (top) at (0,0) {$\Lambda^x$}; \node (bottom) at (0,1) {$\Lambda^x$}; \draw (-0.5,0) -- (0.5,0); \draw (-0.5,1) -- (0.5,1); \node[above] at (0.25,0.5) {$\oplus ZX$}; \node[below] at (0.25,0.5) {$\oplus ZX$}; \end{tikzpicture}$}; \draw (A) -- (B); \node[above] at (1,0.5) {$=$}; \node[below] at (1,0.5) {$=$}; \end{tikzpicture} \tag{358} $$

$$ \begin{tikzpicture}[baseline=(current bounding box.center), scale=0.35] \node (A) at (0,0) {$\begin{tikzpicture}[baseline=-0.5ex, scale=0.7] \node (top) at (0,0) {$\Lambda^x$}; \node (bottom) at (0,1) {$\Lambda^x$}; \draw (-0.5,0) -- (0.5,0); \draw (-0.5,1) -- (0.5,1); \node[above] at (0.25,0.5) {$\oplus ZX$}; \node[below] at (0.25,0.5) {$\oplus ZX$}; \end{tikzpicture}$}; \node (B) at (2,0) {$\begin{tikzpicture}[baseline=-0.5ex, scale=0.7] \node (top) at (0,0) {$\Lambda^x$}; \node (bottom) at (0,1) {$\Lambda^x$}; \draw (-0.5,0) -- (0.5,0); \draw (-0.5,1) -- (0.5,1); \node[above] at (0.25,0.5) {$\oplus ZX$}; \node[below] at (0.25,0.5) {$\oplus ZX$}; \end{tikzpicture}$}; \draw (A) -- (B); \node[above] at (1,0.5) {$=$}; \node[below] at (1,0.5) {$=$}; \end{tikzpicture} \tag{359} $$
(333)
=

<figure><img src="image_1150.png" /></figure>

(312)(311)
=

<figure><img src="image_1151.png" /></figure>

(334)
=

<figure><img src="image_1152.png" /></figure>

*Proof of Equation* (357): The two-qubit case of the equation (where there are no controls above the two bottom qubits) follows from Lemma 7.6. In the other cases (where there is at least one control above), we have

(353)
=

<figure><img src="image_1153.png" /></figure>

<figure><img src="image_1154.png" /></figure>

(309)(347)(311)
=

<figure><img src="image_1155.png" /></figure>

=

<figure><img src="image_1156.png" /></figure>

(299)
=

<figure><img src="image_1157.png" /></figure>

=

<figure><img src="image_1158.png" /></figure>
(353) =

<figure><img src="image_1159.png" /></figure>

Proof of Equation (358):

<figure><img src="image_1160.png" /></figure>

(317) =

<figure><img src="image_1161.png" /></figure>

(355) =

<figure><img src="image_1162.png" /></figure>

(317) =

<figure><img src="image_1163.png" /></figure>

(317)(351) =

<figure><img src="image_1164.png" /></figure>

(351)(354) =

<figure><img src="image_1165.png" /></figure>

(351)(352) =

<figure><img src="image_1166.png" /></figure>

(351)(354) =

<figure><img src="image_1167.png" /></figure>

similar to
the last 4 steps
= =

<figure><img src="image_1168.png" /></figure>

similar again
= =

<figure><img src="image_1169.png" /></figure>
<figure><img src="image_1170.png" /></figure>

<figure><img src="image_1171.png" /></figure>

<figure><img src="image_1172.png" /></figure>

<figure><img src="image_1173.png" /></figure>

<figure><img src="image_1174.png" /></figure>

<figure><img src="image_1175.png" /></figure>

<figure><img src="image_1176.png" /></figure>

Proof of Equation (359):

<figure><img src="image_1177.png" /></figure>

<figure><img src="image_1178.png" /></figure>
<figure><img src="image_1179.png" /></figure>

(2)

<figure><img src="image_1180.png" /></figure>

and

<figure><img src="image_1181.png" /></figure>

=

<figure><img src="image_1182.png" /></figure>

(309)(346)

<figure><img src="image_1183.png" /></figure>

(2)

<figure><img src="image_1184.png" /></figure>

(81)

<figure><img src="image_1185.png" /></figure>
*Proof of Equation (360):* The two-qubit case of the equation (where the ZX gate has no controls) follows from Lemma 7.6. The three-qubit case of the equation (where the ZX gate has exactly one control) is just Equation (129). In the other cases, we have

<figure><img src="image_1186.png" /></figure>

(333) =

<figure><img src="image_1187.png" /></figure>

(112) =

<figure><img src="image_1188.png" /></figure>

(130) =

<figure><img src="image_1189.png" /></figure>

(308) =

<figure><img src="image_1190.png" /></figure>

(112)(130) =

<figure><img src="image_1191.png" /></figure>

(308) =

<figure><img src="image_1192.png" /></figure>

(112)(246) =

<figure><img src="image_1193.png" /></figure>
<figure><img src="image_1194.png" /></figure>

# E Completeness in the General Case

## E.1 Definition of $\Sigma_{a,b,c,d}$ and $\Sigma'_{a,b,c,d}$

We need to provide two words $\Sigma_{a,b,c,d}$ and $\Sigma'_{a,b,c,d}$ over $\mathcal{P}_N$, not containing any generators of the form $(H_{[e,f]}H_{[g,h]})$, such that $[\Sigma'_{a,b,c,d}] = [\Sigma_{a,b,c,d}]^{-1}$, and that there exists $\mu \notin \{0, 1, 3, 2\}$ and a permutation matrix $\sigma$ mapping respectively $|0\rangle, |1\rangle, |3\rangle, |2\rangle$ to $|a\rangle, |b\rangle, |c\rangle, |d\rangle$, such that either $[\Sigma_{a,b,c,d}] = \sigma$ or $[\Sigma_{a,b,c,d}] = \sigma \circ [(-1)_{[\mu]}]$. The following definition fulfils these properties:

**Definition E.1.**

$$ \Sigma_{a,b,c,d} = ((-1)_{[m_0]} X_{[a,i_0]})^{e_0} ((-1)_{[m_1]} X_{[b,i_1]})^{e_1} ((-1)_{[m_2]} X_{[c,i_2]})^{e_2} ((-1)_{[m_3]} X_{[d,2]})^{e_3} $$
$$ \Sigma'_{a,b,c,d} = ((-1)_{[4]} X_{[d,2]})^{e_3} ((-1)_{[m'_2]} X_{[c,i_2]})^{e_2} ((-1)_{[m'_1]} X_{[b,i_1]})^{e_1} ((-1)_{[m'_0]} X_{[a,i_0]})^{e_0} $$

where

$$ \bullet e_3 = \begin{cases} 1 & \text{if } d \neq 2 \\ 0 & \text{if } d = 2 \end{cases} $$

$$ \bullet m_3 = m'_2 = \tau_{d,2}^{e_3} (4) $$

$$ \bullet i_2 = \tau_{d,2}^{e_3} (3) $$

$$ \bullet e_2 = \begin{cases} 1 & \text{if } c \neq i_2 \\ 0 & \text{if } d = i_2 \end{cases} $$

$$ \bullet m_2 = m'_1 = \tau_{c,i_2}^{e_2} (\tau_{d,2}^{e_3} (4)) $$

$$ \bullet i_1 = \tau_{c,i_2}^{e_2} (\tau_{d,2}^{e_3} (1)) $$

$$ \bullet e_1 = \begin{cases} 1 & \text{if } b \neq i_1 \\ 0 & \text{if } b = i_1 \end{cases} $$

$$ \bullet m_1 = m'_0 = \tau_{b,i_1}^{e_1} (\tau_{c,i_2}^{e_2} (\tau_{d,2}^{e_3} (4))) $$

$$ \bullet i_0 = \tau_{b,i_1}^{e_1} (\tau_{c,i_2}^{e_2} (\tau_{d,2}^{e_3} (0))) $$

$$ \bullet e_0 = \begin{cases} 1 & \text{if } a \neq i_0 \\ 0 & \text{if } a = i_0 \end{cases} $$

$$ \bullet m_0 = \tau_{a,i_0}^{e_0} (\tau_{b,i_1}^{e_1} (\tau_{c,i_2}^{e_2} (\tau_{d,2}^{e_3} (4)))) $$

where $\tau_{k,i} : \{0, ..., 2^n - 1\} \to \{0, ..., 2^n - 1\}$ denotes the transposition which exchanges $k$ and $i$, and where we allow $k = i$ in $((-1)_{[m]} X_{[k,i]})^e$ whenever $e = 0$ (in which case this is still equal to the identity).

## E.2 Proof of Lemma 8.4

We proceed by induction on $|G_n^{-1}(x0y) - G_n^{-1}(x1y)|$. Note that by an elementary property of the Gray code,<fnref n="23" /> since $x0y$ and $x1y$ differ by an odd number of bits (namely 1), $|G_n^{-1}(x0y) - G_n^{-1}(x1y)|$ is necessarily odd.

If $|G_n^{-1}(x0y) - G_n^{-1}(x1y)| = 1$, then the result follows directly from Definition 8.3.

If $|G_n^{-1}(x0y) - G_n^{-1}(x1y)| \ge 3$, let $\alpha \in \{0, 1\}$ be such that $G_n^{-1}(x\alpha y) < G_n^{-1}(x\tilde{\alpha}y)$. It follows from the properties of the Gray code that $G_n(G_n^{-1}(x\alpha y) + 1)$ differs from $x\alpha y$ by exactly one bit (which is either in $x$ or in $y$ since $G_n^{-1}(x\alpha y) + 1 \ne G_n^{-1}(x\tilde{\alpha}y)$), and moreover, that $G_n(G_n^{-1}(x\tilde{\alpha}y) - 1)$ differs from $x\tilde{\alpha}y$ by the same bit. We assume that the changing bit is in $x$ (the proof being the same if it is in $y$). In other words, there exist two bitstrings $x_1 \in \{0, 1\}^{k_1}$, $x_2 \in \{0, 1\}^{k_2}$ and a bit $\beta \in \{0, 1\}$ such that $x = x_1\beta x_2$, $G_n(G_n^{-1}(x\alpha y) + 1) = x_1\beta x_2\alpha y$ and $G_n(G_n^{-1}(x\tilde{\alpha}y) - 1) = x_1\beta x_2\tilde{\alpha}y$. Then we have

$$ \begin{align*} D((-1)_{[G_n^{-1}(x0y)]} (-1)_{[G_n^{-1}(x1y)]}) & \\ &= \prod_{k=G_n^{-1}(x\alpha y)}^{G_n^{-1}(x\tilde{\alpha}y)-1} D((-1)_{[k]} (-1)_{[k+1]}) \\ &= D((-1)_{[G_n^{-1}(x\tilde{\alpha}y)-1]} (-1)_{[G_n^{-1}(x\tilde{\alpha}y)]}) \circ \prod_{k=G_n^{-1}(x\alpha y)+1}^{G_n^{-1}(x\tilde{\alpha}y)-2} D((-1)_{[k]} (-1)_{[k+1]}) \circ D((-1)_{[G_n^{-1}(x\alpha y)]} (-1)_{[G_n^{-1}(x\alpha y)+1]}) \\ &= D((-1)_{[G_n^{-1}(x\tilde{\alpha}y)-1]} (-1)_{[G_n^{-1}(x\tilde{\alpha}y)]}) \circ D((-1)_{[G_n^{-1}(x\alpha y)+1]} (-1)_{[G_n^{-1}(x\tilde{\alpha}y)-1]}) \circ D((-1)_{[G_n^{-1}(x\alpha y)]} (-1)_{[G_n^{-1}(x\alpha y)+1]}) \\ &= D((-1)_{[G_n^{-1}(x_1\beta x_2\tilde{\alpha}y)]} (-1)_{[G_n^{-1}(x_1\beta x_2\tilde{\alpha}y)]}) \circ D((-1)_{[G_n^{-1}(x_1\beta x_2\alpha y)]} (-1)_{[G_n^{-1}(x_1\beta x_2\alpha y)]}) \circ D((-1)_{[G_n^{-1}(x_1\beta x_2\alpha y)]} (-1)_{[G_n^{-1}(x_1\beta x_2\alpha y)]}) \end{align*} $$

<footnote n="23">²³Namely, that the parity of a natural integer is always equal to the parity of the number of 1's in its Gray code.</footnote>
<figure><img src="image_1195.png" /></figure>

<figure><img src="image_1196.png" /></figure>

<figure><img src="image_1197.png" /></figure>

<figure><img src="image_1198.png" /></figure>

## E.3 Proof of Lemma 8.5

We proceed by induction on $|G_n^{-1}(x0y) - G_n^{-1}(x1y)|$. Note that for the same reason as in the proof of Lemma 8.4, $|G_n^{-1}(x0y) - G_n^{-1}(x1y)|$ is necessarily odd.

If $|G_n^{-1}(x0y) - G_n^{-1}(x1y)| = 1$, then the result follows directly from Definition 8.3.

If $|G_n^{-1}(x0y) - G_n^{-1}(x1y)| \ge 3$, let $\alpha \in \{0, 1\}$ be such that $G_n^{-1}(x\alpha y) < G_n^{-1}(x\tilde{\alpha}y)$. Note that for any $a, b$, one has $D((-1)_{[a]}X_{[a,b]}) = D((-1)_{[a]}X_{[b,a]})$ and $D((-1)_{[b]}X_{[a,b]}) = D((-1)_{[b]}X_{[b,a]})$. Therefore the result that we want to prove is equivalent to

$$ \mathrm{QC} \vdash D((-1)_{[G_n^{-1}(x\alpha y)]} X_{[G_n^{-1}(x\alpha y), G_n^{-1}(x\tilde{\alpha} y)]}) = \begin{tikzpicture}[baseline=(current bounding box.center)] \node (A) at (0,0) {$\Lambda^x$}; \node (B) at (0,1) {$\Lambda^y$}; \draw[->] (A) -- (B); \draw[->] (-0.5,0) -- (0.5,0); \draw[->] (-0.5,1) -- (0.5,1); \end{tikzpicture} $$
<figure><img src="image_1199.png" /></figure>

<figure><img src="image_1200.png" /></figure>

<figure><img src="image_1201.png" /></figure>

<figure><img src="image_1202.png" /></figure>

<figure><img src="image_1203.png" /></figure>

where is an abuse of notation denoting

<figure><img src="image_1204.png" /></figure>

, where the order of the product is left unspecified.

We also have

$$D((-1)_{[G_n^{-1}(x\hat{\alpha}y)]} X_{[G_n^{-1}(x\alpha y), G_n^{-1}(x\hat{\alpha}y)]})$$

$$= D((-1)_{[G_n^{-1}(x\alpha y)+1]} X_{[G_n^{-1}(x\alpha y), G_n^{-1}(x\alpha y)+1]}) \circ D((-1)_{[G_n^{-1}(x\hat{\alpha}y)-1]} X_{[G_n^{-1}(x\hat{\alpha}y)-1, G_n^{-1}(x\alpha y)]}) \circ$$
$$
\begin{align*}
& D((-1)_{[G_n^{-1}(x\bar{\alpha}y)-1]} X_{[G_n^{-1}(x\alpha y)+1, G_n^{-1}(x\bar{\alpha}y)-1]}) \circ \\
& D((-1)_{[G_n^{-1}(x\bar{\alpha}y)]} X_{[G_n^{-1}(x\bar{\alpha}y)-1, G_n^{-1}(x\bar{\alpha}y)]}) \circ D((-1)_{[G_n^{-1}(x\alpha y)]} X_{[G_n^{-1}(x\alpha y), G_n^{-1}(x\alpha y)+1]}) \\
= & D((-1)_{[G_n^{-1}(x_1\bar{\beta}x_2\alpha y)]} X_{[G_n^{-1}(x_1\beta x_2\alpha y), G_n^{-1}(x_1\bar{\beta}x_2\alpha y)]}) \circ D((-1)_{[G_n^{-1}(x_1\bar{\beta}x_2\bar{\alpha}y)]} X_{[G_n^{-1}(x_1\beta x_2\bar{\alpha}y), G_n^{-1}(x_1\beta x_2\bar{\alpha}y)]}) \circ \\
& D((-1)_{[G_n^{-1}(x_1\beta x_2\bar{\alpha}y)]} X_{[G_n^{-1}(x_1\beta x_2\alpha y), G_n^{-1}(x_1\beta x_2\bar{\alpha}y)]}) \circ D((-1)_{[G_n^{-1}(x_1\beta x_2\alpha y)]} X_{[G_n^{-1}(x_1\beta x_2\alpha y), G_n^{-1}(x_1\beta x_2\alpha y)+1]})
\end{align*}
$$

<figure><img src="image_1205.png" /></figure>

<figure><img src="image_1206.png" /></figure>

<figure><img src="image_1207.png" /></figure>

which completes the proof.

E.4 Proof of Lemma 8.6

More precisely, we are going to prove the equivalent statement that for any raw circuit $C''$ such that $|||C''||| = C$, one has QC $\vdash$
$D(E_{k,l}(|||C''|||)) = id_k \otimes C \otimes id_l$, where $|||C''|||$ (resp. $|||C''|||$) denotes the image of $|||C''|||$ by the canonical projection onto $\mathbf{QC}_{\text{seq}}$ (resp. onto
$\mathbf{QC}$). For simplicity, by abuse of notation we will omit the canonical projection $|| \cdot ||$ in the rest of the proof. We proceed by structural
induction on $C''$.

For any two *n*-qubit raw circuits $C_1, C_2$ such that $k + n + l \ge 3$, one has

$$
D(E_{k,l}(C_2 \circ C_1)) = D(E_{k,l}(C_2)) \circ D(E_{k,l}(C_1))
$$

and for any *n*-qubit raw circuit $C_1$ and *m*-qubit raw circuit $C_3$ such that $k + n + m + l \ge 3$,

$$
D(E_{k,l}(C_1 \otimes C_3)) = D(E_{k+n,l}(C_3)) \circ D(E_{k,l+m}(C_1)).
$$

Hence, it remains the base cases:
$$ \text{Case of } [\overline{\text{---}}]: D(E_{k,l}(\overline{\text{---}})) = D(\epsilon) = (\text{---})^{\otimes n} $$

$$ \text{Case of } \text{---}: D(E_{k,l}(\text{---})) = D(\epsilon) = (\text{---})^{\otimes n} $$

Case of $\text{---}$: If $\ell \ge 1$, then it follows from Definition 8.2 and Lemma 8.4 that

$$ \mathrm{QC} \vdash D(E_{k,l}(\text{---})) = \prod_{\substack{x \in \{0,1\}^k \\ y \in \{0,1\}^{\ell-1}}} \begin{tikzpicture}[baseline=-0.5ex, scale=0.5, every node/.style={font=\scriptsize}] 
    \node (A) at (0,0) {$\Lambda^x$};
    \node (B) at (0,1) {$\Lambda^y$};
    \draw (A) -- (B);
    \node at (0.2,0.5) {$=$};
    \draw (A) -- (0.2,1.5) -- (B) -- (A);
    \draw (B) -- (0.2,-0.5) -- (A) -- (B);
    \node at (0.4,0.5) {$\cdots$};
    \node at (0.4,-0.5) {$\cdots$};
    \node at (0.2,0) {$\cdots$};
    \node at (0.2,1) {$\cdots$};
    \node at (0.2,-0.5) {$\cdots$};
    \node at (0.2,1.5) {$\cdots$};
    \end{tikzpicture} $$

By Equation (335), one can put the factors in an appropriate order, so as to be able to merge them all together using Equation (309). One gets $id_k \otimes \text{---} \otimes id_\ell$, which is the desired result.

If $\ell = 0$, then it follows from Definition 8.2 and Lemma 8.4 that

$$ \mathrm{QC} \vdash D(E_{k,0}(\text{---})) = \prod_{x \in \{0,1\}^{k-1}} \begin{tikzpicture}[baseline=-0.5ex, scale=0.5, every node/.style={font=\scriptsize}] 
    \node (A) at (0,0) {$\Lambda^x$};
    \node (B) at (0,1) {$\Lambda^y$};
    \draw (A) -- (B);
    \node at (0.2,0.5) {$=$};
    \draw (A) -- (0.2,1.5) -- (B) -- (A);
    \draw (B) -- (0.2,-0.5) -- (A) -- (B);
    \node at (0.4,0.5) {$\cdots$};
    \node at (0.4,-0.5) {$\cdots$};
    \node at (0.2,0) {$\cdots$};
    \node at (0.2,1) {$\cdots$};
    \node at (0.2,-0.5) {$\cdots$};
    \node at (0.2,1.5) {$\cdots$};
    \end{tikzpicture} $$

One can similarly merge all the factors together using Equations (335) and (309), so as to get $id_k \otimes \text{---}$, which is the desired result.

Case of $\text{---}$: If $\ell \ge 1$, then it follows from Definition 8.2 and Lemma 8.4 that

$$ \mathrm{QC} \vdash D(E_{k,0}(\text{---})) = \prod_{\substack{x \in \{0,1\}^k \\ y \in \{0,1\}^{\ell-1}}} \begin{tikzpicture}[baseline=-0.5ex, scale=0.5, every node/.style={font=\scriptsize}] 
    \node (A) at (0,0) {$\Lambda^x$};
    \node (B) at (0,1) {$\Lambda^y$};
    \draw (A) -- (B);
    \node at (0.2,0.5) {$=$};
    \draw (A) -- (0.2,1.5) -- (B) -- (A);
    \draw (B) -- (0.2,-0.5) -- (A) -- (B);
    \node at (0.4,0.5) {$\cdots$};
    \node at (0.4,-0.5) {$\cdots$};
    \node at (0.2,0) {$\cdots$};
    \node at (0.2,1) {$\cdots$};
    \node at (0.2,-0.5) {$\cdots$};
    \node at (0.2,1.5) {$\cdots$};
    \end{tikzpicture} $$

By Equation (335), one can put the factors in an appropriate order, so as to be able to merge them all together using Equation (309). One gets $id_k \otimes \text{---} \otimes id_\ell$, which is the desired result.

If $\ell = 0$, then it follows from Definition 8.2 and Lemma 8.4 that

$$ \mathrm{QC} \vdash D(E_{k,0}(\text{---})) = \prod_{x \in \{0,1\}^{k-1}} \begin{tikzpicture}[baseline=-0.5ex, scale=0.5, every node/.style={font=\scriptsize}] 
    \node (A) at (0,0) {$\Lambda^x$};
    \node (B) at (0,1) {$\Lambda^y$};
    \draw (A) -- (B);
    \node at (0.2,0.5) {$=$};
    \draw (A) -- (0.2,1.5) -- (B) -- (A);
    \draw (B) -- (0.2,-0.5) -- (A) -- (B);
    \node at (0.4,0.5) {$\cdots$};
    \node at (0.4,-0.5) {$\cdots$};
    \node at (0.2,0) {$\cdots$};
    \node at (0.2,1) {$\cdots$};
    \node at (0.2,-0.5) {$\cdots$};
    \node at (0.2,1.5) {$\cdots$};
    \end{tikzpicture} $$

One can similarly merge all the factors together using Equations (335) and (309), so as to get $id_k \otimes \text{---}$, which is the desired result.

Case of $\text{---}$: If $\ell \ge 1$, then it follows from Definitions 8.2 and 8.3 that

$$ D(E_{k,l}(-\text{---})) = \prod_{\substack{x \in \{0,1\}^k \\ y \in \{0,1\}^{\ell-1}}} \begin{tikzpicture}[baseline=-0.5ex, scale=0.5, every node/.style={font=\scriptsize}] 
    \node (A) at (0,0) {$\Lambda^x$};
    \node (B) at (0,1) {$\text{---}$};
    \draw (A) -- (B);
    \node at (0.2,0.5) {$=$};
    \draw (A) -- (0.2,1.5) -- (B) -- (A);
    \draw (B) -- (0.2,-0.5) -- (A) -- (B);
    \node at (0.4,0.5) {$\cdots$};
    \node at (0.4,-0.5) {$\cdots$};
    \node at (0.2,0) {$\cdots$};
    \node at (0.2,1) {$\cdots$};
    \node at (0.2,-0.5) {$\cdots$};
    \node at (0.2,1.5) {$\cdots$};
    \end{tikzpicture} $$

By Equation (340), one can put the factors in an appropriate order, so as to be able to merge them all together using Equation (310). One gets $id_k \otimes \text{---} \otimes id_\ell$, which is the desired result.
Decoding Equation (20): $D((-1)_{[a]}(-1)_{[a]}) = \text{id}_n = D(\epsilon)$.

Decoding Equation (21): The result follows from the definition of $D((-1)_{[a]}(-1)_{[b]})$.

Decoding Equation (22): We distinguish depending on the respective order of a, b, c.

* If $a \le b \le c$, then we have

$$ 
\begin{aligned}
D((-1)_{[a]}(-1)_{[b]})(-1)_{[b]}(-1)_{[c]}) &= \left(\prod_{k=a}^{b-1} D((-1)_{[k]})(-1)_{[k+1]}) \circ \prod_{k=b}^{c-1} D((-1)_{[k]})(-1)_{[k+1]})\right) \\
&\stackrel{(335)(336)}{=} \left(\prod_{k=b}^{c-1} D((-1)_{[k]})(-1)_{[k+1]}) \circ \prod_{k=a}^{b-1} D((-1)_{[k]})(-1)_{[k+1]})\right) \\
&= \left(\prod_{k=a}^{c-1} D((-1)_{[k]})(-1)_{[k+1]})\right) \\
&= D((-1)_{[a]}(-1)_{[c]})
\end{aligned}
$$

* If $a \le c \le b$, then we have

$$ 
\begin{aligned}
D((-1)_{[a]}(-1)_{[b]})(-1)_{[b]}(-1)_{[c]}) &= \left(\prod_{k=a}^{b-1} D((-1)_{[k]})(-1)_{[k+1]}) \circ \prod_{k=c}^{b-1} D((-1)_{[k]})(-1)_{[k+1]})\right) \\
&= \left(\prod_{k=c}^{b-1} D((-1)_{[k]})(-1)_{[k+1]}) \circ \prod_{k=a}^{c-1} D((-1)_{[k]})(-1)_{[k+1]}) \circ \prod_{k=c}^{b-1} D((-1)_{[k]})(-1)_{[k+1]})\right) \\
&\stackrel{(335)(336)}{=} \left(\prod_{k=a}^{c-1} D((-1)_{[k]})(-1)_{[k+1]}) \circ \prod_{k=c}^{b-1} \left(D((-1)_{[k]})(-1)_{[k+1]}) \circ D((-1)_{[k]})(-1)_{[k+1]})\right)\right) \\
&= \left(\prod_{k=a}^{c-1} D((-1)_{[k]})(-1)_{[k+1]})\right) \\
&= D((-1)_{[a]}(-1)_{[c]})
\end{aligned}
$$

* The other cases ($b \le a \le c$, $b \le c \le a$, $c \le a \le b$, and $c \le b \le a$) are similar.

Decoding Equation (23):

$$ D((-1)_{[a]}(-1)_{[a+1]}) = \underset{\Lambda^y}{\overset{\Lambda^x}{\searrow}} \stackrel{(355)}{=} \underset{\Lambda^y}{\overset{\Lambda^x}{\searrow}} \underset{\Lambda^x}{\overset{\Lambda^x}{\searrow}} \underset{\Lambda^y}{\overset{\Lambda^y}{\searrow}} = D((-1)_{[a]}X_{[a,a+1]})(-1)_{[a]}X_{[a,a+1]}) $$

where $G_n(a) = x\beta y$ and $G_n(a+1) = x\tilde{\beta}y$, with $\beta \in \{0, 1\}$, $x \in \{0, 1\}^k$, and $y \in \{0, 1\}^{n-k-1}$.

Decoding Equation (24): By the properties of the Gray code, there exist bitstrings x, y, z (of combined length $n-2$) and bits $\alpha, \beta \in \{0, 1\}$ such that $G_n(a) = x\alpha y\beta z$, either $G_n(a+1) = x\alpha y\beta z$ or $G_n(a+1) = x\alpha\tilde{\beta}y\beta z$, and $G_n(a+2) = x\alpha\tilde{\beta}y\beta z$. We treat only the case $G_n(a+1) = x\alpha y\beta z$,
the other case being similar. One has

<figure><img src="image_1208.png" /></figure>

and

<figure><img src="image_1209.png" /></figure>

To derive the equality between the two, up to using Equations (81) and (356), we can assume without loss of generality that $\alpha = 1$ and $\beta = 0$. Moreover, up to using Equation (81), we can assume without loss of generality that $x, y, z$ consist only of 1's, and up to using Equation (353), we can assume without loss of generality that $y$ and $z$ are of length zero. Thus, it suffices to prove the following equation:

<figure><img src="image_1210.png" /></figure>

which is exactly Equation (358).

Decoding Equation (25): The result follows directly from the definitions of $D((-1)_{[a]}X_{[a,c]})$ and $D((-1)_{[a+1]}X_{[a+1,c]})$.

Decoding Equation (26): The result follows directly from the definitions of $D((-1)_{[c]}X_{[a,c]})$ and $D((-1)_{[c]}X_{[a+1,c]})$.

Decoding Equation (27): The result follows directly from the definitions of $D((-1)_{[a]}X_{[a,c]})$ and $D((-1)_{[a+1]}X_{[a+1,c]})$.

Decoding Equation (28): The result follows directly from the definitions of $D((-1)_{[c]}X_{[a,c]})$ and $D((-1)_{[c]}X_{[a+1,c]})$.

Decoding Equation (29):

$$D((-1)_{[a]}X_{[a,a+1]})(-1)_{[a+1]}X_{[a,a+1]}) = \overset{(317)}{=} \mathrm{id}_n = D(\epsilon)$$

where $G_n(a) = x\beta y$ and $G_n(a+1) = x\bar{\beta}y$, with $\beta \in \{0, 1\}$, $x \in \{0, 1\}^k$, and $y \in \{0, 1\}^{n-k-1}$.

Decoding Equation (30): The result follows directly from the definition of $D((-1)_{[c]}X_{[a,b]})$, given that $c \neq a, b$.

Decoding Equation (31): We only treat the case $c > a + 1$, the other case (where $c < a$) being similar. We have

$$D((-1)_{[a+1]}(-1)_{[c]})(-1)_{[a]}X_{[a,a+1]}) = \overset{c-1}{\underset{k=a+1}{\overbrace{\prod}}} D((-1)_{[k]}(-1)_{[k+1]}) \circ D((-1)_{[a]}X_{[a,a+1]})$$
$$= \prod_{k=a+2}^{c-1} D((-1)_{[k]}(-1)_{[k+1]}) \circ D((-1)_{[a+1]}(-1)_{[a+2]}) \circ D((-1)_{[a]}X_{[a,a+1]})$$

By the properties of the Gray code, there exist bitstrings x, y, z (of combined length $n-2$) and bits $\alpha, \beta \in \{0, 1\}$ such that $G_n(a) = x\alpha y\beta z$, either $G_n(a+1) = x\alpha y\beta z$ or $G_n(a+1) = x\alpha y\beta z$, and $G_n(a+2) = x\alpha y\beta z$. We treat only the case $G_n(a+1) = x\alpha y\beta z$, the other case being similar. We have

$$D((-1)_{[a+1]}(-1)_{[a+2]}) \circ D((-1)_{[a]}X_{[a,a+1]}) = \begin{tikzpicture}[baseline=(current bounding box.center), scale=0.7, every node/.style={scale=0.7}]
\node (A1) at (0,0) {$\Lambda^x$};
\node (A2) at (1,0) {$\Lambda^x$};
\node (B1) at (0,1) {$\Lambda^y$};
\node (B2) at (1,1) {$\Lambda^y$};
\node (C1) at (0,-1) {$\Lambda^z$};
\node (C2) at (1,-1) {$\Lambda^z$};
\draw[->] (A1) -- (B1) node right {$\Lambda^x$};
\draw[->] (A2) -- (B2) node right {$\Lambda^x$};
\draw[->] (A1) -- (C1) node right {$\Lambda^y$};
\draw[->] (A2) -- (C2) node right {$\Lambda^y$};
\draw[->] (B1) -- (C2) node left {$\Lambda^z$};
\draw[->] (B2) -- (C1) node left {$\Lambda^z$};
\end{tikzpicture}$$

Up to using Equations (81) and (356), we can do as if the control parametrized by $\beta$ was black, and up to using Equation (315), we can do as if the control parametrized by $\alpha$ was black as well. Then by using Equations (353), (307) and (313), we can swap the controls appropriately so that

<figure><img src="image_1211.png" /></figure>

<figure><img src="image_1212.png" /></figure>

$$\prod_{k=a+2}^{c-1} D((-1)_{[k]}(-1)_{[k+1]}) \circ D((-1)_{[a]}X_{[a,a+1]}) \circ D((-1)_{[a]}(-1)_{[a+1]}) \circ D((-1)_{[a+1]}(-1)_{[a+2]})$$

Finally, note that for any $k \notin \{a-1, a, a+1\}$, $D((-1)_{[k]}(-1)_{[k+1]})$ commutes with $D((-1)_{[a]}X_{[a,a+1]})$ by Equation (346) or (347) (and Equation (307)). Indeed, $G_n(k)$ differs from $G_n(k+1)$ by exactly one bit, and $G_n(a)$ differs from $G_n(a+1)$ by exactly one bit as well. Since $G_n(a), G_n(a+1), G_n(k), G_n(k+1)$ are all distinct, there must be at least one bit where $G_n(k)$ and $G_n(k+1)$ agree, $G_n(a)$ and $G_n(a+1)$ agree, but $G_n(k), G_n(k+1)$ differ from $G_n(a), G_n(a+1)$ (otherwise, we would be able to get two identical vectors by choosing appropriately between $G_n(k)$ and $G_n(k+1)$ and between $G_n(a)$ and $G_n(a+1)$, in contradiction with the fact that all four are distinct). This corresponds to a control of $D((-1)_{[k]}(-1)_{[k+1]})$ being in front of a control of $D((-1)_{[a]}X_{[a,a+1]})$ of the opposite color.
$$ \stackrel{(2)}{=} \mathrm{id}_k \otimes - \otimes \mathrm{id}_l $$

Together with Lemma 8.6 applied to $\mathcal{X}$, and with Equation (81), this implies that the decoding of the right-hand side of Equation (36) can

<figure><img src="image_1213.png" /></figure>

*Decoding Equation (37):* On the one hand, by definition of $D(H_{[a,b]}H_{[c,d]})$ in the case where the tuple $(G_n(a), G_n(b), G_n(c), G_n(d))$ is of

<figure><img src="image_1214.png" /></figure>

On the other hand, with the same argument as for Equation (36), one can show that the decoding of the right-hand side of Equation (37)

<figure><img src="image_1215.png" /></figure>

$$ \textit{Decoding Equation (38): On the one hand, } D(H_{[0,1]}H_{[3,2]}) = \frac{\text{---}}{\text{---}}. $$

On the other hand, for any $a \ge 4$, $G_n(a)$ and $G_n(a+1)$ necessarily have a common bit with value 1 which is not one of the last two. Indeed, otherwise, since they differ by only one bit, at least one of the two would be equal to either 0...000, 0...001, 0...011, or 0...010 (more precisely, if the changing bit was among the last two bits, both $G_n(a)$ and $G_n(a+1)$ would be equal to one of these codes, and if it was among the first $n-2$ bits, exactly one of $G_n(a)$ and $G_n(a+1)$ would be equal to one of these codes). But these are already the Gray codes of 0, 1, 2, 3 respectively.

Consequently, $D((-1)_{[a]}X_{[a,a+1]})$ has a black control in front of one of the white controls of $\text{---}$, and therefore commutes with it by Equation (348), (349) or (350).

*Decoding Equation (39):*

$$ D((-1)_{[0]}(-1)_{[1]}(H_{[0,1]}H_{[3,2]}) = \stackrel{(338)}{=} \frac{\text{---}}{\text{---}} = D((H_{[0,1]}H_{[3,2]})(-1)_{[0]}(-1)_{[1]}) $$
We can then prove the result as follows:

<figure><img src="image_1216.png" /></figure>
(353)(349) =

<figure><img src="image_1217.png" /></figure>

(354) =

<figure><img src="image_1218.png" /></figure>

(360) =

<figure><img src="image_1219.png" /></figure>

(112)(307) =

<figure><img src="image_1220.png" /></figure>

(112)(130) =

<figure><img src="image_1221.png" /></figure>

(112) =

<figure><img src="image_1222.png" /></figure>

(311)(355) =

<figure><img src="image_1223.png" /></figure>

(317) =

<figure><img src="image_1224.png" /></figure>
(274)

<figure><img src="image_1225.png" /></figure>

(309)

<figure><img src="image_1226.png" /></figure>

(307)(338)

<figure><img src="image_1227.png" /></figure>

(354)

<figure><img src="image_1228.png" /></figure>

(81)(315)

<figure><img src="image_1229.png" /></figure>

(311)

<figure><img src="image_1230.png" /></figure>

(315)(81)

<figure><img src="image_1231.png" /></figure>

(309)(354)

<figure><img src="image_1232.png" /></figure>
(274) =

<figure><img src="image_1233.png" /></figure>

(355) =

<figure><img src="image_1234.png" /></figure>

(317) =

<figure><img src="image_1235.png" /></figure>

(361) =

<figure><img src="image_1236.png" /></figure>

Decoding Equation (44):

$$D((X_{[0,3]}X_{[1,2]})(H_{[0,1]}H_{[3,2]}))$$

(361) =

<figure><img src="image_1237.png" /></figure>

(338) =

<figure><img src="image_1238.png" /></figure>

(112) =

<figure><img src="image_1239.png" /></figure>

(111)(360) =

<figure><img src="image_1240.png" /></figure>

(354) =

<figure><img src="image_1241.png" /></figure>

(346) =

<figure><img src="image_1242.png" /></figure>
<figure><img src="image_1243.png" /></figure>

<figure><img src="image_1244.png" /></figure>

<figure><img src="image_1245.png" /></figure>

$$ (361) \quad = \quad D((H_{[0,1]}H_{[3,2]})(X_{[0,3]}X_{[1,2]})) $$

Decoding Equation (45): The decoding of the left-hand side is

<figure><img src="image_1246.png" /></figure>

and the decoding of the left-hand side is

<figure><img src="image_1247.png" /></figure>

If $n = 3$, then the result is given by Equation (146).
If $n \ge 4$, then we have

<figure><img src="image_1248.png" /></figure>

(311) =

<figure><img src="image_1249.png" /></figure>

(337) =

<figure><img src="image_1250.png" /></figure>

(I)

<figure><img src="image_1251.png" /></figure>

(312) =
(337)(344) =

<figure><img src="image_1252.png" /></figure>

(309) =

<figure><img src="image_1253.png" /></figure>

(310) =

<figure><img src="image_1254.png" /></figure>

(277) =

<figure><img src="image_1255.png" /></figure>

(274) =

<figure><img src="image_1256.png" /></figure>

(112) =

<figure><img src="image_1257.png" /></figure>

(308) =

<figure><img src="image_1258.png" /></figure>
(309) =

<figure><img src="image_1259.png" /></figure>

(311) =

<figure><img src="image_1260.png" /></figure>

(112) =

<figure><img src="image_1261.png" /></figure>

(112)(246) =

<figure><img src="image_1262.png" /></figure>

(308)(111) =

<figure><img src="image_1263.png" /></figure>

(112)(130) =

<figure><img src="image_1264.png" /></figure>

(311) =

<figure><img src="image_1265.png" /></figure>
(309) =

<figure><img src="image_1266.png" /></figure>

(308) =

<figure><img src="image_1267.png" /></figure>

(309)(307)(336) =

<figure><img src="image_1268.png" /></figure>

(112) =

<figure><img src="image_1269.png" /></figure>

(309) =

<figure><img src="image_1270.png" /></figure>

(307)(313)(339)(311) =

<figure><img src="image_1271.png" /></figure>

(II)

(307)(313)(339)(340) =

<figure><img src="image_1272.png" /></figure>
(274)
≡

<figure><img src="image_1273.png" /></figure>

(309)
≡

<figure><img src="image_1274.png" /></figure>

(307)(338)(336)
≡

<figure><img src="image_1275.png" /></figure>

(335)
≡

<figure><img src="image_1276.png" /></figure>

(307)(336)(338)
≡

<figure><img src="image_1277.png" /></figure>

(309)
≡

<figure><img src="image_1278.png" /></figure>

(274)
≡

<figure><img src="image_1279.png" /></figure>
same steps as
from (I) to (II),
mirrored,
in reverse order
≡

<figure><img src="image_1280.png" /></figure>

(307)(338)(337)(335)
≡

<figure><img src="image_1281.png" /></figure>

(311)
≡

<figure><img src="image_1282.png" /></figure>

*Decoding Equation (46):* The decoding of the left-hand side is

<figure><img src="image_1283.png" /></figure>

and the decoding of the left-hand side is

<figure><img src="image_1284.png" /></figure>

If $n = 3$, then the result is given by Equation (147).

If $n \ge 4$, then we have

(311)
≡

<figure><img src="image_1285.png" /></figure>
(III)

$$ (307)(337) = $$

<figure><img src="image_1286.png" /></figure>

$$ (312) = $$

<figure><img src="image_1287.png" /></figure>

$$ (337)(307)(345)(344) = $$

<figure><img src="image_1288.png" /></figure>

$$ (342) = $$

<figure><img src="image_1289.png" /></figure>

$$ (310) = $$

<figure><img src="image_1290.png" /></figure>

$$ (277) = $$

<figure><img src="image_1291.png" /></figure>

$$ (311) = $$

<figure><img src="image_1292.png" /></figure>
(338)
≡

<figure><img src="image_1293.png" /></figure>

(309)
≡

<figure><img src="image_1294.png" /></figure>

(274)
≡

<figure><img src="image_1295.png" /></figure>

(112)
≡

<figure><img src="image_1296.png" /></figure>

(308)
≡

<figure><img src="image_1297.png" /></figure>

(309)
≡

<figure><img src="image_1298.png" /></figure>

(311)
≡

<figure><img src="image_1299.png" /></figure>
(112)(307)
≡

<figure><img src="image_1300.png" /></figure>

(112)(130)
≡

<figure><img src="image_1301.png" /></figure>

(308)(111)
≡

<figure><img src="image_1302.png" /></figure>

(307)
≡

<figure><img src="image_1303.png" /></figure>

(309)
≡

<figure><img src="image_1304.png" /></figure>

(307)(313)(339)(336)
≡

<figure><img src="image_1305.png" /></figure>

(311)
≡

<figure><img src="image_1306.png" /></figure>
$^{(309)}$

<figure><img src="image_1307.png" /></figure>

<figure><img src="image_1308.png" /></figure>

$^{(130)}$

<figure><img src="image_1309.png" /></figure>

<figure><img src="image_1310.png" /></figure>

$^{(111)}$

<figure><img src="image_1311.png" /></figure>

<figure><img src="image_1312.png" /></figure>

$^{(313)}$

<figure><img src="image_1313.png" /></figure>

<figure><img src="image_1314.png" /></figure>

$^{(112)}$

<figure><img src="image_1315.png" /></figure>

<figure><img src="image_1316.png" /></figure>

$^{(112)}$

<figure><img src="image_1317.png" /></figure>

<figure><img src="image_1318.png" /></figure>

$^{(111)}$

<figure><img src="image_1319.png" /></figure>

<figure><img src="image_1320.png" /></figure>
<figure><img src="image_1321.png" /></figure>

(311)

<figure><img src="image_1322.png" /></figure>

(112)(111)

<figure><img src="image_1323.png" /></figure>

(311)

<figure><img src="image_1324.png" /></figure>

(309)

<figure><img src="image_1325.png" /></figure>

(308)

<figure><img src="image_1326.png" /></figure>

(309)(307)(336)

<figure><img src="image_1327.png" /></figure>

(IV)
(308)
≡

<figure><img src="image_1328.png" /></figure>

(309)(307)(336)
≡

<figure><img src="image_1329.png" /></figure>

(309)(335)
≡

<figure><img src="image_1330.png" /></figure>

(309)(338)
≡

<figure><img src="image_1331.png" /></figure>

(309)(307)(336)
≡

<figure><img src="image_1332.png" /></figure>

(308)
≡

<figure><img src="image_1333.png" /></figure>

(309)(307)(336)
≡

<figure><img src="image_1334.png" /></figure>
$$ (307)(112) = $$

<figure><img src="image_1335.png" /></figure>

$$ (307)(313)(339) = $$

<figure><img src="image_1336.png" /></figure>

$$ (112)(111) = $$

<figure><img src="image_1337.png" /></figure>

$$ (307)(336)(335)(338) = $$

<figure><img src="image_1338.png" /></figure>

$$ (111)(112) = $$

<figure><img src="image_1339.png" /></figure>

$$ (307)(313)(339) = $$

<figure><img src="image_1340.png" /></figure>

$$ (112)(307) = $$

<figure><img src="image_1341.png" /></figure>
(308)
≡

<figure><img src="image_1342.png" /></figure>

(309)
≡

<figure><img src="image_1343.png" /></figure>

(311)
≡

<figure><img src="image_1344.png" /></figure>

(307)(338)
≡

<figure><img src="image_1345.png" /></figure>

same steps as
from (III) to (IV),
mirrored,
in reverse order

<figure><img src="image_1346.png" /></figure>

(307)(338)(337)(335)
≡

<figure><img src="image_1347.png" /></figure>

(311)
≡

<figure><img src="image_1348.png" /></figure>