singular string of length ($\leq l^{(2)}-1$) in $\nu^{(2)}$ and hence [5] can
be marked in a string in a string (singular or q-singular) of length $l^{(2)}$
in $\nu^{(2)}$ even if [5] cannot be marked in a string of length ($\leq
l^{(2)}-1$), which contradicts $b=\framebox{$6$}$. Hence
$m_{l^{(2)}}^{(2)}\geq 2$. The remaining argument is similar to case (1) and
results in a contradiction. Hence, $\lambda_{2}\geq 2$.

Case 7. $b=\framebox{$7$}$.

Since $\mathrm{wt}(\framebox{$7$})=0$, there is nothing to prove.

Case 8. $b=\framebox{$8$}$.

Since $\mathrm{wt}(\framebox{$8$})=0$, there is nothing to prove.

Case 9. $b=\framebox{$9$}$.

In this case, after $[6]$ was marked in $\nu^{(1)}$ the box marking has
terminated. Since
$\mathrm{wt}(\framebox{$9$})=-2\bar{\Lambda}_{1}+\bar{\Lambda}_{2}$, we must
show that $\lambda_{2}\geq 1$. The proof is analogous to that in Case 4.

Case 10. $b=\framebox{$10$}$.

In this case, after $[6]$ was marked in $\nu^{(2)}$ the box marking has
terminated. Since
$\mathrm{wt}(\framebox{$10$})=3\bar{\Lambda}_{1}-2\bar{\Lambda}_{2}$, we must
show that $\lambda_{1}\geq 3$. We first suppose that $\lambda_{1}=0$. Then we
have that $l^{(1)}\geq 3l^{(2)}$ and $p_{l^{(1)}}^{(1)}=0$ as in Case 2. Note
that all $l^{(1)}$-strings in $\nu^{(1)}$ are singular. If $[6]$ is marked in
a singular string of length $(<l^{(2)})$, then it is obvious that $[7]$ can be
marked in an $l^{(1)}$-string even if [7] cannot be marked in a string of
length ($\leq l^{(1)}-1$). So we assume that $[6]$ is marked in an
$l^{(2)}$-string and the selected $l^{(2)}$-string is singular. If $[4]$ is
marked in a singular string of length $(<l^{(1)})$, it is obvious that $[7]$
can be marked in an $l^{(1)}$-string. If $[4]$ were marked in a singular
string of length $(>3l^{(2)})$, then $[5]$ (and therefore $[6]$) would not be
marked in $\nu^{(2)}$. So it suffices to consider the case when
$l^{(1)}=3l^{(2)}$ and $[4]$ is marked in an $l^{(1)}$-string. We have two
possible ways of box marking in $\nu^{(1)}$.

$\downarrow$$\uparrow$$[4]$$[3]$$3l^{(2)}$and$\downarrow$$\uparrow$$[4]$$[3]$$[2]$$3l^{(2)}$

In both cases, $[7]$ can be marked in the string marked by $[4]$, which
contradicts $b=\framebox{$10$}$. Secondly, we suppose that $\lambda_{1}=1$. As
in Case 2, we have two possible cases.

  * (1)

$m_{l^{(2)}}^{(2)}=1$, $l^{(1)}=3l^{(2)}-1$, and $p_{l^{(1)}}^{(1)}=0$,

  * (2)

$m_{l^{(2)}}^{(2)}\geq 1$, $l^{(1)}\geq 3l^{(2)}$, and $p_{l^{(1)}}^{(1)}=1$.

As before it suffices to consider the case when $[4]$ is marked in an
$l^{(1)}$-string and the selected $l^{(1)}$\- and $l^{(2)}$-strings are
singular. By the same arguments as in Case 2, both cases (1) and (2) result in
a contradiction. Thirdly, we suppose that $\lambda_{1}=2$. As in Case 2, we
have four possible cases.

  * (1)

$m_{l^{(2)}}^{(2)}=1$, $l^{(1)}=3l^{(2)}-2$, and $p_{l^{(1)}}^{(1)}=0$,

  * (2)

$m_{l^{(2)}}^{(2)}=1$, $l^{(1)}=3l^{(2)}-1$, and $p_{l^{(1)}}^{(1)}=1$,

  * (3)

$m_{l^{(2)}}^{(2)}=2$, $l^{(1)}=3l^{(2)}-1$, and $p_{l^{(1)}}^{(1)}=0$,

  * (4)

$m_{l^{(2)}}^{(2)}\geq 1$, $l^{(1)}\geq 3l^{(2)}$, and $p_{l^{(1)}}^{(1)}=2$.

As before it suffices to consider the case when $[4]$ is marked in an
$l^{(1)}$-string and the selected $l^{(1)}$\- and $l^{(2)}$-strings are
singular. By the same arguments as in Case 2, these cases result in a
contradiction. Hence, $\lambda_{1}\geq 3$.

Case 11. $b=\framebox{$11$}$.

In this case, after $[7]$ was marked in $\nu^{(1)}$ or $\nu^{(2)}$ the box
marking has terminated. Since
$\mathrm{wt}(\framebox{$11$})=\bar{\Lambda}_{1}-\bar{\Lambda}_{2}$, we must
show that $\lambda_{1}\geq 1$. Suppose that $\lambda_{1}=0$. Then we have that
$l^{(1)}\geq 3l^{(2)}$ and $p_{l^{(1)}}^{(1)}=0$ as in Case 2. Note that all
$l^{(1)}$-strings in $\nu^{(1)}$ are singular.

  * (1)

$[7]$ is marked in $\nu^{(2)}$.

If [7] is marked in a singular string of length $(<l^{(2)})$, then it is
obvious that [8] can be marked in an $l^{(1)}$-string in $\nu^{(1)}$ ((6) must
have been marked in a string of length $(<3l^{(2)})$). So we assume that [7]
is marked in an $l^{(2)}$-string and that the string is singular. If [6] is
marked in a singular string of length $(<3l^{(2)})$ (i.e., it is not an
$l^{(1)}$-string), it is obvious that [8] can be marked in an
$l^{(1)}$-string. If [6] were marked in a singular string of length
$(>3l^{(2)})$, then [7] would not be marked in $\nu^{(2)}$. So it suffices to
consider the case when $l^{(1)}=3l^{(2)}$ and [6] is marked in an
$l^{(1)}$-string and [7] is marked in an $l^{(2)}$-string and both selected
strings are singular. We have three possible ways of box marking in
$\nu^{(1)}$.

$\downarrow$$\uparrow$$[6]$$[5]$$[3]$$3l^{(2)}$$\downarrow$$\uparrow$$[6]$$[5]$$3l^{(2)}$$\downarrow$$\uparrow$$[6]$$3l^{(2)}$

It is obvious that [8] can be marked on the left of the box marked by [6].
Note that the box marking depicted below

$\downarrow$$\uparrow$$[3]$$[6]$$[5]$$3l^{(2)}$

cannot occur because the $l^{(1)}$-string is a boomerang string for the string
marked by [3] (q-singular) in the above figure (see (BS-4)).

  * (2)

[7] is marked in $\nu^{(1)}$.

If [7] is marked in a string of length $(<l^{(1)})$, then it is obvious that
[8] can be marked in a singular $l^{(1)}$-string. So we assume that [7] is
marked in an $l^{(1)}$-string.

    * (a)

$l^{(1)}=3l^{(2)}$.

We have three possible ways of box marking in $\nu^{(1)}$.

$\downarrow$$\uparrow$$[7]$$[4]$$[3]$$[2]$$3l^{(2)}$$\downarrow$$\uparrow$$[7]$$[4]$$[3]$$3l^{(2)}$$\downarrow$$\uparrow$$[7]$$3l^{(2)}$

The box marking of

$\downarrow$$\uparrow$$[2]$$[7]$$[4]$$[3]$$3l^{(2)}$

cannot occur because $l^{(1)}$-string is a boomerang string for the string
marked by [2] (qq-singular) in the above figure (see (BS-3)).

In all cases, [8] can be marked on the left of the box marked by [7].

    * (b)

$l^{(1)}=3l^{(2)}+1$.

The only possible way of marking [7] is

$\downarrow$$\uparrow$$[7]$$3l^{(2)}+1$

where [4] must have been marked in a string of length $(\leq 3l^{(2)}-2)$ as
otherwise [7] could be marked in the left of the box marked by [4]. It is
obvious [8] can be marked on the left of the box marked by [7].We omit the
details.

    * (c)

$l^{(1)}\geq 3l^{(2)}+2$.

It is obvious that [8] can be marked on the left of the box marked by [7].

Both cases (1) and (2) contradict $b=\framebox{$11$}$ and therefore
$\lambda_{1}\geq 1$.

Case 12. $b=\framebox{$12$}$.

Since $\mathrm{wt}(\framebox{$12$})=-\bar{\Lambda}_{1}$, there is nothing to
prove.

Case 13. $b=\framebox{$13$}$.

In this case, after [9] was marked in $\nu^{(1)}$ the box marking has
terminated. Since
$\mathrm{wt}(\framebox{$13$})=-3\bar{\Lambda}_{1}+\bar{\Lambda}_{2}$, we must
show that $\lambda_{2}\geq 1$. The proof is analogous to that in Case 4.

Case 14. $b=\framebox{$14$}$.

Since $\mathrm{wt}(\framebox{$14$})=-\bar{\Lambda}_{2}$, there is nothing to
prove.

∎

###### Proof of (II).

Case 1. $b=\framebox{$1$}$.

Since $\Delta{p}_{i}^{(2)}=-1$ $(i\geq 1)$ (see (VC-1)), we must show that
$p_{i}^{(2)}\geq 1$ $(i\geq 1)$. If $m_{i}^{(2)}=0$ $(i\geq 1)$, then
$\nu^{(2)}=\emptyset$ so that we assume $m_{j}^{(2)}\neq 0$ for some $j$.
Since the $j$-string is q-singular at best, $p_{j}^{(2)}\geq 1$ so that
$p_{i}^{(2)}\geq 1$ $(i\geq 1)$ by Lemma 5.3. Hence the admissibility in the
new RC is guaranteed.

Case 2. $b=\framebox{$2$}$.

Firstly we show that $p_{i}^{(2)}\geq 1$ $(i\leq i_{1}-1)$. In the proof of
Case 1, we assumed that $m_{j}^{(2)}\neq 0$ for some $j$. In the following we
assume that $i_{1}\geq 2$ and $m_{i}^{(2)}=0$ $(i\leq i_{1}-1)$. Suppose that
$p_{i_{1}-1}^{(2)}=0$. By convexity we have $p_{i}^{(2)}=0$ $(i\leq i_{1})$.
By Eq. (5.6), we have

$0=3m_{3(i_{1}-1)}^{(1)}+2(m_{3(i_{1}-1)-1}^{(1)}+m_{3(i_{1}-1)+1}^{(1)})+m_{3(i_{1}-1)-2}^{(1)}+m_{3(i_{1}-1)+2}^{(1)}$

so that $m_{3i_{1}-1}^{(1)}=\cdots=m_{3i_{1}-5}^{(1)}=0$. Similarly we have
$m_{i}^{(1)}=0$ $(i\leq 3i_{1}-1)$ so that

$\displaystyle p_{3(i_{1}-1)}^{(1)}=$ $\displaystyle-2\sum_{j\geq
3i_{1}}\min(3(i_{1}-1),j)m_{j}^{(1)}+\sum_{j\geq
i_{1}}\min(3(i_{1}-1),3j)m_{j}^{(2)}$ $\displaystyle=$
$\displaystyle-6(i_{1}-1)\sum_{j\geq 3i_{1}}m_{j}^{(1)}+3(i_{1}-1)\sum_{j\geq
i_{1}}m_{j}^{(2)}.$

Since $p_{3(i_{1}-1)}^{(1)}\geq 0$, we have $\sum_{j\geq
3i_{1}}m_{j}^{(1)}\leq\frac{1}{2}\sum_{j\geq i_{1}}m_{j}^{(2)}$ which yields

$p_{i_{1}}^{(2)}-p_{i_{1}-1}^{(2)}=3\sum_{j\geq
3i_{1}}m_{j}^{(1)}-2\sum_{j\geq i_{1}}m_{j}^{(2)}\leq-\frac{1}{2}\sum_{j\geq
i_{1}}m_{j}^{(2)}<0.$

The last inequality is due to the fact that $m_{i_{1}}^{(2)}>0$. Therefore
$p_{i_{1}}^{(2)}<p_{i_{1}-1}^{(2)}=0$, which is a contradiction. Hence
$p_{i_{1}-1}^{(2)}\geq 1$ and therefore $p_{i}^{(2)}\geq 1$ $(i\leq i_{1}-1)$
by convexity. Note that $m_{i}^{(2)}=0$ $(i\leq i_{1}-1)$.

Secondly we show that $p_{3i_{1}}^{(1)}\geq 3$, $p_{3i_{1}-1}^{(1)}\geq 2$,
and $p_{3i_{1}-2}^{(1)}\geq 1$. We assume that
$m_{3i_{1}}^{(1)}=m_{3i_{1}-1}^{(1)}=m_{3i_{1}-2}^{(1)}=0$. Otherwise the
proof is much easier. If $m_{3i_{1}+1}^{(1)}\neq 0$, then $(3i_{1}+1)$-strings
in $\nu^{(1)}$ must be qqq-singular at best so that $p_{3i_{1}+1}^{(1)}\geq
3$. By convexity we have $p_{3i_{1}}^{(1)}\geq 3$, $p_{3i_{1}-1}^{(1)}\geq 2$,
and $p_{3i_{1}-2}^{(1)}\geq 1$. Therefore we further assume that
$m_{3i_{1}+1}^{(1)}=0$. Let us show that $p_{3i_{1}-2}^{(1)}\geq 1$. From Eq.
(5.4), we have

(5.9)
$-p_{3i_{1}-1}^{(1)}+2p_{3i_{1}}^{(1)}-p_{3i_{1}+1}^{(1)}=m_{i_{1}}^{(2)}>0.$

Suppose that $p_{3i_{1}-2}^{(1)}=0$, then
$p_{3i_{1}-1}^{(1)}=p_{3i_{1}}^{(1)}=0$ by convexity. This contradicts Eq.
(5.9). Hence $p_{3i_{1}-2}^{(1)}\geq 1$. This implies $p_{3i_{1}-1}^{(1)}\geq
1$, $p_{3i_{1}}^{(1)}\geq 1$, and $p_{3i_{1}+1}^{(1)}\geq 1$ by convexity.
However, $p_{3i_{1}}^{(1)}=1$ contradicts Eq. (5.9) so that
$p_{3i_{1}}^{(1)}\geq 2$ and hence $p_{3i_{1}-1}^{(1)}\geq 2$ by convexity.
Now we show that $p_{3i_{1}}^{(1)}=2$ is not the case. Suppose that
$p_{3i_{1}}^{(1)}=2$. Then $(p_{3i_{1}-1}^{(1)},p_{3i_{1}+1}^{(1)})=(2,2)$ or
$(3,1)$ by convexity, which contradicts Eq. (5.9). Hence $p_{3i_{1}}^{(1)}\geq
3$. If $m_{j}^{(1)}>0$ for some $j(\geq 3i_{1}+1)$, then $p_{j}^{(1)}\geq 3$
so that $p_{i}^{(1)}\geq 3$ $(i\geq 3i_{1})$ by Lemma 5.4. If not, then
$p_{i}^{(1)}\geq 3$ $(i\geq 3i_{1})$ by convexity. We have shown that
$p_{i}^{(1)}\geq 3$ $(i\geq 3i_{1})$, $p_{3i_{1}-1}^{(1)}\geq 2$,
$p_{3i_{1}-2}^{(1)}\geq 1$, and $p_{i}^{(2)}\geq 1$ $(i\leq i_{1}-1)$.
Therefore, the admissibility of the new RC is guaranteed (see (VC-2)).

Case 3. $b=\framebox{$3$}$.

Although it suffices to show that $p_{i}^{(1)}\geq 1$ $(i\geq 3i_{1})$ when
$i_{2}=3i_{1}-2$ or $i_{2}=3i_{1}-1$ and $p_{i}^{(1)}\geq 1$ $(i\geq i_{2})$
when $i_{2}\geq 3i_{1}$ (see (VC-3)), we show that $p_{i}^{(1)}\geq 2$
$(i\geq\max(i_{2},3i_{1}))$. We assume that $m_{i}^{(1)}=0$ $(i\geq i_{2}+1)$.
Otherwise, the proof is much easier. From Eq. (5.4), we have

(5.10)
$-p_{3i_{1}-1}^{(1)}+2p_{3i_{1}}^{(1)}-p_{3i_{1}+1}^{(1)}=m_{i_{1}}^{(2)}>0.$

  * (1)

$i_{2}=3i_{1}-2$.

From Eq. (5.10), we have $p_{3i_{1}}^{(1)}\geq 1$. The convexity relation
yields $p_{3i_{1}-1}^{(1)}\geq 1$ and $p_{3i_{1}+1}^{(1)}\geq 1$. If
$p_{3i_{1}}^{(1)}=1$, then Eq. (5.10) leads to a contradiction so that
$p_{3i_{1}}^{(1)}\geq 2$.

  * (2)

$i_{2}=3i_{1}-1$.

Since the selected $i_{2}$-string is q-singular, $p_{3i_{1}-1}^{(1)}\geq 1$ so
that Eq. (5.10) yields $p_{3i_{1}}^{(1)}\geq 2$ as in case (1).

  * (3)

$i_{2}\geq 3i_{1}$.

Since the selected $i_{2}$-string is qq-singular, $p_{i_{2}}^{(1)}\geq 2$.

Therefore, $p_{i}^{(1)}\geq 2$ $(i\geq\max(i_{2},3i_{1}))$. In particular,
$\tilde{p}_{i}^{(1)}\geq 1$ $(i\geq\max(i_{2},3i_{1}))$.

Case 4. $b=\framebox{$4$}$.

It is easily verified that $p_{i}^{(1)}+\Delta p_{i}^{(1)}\geq 0$ $(i\leq
i_{3}-1)$. Since $\Delta p_{i}^{(2)}=-1$ $(i\geq i_{3}^{eff})$ we must show
that $p_{i}^{(2)}\geq 1$ $(i\geq i_{3}^{eff})$. Since [4] or [5] are not
marked in $\nu^{(2)}$, $p_{j}^{(2)}\geq 1$ for $j(\geq i_{1}+1)$ such that
$m_{j}^{(2)}\neq 0$. Therefore it suffices to show that
$p_{i_{3}^{eff}}^{(2)}\geq 1$.

  * (1)

$i_{3}^{eff}=i_{1}$.

If $m_{i_{1}}^{(2)}\geq 2$, then $p_{i_{1}}^{(2)}\geq 1$. Assume that
$m_{i_{1}}^{(2)}=1$. In Case 2, we have shown that $p_{i_{1}-1}^{(2)}\geq 1$.
From Eq. (5.6), we have

$\displaystyle-p_{i_{1}-1}^{(2)}+2p_{i_{1}}^{(2)}-p_{i_{1}+1}^{(2)}$
$\displaystyle=$ $\displaystyle
3m_{3i_{1}}^{(1)}+2(m_{3i_{1}-1}^{(1)}+m_{3i_{1}+1}^{(1)})+m_{3i_{1}-2}^{(1)}+m_{3i_{1}+2}^{(1)}-2.$

Suppose that $p_{i_{1}}^{(2)}=0$. Then the left-hand side is smaller than $0$.
On the other hand, the right-hand side is greater than or equal to $0$ because
$m_{3i_{1}}^{(1)}\neq 0$ or $m_{3i_{1}-1}^{(1)}\neq 0$. This is a
contradiction so that $p_{i_{1}}^{(2)}\geq 1$.

  * (2)

$i_{3}^{eff}>i_{1}$.

The argument is similar to case (1) and we have $p_{i}^{(2)}\geq 1$ $(i\geq
i_{3}^{eff})$.

When $i_{2}<i_{3}$, the box-deleted $(i_{3}-1)$-string in $\tilde{\nu}^{(1)}$
is set to be q-singular by (RA-2). Therefore, we must check that

(5.11) $p_{i_{3}-1}^{(1)}+\Delta p_{i_{3}-1}^{(1)}\geq 1.$

  * (1)

$i_{3}=3i_{1}$.

In this case, $i_{2}=3i_{1}-2$. If $m_{3i_{1}-1}^{(1)}=0$, then
$p_{3i_{1}-1}^{(2)}\geq 1$ by convexity because $p_{3i_{1}}^{(1)}\geq 1$. If
$m_{3i_{1}-1}^{(1)}\neq 0$, then $p_{3i_{1}-1}^{(2)}\geq 1$ by the box marking
of this case. By (VC-3), $\Delta p_{3i_{1}-1}^{(1)}=0$. Hence Eq. (5.11) is
satisfied.

  * (2)

$i_{3}\geq 3i_{1}+1$.

As shown in Case 3, $p_{i_{3}-1}^{(1)}\geq 2$ and $\Delta
p_{i_{3}-1}^{(1)}=-1$ by (VC-3). Hence Eq. (5.11) is satisfied.

Case 5. $b=\framebox{$5$}$.

We must show that $p_{i}^{(1)}\geq 2$ $(i\geq 3i_{4})$ and
$p_{3i_{4}-1}^{(1)}\geq 1$. In order to show that $p_{i}^{(1)}\geq 2$ $(i\geq
3i_{4})$, it suffices to show $p_{3i_{4}}^{(1)}\geq 2$ by noting Lemma 5.4.
Furthermore, we must show that $p_{3i_{4}-2}^{(1)}\geq 2$ when the
$i_{3}$-string is type-I with $i_{3}\geq 3i_{1}+2$ and $i_{3}^{eff}=i_{4}$;
$i_{3}=3i_{4}-1$ (see (VC-5)).

Firstly, let us show that $p_{3i_{4}}^{(1)}\geq 2$ and $p_{3i_{4}-1}^{(1)}\geq
1$. If $m_{3i_{4}}^{(1)}\neq 0$, then $p_{3i_{4}}^{(1)}\geq 2$ as otherwise
[5] would be marked in the $3i_{4}$-string in $\nu^{(1)}$. Suppose that
$m_{3i_{4}}^{(1)}=0$. Then, from Eq. (5.4) we have

(5.12)
$-p_{3i_{4}-1}^{(1)}+2p_{3i_{4}}^{(1)}-p_{3i_{4}+1}^{(1)}=m_{i_{4}}^{(2)}\geq
1.$

Hence $p_{3i_{4}}^{(1)}\geq 1$. We show that $p_{3i_{4}}^{(1)}=1$ is not the
case. Suppose that $p_{3i_{4}}^{(1)}=1$. Then,
$(p_{3i_{4}-1}^{(1)},p_{3i_{4}+1}^{(1)})=(0,0),(0,1)$ or $(1,0)$ by Eq. (5.12)
and by convexity. If $p_{3i_{4}+1}^{(1)}=0$, then $m_{3i_{4}+1}^{(1)}=0$ since
otherwise [5] would be marked in the $(3i_{4}+1)$-string if
$m_{3i_{4}+1}^{(1)}\neq 0$. However, the convexity relation

$p_{3i_{4}+1}^{(1)}\geq\frac{1}{2}(p_{3i_{4}}^{(1)}+p_{3i_{4}+2}^{(1)})=\frac{1}{2}(1+p_{3i_{4}+2}^{(1)})$

implies $p_{3i_{4}+1}^{(1)}\geq 1$. This contradicts our assumption so that
$(p_{3i_{4}-1}^{(1)},p_{3i_{4}+1}^{(1)})=(0,1)$. Since [5] is not marked in a
$(3i_{4}-1)$-string, $m_{3i_{4}-1}^{(1)}=0$. From Eq. (5.3), we have

$-p_{3i_{4}-2}^{(1)}+2p_{3i_{4}-1}^{(1)}-p_{3i_{4}}^{(1)}=-p_{3i_{4}-2}^{(1)}-1=0.$

However this does not hold. Hence $p_{3i_{4}}^{(1)}\geq 2$. If
$m_{3i_{4}-1}^{(1)}\neq 0$, then $p_{3i_{4}-1}^{(1)}\geq 1$ because [5] is not
marked in the $(3i_{4}-1)$-string in $\nu^{(1)}$. If not, then
$p_{3i_{4}-1}^{(1)}\geq 1$ by convexity.

Secondly, we show that $p_{3i_{4}-2}^{(1)}\geq 2$ when the $i_{3}$-string is
type-I with $i_{3}\geq 3i_{1}+2$ and $i_{3}^{eff}=i_{4}$; $i_{3}=3i_{4}-1$.

  * (1)

$i_{2}=i_{3}$.

It suffices to consider the case when
$m_{3i_{1}-1}^{(1)}=\cdots=m_{3i_{4}-2}^{(1)}=0$ by noting Lemma 5.4. Since
$p_{3i_{4}-1}^{(1)}\geq 1$, we have $p_{3i_{1}-1}^{(1)}\geq
1,\ldots,p_{3i_{4}-2}^{(1)}\geq 1$ by convexity. We show that
$p_{3i_{4}-2}^{(1)}=1$ is not the case. Suppose that $p_{3i_{4}-2}^{(1)}=1$.
Then $p_{3i_{1}-1}^{(1)}=\cdots=p_{3i_{4}-2}^{(1)}=1$ by Lemma 5.5. Hence

$0=p_{3i_{4}-2}^{(1)}-p_{3i_{4}-3}^{(1)}=-2\sum_{j\geq
i_{3}}m_{j}^{(1)}+\sum_{j\geq i_{4}}m_{j}^{(2)}$

and

$0=p_{3i_{1}}^{(1)}-p_{3i_{1}-1}^{(1)}=-2\sum_{j\geq
i_{3}}m_{j}^{(1)}+\sum_{j\geq i_{1}}m_{j}^{(2)}$

so that $\sum_{j\geq i_{4}}m_{j}^{(2)}=\sum_{j\geq i_{1}}m_{j}^{(2)}$, which
is a contradiction because $m_{i_{1}}^{(2)}\neq 0$. Hence
$p_{3i_{4}-2}^{(1)}\geq 2$.

  * (2)

$i_{2}<i_{3}$.

We first suppose that he box marking in $\nu^{(1)}$ is

$\downarrow$$[2]$$[3]$

Then $p_{3i_{4}-2}^{(1)}=p_{i_{2}}^{(1)}\geq 2$, where the $i_{2}$ (resp.
$i_{3}$)-string is qq-singular (resp. q-singular). Next, suppose that
$i_{2}\leq 3i_{4}$. It suffices to consider the case when $m_{i}^{(1)}=0$
$(i_{2}+1\leq i\leq 3i_{4}-2)$ as in Case (1). If $i_{2}\geq 3i_{1}$, then the
selected $i_{2}$-string is qq-singular and $p_{i_{2}}^{(1)}\geq 2$ so that
$p_{3i_{4}-2}^{(1)}\geq 2$. Therefore we further assume that $i_{2}\leq
3i_{1}-1$. Suppose that $p_{3i_{4}-2}^{(1)}=1$. Then
$p_{i_{2}+1}^{(1)}=\cdots=p_{3i_{4}-2}^{(1)}=1$ by Lemma 5.5. If
$i_{2}=3i_{1}-1$, then $p_{i_{2}}^{(1)}\geq 1$. This is the equality because
the convexity relation

$1=p_{i_{2}+1}^{(1)}\geq\frac{1}{2}(p_{i_{2}}^{(1)}+p_{i_{2}+2}^{(1)})=\frac{1}{2}(p_{i_{2}}^{(1)}+1)$

implies $p_{i_{2}}^{(1)}=1$. Hence,
$p_{3i_{1}-1}^{(1)}=\cdots=p_{3i_{4}-2}^{(1)}=1$. The remaining argument is
the same as in case (a) and we have $p_{3i_{4}-2}^{(1)}\geq 2$.

Case 6. $b=\framebox{$6$}$.

It is easily verified that $p_{i}^{(1)}+\Delta p_{i}^{(1)}\geq 0$ $(i\leq
i_{4}-1)$. We must show that $p_{i_{4}^{eff}}^{(2)}\geq 2$; $p_{i}^{(2)}\geq
2$ $(i\geq i_{4}^{eff})$ follows from this. Furthermore, we must show that
$p_{i_{4}^{eff}-1}^{(2)}\geq 1$ when the $i_{4}$-string is type-II with
$i_{2}=i_{3}=i_{4}$ or $i_{2}<i_{3}=i_{4}$ (see (VC-6)).

Firstly, we show that $p_{i_{4}^{eff}}^{(2)}\geq 2$.

  * (1)

The $i_{4}$-string is type-0/I.

If $m_{i_{4}^{eff}}^{(2)}\neq 0$, then the $i_{4}^{eff}$-string in $\nu^{(2)}$
must be qq-singular at best so that $p_{i_{4}^{eff}}^{(2)}\geq 2$. Now we
assume that $m_{i_{4}^{eff}}^{(2)}=0$. From Eq. (5.6), we have

(5.13) $\displaystyle-
p_{i_{4}^{eff}-1}^{(2)}+2p_{i_{4}^{eff}}^{(2)}-p_{i_{4}^{eff}+1}^{(2)}$
$\displaystyle=$ $\displaystyle
3m_{3i_{4}^{eff}}^{(1)}+2(m_{3i_{4}^{eff}-1}^{(1)}+m_{3i_{4}^{eff}+1}^{(1)})+m_{3i_{4}^{eff}-2}^{(1)}+m_{3i_{4}^{eff}+2}^{(1)}.$

Obviously, $p_{i_{4}^{eff}}^{(2)}\geq 1$ because $m_{3i_{4}^{eff}}^{(1)}$ or
$m_{3i_{4}^{eff}-1}^{(1)}$ is not $0$. We show that $p_{i_{4}^{eff}}^{(2)}=1$
is not the case. Suppose that $p_{i_{4}^{eff}}^{(2)}=1$. If
$m_{i_{4}^{eff}+1}^{(2)}\neq 0$, then $p_{i_{4}^{eff}+1}^{(2)}\geq 2$. If
$m_{i_{4}^{eff}+1}^{(2)}=0$, then $p_{i_{4}^{eff}+1}^{(2)}\geq 1$ by
convexity. In either case, the left-hand side of Eq. (5.13) is smaller than or
equal to $1$ while the right-hand side is greater than or equal to $2$, which
is a contradiction. Hence $p_{i_{4}^{eff}}^{(2)}\geq 2$.

  * (2)

The $i_{4}$-string is type-II.

If $m_{i_{4}^{eff}-1}^{(2)}\neq 0$, then [4] would be marked in $\nu^{(2)}$
contradicting $b=\framebox{$6$}$ so that $m_{i_{4}^{eff}-1}^{(2)}=0$. The
remaining argument is similar to case (1).

Secondly, we show that $p_{i_{4}^{eff}-1}^{(2)}\geq 1$ when the $i_{4}$-string
is type-II with $i_{2}=i_{3}=i_{4}$ or $i_{2}<i_{3}=i_{4}$. We consider two
cases (1) $m_{i_{4}^{eff}-1}^{(2)}=0$ and (2) $m_{i_{4}^{eff}-1}^{(2)}\neq 0$
separately.

  * (1)

$m_{i_{4}^{eff}-1}^{(2)}=0$.

If $m_{i_{4}^{eff}}^{(2)}\neq 0$, then $p_{i_{4}^{eff}}^{(2)}\geq 2$ and
therefore $p_{i_{4}^{eff}-1}^{(2)}\geq 1$ by convexity. This is because [5] is
not marked in the $i_{4}^{eff}$-string in $\nu^{(2)}$. Now, we assume that
$m_{i_{4}^{eff}}^{(2)}=0$. From Eq. (5.6), we have

(5.14) $\displaystyle-
p_{i_{4}^{eff}-1}^{(2)}+2p_{i_{4}^{eff}}^{(2)}-p_{i_{4}^{eff}+1}^{(2)}$
$\displaystyle=$ $\displaystyle
3m_{3i_{4}^{eff}}^{(1)}+2(m_{3i_{4}^{eff}-1}^{(1)}+m_{3i_{4}^{eff}+1}^{(1)})+m_{3i_{4}^{eff}-2}^{(1)}+m_{3i_{4}^{eff}+2}^{(1)}.$

Suppose that $p_{i_{4}^{eff}-1}^{(2)}=0$. Then $p_{i_{4}^{eff}}^{(2)}=0$ by
convexity, which contradicts Eq. (5.14) because
$m_{i_{4}}^{(1)}=m_{3i_{4}^{eff}-2}^{(1)}\neq 0$. Hence
$p_{i_{4}^{eff}-1}^{(2)}\geq 1$.

  * (2)

$m_{i_{4}^{eff}-1}^{(2)}\neq 0$.

The $i_{4}$-string is

$\downarrow$$[4]$$[3]$$[2]$or$\downarrow$$[4]$$[3]$

Since the marking [2] and [3] causes the effective length reduction
$i_{4}^{eff}\rightarrow i_{4}^{eff}-1$, the $(i_{4}^{eff}-1)$-string in
$\nu^{(2)}$ must be q-singular at best. Otherwise [4] would be marked in this
string. Hence $p_{i_{4}^{eff}-1}^{(2)}\geq 1$.

When $i_{3}<i_{4}$, the box-deleted $(i_{4}-1)$-string in $\tilde{\nu}^{(1)}$
is set to be qq-singular by (RA-3), we must check that

$p_{i_{4}-1}^{(1)}+\Delta p_{i_{4}-1}^{(1)}\geq 2.$

Here, $\Delta p_{i_{4}-1}^{(1)}=1$ (see (VC-4)) and it is obvious that
$p_{i_{4}-1}^{(1)}\geq 1$ so that the above inequality is satisfied. When
$i_{2}<i_{3}=i_{4}$, the box-deleted $(i_{4}-2)$-string in $\tilde{\nu}^{(1)}$
is set to be q-singular by (RA-4), we must check that

(5.15) $p_{i_{4}-2}^{(1)}+\Delta p_{i_{4}-2}^{(1)}\geq 1.$

If $i_{4}\geq 3i_{1}+2$, then $p_{i_{4}-2}^{(1)}\geq 2$ (see Case 3) and
$\Delta p_{i_{4}-2}^{(1)}=-1$ (see (VC-3)) so that Eq. (5.15) is satisfied. If
$i_{4}=3i_{1}+1$, then $\Delta p_{i_{4}-2}^{(1)}=\Delta p_{3i_{1}-1}^{(1)}=0$
(see (VC-3)). Let us show that $p_{3i_{1}-1}^{(1)}\geq 1$. Since this is
obvious if $m_{3i_{1}}^{(1)}\neq 0$ or $m_{3i_{1}-1}^{(1)}\neq 0$, we assume
that $m_{3i_{1}}^{(1)}=0$ and $m_{3i_{1}-1}^{(1)}=0$. From Eq. (5.4), we have

$-p_{3i_{1}-1}^{(1)}+2p_{3i_{1}}^{(1)}-p_{3i_{1}+1}^{(1)}=m_{i_{1}}^{(2)}>0$

so that $p_{3i_{1}}^{(1)}\geq 1$ and therefore $p_{3_{1}-1}^{(1)}\geq 1$ by
convexity. Hence, Eq. (5.15) is satisfied.

Case 7. $b=\framebox{$7$}$.

It is not hard to check that $p_{i}^{(1)}+\Delta p_{i}^{(1)}\geq 0$ $(i\leq
i_{5}-1)$.

Case 8. $b=\framebox{$8$}$.

It is is not hard to check that $p_{i}^{(1)}+\Delta p_{i}^{(1)}\geq 0$ $(i\leq
3i_{5}-1)$.

Case 9. $b=\framebox{$9$}$.

It is not hard to check that $p_{1}^{(1)}+\Delta p_{i}^{(1)}\geq 0$ $(i\leq
i_{6}-1)$. Since $\Delta p_{i}^{(2)}=-1$ $(i\geq i_{6}^{eff})$ (see (VC-9)),
we must show that $p_{i_{6}^{eff}}^{(2)}\geq 1$; $p_{i_{6}^{i}}^{(2)}\geq 1$
$(i\geq i_{6}^{eff})$ follows from this. If $m_{i_{6}^{eff}}^{(2)}\neq 0$,
then $p_{i_{6}^{eff}}^{(2)}\geq 1$. So we assume that
$m_{i_{6}^{eff}}^{(2)}=0$. From Eq. (5.6), we have

$\displaystyle-
p_{i_{6}^{eff}-1}^{(2)}+2p_{i_{6}^{eff}}^{(2)}-p_{i_{6}^{eff}+1}^{(2)}$
$\displaystyle=$ $\displaystyle
3m_{3i_{6}^{eff}}^{(1)}+2(m_{3i_{6}^{eff}-1}^{(1)}+m_{3i_{6}^{eff}+1}^{(1)})+m_{3i_{6}^{eff}-2}^{(1)}+m_{3i_{6}^{eff}+2}^{(1)}.$

Suppose that $p_{i_{6}^{eff}}^{(2)}=0$. Then
$p_{i_{6}^{eff}-1}^{(2)}=p_{i_{6}^{eff}+1}^{(2)}=0$ by convexity. The left-
hand side is $0$ while the right-hand side is positive because
$m_{i_{6}}^{(1)}\neq 0$, which is a contradiction. Hence, we have
$p_{i_{6}^{eff}}^{(2)}\geq 1$.

When $i_{5}<i_{6}$, we must check that $p_{i_{6}-1}^{(1)}\geq 1$ because the
box-deleted $(i_{6}-1)$-string in $\tilde{\nu}^{(1)}$ is set to be q-singular
by (RA-1) and $\Delta p_{i_{6}-1}^{(1)}=0$ (see (VC-7)). However this is
obvious because $i_{5}$-string is q-singular and [6] is not marked in a string
of length $i$ $(i_{5}\leq i\leq i_{6}-1)$.

We omit the proof in the case when $b=\framebox{$i$}$ with $10\leq i\leq 14$.
The proof is similar to those in the previous cases. When
$b=\framebox{$\emptyset$}$, there is nothing to prove.

∎

###### Proof of (IV).

Let $\tilde{\nu}=(\tilde{m}_{i}^{(a)})_{(a,i)\in\mathcal{H}_{0}}$.

Case 0. $b=\framebox{$\emptyset$}$.

In this case, $\tilde{m}_{1}^{(2)}=m_{1}^{(2)}-2$ and
$\tilde{m}_{3}^{(1)}=m_{3}^{(1)}-1$.

$\displaystyle\Delta c(\nu)=$ $\displaystyle
3\left((m_{3}^{(1)})^{2}-(\tilde{m}_{3}^{(1)})^{2}\right)+2\sum_{j\neq
3}\min(3,j)(m_{3}^{(1)}-\tilde{m}_{3}^{(1)})m_{j}^{(1)}$ $\displaystyle-$
$\displaystyle
3(m_{3}^{(1)}m_{1}^{(2)}-\tilde{m}_{3}^{(1)}\tilde{m}_{1}^{(2)})-\sum_{j\neq
1}\min(3,3j)(m_{3}^{(1)}-\tilde{m}_{3}^{(1)})m_{j}^{(2)}$
$\displaystyle-\sum_{i\neq
3}\min(i,3)m_{i}^{(1)}(m_{1}^{(2)}-\tilde{m}_{1}^{(2)})$ $\displaystyle+$
$\displaystyle(m_{1}^{(2)})^{2}-(\tilde{m}_{1}^{(2)})^{2}+4\sum_{j\neq
1}\min(1,j)(m_{1}^{(2)}-\tilde{m}_{1}^{(2)})m_{j}^{(2)}$ $\displaystyle-$
$\displaystyle L\sum_{j\geq 1}m_{j}^{(2)}+(L-1)\left(\sum_{j\geq
1}m_{j}^{(2)}-2\right)$ $\displaystyle=$ $\displaystyle\sum_{j\geq
1}m_{j}^{(2)}-\alpha_{1}^{(2)}-2L+1.$

The change of the sum of riggings is

$\Delta|J|=J^{(1,3)}+2J^{(2,1)}=p_{3}^{(1)}+2p_{1}^{(2)},$

where

$p_{3}^{(1)}=-6\sum_{j\geq
3}m_{j}^{(1)}-4m_{2}^{(1)}-2m_{1}^{(1)}+3\sum_{j\geq 1}m_{j}^{(2)}$

and

$p_{1}^{(2)}=L+3\sum_{j\geq
3}m_{j}^{(1)}+2m_{2}^{(1)}+m_{1}^{(1)}-2\sum_{j\geq 1}m_{j}^{(2)}.$

Hence altogether $\Delta c(\nu,J)=-\alpha_{1}^{(2)}+1$.

Case 1. $b=\framebox{$1$}$.

Let $(\tilde{\nu},\tilde{J})=\delta_{b}(\nu,J)$. Then

$\Delta c(\nu)=c(\nu)-c(\tilde{\nu})=-\sum_{i}m_{i}^{(2)}=-\alpha_{1}^{(2)}.$

The riggings are unchanged so that $\Delta c(\nu,J)=-\alpha_{1}^{(2)}$.

In what follows we assume that $i_{1}>1$. The proof in the case when $i_{1}=1$
is similar. We write $\Delta_{i\rightarrow
j}c(\nu)=c(\tilde{\nu}_{i})-c(\tilde{\nu}_{j})$, $\Delta_{i\rightarrow
j}|J|=|\tilde{J}_{i}|-|\tilde{J}_{j}|$, and $\Delta_{i\rightarrow
j}c(\nu,J)=c(\tilde{\nu}_{i},\tilde{J}_{i})-c(\tilde{\nu}_{j},\tilde{J}_{j})$,
where $\delta_{b_{i}}(\nu,J)=(\tilde{\nu}_{i},\tilde{J}_{i})$ and
$\delta_{b_{j}}(\nu,J)=(\tilde{\nu}_{j},\tilde{J}_{j})$ with
$b_{i}=\framebox{$i$}$ and $b_{j}=\framebox{$j$}$ $(i<j)$. We also write
$\Delta_{1}c(\nu,J)=c(\nu,J)-c(\tilde{\nu},\tilde{J})$ where
$(\tilde{\nu},\tilde{J})=\delta_{b}(\nu,J)$ with $b=\framebox{$1$}$.

Case 2. $b=\framebox{$2$}$.

In this case, $\tilde{m}_{i_{1}}^{(2)}=m_{i_{1}}^{(2)}-1$ and
$\tilde{m}_{i_{1}-1}^{(2)}=m_{i_{1}-1}^{(2)}+1$.

$\Delta_{1\rightarrow 2}c(\nu)=-3\sum_{j\geq
3i_{1}}m_{j}^{(1)}-2m_{3i_{1}-1}^{(1)}-m_{3i_{1}-2}^{(1)}+2\sum_{j\geq
i_{1}}m_{j}^{(2)}-1$

and

$\Delta_{1\rightarrow 2}|J|=p_{i_{1}}^{(2)}-(p_{i_{1}-1}^{(2)}+\Delta
p_{i_{1}-1}^{(2)}),$

where $\Delta p_{i_{1}-1}^{(2)}=-1$ by (VC-1). The direct calculation yields

$\Delta_{1\rightarrow 2}|J|=3\sum_{j\geq
3i_{1}}m_{j}^{(1)}+2m_{3i_{1}-1}^{(1)}+m_{3i_{1}-2}^{(1)}-2\sum_{j\geq
i_{1}}m_{j}^{(2)}+1.$

Hence $\Delta_{1\rightarrow 2}c(\nu,J)=0$ so that $\Delta
c(\nu,J)=\Delta_{1}c(\nu,J)+\Delta_{1\rightarrow
2}c(\nu,J)=-\alpha_{1}^{(2)}$.

As easily verified that $\Delta_{i\rightarrow j}c(\nu,J)$ does not contain
terms involving $m_{i}^{(a)}$. Let us write

$\Delta_{i\rightarrow j}|J|=\sum_{a}\sum_{k}(p_{i_{k}}^{(a)}+\Delta
r_{i_{k}}^{(a)}-(p_{i_{k}-n_{k}}^{(a)}+\Delta p_{i_{k}-n_{k}}^{(a)}+\Delta
r_{i_{k}-n_{k}}^{(a)})),$

where $\Delta r_{k}^{(a)}$ is the rigging adjustment for the selected
$i_{k}$-string in $\nu^{(a)}$, i.e., $\Delta r_{i_{k}}^{(a)}=0,-1$, and $-2$
when the selected $i_{k}$-string in $\nu^{(a)}$ is singular, q-singular, and
qq-singular, respectively so that $p_{i_{k}}^{(a)}+\Delta r_{i_{k}}^{(a)}$ is
the value of the rigging of the selected $i_{k}$-string in $\nu^{(a)}$ and
$n_{k}$ is the numbers of box marking in the $i_{k}$-string in $\nu^{(a)}$,
$p_{i_{k}-n_{k}}^{(a)}+\Delta
p_{i_{k}-n_{k}}^{(a)}=\tilde{p}_{i_{k}-n_{k}}^{(a)}$, and $\Delta
r_{i_{k}-n_{k}}^{(a)}$ is the rigging adjustment for the
$\tilde{p}_{i_{k}-n_{k}}^{(a)}$-string in $\tilde{\nu}^{(a)}$ so that
$p_{i_{k}-n_{k}}^{(a)}+\Delta p_{i_{k}-n_{k}}^{(a)}+\Delta
r_{i_{k}-n_{k}}^{(a)}$ is the value of rigging of the box-deleted
$\tilde{p}_{i_{k}-n_{k}}^{(a)}$-string in $\tilde{\nu}^{(a)}$. Then,

$\Delta_{i\rightarrow j}c(\nu,J)=t-\sum_{a}\sum_{k}(\Delta
r_{i_{k}}^{(a)}-(\Delta p_{i_{k}-n_{k}}^{(a)}+\Delta r_{i_{k}-n_{k}}^{(a)})),$

where $t$ is the “constant” term in $\Delta_{i\rightarrow j}c(\nu)$, which
does not contain $m_{l}^{(a)}$.

Case 3. $b=\framebox{$3$}$.

In this case one caution is in order. When $i_{2}^{eff}=i_{1}$, we must
consider the changes of $m_{i_{2}}^{(1)}$, $m_{i_{2}-1}^{(1)}$,
$m_{i_{1}}^{(2)}$, and $m_{i_{1}-1}^{(2)}$ simultaneously in $c(\nu)$.

  * (1)

$i_{2}^{eff}=i_{1}$.

We compute $\Delta_{1\rightarrow 3}c(\nu,J)$.

$\displaystyle\Delta_{1\rightarrow 3}c(\nu,J)=-1$ $\displaystyle+\Delta
r_{i_{1}}^{(2)}-(\Delta p_{i_{1}-1}^{(2)}+\Delta r_{i_{1}-1}^{(2)})$
$\displaystyle+\Delta r_{i_{2}}^{(1)}-(\Delta p_{i_{2}-1}^{(1)}+\Delta
r_{i_{2}-1}^{(1)}),$

where $\Delta p_{i_{1}-1}^{(2)}=-1$ by (VC-1) and $\Delta
r_{i_{1}}^{(2)}=\Delta r_{i_{2}-1}^{(1)}=0$. The values of the rigging
adjustment $\Delta r_{i_{2}}^{(1)}$ and $\Delta p_{i_{2}-1}^{(1)}$ are listed
below. In this table the type is refereed to the type of the $i_{2}$-string.

type | $\Delta r_{i_{2}}^{(1)}$ | $\Delta p_{i_{2}-1}^{(1)}$
---|---|---
0 | $-2$ | $-2$
I | $-1$ | $-1$
II | $0$ | $0$

The values of $\Delta r_{i_{2}}^{(1)}$ is due to the fact that the selected
$i_{2}$-string is singular (resp. q-singular) when it is type-II (resp.
type-I) and is qq-singular when it is type-0. The values of $\Delta
p_{i_{2}-1}^{(1)}$ come from (VC-2). Hence altogether $\Delta_{1\rightarrow
3}c(\nu,J)=0$ so that $\Delta c(\nu,J)=\Delta_{1}c(\nu,J)+\Delta_{1\rightarrow
3}c(\nu,J)=-\alpha_{1}^{(2)}$.

  * (2)

$i_{1}<i_{2}^{eff}$.

Since $\Delta_{1\rightarrow 3}c(\nu,J)=\Delta_{1\rightarrow
2}c(\nu,J)+\Delta_{2\rightarrow 3}c(\nu,J)$, we compute $\Delta_{2\rightarrow
3}c(\nu,J)$.

$\Delta_{1\rightarrow 3}c(\nu,J)=-1+\Delta r_{i_{2}}^{(1)}-(\Delta
p_{i_{2}-1}^{(1)}+\Delta r_{i_{2}-1}^{(1)}),$

where $\Delta r_{i_{2}}^{(1)}=-2$ because the selected $i_{2}$-string in
$\nu^{(1)}$ is qq-singular and $\Delta r_{i_{2}-1}^{(1)}=0$. By (VC-2),
$\Delta p_{i_{2}-1}^{(1)}=-3$. Hence altogether $\Delta_{2\rightarrow
3}c(\nu,J)=0$ so that $\Delta c(\nu,J)=\Delta_{1}c(\nu,J)+\Delta_{1\rightarrow
2}c(\nu,J)+\Delta_{2\rightarrow 3}c(\nu,J)=-\alpha_{1}^{(2)}$.

Case 4. $b=\framebox{$4$}$.

  * (1)

$i_{2}=i_{3}$.

In this case, $\tilde{m}_{i_{3}}^{(1)}=m_{i_{3}}^{(1)}-1$ and
$\tilde{m}_{i_{3}-2}^{(1)}=m_{i_{3}-2}^{(1)}+1$. As before, we must consider
the changes of $m_{i_{3}}^{(1)}$, $m_{i_{3}-2}^{(1)}$, $m_{i_{1}}^{(2)}$ and
$m_{i_{1}-1}^{(2)}$ simultaneously in $c(\nu)$ when $i_{2}\leq 3i_{1}+1$.

    * (a)

$i_{2}\leq 3i_{1}+1$.

We compute $\Delta_{1\rightarrow 4}c(\nu,J)$.

$\displaystyle\Delta_{1\rightarrow 4}c(\nu,J)=t$ $\displaystyle+\Delta
r_{i_{1}}^{(2)}-(\Delta p_{i_{1}-1}^{(2)}+\Delta r_{i_{1}-1}^{(2)})$
$\displaystyle+\Delta r_{i_{3}}^{(1)}-(\Delta p_{i_{3}-2}^{(1)}+\Delta
r_{i_{3}-2}^{(1)}),$

where $t=-2$ when the $i_{3}$-string is type-II $(i_{3}=3i_{1}+1)$ and $t=-1$
otherwise. As before $\Delta r_{i_{1}}^{(2)}=\Delta r_{i_{1}-1}^{(2)}=0$,
$\Delta p_{i_{1}-1}^{(2)}=-1$, and $\Delta r_{i_{3}-2}^{(1)}=0$ because the
box-deleted $(i_{3}-2)$-string is set to be singular. The values of $\Delta
r_{i_{3}}^{(1)}$ and $\Delta p_{i_{3}-2}^{(1)}$ (see (VC-2)) are listed below.
In this table the type is refereed to the type of the $i_{3}$-string. Note
that the $i_{3}$-string is singular when it is type-I ($i_{3}=3i_{1}-1$) while
it is q-singular when the type is not I.

type | $\Delta_{i_{3}}^{rig}$ | $\Delta p_{i_{3}-2}^{(1)}$
---|---|---
0 | $-1$ | $-1$
I | $0$ | $0$
II | $-1$ | $-2$

Hence altogether $\Delta_{1\rightarrow 4}c(\nu,J)=0$ so that $\Delta
c(\nu,J)=\Delta_{1}c(\nu,J)+\Delta_{1\rightarrow
4}c(\nu,J)=-\alpha_{1}^{(2)}$.

    * (b)

$i_{2}\geq 3i_{1}+2$.

Since $\Delta_{1\rightarrow 4}c(\nu,J)=\Delta_{1\rightarrow
2}c(\nu,J)+\Delta_{2\rightarrow 4}c(\nu,J)$, we compute $\Delta_{2\rightarrow
4}c(\nu,J)$.

$\Delta_{1\rightarrow 3}c(\nu,J)=-2+\Delta r_{i_{3}}^{(1)}-(\Delta
p_{i_{3}-2}^{(1)}+\Delta r_{i_{3}-2}^{(1)}),$

where $\Delta p_{i_{3}-2}^{(1)}=-3$ by (VC-2), $\Delta r_{i_{3}-2}^{(1)}=0$,
and $\Delta r_{i_{3}}^{(1)}=-1$ because the selected $i_{3}$-string is
q-singular. Hence $\Delta_{2\rightarrow 4}c(\nu,J)=0$ so that $\Delta
c(\nu,J)=\Delta_{1}c(\nu,J)+\Delta_{1\rightarrow
2}c(\nu,J)+\Delta_{2\rightarrow 4}c(\nu,J)=-\alpha_{1}^{(2)}$.

  * (2)

$i_{2}<i_{3}$.

    * (a)

$i_{3}^{eff}=i_{1}$.

We compute $\Delta_{1\rightarrow 4}c(\nu,J)$. In this case $i_{2}=3i_{1}-2$
and $i_{3}=3i_{1}$. The $i_{2}$-string is singular and the $i_{3}$-string is
q-singular (see (BM-3)); $\Delta r_{i_{2}}^{(1)}=0$ and $\Delta
r_{i_{3}}^{(1)}=-1$.

$\displaystyle\Delta_{1\rightarrow 4}c(\nu,J)=-1$ $\displaystyle+\Delta
r_{i_{1}}^{(2)}-(\Delta p_{i_{1}-1}^{(2)}+\Delta r_{i_{1}-1}^{(2)})$
$\displaystyle+\Delta r_{i_{2}}^{(1)}-(\Delta p_{i_{2}-1}^{(1)}+\Delta
r_{i_{2}-1}^{(1)})$ $\displaystyle+\Delta r_{i_{3}}^{(1)}-(\Delta
p_{i_{3}-1}^{(1)}+\Delta r_{i_{3}-1}^{(1)}),$

where $\Delta r_{i_{1}}^{(2)}=\Delta r_{i_{1}-1}^{(2)}=0$, $\Delta
p_{i_{1}-1}^{(2)}=-1$ as before. By (RA-2) $\Delta r_{i_{2}-1}^{(1)}=-1$ and
$\Delta r_{i_{3}-1}^{(1)}=-1$. By (VC-3) $\Delta p_{i_{2}-1}^{(1)}=0$ and
$\Delta p_{i_{3}-1}^{(1)}=0$. Hence altogether $\Delta_{1\rightarrow
4}c(\nu,J)=0$ so that $\Delta c(\nu,J)=\Delta_{1}c(\nu,J)+\Delta_{1\rightarrow
4}c(\nu,J)=-\alpha_{1}^{(2)}$.

    * (b)

$i_{3}^{eff}>i_{1}$.

We compute $\Delta_{3\rightarrow 4}c(\nu,J)$.

$\Delta_{3\rightarrow 4}c(\nu,J)=-1+\Delta r_{i_{3}}^{(1)}-(\Delta
p_{i_{3}-1}^{(1)}+\Delta r_{i_{3}-1}^{(1)}),$

where $\Delta p_{i_{3}-1}^{(1)}=-1$ by (VC-3) and $\Delta
r_{i_{3}}^{(1)}=\Delta r_{i_{3}-1}^{(1)}=-1$. Hence $\Delta_{3\rightarrow
4}c(\nu,J)=0$ so that $\Delta c(\nu,J)=\Delta_{1}c(\nu,J)+\Delta_{1\rightarrow
3}c(\nu,J)+\Delta_{3\rightarrow 4}c(\nu,J)=-\alpha_{1}^{(2)}$.

The verifications of $\Delta c(\nu,J)=-\alpha_{1}^{(2)}$ for
$b=\framebox{$i$}$ $(5\leq i\leq 14)$ are similar and we omit the details.

∎

Define the following subsets of $B_{0}\otimes B_{0}$.

$\displaystyle S_{0}=$
$\displaystyle\left\\{\framebox{$1$}\otimes\framebox{$1$},\framebox{$14$}\otimes\framebox{$14$}\right\\}\bigsqcup\left\\{\framebox{$i$}\otimes\framebox{$j$}\mathrel{}\middle|\mathrel{}i=1,2;2\leq
j\leq 14\right\\}$
$\displaystyle\bigsqcup\left\\{\framebox{$i$}\otimes\framebox{$j$}\mathrel{}\middle|\mathrel{}i=3,4,6;6\leq
j\leq 14;j\neq 7\right\\}$
$\displaystyle\bigsqcup\left\\{\framebox{$i$}\otimes\framebox{$j$}\mathrel{}\middle|\mathrel{}i=5,8,10;10\leq
j\leq 14\right\\}$
$\displaystyle\bigsqcup\left\\{\framebox{$i$}\otimes\framebox{$j$}\mathrel{}\middle|\mathrel{}i=7,9,11,12,13;j=13,14\right\\},$
$\displaystyle S_{1}=$
$\displaystyle\left\\{\framebox{$2$}\otimes\framebox{$1$}\right\\}\bigsqcup\left\\{\framebox{$i$}\otimes\framebox{$j$}\mathrel{}\middle|\mathrel{}i=3,4,6;1\leq
j\leq 7;j\neq 6\right\\}$
$\displaystyle\bigsqcup\left\\{\framebox{$i$}\otimes\framebox{$j$}\mathrel{}\middle|\mathrel{}i=5,8,10;2\leq
j\leq 9\right\\}$
$\displaystyle\bigsqcup\left\\{\framebox{$i$}\otimes\framebox{$j$}\mathrel{}\middle|\mathrel{}i=7,9,11,12,13;6\leq
j\leq 12;j\neq 7\right\\}$
$\displaystyle\bigsqcup\left\\{\framebox{$14$}\otimes\framebox{$j$}\mathrel{}\middle|\mathrel{}10\leq
j\leq 13\right\\},$

and

$\displaystyle S_{2}=$
$\displaystyle\left\\{\framebox{$i$}\otimes\framebox{$1$}\mathrel{}\middle|\mathrel{}i=5,8,10\right\\}$
$\displaystyle\bigsqcup\left\\{\framebox{$i$}\otimes\framebox{$j$}\mathrel{}\middle|\mathrel{}i=7,9,11,12,13;1\leq
j\leq 7;j\neq 6\right\\}$
$\displaystyle\bigsqcup\left\\{\framebox{$14$}\otimes\framebox{$j$}\mathrel{}\middle|\mathrel{}1\leq
j\leq 9\right\\}.$

The subset $S_{0}$ (resp. $S_{1}$) is $B(2\bar{\Lambda}_{2})$ (resp.
$B(3\bar{\Lambda}_{1})$) in Eq. (2.1) and the subset $S_{2}$ is the disjoint
union of $B(2\bar{\Lambda}_{1})$, $B(\bar{\Lambda}_{2})$, and $B(0)$ in Eq.
(2.1) so that

(5.16) $H(b_{1}\otimes b_{2})=\begin{cases}0&\text{if }b_{1}\otimes b_{2}\in
S_{0},\\\ -1&\text{if }b_{1}\otimes b_{2}\in S_{1},\\\
-2&\text{otherwise}.\end{cases}$

###### Proof of (V).

The proof is reduced to showing the following lemma. ∎

######  Lemma 5.6.

$b_{1}\otimes b_{2}$ belongs to $S_{2}$ or takes the form of
$\framebox{$\emptyset$}\otimes b$ if and only if
$\alpha_{1}^{(2)}-\tilde{\alpha}_{1}^{(2)}=2$, where $b=\framebox{$j$}$ with
$1\leq j\leq 14$ or $j=\emptyset$. $b_{1}\otimes b_{2}$ belongs to $S_{1}$ if
and only if $\alpha_{1}^{(2)}-\tilde{\alpha}_{1}^{(2)}=1$. $b_{1}\otimes
b_{2}$ belongs to $S_{0}$ or takes the form of
$b\otimes\framebox{$\emptyset$}$ with $b=\framebox{$i$}$ with $1\leq i\leq 14$
if and only if $\alpha_{1}^{(2)}-\tilde{\alpha}_{1}^{(2)}=0$.

###### Proof.

Assertions are the direct consequences of the following three lemmas. ∎

######  Lemma 5.7.

If $\alpha_{1}^{(2)}-\tilde{\alpha}_{1}^{(2)}=2$. then $b_{1}\otimes b_{2}$
belongs to $S_{2}$ or it takes the form of $\framebox{$\emptyset$}\otimes b$
with $b=\framebox{$i$}$ with $1\leq i\leq 14$ or $i=\emptyset$.

######  Lemma 5.8.

If $\alpha_{1}^{(2)}-\tilde{\alpha}_{1}^{(2)}=1$. then $b_{1}\otimes b_{2}$
belongs to $S_{1}$. In particular, it does not take the form of
$b\otimes\framebox{$\emptyset$}$ or $\framebox{$\emptyset$}\otimes b$.

######  Lemma 5.9.

If $\alpha_{1}^{(2)}-\tilde{\alpha}_{1}^{(2)}=0$. then $b_{1}\otimes b_{2}$
belongs to $S_{0}$ or takes the form of $b\otimes\framebox{$\emptyset$}$ with
$b=\framebox{$i$}$ with $1\leq i\leq 14$.

We give the proof of Lemma 5.7 only. Proofs of Lemma 5.8 and Lemma 5.9 are
similar.

###### Proof of Lemma 5.7.

Since $\alpha_{1}^{(2)}-\tilde{\alpha}_{1}^{(2)}=2$, $i\geq 5$ but $i\neq 6$.
It is obvious that $b_{1}=\framebox{$\emptyset$}$ is also possible.

Case 1. $b_{1}=\framebox{$\emptyset$}$.

From (VC-$\emptyset$), it is obvious that $1\leq j\leq 14$ or $j=\emptyset$.

Case 2. $b_{1}=\framebox{$5$}$

Since $b_{1}=\framebox{$5$}$, after [4] was marked in $\nu^{(2)}$ the box
marking has terminated. Since $\alpha_{1}^{(2)}-\tilde{\alpha}_{1}^{(2)}=2$,
two strings of length one are marked by [1] and [4] in $\nu^{(2)}$ and they
are deleted in $\tilde{\nu}^{(1)}$. Since $\Delta p_{i}^{(2)}=1$ $(i\geq 1)$
(see (VC-5)), all strings of $\tilde{\nu}^{(2)}$ are not singular so that
$j=1$. In particular $j\neq\emptyset$.

Case 3. $b_{1}=\framebox{$7$}$.

Since $b_{1}=\framebox{$7$}$, after [5] was marked in $\nu^{(1)}$ the box
marking has terminated. We claim that $1\leq j\leq 7$ but $j\neq 6$ and
$j\neq\emptyset$. To do so, we show that neither [4] nor [6] can be marked in
$\tilde{\nu}^{(1)}$. From (VC-7), the changes of vacancy numbers are
summarized as

$\Delta p_{i}^{(1)}$ |  | $\Delta p_{i}^{(2)}$ |
---|---|---|---
$0$ | $(i\geq i_{5})$ | $0$ | $(i\geq i_{5}^{eff})$
|  | $1$ | $(1\leq i\leq i_{5}^{eff}-1)$

Note that strings of length ($\geq i_{5}$) in $\nu^{(1)}$ and therefore in
$\tilde{\nu}^{(1)}$ are q-singular at best. Otherwise [6] would be marked in
$\nu^{(1)}$ contradicting $b_{1}=\framebox{$7$}$. The selected $i_{1}$\- and
$i_{4}$-strings of length one are deleted in $\tilde{\nu}^{(2)}$ so that it is
obvious from the table above that [1] can be marked in a string of length
($\geq i_{5}^{eff}$) if possible. If [1] cannot be marked in
$\tilde{\nu}^{(2)}$, then $b_{2}=\framebox{$1$}$. If [1] is marked in
$\tilde{\nu}^{(2)}$, then the box marking in $\tilde{\nu}^{(1)}$ is possible
only for strings of length ($\geq 3i_{5}^{eff}-2$). Therefore, in order to
mark [4] or [6] there must be a singular string of length $(\geq
3i_{5}^{eff})$ in $\tilde{\nu}^{(1)}$. However, such a string does not exist
because string of length $(\geq i_{5})$ in $\tilde{\nu}^{(1)}$ are q-singular
at best. Hence the claim follows.

Case 4. $b_{1}=\framebox{$8$}$.

Since $b_{1}=\framebox{$8$}$, after [5] is marked in $\nu^{(2)}$ the box
marking has terminated. Two strings of length one are marked by [1] and [5] in
$\nu^{(2)}$. Only one 1-string (the selected $i_{1}$-string) in $\nu^{(2)}$ is
singular. Strings of length $(\geq 2)$ in $\nu^{(2)}$ are not singular.
Otherwise [6] would be marked in $\nu^{(2)}$ contradicting
$b_{1}=\framebox{$8$}$. Since $\Delta p_{i}^{(2)}=0$ $(i\geq 1)$ (see (VC-8)),
all strings of $\tilde{\nu}^{(2)}$ are still not singular so that $j=1$ and
$j\neq\emptyset$.

Case 5. $b_{1}=\framebox{$9$}$.

Since $b_{1}=\framebox{$9$}$, after [6] was marked in $\nu^{(1)}$ the box
marking has terminated. We claim that $1\leq j\leq 7$ but $j\neq 6$ and
$j\neq\emptyset$. To do so, we show that neither [4] nor [6] can be marked in
$\tilde{\nu}^{(1)}$. From (VC-9), we have two cases.

  * (1)

$i_{5}=i_{6}$.

    * (a)

The selected $i_{6}$-string in $\nu^{(1)}$ is type-II.

$\Delta p_{i}^{(1)}$ |  | $\Delta p_{i}^{(2)}$ |
---|---|---|---
2 | $(i\geq i_{6})$ | $-1$ | $(i\geq i_{6}^{eff})$
0 | $(i=i_{6}-1)$ | 0 | $(i=i_{6}^{eff}-1)$
|  | 1 | $(1\leq i\leq i_{6}^{eff}-2)$

Since the selected $i_{1}$\- and $i_{4}$-strings are deleted in
$\tilde{\nu}^{(2)}$, it is obvious from the table above that [1] can be marked
in a string of length ($\geq i_{6}^{eff}-1$) if possible. If [1] cannot be
marked in $\tilde{\nu}^{(2)}$, then $b_{2}=\framebox{$1$}$. If [1] is marked
in $\tilde{\nu}^{(2)}$, then the box marking in $\tilde{\nu}^{(1)}$ is
possible only for strings of length ($\geq 3(i_{6}^{eff}-1)-2=i_{6}-3$).
Therefore, in order to mark [4] or [6] there must a singular string of length
($\geq i_{6}-1$) in $\tilde{\nu}^{(1)}$. However such a string does not exist
in $\tilde{\nu}^{(1)}$. This is shown as follows. The $(i_{6}-1)$-strings
(type-0) in $\nu^{(1)}$ are qq-singular at best. Otherwise [5] would be marked
in an $(i_{6}-1)$-string. Since $\Delta p_{i_{6}-1}^{(1)}=0$ and $\Delta
p_{i}^{(1)}=2$ $(i\geq i_{6})$, strings of length $(\geq i_{6}-1)$ in
$\tilde{\nu}^{(1)}$ are qq-singular at best.

    * (b)

The selected $i_{6}$-string in $\nu^{(1)}$ is type-0/I.

$\Delta p_{i}^{(1)}$ |  | $\Delta p_{i}^{(2)}$ |
---|---|---|---
2 | $(i\geq i_{6})$ | $-1$ | $(i\geq i_{6}^{eff})$
0 | $(i=i_{6}-1)$ | 1 | $(1\leq i\leq i_{6}^{eff}-1)$

From the table above, box marking in $\tilde{\nu}^{(1)}$ is possible only for
strings of length ($\geq 3i_{6}^{eff}-2$). The remaining argument is the same
as in (a).

  * (2)

$i_{5}<i_{6}$.

$\Delta p_{i}^{(1)}$ |  | $\Delta p_{i}^{(2)}$ |
---|---|---|---
2 | $(i\geq i_{6})$ | $-1$ | $(i\geq i_{6}^{eff})$

The changes of vacancy numbers $\Delta p_{i}^{(1)}$ $(i\leq i_{6}-1)$ and
$\Delta p_{i}^{(2)}$ $(i\leq i_{6}^{eff}-1)$ are the same as those in Case 3.
Note that the box-deleted $(i_{6}-1)$-string in $\tilde{\nu}^{(1)}$ is set to
be q-singular by (RA-1). The argument is the same as that in Case 3.

Case 6. $b_{1}=\framebox{$10$}$.

Since $b_{1}=\framebox{$10$}$, after [6] was marked in $\nu^{(2)}$ the box
marking has terminated.

  * (1)

$i_{5}=i_{6}=2$.

The effective length of the $i_{4}$-string in $\nu^{(1)}$ is one. Otherwise,
[7] would be marked on the left of [4] in $\nu^{(1)}$. Therefore, only one
1-string (the $i_{1}$-string) in $\nu^{(2)}$ is singular. Otherwise [5] would
be marked in the singular 1-string in $\nu^{(2)}$ and the box marking would
terminate. Since $\Delta p_{1}^{(2)}=0$ (see (VC-10)), strings of length one
in $\tilde{\nu}^{(2)}$ are not singular. Furthermore, $\Delta p_{i}^{(2)}=2$
$(i\geq 2)$ (see (VC-10)), strings of length ($\geq 2$) in $\tilde{\nu}^{(2)}$
are also not singular. Hence $j=1$. In particular, $j\neq\emptyset$.

  * (2)

$i_{5}=1$ and $i_{6}\geq 2$.

Since strings of length $i$ $(1\leq i\leq i_{6}-1)$ in $\nu^{(1)}$ except the
$i_{1}$-string are not singular and $\Delta p_{i}^{(2)}=0$ $(1\leq i\leq
i_{6}-1)$ (see (VC-10)), strings of length $i$ $(2\leq i\leq i_{6}-1)$ in
$\tilde{\nu}^{(1)}$ are also not singular. Furthermore, $\Delta p_{i}^{(2)}=2$
$(i\geq i_{6})$ (see (VC-10)) so that all strings in $\tilde{\nu}^{(2)}$ are
not singular. Hence $j=1$. In particular, $j\neq\emptyset$.

Case 7. $b_{1}=\framebox{$11$}$.

Since $b_{1}=\framebox{$11$}$, after [7] was marked in $\nu^{(1)}$ or
$\nu^{(2)}$ the box marking has terminated. We claim that $1\leq j\leq 7$ but
$j\neq 6$ and $j\neq\emptyset$. To do so, we show that neither [4] nor [6] can
be marked in $\tilde{\nu}^{(1)}$.

Suppose that [7] is marked in $\nu^{(1)}$. Note that [7] is always marked in
the string which is unmarked so far. The changes of vacancy numbers $\Delta
p_{i}^{(2)}$ in Case 6 are changed to $\Delta p_{i}^{(2)}=1$ only for $i\geq
i_{7}^{eff}$ in this case. Hence $j=1$ and $j\neq\emptyset$ as in Case 6. So
we assume that [6] is always marked in $\nu^{(1)}$ and [7] is always marked in
$\nu^{(2)}$. Furthermore if $i_{6}^{eff}<i_{7}$, then $\Delta p_{i}^{(1)}=-1$
$(i\geq 3i_{7})$, $\Delta p_{3i_{7}-1}^{(1)}=0$, and $\Delta
p_{3i_{7}-2}^{(1)}=1$ from (VC-11). Here we note the fact that [8] cannot be
marked in $\nu^{(1)}$. This fact implies that strings of length $3i_{7}-2$ in
$\nu^{(1)}$ are singular at best, strings of length $3i_{7}-1$ in $\nu^{(1)}$
are q-singular at best, and strings of length ($\geq 3i_{7}$) are qq-singular
at best by the algorithm $\delta_{\theta}$. Hence strings of length ($\geq
3i_{7}-2$) in $\tilde{\nu}^{(1)}$ are q-singular at best. The singularity of
strings of length ($\leq 3i_{7}-3$) is the same as in Case 5. Hence $1\leq
j\leq 7$ but $j\neq 6$ and $j\neq\emptyset$. So we assume that
$i_{6}^{eff}=i_{7}$. From (VC-11), we have following two cases.

  * (1)

$i_{5}=i_{6}$.

$\Delta p_{i}^{(1)}$ |  | $\Delta p_{i}^{(2)}$ |
---|---|---|---
$-1$ | $(i\geq 3i_{7})$ | 1 | $(i\geq i_{7})$
0 | $(i=i_{6}+1)$ | $0$ | $(i=i_{7}-1)$
1 | $(i=i_{6})$ | 1 | $(1\leq i\leq i_{7}-2)$
0 | $(i=i_{6}-1)$ |  |

Note that the $i_{6}$-string must be type-II. Otherwise [8] would be marked on
the left of [6]. Since the selected $i_{1}$\- and $i_{4}$-strings are deleted
in $\tilde{\nu}^{(2)}$, it is obvious that [1] can be marked in a string of
length ($\geq i_{7}-1$). Therefore if [1] is marked in $\tilde{\nu}^{(2)}$,
then the box marking is possible only for strings of length ($\geq
3(i_{7}-1)-2=i_{6}-3$). Therefor, in order to mark [4] or [6] there must be a
singular string of length ($\geq i_{6}-1$) in $\tilde{\nu}^{(1)}$. However,
such a string does not exist in $\tilde{\nu}^{(1)}$. This is shown as follows.
The $(i_{6}-1)$-strings in $\nu^{(1)}$ are qq-singular at best. Otherwise [5]
would be marked in an $(i_{6}-1)$-string. Furthermore, since [8] cannot be
marked in $\nu^{(1)}$, strings of length $3i_{7}-1(=i_{6}+1)$ are q-singular
at best and strings of length ($\geq 3i_{7}$) are qq-singular at best. Hence,
from the table above, strings of length ($\geq i_{6}-1$) are not singular so
that the box marking of [4] or [6] in $\tilde{\nu}^{(1)}$ is impossible.

  * (2)

$i_{5}<i_{6}$.

$\Delta p_{i}^{(1)}$ |  | $\Delta p_{i}^{(2)}$ |
---|---|---|---
$-1$ | $(i\geq 3i_{7})$ | 1 | $(i\geq i_{7})$
0 | $(i=i_{6}+1)$ | $0$ | $(i_{5}^{eff}\leq i\leq i_{7}-1)$
1 | $(i=i_{6})$ | 1 | $(1\leq i\leq i_{5}^{eff}-1)$
0 | $(i=i_{6}-1)$ |  |

Note that the $i_{6}$-string must be type-II. Otherwise [8] would be marked on
the left of [6]. Since the selected $i_{1}$\- and $i_{4}$-strings are deleted
in $\tilde{\nu}^{(2)}$, it is obvious that [1] can be marked in a string of
length ($\geq i_{5}^{eff}$). Therefore if [1] is marked in
$\tilde{\nu}^{(2)}$, then the box marking is possible only for strings of
length ($\geq 3i_{5}^{eff}-2$) and in order to [4] or [6] there must be a
singular string of length ($\geq 3i_{5}^{eff}$). However, such a string does
not exist in $\tilde{\nu}^{(1)}$. This is shown as follows. Strings of length
$i$ ($i_{5}\leq i\leq i_{6}-1$) in $\nu^{(1)}$ are q-singular at best and
their singularity does not change in $\tilde{\nu}^{(1)}$. Furthermore, since
[8] cannot be marked in $\nu^{(1)}$, strings of length $3i_{7}-1(=i_{6}+1)$
are q-singular at best and strings of length ($\geq 3i_{7}$) are qq-singular
at best. Hence from the table above, strings of length ($\geq i_{5}$) in
$\tilde{\nu}^{(1)}$ are not singular so that the box marking of [4] or [6] in
$\tilde{\nu}^{(1)}$ is impossible.

Case 8. $b_{1}=\framebox{$12$}$.

Since $b_{1}=\framebox{$12$}$, after [8] was marked in $\nu^{(1)}$ the box
marking has terminated. We claim that $1\leq j\leq 7$ but $j\neq 6$ and
$j\neq\emptyset$. To do so, we show that neither [4] nor [6] can be marked in
$\tilde{\nu}^{(1)}$.

Firstly, we suppose that [8] is marked in the string which is unmarked so far
in $\nu^{(1)}$, where [7] is marked in $\nu^{(1)}$ or $\nu^{(2)}$. From
(VC-12) (5), $\Delta p_{i}^{(1)}=1$ $(i\geq i_{8})$ and $\Delta p_{i}^{(2)}=0$
$(i\geq i_{8}^{eff})$. The changes of vacancy numbers $\Delta p_{i}^{(1)}$
($i\leq i_{8}-1$) and $\Delta p_{i}^{(2)}$ ($i\leq i_{8}^{eff}-1$) are the
same as those in Case 7. The box-deleted $(i_{8}-1)$-string in
$\tilde{\nu}^{(1)}$ is set to be q-singular by the rule of rigging adjustment
(RA-5). Therefore, the box marking of [4] or [6] in $\tilde{\nu}^{(1)}$ is
impossible as in Case 7.

Secondly, we suppose that [8] is marked in a marked string. From (VC-12), we
have following four cases.

  * (1)

The $i_{7}$-string is in $\nu^{(1)}$ with $i_{8}=i_{7}=i_{4}$
($i_{5}=i_{6}=2$) or the $i_{7}$-string is in $\nu^{(2)}$ with
$i_{4}=i_{7}(=2)$. The box marking in $\nu^{(1)}$ is one of the following
three.

$\downarrow$$[8]$$[7]$$[4]$$[3]$$[2]$or$\downarrow$$[2]$$[8]$$[7]$$[4]$$[3]$or$\downarrow$$[3]$$[2]$$[8]$$[6]$$[5]$

$\Delta p_{i}^{(1)}$ |  | $\Delta p_{i}^{(2)}$ |
---|---|---|---
$1$ | $(i\geq 6)$ | $0$ | $(i\geq 1)$
$2$ | $(i=5)$ |  |
$1$ | $(i=4)$ |  |
$0$ | $(i=3)$ |  |

Since 3-strings in $\nu^{(1)}$ are not singular, it is obvious that the box
marking of [4] or [6] in $\tilde{\nu}^{(1)}$ is impossible.

  * (2)

$i_{8}=i_{7}\neq i_{4}$.

    * (a)

$i_{8}^{eff}=i_{6}$.

$\Delta p_{i}^{(1)}$ |  | $\Delta p_{i}^{(2)}$ |
---|---|---|---
$1$ | $(i\geq 3i_{6})$ | $0$ | $(i\geq i_{6})$

    * (b)

$i_{8}^{eff}>i_{6}$.

$\Delta p_{i}^{(1)}$ |  | $\Delta p_{i}^{(2)}$ |
---|---|---|---
$1$ | $(i\geq i_{8})$ | $0$ | $(i\geq i_{8}^{eff})$

The changes of vacancy numbers $\Delta p_{i}^{(2)}$ $(i\leq i_{6}-1)$ (case
(a)) and $\Delta p_{i}^{(2)}$ $(i\leq i_{8}^{eff}-1)$ (case (b)) are the same
as those in Case 6 so that the box marking in $\tilde{\nu}^{(1)}$ is possible
only for strings of length $(\geq 3i_{6}-2)$ (case (a)) and $(\geq
3i_{8}^{eff}-2)$ (case (b)). It is obvious that the box marking of [4] or [6]
in $\tilde{\nu}^{(1)}$ is impossible.

  * (3)

$i_{8}=i_{6}=i_{5}\neq i_{3}$.

In this case, the $i_{8}$-string is type-I and $i_{8}^{eff}=i_{7}$;
$i_{8}=3i_{7}-1$.

$\Delta p_{i}^{(1)}$ |  | $\Delta p_{i}^{(2)}$ |
---|---|---|---
$1$ | $(i\geq i_{8}+1)$ | $0$ | $(i\geq i_{7}-1)$
$2$ | $(i=i_{8})$ | $1$ | $(1\leq i\leq i_{7}-2)$
$1$ | $(i=i_{8}-1)$ |  |
$0$ | $(i=i_{8}-2)$ |  |

The box marking in $\tilde{\nu}^{(1)}$ is possible only for strings of length
$(\geq 3(i_{7}-1)-2=i_{8}-4)$. Therefore, in order to mark [4] or [6] there
must be a singular string of length ($\geq i_{8}-2$) in $\tilde{\nu}^{(1)}$.
However, such a string does not exist. This is shown as follows. The
$(i_{8}-2)$-strings in $\nu^{(1)}$ are not singular as otherwise [4] would be
marked in this string. Therefore, strings of length ($\geq i_{8}-2$) in
$\tilde{\nu}^{(1)}$ are not singular.

  * (4)

$i_{8}=i_{6}\neq i_{5}$.

In this case, the $i_{8}$-string is type-I and $i_{8}^{eff}=i_{7}$;
$i_{8}=3i_{7}-1$. Note that the box-deleted $(i_{8}-2)$-string is set to be
q-singular by (RA-1).

    * (a)

$i_{3}=i_{5}(=3)$.

$\Delta p_{i}^{(1)}$ |  | $\Delta p_{i}^{(2)}$ |
---|---|---|---
$1$ | $(i\geq i_{8}+1)$ | $0$ | $(i\geq i_{7}-1)$
$2$ | $(i=i_{8})$ | $1$ | $(1\leq i\leq i_{7}-2)$
$1$ | $(i=i_{8}-1)$ |  |

    * (b)

$i_{3}<i_{5}$.

$\Delta p_{i}^{(1)}$ |  | $\Delta p_{i}^{(2)}$ |
---|---|---|---
$1$ | $(i\geq i_{8}+1)$ | $0$ | $(i\geq i_{5}^{eff})$
$2$ | $(i=i_{8})$ | $1$ | $(1\leq i\leq i_{5}^{eff}-1)$
$1$ | $(i=i_{8}-1)$ |  |

In both cases, it is obvious that the box marking of [4] or [6] in
$\tilde{\nu}^{(1)}$ is impossible.

Case 9. $b_{1}=\framebox{$13$}$.

Since $b_{1}=\framebox{$13$}$, after [9] was marked in $\nu^{(1)}$ the box
marking has terminated. We claim that $1\leq j\leq 7$ but $j\neq 6$ and
$j\neq\emptyset$. The proof is very similar to Case 8 and we omit the details.

Case 10. $b_{1}=\framebox{$14$}$.

We claim that $1\leq j\leq 9$ and $j\neq\emptyset$.

  * (1)

$i_{1}<i_{10}$.

    * (a)

$i_{9}^{eff}<i_{10}$.

$\Delta p_{i}^{(1)}$ |  | $\Delta p_{i}^{(2)}$ |
---|---|---|---
$0$ | $(i\geq 3i_{10})$ | 1 | $(i\geq i_{10})$
1 | $(i=3i_{10}-1)$ |  |
2 | $(i=3i_{10}-2)$ |  |
3 | $(i_{9}\leq i\leq 3i_{10}-3)$ |  |

The changes of vacancy numbers $\Delta p_{i}^{(1)}$ $(i\leq 3(i_{10}-1))$ are
the same as in Case 13. Therefore, [4] or [6] can be marked in a singular
string of length ($\geq 3i_{10}$) in $\tilde{\nu}^{(1)}$. If [4] is marked in
such a string, then $j\geq 6$. Although [5] can be marked in a q-singular
string of length ($\geq i_{10}$) in $\tilde{\nu}^{(2)}$, [6] cannot so that
$j=8$ is possible but $j\ngeq 10$. In particular, $j\neq\emptyset$. If [6] is
marked in such a string, then $j\geq 9$. Since [7] cannot be marked in
$\tilde{\nu}^{(2)}$ because strings of length ($\geq i_{10}$) in
$\tilde{\nu}^{(2)}$ are q-singular at best so that $j\ngeq 11$. Hence
altogether we have $1\leq j\leq 9$ and $j\neq\emptyset$.

    * (b)

$i_{9}^{eff}=i_{10}$.

Possible ways of box marking in $\nu^{(1)}$ are following three;

$\downarrow$$[9]$$[8]$$[7]$(i)$[9]$$[8]$(ii)$[9]$(iii)

We omit the unmarked or marked strings except the selected $i_{9}$-string. In
either case, $\Delta p_{i}^{(2)}=1$ $(i\geq i_{10})$. The changes of vacancy
numbers $\Delta p_{i}^{(1)}$ are obtained by overwriting

$-3\chi(i\geq 3i_{10})-2\chi(i=3i_{10}-1)-\chi(i=3i_{10}-2)$

on $\Delta p_{i}^{(1)}$ in Case 13 where $\Delta p_{i}^{(1)}\leq 3$ and in
particular $\Delta p_{i}^{(1)}=3$ $(i\geq i_{9})$. Therefore, $\Delta
p_{i}^{(1)}\leq 0$ for some strings of effective length $i_{10}$ and $\Delta
p_{i}^{(1)}=0$ $(i\geq 3i_{10})$. In case (i), the selected $i_{9}$-string is
type-0 and the box-deleted $(i_{9}-3)$-string in $\tilde{\nu}^{(1)}$ is set to
be singular but [4] or [6] cannot be marked in this string as in Case 9 so
that [4] or [6] can be marked in only a singular string of effective length
($\geq i_{10}$) in $\tilde{\nu}^{(1)}$. The remaining argument is the same as
in case (a) so that $1\leq j\leq 9$ and $j\neq\emptyset$. In cases (ii) and
(iii), the box-deleted $(i_{9}-1)$\- or $(i_{9}-2)$-string is set to be q/qq-
singular by (RA-6) or (RA-7) so that [4] or [6] cannot marked in this string.
Similarly, we have $1\leq j\leq 9$ and $j\neq\emptyset$.

  * (2)

$i_{1}=i_{10}$.

From (VC-14), it is obvious that $j=1$.

∎

##  6\. Verifications of the Rules of Forbidden Box Marking

In this section we verify the rules of forbidden box marking. They are
necessary to ensure Eq. (5.1).

In (BM-2), we excluded the box marking in $\nu^{(1)}$ depicted below.

$\downarrow$$[2]$$[3]$and$\downarrow$$[2]$$[3]$

where $i_{2}^{eff}>i_{1}$ and the $i_{2}$ (resp. $i_{3}$)-string is qq-
singular (resp. singular). Suppose that [4] is not marked in $\nu^{(2)}$.
Then,

$\displaystyle\Delta_{2\rightarrow 4}c(\nu,J)=-2$ $\displaystyle+\Delta
r_{i_{2}}^{(1)}-(\Delta p_{i_{2}-1}^{(1)}+\Delta r_{i_{2}-1}^{(1)})$
$\displaystyle+\Delta r_{i_{3}}^{(1)}-(\Delta p_{i_{3}-1}^{(1)}+\Delta
r_{i_{3}-1}^{(1)}),$

where $\Delta r_{i_{2}}^{(1)}=-2$, $\Delta r_{i_{3}}^{(1)}=0$, $\Delta
r_{i_{2}-1}^{(1)}=0$ and $\Delta r_{i_{3}-1}^{(1)}=-1$ by (RA-2). The changes
of vacancy numbers are $\Delta p_{i_{2}-1}^{(1)}=-3$ and $\Delta
p_{i_{3}-1}^{(1)}=-1$ (see (VC-2) and (VC-3)). Therefore $\Delta_{2\rightarrow
4}c(\nu,J)=-1$ so that $\Delta c(\nu,J)\neq\alpha_{1}^{(2)}$.

We excluded the box marking in $\nu^{(1)}$ depicted below.

$\downarrow$$[2]$$[3]$

where $i_{2}^{eff}>i_{1}$ and the $i_{2}$ (resp. $i_{3}$)-string is qq-
singular (resp. q-singular). Suppose that [4] is marked in $\nu^{(2)}$ with
$i_{3}^{eff}=i_{4}$. Then,

$\displaystyle\Delta_{2\rightarrow 5}c(\nu,J)=-1$ $\displaystyle+\Delta
r_{i_{2}}^{(1)}-(\Delta p_{i_{2}-1}^{(1)}+\Delta r_{i_{2}-1}^{(1)})$
$\displaystyle+\Delta r_{i_{3}}^{(1)}-(\Delta p_{i_{3}-1}^{(1)}+\Delta
r_{i_{3}-1}^{(1)})$ $\displaystyle+\Delta r_{i_{4}}^{(2)}-(\Delta
p_{i_{4}-1}^{(2)}+\Delta r_{i_{4}-1}^{(2)}),$

where $\Delta r_{i_{2}}^{(1)}=-2$, $\Delta r_{i_{3}}^{(1)}=-1$, $\Delta
r_{i_{4}}^{(2)}=0$, $\Delta r_{i_{2}-1}^{(1)}=0$, $\Delta r_{i_{3}-1}^{(1)}=0$
by (RA-2), and $\Delta r_{i_{4}-1}^{(2)}=0$, The changes of vacancy numbers
$\Delta p_{i_{2}-1}^{(1)}$ and $\Delta p_{i_{3}-1}^{(1)}$ are obtained by
overwriting

$-3\chi(i\geq 3i_{5})-2\chi(i=3i_{5}-1)-\chi(i=3i_{5}-2)$

on the values of previous case. That is, $\Delta p_{i_{2}-1}^{(1)}=-4$ and
$\Delta p_{i_{3}-1}^{(1)}=-3$. By (VC-3) and (VC-4) $\Delta
p_{i_{4}-1}^{(2)}=0$. Therefore, $\Delta_{2\rightarrow 5}c(\nu,J)=3$ so that
$\Delta c(\nu,J)\neq\alpha_{1}^{(2)}$.

We excluded the box marking (a) in $\nu^{(1)}$ depicted below and prescribe
that the box marking must be (b).

$\downarrow$$[2]$$[4]$$[3]$(a)$\downarrow$$[4]$$[3]$$[2]$(b)

where $i_{2}^{eff}>i_{1}$ and the first (resp. second) string is singular
(resp. qq-singular). It is easily verified that (a) and (b) yields the same RC
unless [5] and [6] are marked in $\nu^{(2)}$ with $i_{5}=i_{6}$ and
$i_{4}^{eff}=i_{6}$. The changes of vacancy numbers in (a) are $\Delta
p_{i_{2}-1}^{(1)}=-3$ and $\Delta p_{i_{4}-2}^{(1)}=-1$ (see (VC-2) and
(VC-3)). Now suppose that [5] and [6] are marked in $\nu^{(2)}$ with
$i_{5}=i_{6}$ and $i_{4}^{eff}=i_{6}$. In case (a), the new vacancy changes
$\Delta p_{i_{2}-1}^{(1)}$ and $\Delta p_{i_{4}-2}^{(1)}$ are obtained by
overwriting

$\displaystyle-6\chi(i\geq 3i_{6})-5\chi(i=3i_{6}-1)-4\chi(i=3i_{6}-2)$
$\displaystyle-3\chi(i=3i_{6}-3)-2\chi(i=3i_{6}-4)-\chi(i=3i_{6}-5)$

on the old ones. That is, $\Delta p_{i_{2}-1}^{(1)}=-6$ and $\Delta
p_{i_{4}-2}^{(1)}=-5$. $\Delta p_{i_{6}-2}^{(2)}=0$, which is unchanged (see
(VC-3)). $\Delta_{2\rightarrow 10}c(\nu,J)$ is computed to be

$\displaystyle-2$ $\displaystyle+\Delta r_{i_{2}}^{(1)}-(\Delta
p_{i_{2}-1}^{(1)}+\Delta r_{i_{2}-1}^{(1)})$ $\displaystyle+\Delta
r_{i_{4}}^{(1)}-(\Delta p_{i_{4}-2}^{(1)}+\Delta r_{i_{4}-2}^{(1)})$
$\displaystyle+\Delta r_{i_{6}}^{(2)}-(\Delta p_{i_{6}-2}^{(2)}+\Delta
r_{i_{6}-2}^{(2)}),$

where $\Delta r_{i_{2}}^{(1)}=-2$ and other rigging adjustments are zero so
that $\Delta_{2\rightarrow 10}c(\nu,J)=7$. In this case, we must do the
following box marking in $\nu^{(1)}$.

$\downarrow$$[9]$$[8]$$[7]$$[4]$$[3]$$[2]$

Suppose that $i_{6}\geq i_{1}+2$. Then $\Delta p_{i_{9}-6}^{(1)}=-3$ and
$\Delta p_{i_{6}-2}^{(2)}=1$,

$\Delta_{2\rightarrow 13}c(\nu,J)=-2-\Delta p_{i_{9}-6}^{(1)}-\Delta
p_{i_{6}-2}^{(2)}=0$

so that $\Delta c(\nu,J)=\Delta_{1}c(\nu,J)+\Delta_{1\rightarrow
2}c(\nu,J)+\Delta_{2\rightarrow 13}c(\nu,J)=-\alpha_{1}^{(2)}$. The
computation in the case when $i_{6}=i_{1}+1$ is similar. In this case, we must
consider the changes of $m_{i_{1}}^{(2)}$ and $m_{i_{1}-1}^{(2)}$ as well.

We also excluded the box marking (a) in $\nu^{(1)}$ depicted below and
prescribed that the box marking must be (b).

$\downarrow$$[3]$$[2]$$[4]$(a)$\downarrow$$[4]$$[3]$$[2]$(b)

where $i_{2}^{eff}>i_{1}$ and the first (resp. second) string is singular
(resp. q-singular). The verification of this case is similar.

In (BM-3), we excluded the following box marking (see Example 4.4).

$\downarrow$$[2]$$[3]$

where the $i_{2}$-string is qq-singular and the $i_{3}$-string is singular.
Suppose that [4] is not marked in $\nu^{(2)}$. Then

$\Delta_{3\rightarrow 4}c(\nu,J)=-1+\Delta r_{i_{3}}^{(1)}-(\Delta
p_{i_{3}-1}^{(1)}+\Delta r_{i_{3}-1}^{(1)}),$

where $\Delta r_{i_{3}}^{(1)}=0$, $\Delta r_{i_{3}-1}^{(1)}=-1$ by (RA-2), and
$\Delta p_{i_{3}-1}^{(1)}=-1$ (see (VC-3)) so that $\Delta_{3\rightarrow
4}c(\nu,J)=1$, which yields $\Delta c(\nu,J)\neq\alpha_{1}^{(2)}$.

In (BM-4), we excluded the following box marking.

$\downarrow$$[3]$$[4]$

where $i_{3}^{eff}>i_{1}$ and the first (resp. second) string is singular
(resp. q-singular). Suppose that [5] is marked in a q-singular string in
$\nu^{(2)}$ with $i_{4}^{eff}=i_{5}$. Then

$\displaystyle\Delta_{3\rightarrow 8}c(\nu,J)=-1$ $\displaystyle+\Delta
r_{i_{3}}^{(1)}-(\Delta p_{i_{3}-1}^{(1)}+\Delta r_{i_{3}-1}^{(1)})$
$\displaystyle+\Delta r_{i_{4}}^{(1)}-(\Delta p_{i_{4}-1}^{(1)}+\Delta
r_{i_{4}-1}^{(1)})$ $\displaystyle+\Delta r_{i_{5}}^{(2)}-(\Delta
p_{i_{5}-1}^{(2)}+\Delta r_{i_{5}-1}^{(2)}),$

where $\Delta r_{i_{3}}^{(1)}=-1$, $\Delta r_{i_{4}}^{(1)}=0$, $\Delta
r_{i_{5}}=-1$, $\Delta r_{i_{3}-1}=-1$ by (RA-3), $\Delta r_{i_{4}-1}^{(1)}=0$
by (RA-3), and $\Delta r_{i_{5}-1}^{(2)}=0$. If [5] is not marked in
$\nu^{(2)}$, then $\Delta p_{i_{3}-1}^{(1)}=-1$ and $\Delta
p_{i_{4}-1}^{(1)}=1$. When [5] is marked in $\nu^{(2)}$ with
$i_{4}^{eff}=i_{5}$, the new vacancy changes in $\nu^{(1)}$ are obtained by
overwriting

$-3\chi(i\geq 3i_{5})-2\chi(i=3i_{5}-1)-\chi(i=3i_{5}-2)$

on old ones. Therefore, $\Delta p_{i_{3}-1}^{(1)}=-2$ and $\Delta
p_{i_{4}-1}^{(1)}=-1$. $\Delta p_{i_{5}-1}^{(2)}=-1$, which is unchanged (see
(VC-3) and (VC-4)). Hence, $\Delta_{3\rightarrow 8}c(\nu,J)=2$, which yields
$\Delta c(\nu,J)\neq\alpha_{1}^{(2)}$.

In (BM-5), we excluded the following box marking.

$\downarrow$$[5]$(a)$\downarrow$$[5]$(b)$\downarrow$$[5]$$[3]$(c)

where $i_{5}^{eff}=i_{4}$. In cases (a) and (b),

$\displaystyle\Delta_{5\rightarrow 7}c(\nu,J)=-1$ $\displaystyle+\Delta
r_{i_{4}}^{(2)}-(\Delta p_{i_{4}-1}^{(2)}+\Delta r_{i_{4}-1}^{(2)})$
$\displaystyle+\Delta r_{i_{5}}^{(1)}-(\Delta p_{i_{5}-1}^{(1)}+\Delta
r_{i_{5}-1}^{(1)}),$

where $\Delta r_{i_{4}}^{(2)}=0$, $\Delta r_{i_{5}}^{(1)}=-1$, $\Delta
r_{i_{4}-1}^{(2)}=0$, and $\Delta r_{i_{5}-1}^{(1)}=0$. In case (a) , the box
marking by [4] and [5] does not affect the change of vacancy number of the
$(i_{5}-1)$-string in $\nu^{(1)}$ and that of the $(i_{4}-1)$-string in
$\nu^{(2)}$ so that $\Delta p_{i_{4}-1}^{(2)}=-1$ and $\Delta
p_{i_{5}-1}^{(1)}=-1$ (see (VC-4)) and therefore $\Delta_{5\rightarrow
7}c(\nu,J)=-2$, which yields $\Delta c(\nu,J)\neq\alpha_{1}^{(2)}$. In case
(b), $\Delta p_{i_{4}-1}^{(2)}=-1$ (see (VC-4)) and $\Delta
p_{i_{5}-1}^{(1)}=0$ (see (VC-5)) so that $\Delta_{5\rightarrow
7}c(\nu,J)=-1$, which yields $\Delta c(\nu,J)\neq\alpha_{1}^{(2)}$. In case
(c),

$\displaystyle\Delta_{3\rightarrow 7}c(\nu,J)=-1$ $\displaystyle+\Delta
r_{i_{4}}^{(2)}-(\Delta p_{i_{4}-1}^{(2)}+\Delta r_{i_{4}-1}^{(2)})$
$\displaystyle+\Delta r_{i_{5}}^{(1)}-(\Delta p_{i_{5}-2}^{(1)}+\Delta
r_{i_{5}-2}^{(1)}),$

where $\Delta r_{i_{4}}^{(2)}=0$, $\Delta r_{i_{5}}^{(1)}=-1$, $\Delta
r_{i_{4}-1}^{(2)}=0$, and $\Delta r_{i_{5}-2}^{(1)}=0$. Since the box marking
by [3], [4], and [5] and the deletion these boxes do not affect the change of
the vacancy number of the $(i_{5}-2)$-string in $\tilde{\nu}^{(1)}$ so that
$\Delta p_{i_{5}-2}^{(1)}=-1$ (see (VC-2)). $\Delta p_{i_{4}-1}^{(2)}$ is
obtained by overwriting

$-2\chi(3i\geq i_{5})-\chi(3i=i_{5}-1)+2\chi(i\geq i_{4})$

on $\Delta p_{i_{4}-1}^{(2)}$ in (VC-2). Again it is unchanged and $\Delta
p_{i_{4}-1}^{(2)}=0$. Hence $\Delta_{3\rightarrow 7}c(\nu,J)=-1$, which yields
$\Delta c(\nu,J)\neq\alpha_{1}^{(2)}$.

In (BM-8), we excluded the following box marking in $\nu^{(1)}$.

$\downarrow$$[9]$$[8]$$[7]$$[4]$$[3]$

where $i_{9}^{eff}=i_{6}$ ($i_{9}=3i_{6}-1$) (see Example 4.10).

$\displaystyle\Delta_{3\rightarrow 13}c(\nu,J)=-1$ $\displaystyle+\Delta
r_{i_{6}}^{(2)}-(\Delta p_{i_{6}-2}^{(2)}+\Delta r_{i_{6}-2}^{(2)})$
$\displaystyle+\Delta r_{i_{9}}^{(1)}-(\Delta p_{i_{9}-5}^{(1)}+\Delta
r_{i_{9}-5}^{(1)}),$

where $\Delta r_{i_{6}}^{(2)}=\Delta r_{i_{6}-2}^{(2)}=\Delta
r_{i_{9}}^{(1)}=\Delta r_{i_{9}-5}^{(1)}=0$. The change of vacancy number
$\Delta p_{i_{9}-5}^{(1)}$ is obtained by overwriting

$\displaystyle 10\chi(i\geq i_{9})+8\chi(i=i_{9}-1)+\cdots+2\chi(i=i_{9}-4)$
$\displaystyle-6\chi(i\geq 3i_{6})-5\chi(i=3i_{6}-1)-\cdots-\chi(i=3i_{6}-5)$

on $\Delta p_{i_{9}-5}^{(1)}$ in (VC-3) so that $\Delta p_{i_{9}-5}^{(1)}=-1$.
The change of vacancy number $\Delta p_{i_{6}-2}^{(2)}$ is obtained by
overwriting

$\displaystyle-5\chi(3i\geq i_{4})-4\chi(3i=i_{9}-1)-\cdots-2\chi(3i=i_{9}-4)$
$\displaystyle+4\chi(i\geq i_{6})+2\chi(i=i_{6}-1)$

on $\Delta p_{i_{6}-2}^{(2)}$ in (VC-3) so that $\Delta p_{i_{6}-2}^{(2)}=0$.
Hence $\Delta_{3\rightarrow 13}c(\nu,J)=-1$, which yields $\Delta
c(\nu,J)\neq\alpha_{1}^{(2)}$.

The verification of the following forbidden box marking in $\nu^{(1)}$

$\downarrow$$[9]$$[8]$$[6]$$[5]$or$\downarrow$$[9]$$[8]$$[6]$

with $i_{9}^{eff}=i_{7}$ (see Example 4.11 or 4.12) is similar.

In (BM-9), we excluded the following box marking in $\nu^{(1)}$.

$\downarrow$$[8]$$[9]$

where the first (resp. second) string is singular (resp. q-singular). If [7]
is marked in $\nu^{(1)}$, then $i_{7}\leq i_{8}-2$ and if [7] is marked in
$\nu^{(2)}$, then $i_{8}^{eff}>i_{7}$ (see the preferential rule of (BM-8)
(2)). Suppose that [10] is marked in a singular string in $\nu^{(2)}$ with
$i_{9}^{eff}=i_{10}$. Then,

$\displaystyle\Delta_{11\rightarrow 14}c(\nu,J)=-1$ $\displaystyle+\Delta
r_{i_{8}}^{(1)}-(\Delta p_{i_{8}-1}^{(1)}+\Delta r_{i_{8}-1}^{(1)})$
$\displaystyle+\Delta r_{i_{9}}^{(1)}-(\Delta p_{i_{9}-1}^{(1)}+\Delta
r_{i_{9}-1}^{(1)})$ $\displaystyle+\Delta r_{i_{10}}^{(2)}-(\Delta
p_{i_{10}-1}^{(2)}+\Delta r_{i_{10}-1}^{(2)}),$

where $\Delta r_{i_{8}}^{(1)}=-1$, $\Delta r_{i_{9}}^{(1)}=0$, $\Delta
r_{i_{10}}^{(2)}=0$, $\Delta r_{i_{8}-1}^{(1)}=-1$ by (RA-7), $\Delta
r_{i_{9}-1}^{(1)}=0$ by (RA-7), and $\Delta r_{i_{10}-1}^{(2)}=0$. The changes
of vacancy numbers $\Delta p_{i}^{(1)}$ are obtained by overwriting

$-3\chi(i\geq 3i_{10})-2\chi(i=3i_{10}-1)-\chi(i=3i_{10}-2)$

on $\Delta p_{i}^{(1)}$ in (VC-13) so that $\Delta p_{i_{8}-1}^{(1)}=-2$ and
$\Delta p_{i_{9}-1}^{(1)}=-1$. By (VC-12) and (VC-13), $\Delta
p_{i_{10}-1}^{(2)}=0$. Hence, $\Delta_{11\rightarrow 14}c(\nu,J)=2$, which
yields $\Delta c(\nu,J)\neq\alpha_{1}^{(2)}$.

The verifications of rules of boomerang strings are similar and we omit the
details.

### Acknowledgements

The author would like to express his gratitude to Professor Masato Okado for
helpful discussions. He also would like to thank Professor Travis Scrimshaw
for helpful comments. This work was partly supported by Osaka City University
Advanced Mathematical Institute (MEXT Joint Usage/Research Center on
Mathematics and Theoretical Physics JPMXP0619217849).

## References

  * [1] G. Benkart, I. Frenkel, S.-J. Kang, and H. Lee, Level 1 perfect crystals and path realizations of basic representations at $q=0$, Int. Math. Res. Not. 2006 10312 (2006).
  * [2] D. Bump and A. Schilling, “Crystal Bases: Representations and Combinatorics,” World Scientific Press (2017).
  * [3] L. Deka and A. Schilling, New fermionic formula for unrestricted Kostoka polynomials, J. Combin. Theory Ser. A 113 1435–1461 (2006).
  * [4] G. Hatayama, A, Kuniba, M. Okado, T. Takagi, and Y. Yamada, Remarks on fermionic formula. Contemp. Math. 248 243–291 (1999).
  * [5] G. Hatayama, A. Kuniba, M. Okado, T. Takagi, and Z. Tsuboi, Paths, crystals and fermionic formulae, Prog. Math. Phys. 23 205–272 (2002).
  * [6] D. Hernandez, Kirillov-Reshetikhin conjecture: the general case, Int. Math. Res. Not. 1 149–193 (2010).
  * [7] V. G. Kac, “Infinite-dimensional Lie algebras,” Cambridge University Press, Cambridge, third edition (1990).
  * [8] M. Kashiwara, Crystalizing the $q$-analogue of universal enveloping algebras, Comm. Math. Phys. 133 249–260 (1990).
  * [9] M. Kashiwara, On crystal bases of the $q$-analogue of universal enveloping algebras, Duke Math. J. 63 465–516 (1991).
  * [10] S.-J. Kang, M. Kashiwara, K. C. Misra, T. Miwa, T. Nakashima, and A. Nakayashiki, Affine crystals and vertex models, Int. J. Mod. Phys. A7 (suppl. 1A) 449–484 (1992).
  * [11] S.-J. Kang, M. Kashiwara, K. C. Misra, T. Miwa, T. Nakashima, and A. Nakayashiki, Perfect crystals of quantum affine Lie algebras, Duke Math. J. 68 499–607 (1992).
  * [12] S. Kerov, A. Kirillov, and N. Reshetikhin, Combinatrics, the Bethe anzats and representations of the symmetric group, Zap. Nauchn. Sem. (LOMI) 155 5–64 (1986). (English translation: J. Sov. Math. 41 916–924 (1988).)
  * [13] A. Kirillov, and N. Reshetikhin, The Bethe ansatz and the combinatrics of Young tableaux, J. Sov. Math. 41 925–955 (1988).
  * [14] A. Kirillov, A. Schilling, and M. Shimozono, A bijection between Littlewood-Richardson tableaux and rigged configurations, Selecta Math. (N. S.) 8 67–135 (2002).
  * [15] A. Kuniba, T. Nakanishi, and Z. Tsuboi, The canonical solutions of the Q-systems and the Kirillov-Reshetikhin conjecture, Comm. Math. Phys. 227 155–190 (2002).
  * [16] A. Lascoux and M. Sch$\ddot{\mathrm{u}}$tzenberger, Sur une conjecture de H.O. Foulkes, CR Acad. Sci. Paris 286A 323–324 (1978).
  * [17] K. C. Misra, M. Mohamad, and M. Okado, Zero action on perfect crystals for $U_{q}(G_{2}^{(1)})$, SIGMA 6 022 (2010).
  * [18] K. C. Misra, M. Okado, and E. A. Wilson, Soliton cellular automata associated with $G_{2}^{(1)}$ crystal base, J. Math. Phys. 53 013510 (2012).
  * [19] Mahathir bin Mohamad, Soliton cellular automata constructed from a $U_{q}(\mathfrak{g})$-crystal $B^{n,1}$ and Kirillov-Reshetikhin type bijection for $U_{q}(E_{6}^{(1)})$-crystal $B^{6,1}$, Thesis (Ph.D.)-Osaka University (2012).
  * [20] K. Naoi, Fusion products of Kirillov-Reshetikhin modules and the $X=M$ conjecture, Adv. Math. 231 1546–1571 (2012).
  * [21] H. Nakajima, $t$-analogue of the $q$-characters of Kirillov-Reshetikhin modules of quantum affine algebras, Represent. Theory 7 259–274 (2003).
  * [22] A. Nakayashiki and Y. Yamada, Kostka polynomials and energy functions in solvable lattice models, Selecta Math. (N. S.) 3 547–599 (1997).
  * [23] M. Okado, R. Sakamoto, and A. Schilling, Affine crystal structure on rigged configurations of type $D_{n}^{(1)}$, J. Algebraic Combin. 37 571–599 (2013).
  * [24] M. Okado, R. Sakamoto, A. Schilling, and T. Scrimshaw, Type $D_{n}^{(1)}$ rigged configuration bijection, J. Algebraic Combin. 46 341–401 (2017).
  * [25] M. Okado and N. Sano, KKR type bijection for the exceptional affine algebra $E_{6}^{(1)}$, In “Algebraic groups and quantum groups,” volume 565 of Contemp. Math. 227–242. Amer. Math. Soc., Providence, RI (2012).
  * [26] M. Okado, A. Schilling, and M. Shimozono, A crystal to rigged configuration bijection for nonexceptional affine algebras, In “Algebraic combinatorics and quantum groups,” Edited by N. Jing, World Sci. Publ. 85–124 (2003).
  * [27] M. Okado, A. Schilling, and M. Shimozono, Virtual crystals and fermionic formulas of type $D_{n+1}^{(2)}$, $A_{2n}^{(2)}$, and $C_{n}^{(1)}$, Represent. Theory 7 101–163 (electronic) (2003).
  * [28] M. Okado, A. Schilling, and M. Shimozono, Virtual crystals and Kleber’s algorithm, Comm. Math. Phys. 238 187–209 (2003).
  * [29] M. Okado, A. Schilling, and T. Scrimshaw, Rigged configuration bijection and the proof of the $X=M$ conjecture for nonexceptional affine types, J. Algebra 516 1–37 (2018).
  * [30] R. Sakamoto, Rigged configurations and Kashiwara operators, SIGMA 10 028 (2014).
  * [31] A. Schilling, A bijection between type $D_{n}^{(1)}$ crystals and rigged configurations, J. Algebra 285 292–334 (2005).
  * [32] A. Schilling and M. Shimozono, $X=M$ for symmetric powers, J. Algebra 295 562–610 (2006).
  * [33] A. Schilling and T. Scrimshaw, Crystal structure on rigged configurations and the filling map for nonexceptional affine types, Electon. J. Combin. 22 $\sharp$P1.73 (2015).
  * [34] T. Scrimshaw, A crystal to rigged configuration bijection and the filling map for type $D_{4}^{(3)}$, J. Algebra 448C 294–349 (2016).
  * [35] T. Scrimshaw, Uniform description of the rigged configuration bijection, Selecta Math. (N. S.) 26 42 (2020).