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Proof.
First we enumerate the disjoint sets in
A
(
L
)+
E
1
. By equation (33) we
have
2
n
2
n−r
1
+1
<L<
2
n
2
n−r
1
.
−
−
(38)
Using Theorem 3 and Lemma 5, from equation (38) we have
2
n−r
1
+1
(
A
(
L
)+
E
u
)
∩
(
A
(
L
)+
E
v
)=
∅
,
0
≤
u<v
≤
−
1
,
(39)
,
2
n−r
1
+1
and for
u
=0
,
···
−
1,
,
2
r
1
−
1
A
(
L
)+
E
u
=
A
(
L
)+
E
u
+
t
2
n
−
r
1
+1
,
t
=0
,
···
−
1
.
(40)
Thus, from equation (39) there are 2
n−r
1
+1
disjoint sets
A
(
L
)+
E
i
,
E
i
∈
D
1
(
L
),
in
A
(
L
)+
E
1
. To obtain the disjoint sets in
A
(
L
)+
E
2
,weonlyhavetoenumerate
the disjoint sets in
A
(
L
)+
D
2
(
L
) because from equation (40) we have
A
(
L
)+
2
n−r
1
+1
E
i,j,i
+
u
2
n
−
r
1
+1
,j
+
v
2
n
−
r
1
+1
=
A
(
L
), for 0
≤
i<j
≤
−
1, 0
≤
u, v
≤
2
r
1
−
1
1.
From Theorem 3, we know that
−
A
(
L
)+
E
i,j
=
A
(
L
)+
E
k,l
if and only if there
exists a sequence
S
∈A
(
L
) such that the new sequence
S
+
E
i,j,k,l
∈A
(
L
).
Hence we observe that doubly counted sets in
A
(
L
)+
D
2
(
L
) arise if only if there
2
n−r
1
+1
exist integers
i
,
j
,
k
,and
l
,0
1, that are in the second
form of Theorem 2. Exactly all such
i
,
j
,
k
,and
l
are derived as part of the
proof of Theorem 2 and are enumerated in equation (25). Here we repeat the
enumeration for clarity. Integers
i
,
j
,
k
,and
l
,0
≤
i, j, k, l
≤
−
2
n−r
1
+1
≤
i, j, k, l
≤
−
1, that
are in the second form of Theorem 2 are
i
=
u
+
g
1
2
n−r
2
,
j
=
u
+
g
2
2
n−r
2
,
k
=
i
+2
n−r
1
,
l
=
j
+2
n−r
1
,
(41)
where
2
n−r
2
2
r
2
−r
1
≤
u
≤
−
≤
g
1
<g
2
≤
−
1
.
0
1and0
(42)
From equations (41) and (42) there are exactly 2
n−r
2
2
r
2
−
r
1
2
distinct pairs
2
n−r
1
+1
i
,
j
and hence distinct sets
{
i, j, k, l
}
such that 0
≤
i, j, k, l
≤
−
1and
A
(
L
)+
E
i,j,k,l
=
(
L
).
Hence for all settings of
i
and
j
in equation (41) we have the set equalities
A
A
(
L
)+
E
i,j
=
A
(
L
)+
E
i
+2
n
−
r
1
,j
+2
n
−
r
1
(43)
and
A
(
L
)+
E
i,j
+2
n
−
r
1
=
A
(
L
)+
E
i
+2
n
−
r
1
,j
.
(44)
,
2
n−r
2
1, we have 2
r
2
−r
1
Also, for each
u
=0
,
···
−
−
1 set equalities
i
=
u
+
t
2
n−r
2
A
(
L
)+
E
u,u
+2
n
−
r
1
=
A
(
L
)+
E
i,i
+2
n
−
r
1
,
where
(45)
2
r
2
−r
1
for 1
1.
By the settings of
i
and
j
in equation (41), the set equalities in equations
(43) and (44) result in 2
n−r
2
(2
≤
t
≤
−
·
2
r
2
−
r
1
2
) doubly counted sets in
A
(
L
)+
D
2
(
L
)
2
,
2
n−r
2
enumerated as
A
(
L
)+
D
u
(
L
),
u
=0
,
···
−
1. Similarly, the set equalities
