Proof of Theorem 1. By Lemma 7, for any vertex
,
. Since
,
by Lemma 4. In the following, we will prove
. Take any three distinct vertices
, and
z in
and let
. If we can prove that there are
internally edge disjoint
S-trees in
, we are done.
Case 1. for some .
Without loss of generality, let . By Lemma 3, there exist internally edge disjoint S-trees in . Without loss of generality, suppose , and by Lemma 7(4).
If , we can assume . By Lemma 7(4), is connected. Hence, there exists a tree containing , and in . Take . Since is connected, there exists a tree containing , and in . Take . Then, are internally edge disjoint S-trees. Thus, .
If , without loss of generality, noting that by Lemma 7(3), let . By Lemma 7(5), . By Lemma 7(3)(4), we can let . Since is connected, there exists a tree containing , and in . Take . Since is connected, there exists a tree containing , and in . Take . Then, are internally edge disjoint S-trees. Thus, .
By symmetry and , if for some , we can also obtain .
Case 2. for some . for some and or for some .
Without loss of generality, we let . By Lemmas 2 and 5, there exist s internally disjoint paths from x to y in . Let , such that for . In the following, we will show that for any two distinct vertices and with , lie in distinct for . Without loss of generality, let , , and . Then, , , , , , . By and the definition of , we can show that lie in different for , where and . This implies that lie in distinct for .
Subcase 2.1. for some .
Let . We know that or . Without loss of generality, let . Suppose and for .
Subcase 2.1.1. for some .
Without loss of generality, let . Then, for by the above discussion. We can let or .
First, we consider
(
Figure 4). By Lemma 7(3),
or
. Without loss of generality, let
. Suppose
. Take
s vertices
in
, such that
for
. Let
. By Lemma 6, there exist
s internally disjoint paths
from
z to
Z in
. Let
be the path from
z to
for
. Since
is connected, there exists a tree
containing
and
in
for
. Take
for
. Since
is connected, there exists a tree
containing
, and
in
. Take
. Then,
are
internally edge disjoint
S-trees. Thus,
.
Now, we consider .
If , then . Let . Taking to be the same as above, since is connected, there exists a tree containing and in . Take . Then, are internally edge disjoint S-trees. Thus, .
Let
(
Figure 5). By Lemma 7(4),
. Suppose
. Take
vertices
in
, such that
for
. Let
. By Lemma 6, there exist
s internally disjoint paths
from
z to
Z in
. Let
be the path from
z to
for
and
be the path from
z to
. Since
is connected, there exists a tree
containing
and
in
for
. Take
for
. Noting that
, then
. Since
is connected, there exists a tree
containing
and
in
. Take
and
. Then,
are
internally edge disjoint
S-trees. Thus,
.
Subcase 2.1.2. for each .
By Lemma 7(4), we can show for . Without loss of generality, let .
First, we let . By Lemma 7(3), or . Without loss of generality, let . Suppose . Take s vertices in , such that for . Let . By Lemma 6, there exist s internally disjoint paths from z to Z in . Let be the path from z to for . Since is connected, there exists a tree containing and in for . Take for . Since is connected, there exists a tree containing and in . Take . Then, are internally edge disjoint S-trees. Thus, .
Now, we let .
If , then . We can let . Taking to be the same as above, since is connected, there exists a tree containing and in . Take . Then, are internally edge disjoint S-trees. Thus, .
If
. By Lemma 7(3), suppose
, where
is the hypercube neighbor of
. By Lemma 7(4),
. Without loss of generality, let
. Take
vertices
in
, such that
for
. Let
. By Lemma 6, there exist
s internally disjoint paths
from
z to
Z in
. Let
be the path from
z to
for
and
be the path from
z to
. Since
is connected, there exists a tree
containing
and
in
for
. Take
for
. If
(
Figure 6), noting that
, since
is connected, there exists a tree
containing
and
in
. Take
. Since
is connected, there exists a tree
containing
and
in
. Take
. Then,
are
internally edge disjoint
S-trees. If
(
Figure 7), since
, then
by Lemma 7(3). Since
, we can partition
into
and
, such that
and
. In
, there exists a spanning tree
containing
. Since
for
, there exists a vertex
, such that
for
by Lemma 7(4). Let
. Similarly, there exists a spanning tree
containing
in
. Since
for
, there exists a vertex
, such that
for
by Lemma 7(4). Let
. Since
is connected, there exists a tree
containing
and
. Take
. Since
is connected, there exists a tree
containing
and
. Take
. Then,
are
internally edge disjoint
S-trees. Thus,
.
By symmetry and , if for some and some , we can also obtain .
Subcase 2.2. for some .
Without loss of generality, we let
(
Figure 8), and suppose
for
. Then,
or
or
or
for
. Without loss of generality, let
for
. Suppose
. If
for some
, then
for some
. Then,
for
. If
for each
, then
for
by Lemma 7(4). Without loss of generality, let
for
. Suppose
. Choose
s vertices
in
, such that
for
. Denote
. By Lemma 6, there exist
s internally disjoint paths
from
z to
Z in
. Let
be the path from
z to
for
. Since
is connected, there exists a tree
containing
and
in
for
. Take
for
. Since
is connected, there exists a tree
containing
and
in
. Take
. Then,
are
internally edge disjoint
S-trees. Thus,
.
By symmetry and , if for some and , we can also obtain .
Case 3. , and for some with and some .
Without loss of generality, let .
Subcase 3.1. .
By Lemma 7(3), without loss of generality, let , .
We first consider or . Without loss of generality, let . By Lemma 7(3), we can let and or . Suppose . Then, put . Choose in , such that for . Denote . Choose in , such that for . Denote . Choose in , such that for . Denote . By Lemma 6, there exist s paths from x to X in , s paths from y to Y in , s paths from z to Z in . Let be the paths from x to , from y to , and from z to , respectively, for . Since is connected, there exists a tree containing , and in for . Take for . Since is connected, there exists a tree containing in . Take . Then, are internally edge disjoint S-trees. Thus, .
Now, we consider
and
(
Figure 9). Since
and
, we can partition
into
and
, such that
,
and
,
. Similarly, we partition
into
and
, such that
,
and
. By Lemma 7(4), we can let
and
. Choose
in
such that
for
. This can be performed since
with
. Let
for
. Denote
. Choose
in
, such that
for
. Without loss of generality, for simplicity of description, we can let
and
for
. Note that
and
for
. Denote
. Choose
such that
for
. Denote
. By Lemma 6 and
, there exist
paths
from
x to
X in
,
paths
from
y to
Y in
,
s paths
from
z to
Z in
. Let
be the paths from
x to
, from
y to
, and from
z to
, respectively, for
and
be the path from
z to
. Since
is connected, there exists a tree
containing
and
in
. Take
. Since
is connected for
, there exists a tree
containing
and
in
for
. Take
for
. Since
is connected, there exists a tree
containing
and
in
. Take
. Let
u be the neighbor of
x in
and
v be the neighbor of
y in
. Suppose that
is a spanning tree of
and
is a spanning tree of
. Take
. Then,
are
internally edge disjoint
S-trees. Thus,
.
Subcase 3.2. or .
Without loss of generality, let . Suppose . By Lemma 7(3), or , or . Without loss of generality, we can let , . Choose , such that for . Suppose for . Denote . Choose , such that for . Denote . Choose , such that for . Denote . By Lemma 6, there exist s paths from x to X in , s paths from y to Y in , s paths from z to Z in . Let be the paths from x to , from y to , and from z to , respectively, for . Note that if for some , we regard as the vertex y. Since is connected, there exists a tree containing and in for . Take for . Since is connected, there exists a tree containing in . Take . Then, are internally edge disjoint S-trees. Thus, .
By symmetry and , if for some with and some , we can also obtain .
Case 4. , and for some with .
Let
, and
(
Figure 10). Without loss of generality, suppose
. Choose
, such that
for
. Let
,
and
. By Lemma 6, there exist
s paths
from
x to
X in
,
s paths
from
y to
Y in
,
s paths
from
z to
Z in
. Let
be the paths from
x to
, from
y to
, and from
z to
, respectively, for
. Since
is connected, there exists a tree
containing
and
in
for
. Take
for
. Since
is connected, there exists a tree
containing
and
in
. Take
. Then,
are
internally edge disjoint
S-trees. Thus,
.
By symmetry and , if for some with , we can also obtain .
We have completed the proof. □