In this section, we give the proofs of main results by inductions. For a k-tree , if or , then is a k or (k + 1)-clique, that is, . Thus, all of the theorems are true and we will only consider the case when below.
Proof of Theorem 1. For , let , be a k-clique and for . Just by Definition 3, we can get that for , , then ; for and , , then . Thus, we can get .
For , we will proceed it by induction on . If , let be the simplicial elimination ordering of . By Lemma 3, we can get that and for . Thus, . Assume that Theorem 1 is true for a k-path with at most vertices, where . Let be a k-path such that , and be the simplicial elimination ordering of . Set , then is the simplicial elimination ordering of and for any edge , or is the distance of and in or , respectively.
Let . If we can show that by adding to , , then Theorem 1 is true.
Set
,
and
. By Definition 2 and Lemma 3, we have
,
and
, that is,
and
. Thus,
and
. Similarly, for any edge
with
, we have
, that is,
; For
with
, we have
, that is,
. Thus, we can get that
For
, set
,
,
, and
. For
and by Lemma 3, we have
,
and
, that is,
and
. Thus,
and
. similarly, for any edge
with
, we have
, that is,
; for
with
, we have
, that is,
. Thus, we can get that
Next, we consider the edges in the
-clique
. For any edge
with
, we have
, that is,
. For any edge
with
, by Lemma 3, we can obtain that
,
and when
,
, that is,
,
and
. Similarly, we get that for
and
,
Thus, if , then with and with , that is, ; similarly, we can obtain that
For any edge
with
, by Lemma 3, we can obtain that
and
, that is,
and
. Similarly, we get that for
and
,
Thus, if , then for and with , that is, ; similarly, we have .
Set
, if
with
x or
, by Lemma 2, we have
. Thus,
Thus, , for and Theorem 1 is proved. □
Proof of Theorem 2. We will proceed with it by induction on . If , by Definition 4, we have is also a k-path, that is, . If , assume that Theorem 2 is true for the k-spiral with at most vertices, we will consider with n vertices. Let be a k-spiral with and such that is the simplicial ordering of , where with and for . For any edge , or is the distance of and in or , respectively.
For , let . If we can show that by adding v to , , then Theorem 2 is true.
Set and let , by Lemma 4, we have and for , that is, ; for , that is, . Set , . By Lemma 4, we have and , that is, and . Similarly, for with , we have , that is, ; for with , we have and , that is, . Set . By Lemma 4, we have , that is, . Similarly, for with , we have , that is, .
Set , by Lemma 4, we have for and for . Thus, . Similarly, . In addition, by Lemma 4, we have for , and for . Thus, . Similarly, we have . Set , since , we have . Set , by Lemma 4, we have for , for , for . Thus, for and for .
Set
, if
with
x or
, by Lemma 2, we have
. Thus,
and Theorem 2 is proved.
For , let . If we can show that by adding v to , , then Theorem 2 is proved.
Set , by Lemma 4, we have for , for and for . Thus, and . Set .
By Lemma 4, we can get that and , that is, and . Similarly, for with , we have , that is, ; for with , we have and , that is, . Set . By Lemma 4, we can get that , that is, . Similarly, for with , we have , that is, .
Set , by Lemma 4, we have , for and for . Thus, . Similarly, we have . Set , since , we have . Set , by Lemma 4, we have for and for . Thus, for and for .
Set
, if
with
x or
, by Lemma 2, we have
. Thus,
and Theorem 2 is proved. □