Proof. If T has a strong stem or adjacent stems, then by Propositions 2 and 5 we have Suppose next that T has no adjacent stems and no strong stem. Then, T is a healthy spider. Let and let . Since , we have . We consider the four cases.
- Case 1.
Both x and y are leaves.
Without a loss of generality, assume that and . Observe that . Let be obtained from by subdividing the edges with the vertices , and , respectively. We show that . Let D be a -set. If , then we must have for each and , which leads to . Assume now that . If v is paired with or , then we must have for each and implying that . Assume next that v is not paired with or . Without a loss of generality, assume that v is paired with . Thus, for each and yielding . Thus, and, by Proposition 2, we obtain
- Case 2.
x is a leaf and y is a stem.
Without a loss of generality, assume that and . Then, it is clear that . Let be obtained from by subdividing the edges and e with the vertices and , respectively. As in Case 1, we can see that . Hence, and, by Proposition 2, we have
- Case 3.
x is a leaf and .
Without a loss of generality, assume that . Clearly . Let be obtained from by subdividing the edges with . We show that . Let D be a -set. To paired-dominate , we must have and to paired-dominate we must have . Since is even, we obtain . Hence, and, by Proposition 2, we have
- Case 4.
Both x and y are stems.
In this case, has adjacent stems, and we deduce from Propositions 2 and 5 that □
Proof. As in the proof of Lemma 1, we may assume that T has no strong stem and no adjacent stems. Let be a diametral path in T. By the assumption, the components of are or a healthy spider. It is easy to see that . Let and be the components of containing and , respectively. If , then we deduce from Propositions 2 and 7 that Assume that . Suppose that . If , then it is easy to see that subdividing the edges increases the paired-domination number of T and so and the result follows by Proposition 2. Hence, we assume that . Let , , and let . We consider the four cases.
- Case 1.
Both x and y are leaves.
If x and y are the leaves of (resp., ), then, clearly, and by Propositions 4 and 11, we have If x is a leaf of and y is a leaf of , then, we deduce from Propositions 6 and 11 that
- Case 2.
x is a leaf and y is a stem.
Assume first that are the vertices of . We may assume that and . Then, the set is a paired-dominating set of of size and so . If x and y are the vertices of , then similarly we have . Assume second that and . Without a loss of generality, we can suppose that and . Then, the set is a paired-dominating set of with cardinality , and thus . Applying Propositions 4 and 11 we obtain
- Case 3.
x is a leaf and .
Without loss of generality, we may assume that . If , then the set is a paired-dominating set of of size and so . If , then the set is a paired-dominating set of of size , and thus . Using Propositions 4 and 11, we obtain
- Case 4.
Both x and y are stems.
In this case, has adjacent stems, and we deduce from Propositions 2 and 5 that □
Proof. If T has a strong stem or adjacent stems, then, by Propositions 2 and 5, we have Hence, we assume that T has no strong stem and no adjacent stems. It follows that . According to Lemmas 1 and 2, we may assume that . If x or y is a stem, then by Propositions 9 and 11, we have Hence, we assume that neither x nor y is a stem. If x (resp., y) is a leaf with support vertex (resp., ) of degree 2, respectively, then for , is a path in T with , and therefore by Propositions 6 and 11, we have Thus, we may assume that, if both x and y are leaves, then the stem of x or y is of at least degree 3.
Let be a diametral path such that is minimized. By assumption, the trees and (the components of containing and , respectively) are or a healthy spider. Let and let be the leaf adjacent to for each i. Root T is at . We consider three cases.
- Case 1.
.
We distinguish the following subcases.
Subcase 1.1. is a stem.
By our assumption, T has no strong stem. Let w be a unique leaf adjacent to . First, let . Let be obtained from by subdividing the edges and with vertices and , respectively for each i and the edges and with vertices r and s, respectively. We show that . Let P be a -set. To paired-dominate , we may assume that for each i. On the other hand, to paired-dominate w and , we must have .
If
or
and
is paired with
r or
s, then the set
is a PD-set of
with cardinality of less than
P. Assume that
and
is paired with a vertex not in
. Then, we have
, and clearly the set
is a PD-set of
with cardinality of less than
P. Since the number of subdivided edges is at most
, we deduce from Proposition 11 that
.
Now, let . Let . Note that each component of has at least two vertices. Let be obtained from by subdividing and with vertices and , respectively for each i, the edge with vertex for each j, and the edge e with vertex q. By our assumptions, it is not difficult to verify that the number of subdivided edges is at most . We next show that . Let P be a -set, and let F be the set of all edges in whose subdivision vertices are in P. To paired-dominate , we may assume that for each i.
On the other hand, to paired-dominate , we must have . If or and is not paired with , then and the set is a PD-set of obtained from by subdividing all edges in F with cardinality of less than P. If and is paired with , then the set is a PD-set of obtained from by subdividing the edges in F with cardinality of less than P. By Proposition 1, we have for each . Therefore, in either case, we have . It follows from Proposition 11 that .
Subcase 1.2. is not a stem and there is a path in T such that and .
If e is incident to , then it follows from Propositions 9 and 11 that . Hence, we assume that . First, let . Let be obtained from by subdividing the edges and with the vertices and , respectively, for each i and the edges and with the vertices r and s, respectively. By our assumptions, it is not difficult to verify that the number of subdivided edges is at most . As above, we can see that , and thus . We deduce from Proposition 11 that .
Now, let . We distinguish the following situations.
- (1.2.1)
.
Let . Note that each component of has order at least two. Let be obtained from by subdividing the edges and with the vertices and , respectively, for each i, the edge with vertex for each j, and the edges and e with the vertices and , respectively. By our assumptions, it is not difficult to verify that the number of subdivided edges is at most . We show that . Let P be a -set, and let F be the set of all edges in whose subdivision vertices are in P. To paired-dominate , we may assume that for each i. On the other hand, to paired-dominate , we must have .
If or and is not paired with , then , and the set is a PD-set of obtained from by subdividing the edges F with cardinality of less than P. If and is paired with , then the set is a PD-set of obtained from by subdividing the edges in F with cardinality of less than P. By Proposition 1, we have for each . Therefore, in either case, we have . It follows from Proposition 11 that .
- (1.2.2)
.
Then, . Let , and let be obtained from by subdividing the edges and with the vertices and , respectively, for each i, the edge with vertex for each j, and the edges and e. By our assumptions, it is not difficult to verify that the number of subdivided edges is at most . Using an argument similar to that described in (1.2.1), we can see that .
Subcase 1.3. All children of except have depth 3.
Considering the above cases and subcases and the choice of diametral path, we may assume that each component of has at least three vertices. Let }, and let be obtained from by subdividing the edges and with the vertices and , respectively, for each i and the edges with vertex for each j. By our assumptions, it is not difficult to verify that the number of subdivided edges is at most . We show that . Let P be a -set. To paired-dominate , we may assume that for each i. On the other hand, to paired-dominate , we must have .
Let F be the set of subdivided edges incident to whose subdivision vertices belong to P. If or and its partner is in , then we have , and the set is a PD-set of obtained from by subdividing the edges of F with cardinality of less than P.
If , and its partner is , then the set is a PD-set of obtained from by subdividing the edges of F with cardinality of less than P. By Proposition 1, we have for each . Therefore, in either case, we have . It follows from Proposition 11 that .
- Case 2.
.
By our earlier assumptions, T has no strong stem, neither x nor y is a stem, and if both x and y are leaves, then the stem of x or y is of at least degree 3. Therefore, we may assume without a loss of generality that and . Let be obtained from by subdividing the edges , and e by the subdivision vertices , and , respectively. We show that . Let P be a -set. If , then we must have , and clearly is a PD-set of with cardinality of less than P.
If and its partner is not or , then , and hence is a PD-set of with cardinality of less than P. If and its partner is or , then , and hence is a PD-set of with cardinality of less than P. Therefore, we have . It follows from Proposition 11 that .
- Case 3.
One of x and y belongs to , and the other does not belong to .
Assume, without a loss of generality, that and . By our earlier assumption, x is not a stem. We distinguish two subcases.
Subcase 3.1.x is a leaf of .
Assume without a loss of generality that . If or y is a leaf, then it follows from Proposition 6 that , and the result follows from Proposition 11. Let , and y is not a leaf. By our earlier assumption, y is not a stem, and hence each component of has at least two vertices. Note that the component of containing has at least five vertices and, thus, . Let be obtained from by subdividing the edges and by the subdivision vertices and , respectively, and all edges incident to y in T. We show that . Let P be a -set, be the set of subdivided edges incident to y whose subdivision vertices are in P and let be obtained from by subdividing the edges in .
First, we assume that and y is paired with . Then, clearly . We may assume that , otherwise , and hence we may consider as a -set. If is paired with a vertex other than , then is a PD-set of with cardinality of less than P. If is paired with , then and is also a PD-set of with cardinality of less than P.
Second, assume that and y is paired with a subdivision vertex other than . Then, . As above, we may assume that . If is paired with a vertex other than , then is a PD-set of with cardinality of less than P. If is paired with , then and hence is a PD-set of obtained from T by subdividing the edges in , with cardinality of less than P.
Finally, we assume that . Then, y must be dominated by a subdivision vertex. As above, we may assume that . We consider the following.
- (3.1.1)
.
Then, and are partners and to paired-dominate , we must have . If is paired with , then , and hence is a PD-set of with cardinality of less than P. If is paired with a vertex other than , then , and hence is a PD-set of with cardinality of less than P.
- (3.1.2)
.
Then, we must have . Let be a subdivision vertex that dominates y, and let be obtained from by subdividing the edges . If is paired with , then , and hence is a PD-set of with cardinality of less than P. If is paired with a vertex other than , then is also a PD-set of with cardinality of less than P.
In all cases, it follows from Proposition 1 that for each . Therefore, we have . It follows from Proposition 11 that .
Subcase 3.2..
By assumption, y is not a support vertex. We consider the following situations.
- (3.2.1)
.
Let be the stem of y and be the parent of . Let be obtained from by subdividing the edges and with the vertices , respectively, and the edge with vertex for each i. Let F be the set of all subdivided edges. Note that by the choice of the diametral path , it is not difficult to check that . We show that . Let P be a -set. To paired-dominate , we must have .
If and is the partner of , then is a PD-set of obtained from by subdividing the edges in with cardinality of less than P. If and is the partner of , then is a PD-set of obtained from by subdividing the edge in with cardinality of less than P.
In the following, we may assume that . To paired-dominate , we may assume that , and, to paired-dominate , we must have . First, we assume that y and are partners. If , then is a PD-set of obtained from by subdividing the edges in with cardinality of less than P.
If , then is a PD-set of obtained from by subdividing the edges in with cardinality of less than P. Second, we assume that y and are partners. If , then and , is a PD-set of obtained from by subdividing the edges in with cardinality of less than P. If , then is also a PD-set of with cardinality of less than P.
In all cases, it follows from Proposition 1 that for each . Therefore, we have . It follows from Proposition 11 that .
- (3.2.2)
There is a path in T satisfying that and .
Let be obtained from by subdividing the edge with vertex for each i, the edges and with the vertices and , respectively, and the other edges incident to y. Let F be the set of all subdivided edges. We note that . Let P be a -set. To paired-dominate and , we must have and . To dominate , we must have or .
First, assume that . If y is paired with , then and is a PD-set of obtained from by subdividing the edges with cardinality of less than P. If y is paired with , then is a PD-set of obtained from by subdividing the edges with cardinality of less than P. If y is partner with a subdivision vertex , then is a PD-set of obtained from by subdividing the edges with cardinality of less than P.
Second, assume that . If is paired with a subdivision vertex , then is a PD-set of obtained from by subdividing the edges with cardinality of less than P. If is paired with , then is a PD-set of obtained from by subdividing the edges with cardinality of less than P.
In all cases, it follows from Proposition 1 that for each . Therefore, we have . It follows from Proposition 11 that .
- (3.2.3)
There is no path in T with and .
Let be the shortest -path in T such that is a leaf of T and . By our assumption, T has no strong stem. Thus, we have . Let be the component of containing . Note that . Now, using the argument applied in Case 1, we can see that
This completes the proof.
□