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Axioms
, Volume 12 (5) – Apr 23, 2023

/lp/multidisciplinary-digital-publishing-institute/spanning-k-ended-tree-in-2-connected-graph-TK2C5Vm6VM

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axioms Article 1, 2 Wanpeng Lei * and Jun Yin School of Mathematics and Statistics, Taiyuan Normal University, Jinzhong 030619, China School of Computer Science, Qinghai Normal University, Xining 810008, China; 2013042@qhnu.edu.cn * Correspondence: wplei@tynu.edu.cn Abstract: Win proved a very famous conclusion that states the graph G with connectivity k(G), independence number a(G) and a(G) k(G) + k 1 (k 2) contains a spanning k-ended tree. This means that there exists a spanning tree with at most k leaves. In this paper, we strengthen the Win theorem to the following: Let G be a simple 2-connected graph such that jV(G)j 2k(G) + k, a(G) k(G) + k (k 2) and the number of maximum independent sets of cardinality k + k is at most n 2k k + 1. Then, either G contains a spanning k-ended tree or a subgraph of K _ ((k + k 1)K [ K ). n2kk+1 Keywords: connectivity; independence number; k-ended tree; maximum independent set MSC: 05C10 1. Introduction Notation regarding graph theory is not covered in this paper. We refer the reader to [1]. Let G = (V(G), E(G)) be a graph satisfying vertex set V(G) and edge set E(G). We denote the set of vertices adjacent to v in G as N(v). We write N(X) = N(x) for x2X X V(G). We also denote the subgraph of G induced by S as G[S] for S V(G). Let H and H be two subgraphs of G which vertex disjoint, and P be a path of G. A path xPy in G with end vertices x, y 2 V(G) is called a path from H to H if V(xPy)\ V( H ) = fxg 1 2 1 and V(xPy)\ V( H ) = fyg. (x, U)-path is a path from fxg to a vertex set U. We write an (x, U)-fan of width k for F G if F is a union of (x, U)-paths P , P , . . . , P , where 1 2 k Citation: Lei, W.; Yin, J. Spanning V(P )\ V(P ) = fxg for i 6= j. Let G and G be two subgraphs of G. We denote by xG i j 1 1 k-Ended Tree in 2-Connected Graph. (G x, respectively) the Hamilton path of G[fxg[ V(G )], which starts at x (terminates at x, 1 1 Axioms 2023, 12, 411. https:// respectively). We denote by xG y the Hamilton path of G[V(G )[fx, yg], which starts at 1 1 doi.org/10.3390/axioms12050411 x and terminates at y. We denote by G xG the Hamilton path of G[V(G )[fxg[ V(G )]. 1 2 1 2 A nontrivial graph, G, is considered k-connected if the maximum number of pairwise Academic Editor: Elena Guardo internally disjoint xy-paths for any two distinct vertices, x and y, is greater than or equal Received: 16 March 2023 to k. A trivial graph is considered 0-connected or 1-connected, but it is not considered k- Revised: 16 April 2023 connected for any k greater than 1. The connectivity, k(G), of G is deﬁned as the maximum Accepted: 21 April 2023 value of k for which G is k-connected. Published: 23 April 2023 If a graph contains a Hamilton path, then the graph is said traceable, and if a graph contains a Hamilton cycle, then the graph is said hamiltonian. The sufﬁcient conditions under which a graph can be traceable involving connectivity (k(G)) and independence number (a(G)) were given by Chvátal and Erdos ˝ in 1972. Copyright: © 2023 by the authors. Licensee MDPI, Basel, Switzerland. Theorem 1. (Chvátal and Erdos, ˝ [2]) If a graph G withjV(G)j 3 satisfies the conditions a(G) This article is an open access article k(G), a(G) k(G) + 1, respectively, then G is Hamiltonian and traceable, respectively. distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (https:// Theorem 1 has been extended in various directions, as documented in previous stud- creativecommons.org/licenses/by/ ies [3–8]. For recent results, see [9–12]. Fouquet and Jolivet [13] conjecture whether a 4.0/). Axioms 2023, 12, 411. https://doi.org/10.3390/axioms12050411 https://www.mdpi.com/journal/axioms Y Kk1 Kk2 M2(k2) Kn-k1- k2 F00( k 1, k 2) Axioms 2023, 12, 411 2 of 25 graph’s circumference can have a best possible lower bound when its independence num- ber exceeds its connectivity. This has been proved by Suil O et al. Theorem 2. (Suil O et al., [14]) If G is a simple graph such that jV(G)j = n and a(G) k(G), k(G)(n a(G) k(G)) then G contains a cycle with length of at least . a(G) The number of maximum independent sets of H for a subgraph H G is denoted by m( H). In their study [15], Chen et al. presented the following theorem that generalizes Theorem 1 by bound m(G). Speciﬁcally, the authors demonstrated that expanding the independence number (i.e., a(G) k(G) + 2) slightly and bounding m(G) does not alter the traceability of G. It is worth noting that K represents a complete graph with s vertices, while K is the complement of K . Additionally, the join G_ H of disjoint graphs G and H s s is obtained by joining each vertex of G to each vertex of H in the graph G + H. In the following, we construct three graphs which are excluded. Let H (k ) be i i a copy of K where i = 1, 2. The graph F (k , k ) is deﬁned as ( H (k ) _ H (k )) [ k 0 1 2 1 1 2 2 K [ M (k ), where n k k k and M (k ) is a matching of cardinality k nk k 1 2 1 2 2 1 2 2 1 2 between H (k ) and K . If n k k k , then F (k , k ) is obtained from 2 2 nk k 1 2 2 11 1 2 F (k , k ) by joining exactly two (nonadjacent) vertices of H (k )) or by joining all ver- 0 1 2 2 2 tices of V(K )n V( M (k )) and some ﬁxed vertex w 2 H (k ). Let F (k , k ) be nk k 1 2 0 2 2 00 1 2 1 2 the graph ( H (k )_ H (k ))[ K [ M (k ), where n k k k and M (k ) is 1 1 2 2 2 2 1 2 2 2 2 nk k 1 2 a matching of cardinality n k k between K and H (k ). Deﬁne the graph 1 2 nk k 2 2 1 2 F (k , k ) = K _ (K [ K ); see Figure 1. 2 2 1 k k nk k 2 1 1 2 Kk2 Kk1 Kk2 M1(k2) Kn-k1- k2 Kn-k1- k2 Kk1 F2( k 1, k 2) F0( k 1, k 2) Figure 1. F (k , k ), F (k , k ) and F (k , k ). 0 1 2 00 1 2 2 1 2 Theorem 3. (Chen et al., [15]) Let G be a 2-connected graph with jV(G)j 2k (G), k(G) = k, a(G) k + 1 and m(G) n 2k. Then, either G is Hamiltonian or F (k, k) G F (k, k), 11 2 where F (k, k) and F (k, k) are two graphs deﬁned above. 11 2 Theorem 4. (Chen et al., [15]) Let G be a connected graph with jV(G)j 2k (G), k(G) = k, a(G) k + 2 and m(G) n 2k 1. Then either G is traceable or F (k + 1, k) G K _ ((k + 1)K [ K ), where F (k + 1, k) is the graph deﬁned above. 1 n2k1 11 A Hamilton path is viewed as a spanning tree with exactly two leaves. This perspective allows for the generalization of sufﬁcient conditions for a graph to be traceable to those for a spanning tree with at most k leaves. A tree is called a k-ended tree if it has at most k leaves. Our focus now shifts to spanning k-ended trees. Clearly, if s t, then a spanning s-ended tree is also a spanning t-ended tree. Theorem 1 demonstrates that each graph G such that a(G) k(G) + 1 is traceable. In [16], Win proved the following theorem, which generalizes Theorem 1. Theorem 5. (Win, [16]) Let G be a connected graph and let k 2 be an integer. If a(G) k(G) + k 1, then G contains a spanning k-ended tree. In [17], Lei et al. extend Theorem 5 in cases when k(G) = 1 to the following direction. Axioms 2023, 12, 411 3 of 25 Theorem 6. (Lei et al., [17]) Let k 3 and G be a connect graph with jV(G)j 2k + 2, a(G) 1 + k and m(G) n 2k 2. Then G contains a spanning k-ended tree. In [15], Chen et al. generalize Theorem 1 by bound m(G). The authors demonstrated that expanding the independence number (i.e., a(G) k(G) + 2) slightly and bounding m(G) does not alter the traceability of G. In this paper, our focus will be on the existence of spanning k-ended tree. We will work on extending Theorem 5 to a more general case. A natural question is whether expanding the independence number can alter the existence of the spanning k-ended tree. In the following section, we introduce the k-ended system, which is an important tool for studying the k-ended tree. k-Ended System If there exists a set of paths and cycles where the elements are pairwise vertex-disjoint, we refer to it as a system. This system is often viewed as a subgraph. LetS be a system in a graph. For S 2 S , we put f (S) = 2 if S is a path of order at least 3 and f (S) = 1 otherwise (i.e., S is a vertex, an edge or a cycle). We write V(S) = V(S) and f (S) = f (S). If S2S S2S f (S) k,S is called a k-ended system. Moreover, we callS a spanning k-ended system of G, if V(S) = V(G). Let S = fS 2 S : f (S) = 2g, S = fS 2 S : f (S) = 1g. Then, S = S [S , V(S) = V(S). P C S2S Additionally, V(S ) and V(S ) can be deﬁned in a similar manner. We use jSj, jS j P C P andjS j to represent the number of elements inS ,S andS , respectively. For each S 2 S , C P C we assign an orientation denoted by the symbol <, where x < y if x precedes y in the ! ! orientation. Let S be the orientation of S 2 S and let S be the reverse orientation of S for S 2 S . By assigning an orientation to each S 2 S , we identify S as a system with an orientation, where each element is ordered relative to the others. Let S be a system with a deﬁned orientation. For any P 2 S , we deﬁne v (P) and P L v (P) as the two end-vertices of P such that v (P) < v (P). Additionally, for each C 2 S , R L R C we select an arbitrary vertex v within C. These deﬁnitions will be used in subsequent analyses. Then deﬁne [ [ End(S ) = fv (P), v (P)g, End(S ) = fv g, End(S) = End(S )[ End(S ). P L R C C P C P2S C2S P C For S 2 S and x 2 V(S), we write the ﬁrst, second and ith predecessor (successor, i + ++ i+ respectively) of x as x , x and x (x , x and x , respectively). For convenience, we + + write x = x = x for K = x and y = x = x for K = xy. 1 2 For P 2 S , if x = v (P) (x = v (P), respectively), we have only the predecessor P R L of v (P) (successor of v (P), respectively). For fx, yg V(P), we denote by the section R L + 2+ 3+ s+ P(x, y) a path x x x ...x (= y ) of consecutive vertices of P and denote by the section + 2+ s+ P[x, y] a path xx x ...x (= y) of consecutive vertices of P. Moreover, if x = y, then the section P[x, y] is trivial. The following lemma illustrates the importance of k-ended systems for spanning k-ended trees. Lemma 1. (Win, [16]) Let k 2 be an integer and let G be a connected simple graph. If G contains a spanning k-ended system, then G also contains a spanning k-ended tree. Axioms 2023, 12, 411 4 of 25 A k-ended system S in G is considered a maximal k-ended system if there is no other ˆ ˆ k-ended system S in G satisfying V(S) V(S). The following lemma presents some useful properties of k-ended systems. It is important to note that two distinct elements of S are connected by a path in G V(S) if there exists a path in G whose end-vertices are in elements of S and whose inner vertices are not all contained in V(S). It is worth mentioning that a path may not have any inner vertex. Lemma 2. (Akiyama and Kano, [18]) Let k 2 be an integer and G be a connected simple graph. Assume that G does not contain a spanning k-ended system and let S be a maximal k-ended system of G satisfying the cardinality of the maximum value of S subject to the maximum value of V(S). Then the following characterizations are true. (i) There is no path connecting two distinct elements ofS whose inner vertices are in V(G)n V(S). (ii) There is no path connecting an element of S and one end-vertex of an element of S whose inner vertices are in V(G)n V(S). (iii) There is no path connecting an end-vertex of an element of S and an end-vertex of another element of S whose inner vertices are in V(G)n V(S). (iv) There are no two internally disjoint paths Q and Q connecting two distinct 1 2 elements of S whose inner vertices are in V(G)n V(S) with jV(Q )\ V(Q )j = 1. C 1 2 2. Methods In this paper, our focus will be on the existence of spanning k-ended tree. We will work on extending Theorem 5 to a more general case. We tried to prove that it does not change the existence of spanning k-ended tree if we expand the independent numbers a little bit and bound m(G). The proof will follow an approach similar to Theorem 6, but with additional considerations for the increased connectivity of the graph. Our proof follows a method of contradiction. We primarily utilize the crucial tool of the maximal k-ended system, as mentioned above, to derive contradictions. The subsequent section is the crucial property of the maximal k-ended system which we obtained. This property plays a pivotal role in our proof. Important Properties of Maximal k-Ended System In this section, for convenience, we assume the following: Let k 2 and G be a graph with jV(G)j > 2k(G) + k, k(G) = k 2, a(G) = k + k and m(G) n 2k k + 1. Suppose that there is no spanning k-ended system in G and let S be a k-ended system of G satisfying the following: (I) The cardinality of the set V(S) is maximized. (II) The cardinality of S is maximized subject to condition ( I). ThenS is a set of subgraphs of G satisfying the hypothesis of Lemma 2. Let H = G V(S). Then jV( H)j 1. Let w 2 (V(G) V(S). The following lemma is easily obtained from the selection of S and we omit the proof. Lemma 3. The following characterizations are true. (1) For any P 2 S , v (P) and v (P) are not in N( H). P L R (2) For any C 2 S such that C K , N( H)\ V(C) = Æ. C 1 By the Fan Lemma, there exists a (w, V(S))-fan L with width k. For S 2 S with V(S)\ V(L) 6= Æ, let V(S)\ V(L) = fu , , u g (where u , , u are the ver- S,1 S,t S,1 S,t S S tices of S along the direction of S) and L be the path of L between w and u . Then S,i S,i U = fV(S)\ V(L)g and L = fL , , L g is the set of paths between w and S. S S,1 S,t S2S + + + Denote U = fu : u 2 Ug and U = fu : u 2 Ug. By Lemma 3(1),(2), U and U are well deﬁned and hence jU j = jU j = jUj. The proof of the following lemmas can be easily obtained from the choice of S and we will omit it. Axioms 2023, 12, 411 5 of 25 Lemma 4. A graph G cannot have a k -ended system T that includes all vertices in a k-ended system S , where k < k. Lemma 5. The following characterizations are true: (1) Both U and End(S) are independent sets of G. (2) N(w)\ U = Æ. (3) j fC 2 S : V(C)\ U 6= Æg j 1. (4) Let x 2 U \ V(P), where P 2 S . Then N(x)\ (End(S )nfv (P)g) = Æ. Furthermore, P P R if x 6= u , then N(x)\ End(S ) = Æ. P,t Lemma 6. Let v 2 End(S) and S 2 S with v 2 V(S ). Then the following statements 1 1 1 1 are true: (1) [(N(v ) \ V(S)) [ (N(v ) \ V(S)) ] \ N(v) = Æ for any v 2 V(S fS, S g) [ 1 1 C 1 End(S fS, S g) and S 2 S fS g. 1 1 (2) If v = v (P) (v = v (P), respectively), then (N(v )\ V(S )) \ N(v) = Æ (N(v )\ 1 L 1 R 1 1 1 (N(v)\ V(S )) = Æ, respectively) for any v 2 V(S )[ V( H)[ (End(S)nfv g). 1 C 1 Let Y be an independent set of G with size k + k. Then the following lemma holds. 0 0 Lemma 7. Let S belong to V(G) which satisﬁes S \ Y having precisely one vertex, denoted as z, 0 0 0 i.e, S \ Y = fzg. If N(x)\ Y = fzg for each x 2 S nfzg, then G[S ] forms a clique. Proof of Lemma 7. We begin by assuming the opposite and using a proof by contradiction. Suppose that x x 2 / E(G) for some pair of vertices x and x in S , where x 6= x . Then, 1 2 1 2 1 2 (Ynfzg)[fx , x g forms an independent set of G with a size of k + k + 1. This contradicts 1 2 the fact that a(G) = k + k. Hence, G[S ] is a clique. For convenience, suppose that x is an element of V(P). For each P 2 S , we deﬁne Q(x, P) as follows: v (P), if x = v (P), < R R xv (P), if x = v (P), Q(x, P) = R R x P v (P)x, if xv (P) 2 E(G) (x 6= v (P), x 6= v (P)). R R R R Similarly, we deﬁne Q(x, P) as follows: x = v (P), if x = v (P), L L xv (P), if x = v (P), Q(x, P) = L L x P v (P)x, if xv (P) 2 E(G) (x 6= v (P), x 6= v (P)). L L L L Therefore, f (Q(x, P)) = 1 and f (Q(x, P)) = 1. We say that C(G ) is a spanning subgraph of G satisfying f (C(G )) = 1 if G G. 0 0 0 Some properties ofS are described in the following lemmas, as proved in Appendix B. Lemma 8. U V(S ) and j S j= 1. P P By Lemma 8, S = fPg (say). Then U = V(P)\ V(L) = fu , , u g and t = k. P P,1 P,t P For convenience, denote that U = fu , u , , u g and 1 2 k + + End(S)[ U [fwg, if End(S)\ U 6= Æ, X = + + + (End(S)[ U [fwg)nfu g, i f End(S)\ U = Æ. Lemma 9. The following statements hold. Axioms 2023, 12, 411 6 of 25 + + (1) N(u )\ End(S ) 6= Æ or f (Q(u , P)) = 1 and N(u )\ End(S ) 6= Æ or f (Q(u , C C k k 1 1 P)) = 1. (2) If f (Q(u , P)) 6= 1 and f (Q(u , P)) 6= 1, then there exist at least two elements C, 0 + C 2 S such that u v 0 2 E(G), u v 2 E(G). C k C C (3) X forms an independent set of G with size k + k. + 2+ 2+ (4) If N(u )\ End(S ) 6= Æ and u 6= v (P), then u v (P) 2 E(G). k C k R k R (5) Let y 2 V(P) satisfy jV(P[y , v (P)])j 1, yv (P) 2 E(G) and V(P[y, v (P)])\ U = R R R Æ. Then G[V(P[y , v (P)])] forms a clique. Additionally, if the intersection of N(y) and X is fv (P)g, then the graph G[V(P[y, v (P)])] forms a clique. R R (6) G[V(C)] forms a clique for any C 2 S . Furthermore, N(x) \ X = fv g for any C C x 2 V(C)nfv g. 3. Results and Discussion In [17], the authors provide a novel extension by imposing a limit on the maximum number of independent sets, although the limit is not sharp. Note that G has no spanning k + 1 k -ended tree for each G 2 fF (k , k ), F (k , k ), F (k , k )g. In this paper, we 1 2 0 1 2 00 1 2 2 1 2 extend Theorem 5 to the case where k(G) 2 and the bound on the number of maximum independent sets is already sharp. Theorem 7. Let k 2 and G be a graph with jV(G)j 2k(G) + k , k(G) = k 2, a(G) k + k and m(G) n 2k k + 1. Then G contains a spanning k-ended tree, unless either F (k + k 1, k) G F (k + k 1, k) for n > 3k + k 1, or F (k + k 1, k) G 0 2 00 F (k + k 1, k) for 2k + k n 3k + k 1. Note that a spanning tree having exactly two leaves is called a Hamilton path. Then, we can immediately obtain the following result. Corollary 1. Let G be a graph with jV(G)j 2k(G) + 2, k(G) = k 2, a(G) k + 2 and m(G) n 2k 1. Then G is traceable, unless either F (k + 1, k) G F (k + 1, k) for 0 2 n > 3k + 1, or F (k + 1, k) G F (k + 1, k) for 2k + 2 n 3k + 1. 00 2 In the case of 2-connected, the bounds of jV(G)j can do better. Clearly, Corollary 1 im- proves the result of Theorem 4. It demonstrated that expanding the independence number slightly and bounding m(G) also does not alter the traceability in highly connected graphs. If F (k + k 1, k) G F (k + k 1, k) for n > 3k + k 1, or F (k + k 1, k) 0 2 00 G F (k + k 1, k) for 2k + k n 3k + k 1, then m(G) = n 2k k + 1. Hence, we can obtain the following result immediately. Corollary 2. Let k 2 and G be a graph of order n 2k(G) + k such that k(G) = k 2, a(G) k + k and m(G) n 2k k. Then G contains a spanning k-ended tree. 4. Proof of Theorem 7 In this section, we employ the same terminology and notation in Section 2. Proof of Theorem 7. Let k 2 and G be a graph with jV(G)j > 2k(G) + k, k(G) = k 2, a(G) k + k and m(G) n 2k k + 1. We begin by assuming the opposite and using a proof by contradiction. Suppose that G does not have a spanning k-ended tree. This assumption, along with Theorem 5, implies the following equation: a(G) = k(G) + k. (1) Thus, by Lemma 1, G cannot have a spanning k-ended system. We select a maximal k- ended system S of G that satisﬁes conditions (I) and (II) outlined in Section 2. Deﬁne H = G V(S). Clearly jV( H)j 1. Let w 2 (V(G) V(S)). Axioms 2023, 12, 411 7 of 25 We will show that F (k + k 1, k) G F (k + k 1, k) for n > 3k + k 1, or 0 2 F (k + k 1, k) G F (k + k 1, k) for 2k + k n 3k + k 1. 00 2 Fact 1. m(G) n 2k k + 1. Proof of Fact 1. We consider a (w, V(S))-fan L in Section 2. By Lemma 8, we choose U = fu , u , , u g and S = fPg in Section 2. 1 2 k P We consider a (w, V(S))-fanL in Section 2. By Lemma 8, we choose U = fu , u , , u g 1 2 k and S = fPg in Section 2. Claim 1. N(u )\ V(S ) = Æ and N(u )\ V(S ) = Æ. C C Proof of Claim 1. Using symmetry, we can focus on proving that N(u )\ V(S ) = Æ. By k C + + 0 contradiction, suppose that N(u )\ V(S ) 6= Æ; say v 2 N(u )\ V(S ) and v 2 V(C ). C C k k By Lemma 2(ii), u 6= v (P). k R Denote 3+ + V(P[u , v (P)]), if u v (P) 2 / E(G), k R k R A = 2+ + V(P[u , v (P)]), if u v (P) 2 E(G). R R k k + 2+ If u v (P) 2 / E(G), then, by Lemma 9(4), u v (P) 2 E(G). Therefore, by Lemma 9(5), R R k k 3+ + G[V(P[u , v (P)])] forms a clique. If u v (P) 2 E(G), then, according to Lemma 9(5), k R k R 2+ G[V(P[u , v (P)])] forms a clique. Hence, G[ A] f orms a clique. (2) As G is a connected graph, N( A)\ V(G A) 6= Æ. For y 2 N( A)\ V(G A), there exists a vertex x 2 A with xy 2 E(G). We will show that + 2+ y 2 fu , u g. (3) k k + 2+ By contradiction, suppose that y 2 / fu , u g. By Lemma 6(2) and (2), y 2 / V(S )[ V( H). k k We will examine the following two scenarios to reach a contradiction: • Suppose that y 2 V(P[v (P), u ]). By Lemma 9(1), f (Q(u , P)) = 1 or N(u )\ 1 1 1 V(S ) 6= Æ. We will show that u v (P) 2 E(G). If f (Q(u , P)) = 1, then, by sym- C L 1 1 metry and Lemma 9(5), u v (P) 2 E(G). If f (Q(u , P)) 6= 1, then, by Lemma 9(1), 1 1 N(u ) \ V(S ) 6= Æ. By symmetry and Lemma 9(4), u v (P) 2 E(G). Then, C L 1 1 by symmetry and Lemma 9(5) and (2), either the set of vertices of the subgraph ! ! 0 + 0 C u P x v (P) P xy P u L w and Q(y , P) is equal to V(P[ C )[fwg, which con- R k k ! ! 0 + 0 tradicts (I); or C u P x v (P) P xv (P) P u in G is equal to V(P[ C ), which con- k R L tradicts Lemma 4. • Assume that y 2 V(P[u , u ]). Then, by Lemma 9(1)(2) and (2), the set of vertices of the subgraph ! ! ! 0 + C u P x v (P) P xy P u L wL u P y , if y 2 V(P(u , u )), < R k k 1 1 1 k ! ! 0 + Q = 1 C u P x v (P) P xu P u L w, if y = u , R 1 k k 1 0 + C u P x v (P) P xu P u L w, if y = u , k R k 1 1 k and Cu Q v (Q), if f (Q(u , Q)) 6= 1, Q,1 Q,1 Q = Q(u , Q), if f (Q(u , Q)) = 1. Q,1 Q,1 0 0 is equal to V(P[ C[ C )[fwg or V(P[ C )[fwg, which contradicts (I). Axioms 2023, 12, 411 8 of 25 This contradiction shows that (3) holds. + + 2+ + If y = u and u v (P) 2 / E(G), then, u v (P) 2 E(G). By (2), G[V(P[u , v (P)])] k k R k R k R + 0 has a cycle C = v (P) P xu P x v (P). By (2), we structure a new path P such that k R R ! ! 0 + 2+ P = v (P) P u x P v (P)x P u by rearranging the order of the vertices in P. Then L k R k 0 2+ + 0 + 0 v (P ) = u . It is easy to verify that G[V(P[u , v (P)])] = G[V(P [u , v (P )])]. Note R R R k k k + 0 0 0 that u v (P ) 2 E(G). By Lemma 9(5), the subgraph G[V(P [x, v (P )])] forms a clique. R R 0 0 + 0 + 0 0 0 Let A = V(P [u , v (P )])nfu g = V(P [x, v (P )])]. Since G is connected, N( A )\ R R k k 0 0 0 0 0 0 0 V(G A ) 6= Æ. For y 2 N( A )\ V(G A ), there exists a vertex x 2 A with x y 2 E(G). 0 + 0 0 0 By the proof of (3), y = u . That meansjN( A )\ V(G A )j = 1, contradictingjN( A )\ 0 2+ + V(G A )j k 2. Therefore, by (3), we have either y = u and u v (P) 2 / E(G) k k + + or y = u and u v (P) 2 E(G). Then, jN( A)\ V(G A)j = 1, contradicting jN( A)\ k k R V(G A)j k 2. This contradiction indicates that Claim 1 is true. According to Claim 1 and Lemma 9(1), f (Q(u , P)) = 1 and f (Q(u , P)) = 1. Denote + + End(S)[ U [fwg, if End(S)\ U 6= Æ, X = + + + (End(S)[ U [fwg)nfu g, otherwise(i.e., i f End(S)\ U = Æ). By Lemma 9(3), X is an independent set of G with size k + k. Thus, N(v)\X 6= Æ f or any v 2 V(G)nX . (4) Claim 2. G[V(P[u , v (P)])] and G[V(P[v (P), u ])] form cliques. R L Proof of Claim 2. By virtue of symmetry, we may restrict our consideration to prove that + + G[V(P[u , v (P)])] forms a clique. As G[V(P[u , v (P)])] is connected, we can assume R R k k + + that jV(P[u , v (P)])j 3. According to Lemma 5(1) (4) and Claim 1, N(u )\ (X n k k + + fv (P)g) = Æ. By (4), N(u )\X = fv (P)g. By Lemma 9(5), G[V(P[u , v (P)])] forms R R R k k a clique. Claim 3. N(V( H))\ V(S ) = Æ. Proof of Claim 3. By contradiction, suppose that N(V( H)) \ V(S ) 6= Æ; say x 2 N(V( H))\ V(C) for some C 2 S . This implies that there is a vertex v 2 V( H) with e = vx. By Lemma 8, v 6= w. We will show that v 2 / V(L). (5) Suppose, by way of contradiction, that v 2 V(L ) for some i 2 f1, , kg. Suppose i 0 that v 2 V(L ) [ V(L ). By symmetry, we may only think of v 2 V(L ). Then, by 1 1 Claim 2, v (P) P u L vC and Q(u , P) cover V(P)[ V(C)[fvg, contradicting (I). There- R 1 1 fore, v 2 V(L ) for some i 2 f2, , k 1g. Then, by Claim 2, the set of vertices of the i 0 subgraph v (P) P u L wL vC and Q(u , P) in G is equal to V(P)[ V(C)[fw, vg, which L k k i again contradicts (I). Thus, we have shown that (5) holds. Next, we will prove that vw 2 E(G). (6) By contradiction, suppose that vw 2 / E(G). Note that x 2 V(C). We consider the neigh- + + bourhood of the vertex x . According to Lemma 2(i)(ii), N(x )\ (End(S)nfv g) = Æ. + + If x and u 2 V(P) for some i 2 f1, , k 1g are adjacent in G, then, by Claim 2 and (5), v (P) P u L wL u P u Cv and Q(u , P) in G covers V(P)[ V(C)[fw, vg, contradicting L k k i i k + + + + (I). Hence, N(x )\ (U nfu g) = Æ. If v and u 2 V(P) for some i 2 f1, , k 1g ! ! + + are adjacent in G, then, by Claim 2 and (5), v (P) P u L wL u P u vC and Q(u , P) in L k k i i k + + G covers V(P)[ V(C)[fw, vg, contradicting (I). Hence, N(v)\ (U nfu g) = Æ. Note k Axioms 2023, 12, 411 9 of 25 + + + that vw 2 / E(G). Therefore, by Lemma 2(i)(ii), fv (P), v (P), x , w, vg[ (U nfu g)[ L R (End(S )nfv g) forms an independent set of size k + k + 1. This contradicts the fact that C C a(G) = k + k and thus establishes that (6) holds. Then, by Claim 2, (5) and (6), the set of vertices of the subgraph v (P) P u L wvC k k and Q(u , P) is equal to V(P [ C) [ fw, vg. This contradicts (I) and establishes that N(V( H))\ V(C) = Æ. Claim 4. N(V(C))\ (V(G)n V(C)) = U for each element C 2 S . Proof of Claim 4. Since G is connected, N(V(C))\ (V(G)n V(C)) 6= Æ for any C 2 S . For z 2 N(V(C))\ (V(G)n V(C)), there exists a vertex x 2 V(C) with xz 2 E(G). Accord- ing to Lemma 2 (i), (ii), z 2 / V(S nfCg). By Claim 3, z 2 / V( H). This implies that z 2 V(P). (7) Next, we will show that z 2 U. By contradiction, suppose that z 2 / U. By Lemma 6(2) and Claim 2, z does not belong to V(P(u , v (P)]) [ V(P[v (P), u )). To arrive at a k R L contradiction, we will examine the following three scenarios using (7): + + • Suppose that z 2 fu , u g. Then N(u ) \ V(S ) 6= Æ or N(u ) \ V(S ) 6= Æ, k k C C 1 1 contradicting Claim 1. + + + • Suppose that z 2 U nfu g or U nfu g. By symmetry, we consider that z 2 U n + + fu g say z = u for some i 2 f1, , k 1g. Then, by Claim 2, Cu P u L wL u P k k i i i i v (P) and Q(u , P) cover V(P[ C)[fwg, contradicting (I). L k 2+ 2 • Suppose that z 2 V(P[u , u ]) for some i 2 f1, , k 1g. We consider the i i+1 + + neighbourhood of the vertex z . We claim that N(z )\X = fv g. Suppose oth- erwise that there exists a vertex y 2 N(z )\X such that y 6= v . By Lemma 6(1), y 2 / End(S )nfv g. According to the deﬁnition of X , we will examine the following C C two scenarios to reach a contradiction. – Assume that y 2 U \X ; say y = u for some j 2 f1, , k 1g. If j > i, then, ! ! by Claim 2, the set of vertices of the subgraph v (P) P u z P u L wL u P zC R j j 1 1 and Q(u , P) is equal to V(P[ C)[fwg, which contradicts (I). If j i, then, ! ! by Claim 2, the set of vertices of the subgraph v (P) P u L wu P z u P zC[ L j j k Q(u , P) is equal to V(P[ C)[fwg, which again contradicts (I). – Assume that y 2 End(S ). By Lemma 6(2), y = v (P). Then, by Claim 2, the set P R of vertices of the subgraph u P z v (P) P u L wL u P zC[ Q(u , P) is equal k k k R 1 1 to V(P[ C)[fwg, which contradicts (I). + + This contradiction establishes that N(z )\X fv g. By (4), N(z )\X = fv g. If C C x 6= v and jV(C)j > 1 or jV(C)j = 1, then, by Lemma 9(6), the set of vertices of the ! ! subgraph v (P) P zCz P v (P) is equal to V(P[ C), which contradicts Lemma 4. If L R x = v andjV(C)j > 1, then, according to Lemma 9(6), N(v )\X = fv g. Note that C C + + + + + N(z )\X = fv g. If z v 2 / E(G), then, by Lemma 9(6), (X nfv g)[fz , v g C C C C would be an independent set of cardinality k + k + 1, contradicting (1). Therefore, ! ! + + z v 2 E(G). Then the set of vertices of the subgraph v (P) P zCz P v (P) is equal L R to V(P[ C), which contradicts Lemma 4. This contradiction establishes that z 2 U. Since jN(V(C))\ (V(G)n V(C))j k and jUj = k, N(V(C))\ (V(G)n V(C)) = U for any element C 2 S . Claim 5. Let C 2 S with jV(C)j > 1. For any two disjoint vertices u , u 2 U, there exist two C i j 0 0 disjoint vertices v, v 2 V(C) such that u v, u v 2 E(G). i j Axioms 2023, 12, 411 10 of 25 Proof of Claim 5. We establish Claim 5 by contradiction. Suppose that either N(u )\ V(C) = N(u )\ V(C) = Æ or N(u )\ V(C) = N(u )\ V(C) = fvg and v 2 V(C) for j i j 0 0 0 some u , u 2 U. i j 0 0 If N(u )\ V(C) = N(u )\ V(C) = Æ, then N(V(C))\ (V(G)n V(C)) 6= U, con- i j 0 0 tradicting Claim 4. Now suppose that N(u ) \ V(C) = N(u ) \ V(C) = fvg. Let i j 0 0 ˆ ˆ U = (Unfu , u g)[fvg and C = C v. Since jV(C)j > 1, C 6= Æ. Then, by hypothesis i j 0 0 ˆ ˆ ˆ ˆ and Claim 4, N(V(C))\ (V(G)n V(C)) U. However, jUj = k 1, contradicting the hypothesis that G is k-connected. These contradictions establish that Claim 5 is true. Claim 6. G[V( H)] forms a clique. Proof of Claim 6. We will only focus on the case where jV( H)j 2. For every vertex v 2 V( H)nfwg, N(v)\X 6= Æ. We assume that there is at least one vertex x 2 N(v)\X with x 6= w. By Claim 3, x is not an element of End(S ). By Lemma 3(1), x 2 / End(S ). C P + + Then, x 2 X \ U ; say x = u for some i 2 f1, , k 1g. Then, by Claims 2, 4 and 5, there exist a path Q and Q(u , P) cover V(P)[ V(C)[fvg, see Figure 2, contradicting (I). This contradiction shows that N(v)\X fwg. By (4), N(v)\X = fwg for every vertex v 2 V( H)nfwg. Let S = V( H). Then, according to Lemma 7, G[V( H)] forms a clique. v (P) i u v (P) L i Figure 2. vu 2 E(G) (i 6= k). Denote A = V(P[v (P), u ]) and A = V(P[u , v (P)]). L 2 R 1 k Claim 7. The following two statements are true. (1) N( A )\ (G A ) = U for i 2 f1, 2g; i i (2) N(V( H))\ V(S) = U. Proof of Claim 7. We will prove the ﬁrst statement. By symmetry, we have only proved that N( A )\ (G A ) = U. Let C = G[ A ]. By Claim 2(1), f (C ) = 1. We pick an 2 2 2 element C 2 S , by Claim 4, a new path Q = v (P) P u C would be obtained. We structure C L a new system S such that S = S [S , S = fQg and S = (S nfCg)[fC g. It C P P C is easy to verify that V(S ) = V(S), jS j = jS j and jS j = jS j. Hence, S is also a C P C P k-ended system satisfying (I), (II). Then, by Claim 4, N( A )\ (G A ) = U. 2 2 Next, we need to prove the second statement. The proof here is similar to Claim 4. (For details, see Appendix A.) Claim 8. Suppose that jV( H)j > 1. For any two disjoint vertices u , u 2 U, there exist two i j 0 0 disjoint vertices v, v 2 V( H) such that u v, u v 2 E(G). i j Proof of Claim 8. The proof here is similar to Claim 5. (For details, see Appendix A.) Claim 9. u u 2 E(G) for each i 2 f1, , k 1g. i+1 i Proof of Claim 9. Since G[V(P[u , u ])] is connected, we only need to focus on the case i i+1 + + where jV(P[u , u ])j 3. By contradiction, suppose that u u 2 / E(G) for some i i+1 i +1 i 0 0 Axioms 2023, 12, 411 11 of 25 i 2 f1, , k 1g. By (4), there exists at least one vertex y 2 N(u )\X satisfying i +1 y 6= u . By Claim 4, y 2 / End(S ). We will examine the following two scenarios to reach a contradiction, based on the deﬁnition of X : + + • Assume that y 2 X \ (U nfu g); say y = u for some j 2 f1, , k 1gnfi g. If i j ! ! j > i , then, by Claim 4, the set of vertices of the subgraph v (P) P u u P u L w 0 L k k i +1 j ! ! L u P u C[ Q(u , P) is equal to V(P[ C)[fwg, which contradicts (I). If j < i , i +1 i +1 j 0 0 0 then, by Claims 4 and 5, the set of vertices of the subgraph v (P) P u L wL u P u C k k L j j i +1 ! ! ! + + u P u u P u [ Q(u , P) is equal to V(P[ C)[fwg, which again contradicts (I). i k 0 i +1 j i 0 0 • Assume that y 2 End(S ). If y = v (P), then v (P) P u L wL u P v (P)u P L R i +1 i +1 i i L 0 0 0 0 i +1 P u covers V(P)[fwg, which contradicts (I). Therefore, y = v (P). Then v (P) R L P u v (P) P u L w covers V(P)[fwg, which again contradicts (I). R i +1 i +1 i +1 0 0 + + This contradiction demonstrates that N(u )\X fu g. By (4), N(u )\X = fu g. i+1 i i+1 i By Claim 9, it holds that u u 2 E(G) for every i 2 f1, , k 1g. Let i+1 i + + C = G[E(P[u , u ][fu u g)] for every i 2 f1, , k 1g. i i+1 i i+1 Claim 10. For each section P[u , u ], the following two statements are true. i i+1 • G[V(P[u , u ])] forms a clique; i i+1 + + • N(V(P[u , u ]))\ (V(G)n V(P[u , u ])) = U. i i+1 i i+1 Proof of Claim 10. We pick an element C 2 S ; by Claims 4 and 5, a new path ! ! Q = v (P) P u Cu P v (P) would be obtained. We structure a new system S such i L i i+1 R i that S = S [ S , S = fQ g and S = (S n fCg) [ fC g. It is easy to verify i iC iP iP i iC C i that V(S ) = V(S), jS j = jS j and jS j = jS j. Hence, S is also a k-ended sys- i iC C iP P i tem satisfying (I), (II). According to Lemma 9(6), G[V(C )] forms a clique; by Claim 4, N(V(C ))\ (V(G)n V(C )) = U. Claim 10 is proved. i i By Claims 2–10 and Lemma 9(6), w(G U) = k + k and every component of G U forms a clique. Then, k1 m(G) = jV( H)jjV(P[v (P), u ])jjV(P[u , v (P)])j jV(P[u , u ])j Õ Õ L R 1 i i+1 i=1 C2S jV(C)j 1 1 1 1[n 2k k + 1] = n 2k k + 1. This completes the proof of | {z } k+k Fact 1. Fact 2. jV( H)j = 1 and m(G) = n 2k k + 1. Proof of Fact 2. By Fact 1 and the condition of Theorem 5, m(G) = n 2k k + 1. Then, we will show that jV( H)j = 1. By contradiction, suppose that jV( H)j 2. Then jV(P[u , u ])j = 1 for any i i+1 i 2 f1, , k 1g. Otherwise, m(G) > n 2k k + 1, contradicting m(G) = n 2k k + 1. Let x = V(P[u , u ]) for some i 2 f1, , k 1g. By Claims 7(2) and 8, i i +1 0 0 ! ! v (P) P u Hu P v (P) in G cover (V(P)nfxg)[ V( H), which contradicts (I). This L i i +1 R 0 0 contradiction shows that jV( H)j = 1. Denote + + + + G[V(P[u , u ])], if r 2 U nfu g, say r = u , i i+1 i G[V(P[v (P), u ])], if r = v (P), L L J(r) = > G[V(P[u , v (P)])], if r = v (P), k R R G[V(C)]. if r = v . C Axioms 2023, 12, 411 12 of 25 Finally, we need to prove that G is isomorphic to one of those graphs F with F (k + k 1, k) F F (k + k 1, k) or F (k + k 1, k) F F (k + k 1, k). De- 0 2 00 2 + + note R = End(S)[ (U nfu g). By Fact 2, jV(S)j = n 1 and there exists at most one vertex r 2 R such that j J(r )j 2. Then j J(r )j = n 2k k + 1 = m(G). Let 0 0 0 W (G) = fwg[ (Rnfr g). It follows that W (G) is an independent set of G with a car- 1 1 dinality of k + k 1. Additionally, W (G)[fxg is a maximum independent set of G for any vertex x 2 J(r ). By Claims 4, 7 and 10, y 2 Rnfr g is not adjacent to any vertex in 0 0 J(r )[fwg; it should be adjacent to u for all i 2 f1, , kg. Now let H = G[W (G)] and 0 i 1 1 H = G[U]. This implies that F (k + k 1, k) G F (k + k 1, k) and n > 3k + k 1 2 0 2 or F (k + k 1, k) G F (k + k 1, k) and 2k + k n 3k + k 1 (note that 00 2 J(r ) = K ), which completes the proof of Theorem 7. 0 n2kk+1 5. Conclusions We demonstrats that it does not change the existence of spanning k-ended tree if we expand the independent numbers a little bit and bound m(G). Therefore, we generalize Theorem 5 and the bound on the number of maximum independent sets is already sharp. Note that a Hamilton path is viewed as a spanning tree with exactly two leaves; in other words, a Hamilton path is a spanning 2-ended tree. Hence, our results extend Theorem 4, which has signiﬁcant implications for traceability and the existence of spanning trees. Moreover, we extend Theorem 5 to the case where k(G) 2. This extension has important implications for the study of independent sets in highly connected graphs. The proof of the results is currently too complex and difﬁcult. We hope to ﬁnd a more clever and concise proof technique for Theorem 7 in the future. Author Contributions: Conceptualization, W.L.; methodology, W.L.; validation, W.L.; formal analysis, W.L. and J.Y.; investigation, W.L.; writing—original draft preparation, W.L.; writing—review and editing, W.L.; supervision, W.L.; project administration, W.L. and J.Y. All authors have read and agreed to the published version of the manuscript. Funding: This work was supported by the Natural Science Funds of China (No. 11801296), the Gen- eral project of the Basic Research Program of Shanxi Province (Free exploration) (No. 202103021224303), the Shanxi Province Higher Education Reform and Innovation Project (No. J2021552) and the Shanxi Province Higher Education Science and Technology Innovation Project (No. 2020L0510). Institutional Review Board Statement: Not applicable. Informed Consent Statement: Not applicable. Data Availability Statement: Not applicable. Acknowledgments: The authors are greatly indebted to the Referees for their careful reading of the manuscript and invaluable comments and suggestions, which improved the paper. Conﬂicts of Interest: The authors declare no conﬂict of interest. Appendix A. Some Proofs of Claims of Theorem 7 Proof of Claim 7(2). Since G is connected, N(V( H))\ V(S) 6= Æ. For z 2 N(V( H))\ (V(G)n V( H)), there exists a vertex v 2 V( H) with vz 2 E(G). By Claim 3, z 2 / V(S ). This implies that z 2 V(P). We will prove that z 2 U. Suppose, by way of contradiction, that z 2 / U. We will examine the following three scenarios to reach a contradiction. • Assume that z 2 V(P(u , v (P)]) or V(P[v (P), u )). By symmetry, it would there- k R L fore sufﬁce to consider that z 2 V(P(u , v (P)]). (By) Claim 2(1), the set of vertices of the subgraph v (P) P z v (P) P zv is equal to V(P)[fvg, which contradicts (I). L R • Assume that z 2 U or U . By symmetry, it would therefore sufﬁce to think about ! ! + + + z 2 U ; say z = u for some i 2 f1, , kg. If v = w, then v (P) P u L wu P v (P) L i i R i i covers V(P)[fwg, contradicting (I). If v 6= w, then, by Claim 6, the set of vertices of ! ! the subgraph v (P) P u Hu P v (P) is equal to V(P[ H), which contradicts (I). L i R i Axioms 2023, 12, 411 13 of 25 2+ 2 • Assume z belongs to V(P[u , u ]) for some i 2 f1, , k 1g. We consider i i+1 + + + the neighbourhood of the vertex z . By (4), N(z )\X 6= Æ; say y 2 N(z )\X . By Claim 4, y 2 / End(S ). We will consider the following two cases to obtain a contradiction. + + – Assume that y 2 X \ U ; say y = u for some j 2 f1, , k 1g. Suppose, ﬁrst, that j > i. If v 6= w, then, by Claim 6, the set of vertices of the subgraph ! ! + + v (P) P z Hu P z u P v (P) is equal to V(P [ H), which contradicts (I). If L j R ! ! v = w, then the set of vertices of the subgraph v (P) P zwL u P z u P v (P) is L j j R equal to V(P)[fwg, which also contradicts (I). Suppose, now, that j i. If v 6= w, ! ! then, by Claim 6, v (P) P u Hz P u z P v (P) covers V(P)[ V( H), which con- L j R ! ! tradicts (I). If v = w, then v (P) P u wz P u z P v (P) covers V(P) [ fwg, L R which also contradicts (I). – Assume that y 2 End(S ). Let us take y = v (P) without loss of generality. If v 6= w, P R then, by Claim 6, the set of vertices of the subgraph v (P) P z Hu P z v (P) P L k R u is equal to V(P[ H), which contradicts (I). If v = w, then v (P) P zwL u k k k L + + P z v (P) P u covers V(P)[fwg, which also contradicts (I). This contradiction shows that N(z )\X = Æ, contradicting (4). This contradiction shows z 2 U. Since jN(V( H))\ (V(G)n V( H))j k and jUj = k, N(V( H))\ (V(G)n V( H)) = U. Proof of Claim 8. By contradiction, suppose that either N(u )\ V( H) = N(u )\ V( H) = Æ i j 0 0 or N(u )\ V( H) = N(u )\ V( H) = fvg and v 2 V( H) for some u , u 2 U. i j i j 0 0 0 0 Suppose ﬁrst that N(u )\ V( H) = N(u )\ V( H) = Æ. Then N(V( H))\ (V(G)n i j 0 0 V( H)) 6= U, contradicting Claim 7(2). Suppose now that N(u ) \ V( H) = N(u ) \ i j 0 0 ˆ ˆ V( H) = fvg. Let U = (U n fu , u g) [ fvg and H = H v. Since jV( H)j > 1, i j 0 0 ˆ ˆ ˆ ˆ H 6= Æ. Then, by hypothesis and Claim 7(2), N(V( H)) \ (V(G) n V( H)) U. How- ever, jUj = k 1, contradicting the hypothesis that G is k-connected. These contradictions show that Claim 8 holds. Appendix B. Proof of Lemmas 8 and 9 In this section, we employ the same terminology and notation in Section 2. In order to prove Lemmas 8 and 9, we ﬁrst do some preparatory work. + + Denote X = End(S)[ U [fwg and X = End(S)[ U [fwg. If U\ V(S ) 6= Æ, then, by Lemma 5(3), jfC : C 2 S and U\ V(C) 6= Ægj = 1; say C 2 S . C U C Lemma A1. (Akiyama and Kano, [18]) The following statements are true. (1) If U\ V(S ) 6= Æ, then X nfv g forms an independent set of G with a size of k + k. C C + + (2) If U V(S ) and End(S)\ U 6= Æ, then End(S)\ U = fv (P)g for some P 2 S P R P and X forms an independent set of G with a size of k + k. (3) If U V(S ) and End(S)\ U = Æ, then: (i) The set X does not include four distinct vertices x , x , x , x withfx x , x x g E(G); 1 2 3 4 1 2 3 4 (ii) G[X ] is triangle-free; (iii) U is an independent set of G. Lemma A2. Suppose that U V(S ). The following statements are true. + + + (1) If End(S)\ U = Æ, then G[X ] has exactly one nontrivial component denoted by S(X ) + + + + + such that S(X ) is a star with S(X ) = (V(S(X ))\ U )_ (V(S(X ))\ End(S)); (2) If End(S)\ U = Æ, then G[X ] has exactly one nontrivial component denoted by S(X ) such that S(X ) is a star with S(X ) = (V(S(X ))\ U )_ (V(S(X ))\ End(S)). Axioms 2023, 12, 411 14 of 25 Proof of Lemma A2. By symmetry, it would therefore sufﬁce to show that (1) is true. By Lemmas 3 and 5(1), End(S)[fwg is an independent set of G. By Lemma 5(1)(2), U [fwg is an independent set of G. Since jX j = k + k + 1, there must exist some edges between + + U and End(S). By Lemma A1(3)(i)(ii), G[X ] has exactly one nontrivial component + + S(X ) and S(X ) is a star. Remark A1. S(X ) and S(X ) always denote the stars in Lemma A2 in the following. From Lemma A3 to Lemma A6, for convenience, we assume U V(S ) and End(S)\ U = Æ. Lemma A3. Let C 2 S and P 2 S . Then for each vertex v 2 V(C) with vu 2 E(G) for C P P,i + + some i 2 f1, , t g, it holds that N(v )\ (U nfu g) = Æ. P,i + + Proof of Lemma A3. By contradiction, suppose that N(v )\ (U nfu g) 6= Æ. Then P,i + + + + there exists a vertex x 2 N(v )\ (U nfu g); say x = u for some j 2 f1, , t 0g. Sup- 0 P P,i P ,j pose ﬁrst that P = P. If i < j, then the set of vertices of the subgraph v (P) P u L wL L P,i P,i P,j + + u P u Cu P v (P) is equal to V(P[ C)[fwg, which contradicts Lemma 4. If i > j, P,j R P,i P,j ! ! + + then v (P) P u L wL u P u Cu P v (P) covers V(P [ C) [ fwg, contradicting L R P,j P,j P,i P,i P,j P,i 0 0 0 Lemma 4. Now suppose that P 6= P. Then v (P) P u L wL 0 u 0 P v (P ) and L P,i P,i P ,j P ,j L + + 0 0 0 v (P) P u Cu P v (P ) cover V(P [ P [ C) [ fwg, contradicting Lemma 4. These R 0 R P,i P ,j contradictions show that Lemma A3 holds. + + + Lemma A4. The cardinality of the set V(S(X )) \ U is equal to one and jV(S(X )) \ End(S)j 1. + + + Proof of Lemma A4. By Lemma A2(1),jV(S(X ))\ U j 1 andjV(S(X ))\ End(S)j 1. + + In other words, we need to prove that jV(S(X ))\ U j = 1. + + By contradiction, suppose that jV(S(X )) \ U j 6= 1. Then, by Lemma A2(1), + + + + there exists a vertex x 2 End(S) such that xu , xu 2 E(G) and u 6= u for 0 0 P,i P ,j P,i P ,j some i 2 f1, , t g and some j 2 f1, , t 0g. By Lemma 5(4), x 2 / End(S ). Then, P P P x 2 End(S ); say x = v . By Lemma A2(1), X nfv g is an independent set of G with C C C size k + k. By Lemma A3, jV(C)j 2. We consider the neighbourhood of the vertex + + + + + + + v . By Lemma A3, N(v )\ (U nfu g) = Æ and N(v )\ (U nfu g) = Æ. Then, C C P,i C P ,j + + N(v )\ U = Æ. According to Lemma 2 (i)(ii), N(v )\ (End(S)nfv g) = Æ. Hence, C C + + (X nfv g)[fv g forms an independent set with a cardinality of k + k + 1, which con- tradicts a(G) = k + k. This contradiction show that Lemma A4 holds. + + Remark A2. If U V(S ) and End(S) \ U = Æ, then, by Lemma A4, jV(S(X )) \ + + + U j = 1; say V(S(X ))\ U = fu g for some P 2 S and some i 2 f1, , t g. Denote P P P,i + + + + X = X nfu g. Then, by Lemmas A1(3), A2(1) and A4, X is an independent set of G with i P,i i size k + k. If End(S )\ V(S(X )) 6= Æ, then, by Lemmas 5(4) and A4, + + + u = u and u v (P) 2 E(G). (A1) P,i P,t P,t P P Lemma A5. Let C 2 S . Then, for some P 2 S , the following statements are true. C P + + + (1) If v u 2 E(G) for some i 2 f1, , t g, then N(x)\ (U nfu g) = Æ for each C P P,i P,i x 2 V(C)nfv g; + + + (2) If S(X ) = u v for some i 2 f1, , t g, then u is adjacent to all vertices in C. C P P,i P,i + t+ Proof of Lemma A5. First, we will show that (1) holds. LetjV(C)j 2 and C = v v v C C (jV(C)j1)+ v v . We prove Lemma A5(1) by induction on t. Note that U V(S ), C P + + + + + End(S)\ U = Æ and v u 2 E(G). According to Lemma A3, N(v )\ (U nfu g) = Æ. P,i C P,i Axioms 2023, 12, 411 15 of 25 This implies that Lemma A5(1) holds for t = 1. Next, we assume that Lemma A5(1) holds t + 0 + for all positive integers t t . Then N(v )\ (U nfu g) = Æ. We need to prove that C P,i t + it holds for t = t + 1. By Lemma 2(i)(ii), N(v )\ (End(S)nfv g) = Æ. Note that + + + + + v u 2 E(G). By Lemma A4, V(S(X ))\ U = fu g. Then X is an independent set P,i P,i i t + t + 0 + 0 of G with size k + k. Hence, v v 2 E(G). Otherwise, X [fv g is an independent set C i C of G with size k + k + 1, contradicting a(G) = k + k. t + t + + + 0 0 We claim that G[fv , v , . . . , v g] forms a clique. Since G[fv , v , . . . , v g] is con- C C C C C C t + + 0 nected, we only need to focus on the case when jV(G[fv , v , . . . , v g])j 3. By contra- C C t + t + t + t + t + 1 2 1 2 0 diction, suppose that v v 2 / E(G) for some pair of vertices v , v 2 fv , v , . . . , v g C C C C C C t + t + t + t + 1 2 1 2 with v 6= v , then (X nfv )g)[fv , v g is an independent set of G with size i C C C C C t + k + k + 1, contradicting a(G) = k + k. Hence, G[fv , v , . . . , v g] is a clique. C C t + t + + 0 + 0 By our claim, G[fv , . . . , v g] contains a subgraph C(G[fv , . . . , v g]) such that C C C C t + t + t + + + + 0 0 0 f (C(G[fv , . . . , v g])) = 1 and V(G[fv , . . . , v g]) = V(C(G[fv , . . . , v g])). C C C C C C Next, we will show that Lemma A5(1) holds for t = t + 1. By contradiction, suppose (t +1)+ (t +1)+ 0 + + 0 + that N(v )\ (U nfu g) 6= Æ. Then there exists a vertex x 2 N(v )\ (U n C P,i C + + fu g); say x = u for some j 2 f1, , t g. Suppose ﬁrst that P = P. Then, by our P,i P ,j claim, the set of vertices of the subgraph ! ! ! (t +1)+ + 0 + v (P) P u L wL u P u v C v u P v (P), if j < i, L P,j P,j P,i P,i C R P,j C P,i Q = 2 ! ! (t +1)+ + 0 + v (P) P u L wL u P u v C v u P v (P), if j > i, L P,i P,i P,j P,j C R P,i C P,j t + and C(G[fv , . . . , v g]) is equal to V(P[ C)[fwg, which contradicts (I). Now suppose C C that P 6= P. Then, by our claim, the set of vertices of the subgraph v (P) P u L wL 0 u 0 L P,i P,i P ,j P ,j (t +1)+ t + + + + 0 0 0 0 0 0 P v (P ), v (P) P u v C v u P v (P ) and C(G[fv , . . . , v g]) is equal to V(P L R C 0 R P,i C P ,j C C [ P [ C)[fwg, which contradicts (I). These contradictions show that Lemma A5(1) holds for t = t + 1. Thus, Lemma A5(1) is proved. + + + + + Now, we start to prove Lemma A5(2). If S(X ) = u v , then X nfu g and X n P,i P,i + + fv g are two independent sets of G. For any x 2 V(C)nfv g, N(x)\ (U nfu g) = Æ C C P,i by Lemma A5(1). According to Lemma 2(i)(ii), N(x)\ (End(S)nfv g) = Æ. Hence, + + xu 2 E(G). Otherwise, (X nfv g)[fxg is an independent set of G with size k + k + 1, P,i contradicting a(G) = k + k. Lemma A5(2) is proved. Remark A3. Suppose U V(S ) and End(S) \ U = Æ. For some P 2 S and some P P + + + i 2 f1, , t g, if S(X ) = u v for C 2 S , then, by Lemma A5(2), G[V(C)[fu g] C C P,i P,i + + contains a spanning subgraph C(G[V(C)[fu g]) with f (C(G[V(C)[fu g])) = 1. P,i P,i For some P 2 S and i 2 f1, 2, . . . , t g, we consider the following conﬁguration. P P + + (i) S(X ) = u v , for some C 2 S . C C P,i + 0 (ii) G[fu , v , v 0g] S(X ), for some fC, C g S . C C P,i 0 + + + (iii) For given P 2 S nfPg and C 2 S , S(X ) = u v and fu v , xv g E(G) P C C C C P,i P,i for some x 2 V(P ). 0 0 + (iv) For some P 2 S n fPg and fC, C g S , fu v , xv 0g E(G) for some P C C C P,i x 2 V(P ). Then we denote C(G[V(C)[fu g]), if (i) occurs, P,i Q = + 0 Cu C , if (ii) occurs, P,i Axioms 2023, 12, 411 16 of 25 0 0 v (P) P u Cx P v (P ), if (iii) occurs, R L P,i Q = + 0 0 0 Cu P v (P) and C x P v (P ), if (iv) occurs, R L P,i + + + + Lemma A6. Suppose that U \ V(S(X )) = fu g and End(S )\ V(S(X )) 6= Æ for some P,i 2+ 2+ P 2 S and i 2 f1, , t g. If u 2 / fv (P), u g, then u v (P) 2 E(G). P P R P,i+1 R P,i P,i 2+ + Proof of Lemma A6. By contradiction, suppose that u v (P) 2 / E(G). Note that X is an P,i i 2+ + independent set of G with size k + k. Then there exists at least one vertex x 2 N(u )\ X P,i i with x 6= v (P). Recall that End(S )\ V(S(X )) 6= Æ, we assume that v 2 End(S )\ R C C C V(S(X )). In other words, v u 2 E(G). By Lemma 6(1)(2), x 2 / End(S )nfv (P)g. C P R P,i According to the deﬁnition of X , we will consider the following three cases in order to arrive at a contradiction. + 2+ • Assume that x 2 End(S ). According to Lemma 6(1), N(u )\ End(S ) = N(u )\ C C P,i P,i End(S ) = fv g, then S(X ) = v u . By (A1) and Lemma 6(2), End(S ) \ C C C P P,i + + V(S(X )) = Æ. Note that End(S )\ V(S(X )) 6= Æ. By Lemma A5(2), the set ! ! + 2+ of vertices of the subgraph v (P) P u Cu P v (P) in G is equal to V(P[ C), which L R P,i P,i contradicts Lemma 4. + + + • Assume that x 2 (X \ U )\ V(P); say x = u for some j 2 f1, , t gnfig. i P,j If End(S ) \ V(S(X )) 6= Æ, then, by (A1), the set of vertices of the subgraph + 2+ + u v (P) P u u P u L wL u P v (P) in G is equal to V(P)[fwg, which R P,t P,t P,j P,j L P,t P,t P,j P P P P contradicts Lemma 4. Therefore, End(S )\ V(S(X )) = Æ. Note that End(S )\ P C V(S(X )) 6= Æ. Then (i) or (ii) occurs. By Lemma A5(2), 2+ + v (P) P u u P u L wL u P v (P), if j < i, R L P,i P,i P,j P,j P,i P,j Q = 3 ! + 2+ v (P) P u u P u L wL u P v (P), if j > i, R P,j P,j P,i P,i L P,j P,i Q cover V(P[ C)[fwg or V(P[ C[ C )[fwg, which contradicts (I). + 0 0 • Suppose that x 2 U \ V(P ) for P 2 S nfPg; say x = u for some j 2 f1, , t 0g. P 0 P ,j If End(S ) \ V(S(X )) 6= Æ, then, by (A1), the set of vertices of the subgraph + 2+ + 0 0 0 0 Cu v (P) P u u P v (P )[ v (P) P u L wL 0 u 0 P v (P ) is equal to V R 0 R L P,t P,t P ,j P ,j L P,t P,t P ,j P P P P 0 + (P[ P [ C)[fwg, which contradicts Lemma 4. Therefore, End(S )\ V(S(X )) = Æ. Note that End(S )\ V(S(X )) 6= Æ. Then (i) or (ii) occurs. By Lemma A5(2), the set 2+ + 0 0 0 of vertices of the subgraph v (P) P u u P v (P )[ v (P) P u L wL 0 u 0 P R 0 R L P,i P,i P ,j P ,j P,i P ,j 0 0 0 0 v (P )[ Q is equal to V(P[ P [ C)[fwg or V(P[ P [ C[ C )[fwg, which con- tradicts (I). 2+ + 2+ + The contradiction indicates that N(u )\ X fv (P)g. Note that N(u )\ X 6= Æ. P,i i P,i i 2+ + Therefore, N(u )\ X = fv (P)g. Lemma A6 holds. P,i i Lemma A7. If U V(S ), then there exists exactly one path Q 2 S such that N(u )\ P P Q,t End(S ) 6= Æ or f (Q(u , Q)) = 1. Q,t Proof of Lemma A7. First, we claim that there is a path Q 2 S with N(u )\ End(S ) 6= Æ P C Q,t or f (Q(u , Q)) = 1. To prove this claim, we will consider the following two cases. Q,t • Suppose that U V(S ) and End(S)\ U 6= Æ. Then, by Lemma A1(2), there is a + + path Q 2 S with u = v (Q), i.e., f (Q(u , Q)) = 1. P R Q,t Q,t Q Q • Assume that U V(S ) and End(S)\ U = Æ. According to Lemma A4, End(S)\ + + V(S(X )) 6= Æ. Suppose ﬁrst that End(S ) \ V(S(X )) 6= Æ. By Lemmas 5(4) and A4, there is a path Q 2 S satisfying End(S ) \ V(S(X )) = fv (Q)g and P P R Axioms 2023, 12, 411 17 of 25 + + + + U \ V(S(X )) = fu g, i.e., f (Q(u , Q)) = 1. Now, suppose that End(S )\ Q,t Q,t Q Q + + + V(S(X )) = Æ, i.e., End(S )\ V(S(X )) 6= Æ; say v 2 End(S )\ V(S(X )). By C C C + + + + Lemma A4, jV(S(X ))\ U j = 1; say V(S(X ))\ U = fu g for some Q 2 S Q,i and i 2 f1, , t g. It should be noted that X nfu g forms an independent set Q,i + + of G with a cardinality of k + k. By Lemmas A2 and A4, N(v ) \ U = fu g. Q,i + + + + We will show that u = u . Suppose otherwise that u 6= u . By (A1), Q,i Q,t Q,i Q,t Q Q + + End(S )\ V(S(X )) = Æ. Note that u v 2 E(G). Then (i) or (ii) occurs. By Q,i Lemmas A5(2) and A6, the set of vertices of the subgraph ! ! v (Q) Q u L wL u Q v (Q), if jV(Q[u , u ])j = 1, L Q,i Q,i Q,i+1 Q,i+1 R Q,i Q,i+1 Q = ! ! + 2+ + u Q v (Q)u Q u L wL u Q v (Q), if jV(Q[u , u ])j > 1 R Q,t Q,t Q,i Q,i L Q,t Q Q Q,i Q,i Q,i+1 [Q in G is equal to V(Q[ C)[fwg or V(Q[ C [ C)[fwg, which contradicts (I). + + + The contradiction indicates that u = u . Then N(u )\ End(S ) 6= Æ. Q,i Q,t Q,t Q Q Hence, our claim is proved. Now, we will prove Lemma A7. We begin by assuming the opposite and using a proof by contradiction. Suppose that there exists another path P 2 S with N(u )\ P,t End(S ) 6= Æ or f (Q(u , P)) = 1. To arrive at a contradiction, we consider the following P,t two cases: + + + + • Assume that End(S) \ U 6= Æ. If u = v (P) and u = v (Q), then X R R P,t Q,t P Q does not form an independent set of G with a cardinality of k + k, contradicting + + Lemma A1(2). Therefore, u 6= v (P) or u 6= v (Q). Without loss of generality, R R P,t Q,t P Q + + + suppose that u 6= v (P), then u v (P) 2 E(G) or N(u )\ End(S ) 6= Æ. R R C P,t P,t P,t P P P Hence, there exist two adjacent vertices in X , contradicting Lemma A1(2). + + + • Assume that End(S)\ U = Æ. Then u 6= v (Q) and u 6= v (P). Then, R R Q,t P,t Q P G[X ] has at least two stars, contradicting Lemma A2(1). This statement indicates that Lemma A7 is true. Let X nfv g, if U\ V(S ) 6= Æ, C C + + + X nfu g, if U V(S ) and U \ End(S) = Æ, X = Q,t + + X , if U V(S ) and U \ End(S) 6= Æ. Then, by Lemmas A1, A4 and A7, X forms an independent set of G satisfying size k + k and N(v)\ X 6= Æ f or any v 2 V(G)n X. (A2) Otherwise, there is a vertex v 2 V(G)n X satisfying X[fv g being an independent set of 0 0 G with size k + k + 1, which contradicts a(G) = k + k. Lemma A8. Suppose that U V(S ). The following two statements are true. + + (1) S contains exactly one path Q such that N(u )\ End(S ) 6= Æ or f (Q(u , Q)) = 1 P C Q,t Q,t Q Q and N(u )\ End(S ) 6= Æ or f (Q(u , Q)) = 1; Q,1 Q,1 (2) If f (Q(u , Q)) 6= 1 and f (Q(u , Q)) 6= 1, then there exist at least two elements C, Q,t Q,1 C 2 S with u v 0 2 E(G), u v 2 E(G). C C C Q,t Q,1 Proof of Lemma A8. By symmetry and Lemma A7, S has exactly one path P (say) such that N(u ) \ End(S ) 6= Æ or f (Q(u , P)) = 1. First, we will show Q = P. By P,1 P,1 Axioms 2023, 12, 411 18 of 25 ! ! contradiction, suppose that Q 6= P. Denote Q = v (Q) Q u L wL u P v (P). To 5 L Q,t Q,t P,1 P,1 R Q Q arrive at a contradiction, we consider the following three cases using Lemma A7: • Assume that f (Q(u , Q)) = 1 and f (Q(u , P)) = 1. Then the set of vertices of Q,t P,1 the subgraph Q [ Q(u , Q)[ Q(u , P) in G is equal to V(Q[ P)[fwg, which Q,t P,1 contradicts (I). • Assume that either f (Q(u , Q)) 6= 1 and f (Q(u , P)) = 1 or f (Q(u , P)) 6= 1 Q,t P,1 P,1 ! ! + + and f (Q(u , Q)) = 1. By symmetry, suppose that f (Q(u , Q)) 6= 1 and Q,t Q,t Q Q f (Q(u , P)) = 1. According to Lemma A7, N(u )\ End(S ) 6= Æ; say v 0 2 P,1 Q,t + + 0 N(u )\ End(S ). Then the set of vertices of the subgraph Q [ v (Q) Q u C [ 5 R Q,t Q,t Q Q Q(u , P) in G is equal to V(Q[ P[ C )[fwg, which contradicts (I). P,1 • Suppose that f (Q(u , Q)) 6= 1 and f (Q(u , P)) 6= 1. Applying symmetry and Q,t P,1 using Lemma A7, N(u ) \ End(S ) 6= Æ and N(u ) \ End(S ) 6= Æ. Then C C Q,t P,1 (iii) or (iv) occurs. By Lemma A5(2), Q and Q in G cover V(Q[ P[ C)[fwg or 5 ? V(Q[ P[ C [ C)[fwg, contradicting Lemma 4 or (I). This contradiction shows that Lemma A8(1) holds. Next, we will demonstrate Lemma A8(2). By Lemma A8(1), N(u )\ End(S ) 6= Æ Q,t and N(u )\ End(S ) 6= Æ. We begin by assuming the opposite and using a proof by Q,1 contradiction. Suppose that there is precisely one element C 2 S with u v 2 E(G) C C Q,t and u v 2 E(G). Note that f (Q(u , Q)) 6= 1 and f (Q(u , Q)) 6= 1. By Lemma A4, Q,1 Q,t Q,1 + + S(X ) = u v and S(X ) = u v . Then, by Lemmas A5(2) and A6, the set of C C Q,t Q,1 ! ! + 2+ vertices of the subgraph v (Q) Q u Cu Q u L w[ Q(u , Q) in G is equal to L Q,1 Q,1 Q,1 Q,t Q,t Q Q V(Q[ C)[fwg, which contradicts (I). This contradiction demonstrates that Lemma A8(2) is true. + + Remark A4. If f (Q(u , Q)) 6= 1, then, by Lemma A7, N(u )\ End(S ) 6= Æ; say v 0 2 C C Q,t Q,t Q Q N(u )\ End(S ). If f (Q(u , Q)) 6= 1, then, by Lemma A8(1), N(u )\ End(S ) 6= Æ; C C Q,t Q,1 Q,1 say v 2 N(u )\ End(S ). If f (Q(u , Q)) 6= 1 and f (Q(u , Q)) 6= 1, then, by C C Q,1 Q,t Q,1 Lemma A8(2), V(C)\ V(C ) = Æ. For convenience, we denote ! ! 0 + + C u Q v (Q), if f (Q(u , Q)) 6= 1, Q,t Q,t Q Q Q = 6 ! ! + + Q(u , Q) if f (Q(u , Q)) = 1. Q,t Q,t Q Q Cu Q v (Q), if f (Q(u , Q)) 6= 1, Q,1 Q,1 Q = Q(u , Q), if f (Q(u , Q)) = 1. Q,1 Q,1 Let S nfC g, if U\ V(S ) 6= Æ, U C S = S nfQg, if U\ V(S ) = Æ, 0 0 0 0 and S = S \S , S = S \S . C P C P Lemma A9. Let C 2 S and x 2 V(C)nfv g. It follows that N(x)\ X = fv g. C C C Axioms 2023, 12, 411 19 of 25 Proof of Lemma A9. First, we assert that for each vertex x 2 V(C)nfv g, N(x)\ (X\ U ) = Æ. To prove this, we will use a proof by contradiction. Assume that N(x) \ + + (X \ U ) 6= Æ. According to Lemma 2(i), N(x) \ (V(S ) \ U ) = Æ. Then, there + + is at least one vertex u 2 N(x) \ (X \ V(S ) \ U ) for some P 2 S and some P P P,i i 2 f1, , t g. Assuming U \ V(S ) 6= Æ. Then, the set of vertices of the subgraph ! ! v (P) P u L wG[V(L )[ V(C )][ Cu P v (P) in G is equal to V(P[ C [ C)[ L P,i P,i C ,1 U R U U P,i fwg, which contradicts (I). Therefore, U V(S ). Suppose ﬁrst that P = Q. If N(u )\ Q,t End(S ) 6= Æ, (say v 0 2 N(u )\ End(S )) and C = C , then the set of vertices of the C C Q,t ! ! + + subgraph v (Q) Q u L wL u Q u Cu Q v (Q) in G is equal to V(Q[ C)[ L R Q,i Q,i Q,t Q,t Q Q Q,i Q,t fwg or V(Q[ C)[fwg; see Lemma 4. Otherwise, by Lemma A7, v (Q) Q u L L Q,i Q,i + 0 wL u Q u C and Q in G cover V(Q[ C[ C )[fwg or V(Q[ C)[fwg, contra- Q,t Q,t 6 Q Q Q,i 0 0 dicting (I). Suppose now that P 6= Q, i.e., P 2 S . Let Q = v (P) P u L L P,i P,i + + wL u Q v (Q). If N(u )\ End(S ) 6= Æ, (say v 0 2 N(u )\ End(S )) and Q,t Q,t L C C C Q Q Q,t Q,t Q Q 0 0 + + C = C , then the set of vertices of the subgraph Q [ v (Q) Q u Cu P v (P) in G is R R Q,t P,i equal to V(P[ Q[ C)[fwg, which contradicts Lemma 4. Otherwise, by Lemma A7, 0 + 0 Q , Q and Cu P v (P) in G cover V(P [ Q [ C [ C )[fwg or V(P [ Q [ C)[fwg, 6 R P,i contradicting (I). These contradictions show that our claim holds. According to Lemma 2(i)(ii), N(x)\ (End(S)nfv g) = Æ. Combining this with our claim, we arrive at N(x)\ (Xnfv g) = Æ. By (A2), N(x)\ X = fv g. C C Lemma A10. For any C 2 S , G[V(C)] forms a clique. Proof of Lemma A10. As G[V(C)] is connected, we only need to focus on the case when jV(C)j 3. It is worth noting that V(C)\ X = fv g. According to Lemma A9, N(x)\ X = fv g C C for every vertex x 2 V(C)nfv g. Let S = V(C). Then, according to Lemma 7, G[V(C)] forms a clique. Lemma A11. Suppose that V(S )\ U = Æ. The following two statements are true. (1) Let P 2 S and y 2 V(P) such that yv (P) 2 E(G), jV(P[y , v (P)])j 1 and P R R V(P[y, v (P)])\ U = Æ. Then G[V(P[y , v (P)])] forms a clique. Moreover, if N(y)\ R R X = fv (P)g, then G[V(P[y, v (P)])] also forms a clique; R R (2) Let P 2 S and x 2 V(P) such that xv (P) 2 E(G), jV(P[v (P), x ])j 1 and P L L V(P[v (P), x])\ U = Æ. Then G[V(P[v (P), x ])] forms a clique. Moreover, if N(x)\ L L X = fv (P)g, then G[V(P[v (P), x])] also forms a clique. L L Proof of Lemma A11. By virtue of symmetry, we may restrict our consideration to demon- strate the truth of (1). As G[V(P[y , v (P)])] is connected, it is sufﬁcient to focus on the + + case where jV(P[y , v (P)])j 3. Suppose that there is at least one vertex v 2 N(y )\ X + 0 with v 6= v (P). According to Lemma 6(1)(2), v 2 U \ X\ V(S ). Note that V(S )\ R P U = Æ. Then U V(S ). We assume that v = u for some j 2 f1, , t 1g. P Q Q,j ! ! + + If P 6= Q, then the set of vertices of the subgraph v (P) P yv (P) P y u Q v (Q) [ L R R Q,j v (Q) Q u L w in G is equal to V(Q[ P)[fwg, which contradicts (I). If P = Q, then L Q,j Q,j ! ! + + v (Q) Q u L wL u Q u y Q v (Q) y Q u in G covers V(Q)[fwg, contra- L Q,j Q,j Q,t Q,t R Q Q Q,j Q,t dicting (I). Therefore, we have N(y )\ X = fv (P)g, which is a contradiction. According to (A2), N(y )\ X = fv (P)g. (A3) + 0 + Note that V(P[y , v (P)])\ X = fv (P)g. Let S = V(P[y , v (P)]). According to R R R Lemma 7, it would therefore sufﬁce to show that the following characterization holds, 0 0 + N(y )\ X = fv (P)g for every vertex y 2 V(P[y , v (P))). (A4) R R Axioms 2023, 12, 411 20 of 25 We apply (A3) repeatedly to obtain (A4). Next, we will demonstrate that if N(y)\ X = fv (P)g and V(P[y, v (P)])\ U = Æ, R R then G[V(P[y, v (P)])] forms a clique. Since G[V(P[y, v (P)])] is connected, we can as- R R sume that jV(P[y, v (P)])j 3. It is important to note that N(y)\ X = fv (P)g, which R R combined with (A4) implies that N(x)\ X = fv (P)g for every vertex x 2 V(P[y, v (P))). R R Let S = V(P[y, v (P)]). According to Lemma 7, we can conclude that G[V(P[y, v (P)])] R R forms a clique. Denote 0 + + T (P) :=fx 2 V(P) : P 2 S , f (P[v (P), x]) = 1, V(P[v (P), x ])\ U = Æ, x 6= 1 L L v (P)g; 0 0 + T (P) :=fx 2 V(P) : P 2 S , N(x) \ End(S ) 6= Æ, V(P[v (P), x ]) \ U = Æ, 2 L P C x 6= v (P)g. Remark A5. If x 2 T (P), then, according to the deﬁnition of T (P), there is at least one vertex 2 2 v 2 N(x)\ End(S ). Let Q(x, P), if x 2 T (P), Q = Cx P v (P), if x 2 T (P) (say v 2 N(x)\ End(S )). L 2 C 0 0 0 Lemma A12. Let P 2 S , x 2 V(P) and P 2 S nfPg. Then the following three characteriza- P P tions are true: + + 0 (1) If x 2 T (P)[ T (P), then N(x )\ (U \ V(P )) = Æ; 1 2 + 0 (2) If x 2 T (P), then N(x )\ (Xn (End(S )[fv (P), v (P)g)) = Æ ; 1 L R + 0 (3) If x 2 T (P), then N(x )\ (Xn (End(S )[fv (P)g)) = Æ . 2 R Proof of Lemma A12. First, we will prove Lemma A12(1). We begin by assuming the + + 0 + + 0 opposite, i.e., N(x )\ (U \ V(P )) 6= Æ; say u 2 N(x )\ (U \ V(P )) for some P ,i 0 0 + i 2 f1, , t 0g. Denote Q = v (P ) P u x P v (P). To arrive at a contradiction, we 9 R 0 R P ,i will consider the following two situations. 0 0 • Suppose that U \ V(S ) 6= Æ. Then Q , v (P ) P u 0 L 0 wG[V(L ) [ V(C )] C 9 L C ,1 U P ,i P ,i 0 0 and Q in G cover V(P [ P) [ C ) [ fwg or V(P [ P [ C [ C ) [ fwg, which 8 U U contradicts (I). • Assume that U V(S ). Then, by Lemma A7, there exists exactly one path Q 2 S such P P + + + that N(u )\ End(S ) 6= Æ (say v 0 2 N(u )\ End(S )) or f (Q(u , Q)) = 1. C C C Q,t Q,t Q,t Q Q Q 0 0 Denote Q = v (P ) P u 0 L 0 wL u Q v (Q). To arrive at a contradiction, we 10 L P ,i P ,i Q,t Q,t L Q Q differentiate between the following two cases: – Assume that x 2 T (P). Then, by Lemma A7, the set of vertices of the subgraph 0 0 0 Q [ Q [ Q(x, P)[ Q is equal to V(Q[ P [ P[ C )[fwg or V(Q[ P [ P)[ 9 10 6 fwg, which contradicts (I). – Assume that x 2 T (P). Then f (Q(u , Q)) 6= 1. Otherwise the set of vertices Q,t + 0 of the subgraph Q [ Q [ Cx P v (P)[ Q(u , Q) is equal to V(Q[ P [ P[ 10 L Q,t C)[fwg, which contradicts (I). Then, by Lemma A7, N(u )\ End(S ) 6= Æ. Q,t Then (iii) or (iv) occurs. By Lemma A5(2), the set of vertices of the subgraph 0 0 0 0 Q [ Q [ Q in G is equal to V(Q[ P [ P[ C )[fwg or V(Q[ P [ P[ C [ 9 10 ? C)[fwg, which contradicts Lemma 4 or (I). This statement indicates that Lemma A12(1) is true. + 0 Next, we assert that if x 2 T (P)[ T (P), then N(x )\ (Xn (End(S )[fv (P), v (P) 1 2 L R g)) = Æ. + 0 Assuming a contradiction, let us suppose that N(x )\ (Xn (End(S )[fv (P), v (P) L R + 0 g)) 6= Æ; say z 2 N(x ) \ (X n (End(S ) [ fv (P), v (P)g)). By Lemma 6(1), z 2 / L R C Axioms 2023, 12, 411 21 of 25 End(S )nfv (P), v (P)g. To derive a contradiction, we differentiate between the fol- P L R lowing two cases based on the deﬁnition of X: • Assuming z 2 U \ V(S ); say z = u for some i 2 f1, , t g. Then the C C C ,i U set of vertices of the subgraph Q [ v (P) P x u C u L w in G is equal to 8 R U C ,i C ,i C ,i U U V(P)[ V(C)[ V(C )[fwg or V(P)[ V(C )[fwg, which contradicts (I). U U • Assuming z 2 (X\ U )\ V(S ). To arrive at a contradiction by Lemma A12(1), we differentiate between the following two cases: – Assume that U\ V(S ) 6= Æ; say z = u for some j 2 f1, , t g. Then the set C P P,j of vertices of the subgraph v (P) P u x P u L wG[V(L )[ V(C )] and R P,j P,j C ,1 U P,j U Q is equal to V(P[ C[ C )[fwg or V(P[ C )[fwg, which contradicts (I). 8 U U – Assume that U V(S ). Let + + > v (P) P u x P u L wL u Q v (Q), if z = u for some R P,j P,j Q,t Q,t L > P,j Q Q P,j j 2 f1, , t g, Q = 11 ! + + v (P) P x u Q u L wL u Q v (Q), if z = u for some > R Q,t Q,t Q,j Q,j L Q,j Q Q Q,j j 2 f1, , t 1g. Assume that x 2 T (P). Then, by Lemma A7, the set of vertices of the sub- graph Q [ Q [ Q(x, P) is equal to V(Q[ P[ C )[fwg or V(Q[ P)[fwg, which contradicts (I). Now suppose that x 2 T (P). Then f (Q(u , Q)) 6= 1. Q,t Otherwise, the set of vertices of the subgraph Q [ Q(u , Q)[ Cx P v (P) 11 L Q,t is equal to V(Q [ P [ C) [ fwg, which contradicts (I). Then, by Lemma A7, N(u )\ End(S ) 6= Æ. Then (iii) or (iv) occurs. By Lemma A5(2), Q and C 11 Q,t Q in G cover V(Q[ P[ C)[fwg or V(Q[ P[ C[ C )[fwg, contradicting Lemma 4 or (I). This contradiction demonstrates the validity of our claim. Therefore, Lemma A12(2) is true. Final, we will prove Lemma A12(3). By our claim, if x 2 T (P), then N(x )\ (Xn 0 + (End(S ) [ fv (P), v (P)g)) = Æ. Hence, we only prove that x v (P) 2 / E(G). By L R L + + contradiction, suppose that x v (P) 2 E(G). Then Cx P v (P)x P v (P) in G covers L L R V(P[ C), which contradicts Lemma 4. This contradiction demonstrates that Lemma A12(3) is true. 0 + Lemma A13. Let P 2 S and x 2 V(P) with V(P(x , v (P)])\ U 6= Æ. If x 2 T (P)[ R 1 T (P), then x v (P) 2 / E(G). 2 R Proof of Lemma A13. By contradiction, suppose that x v (P) 2 E(G). To arrive at a contradiction, we differentiate between the following two cases: • Assume that U\ V(S ) 6= Æ. Then the set of vertices of the subgraph G[V(L )[ C C ,1 + + V(C )]wL u P x v (P) P u and Q in G cover V(P[ C[ C )[fwg or V(P[ U P,1 P,1 R 8 U P,1 C )[fwg, which contradicts (I). • Suppose that U V(S ). Let Q = u P x v (P) P u L wL u Q v (Q). P 12 R P,1 P,1 Q,t Q,t L P,1 Q Q Suppose ﬁrst that x 2 T (P). Then, by Lemma A7, the set of vertices of the subgraph Q [ Q [ Q(x, P) is equal to V(Q[ P[ C )[fwg or V(Q[ P)[fwg, which con- 12 6 tradicts (I). Now suppose that x 2 T (P). Then f (Q(u , Q)) 6= 1. Otherwise, the Q,t set of vertices of the subgraph Q [ Q(u , Q)[ Cx P v (P) is equal to V(Q[ P[ 12 L Q,t C)[fwg, which contradicts (I). Then, by Lemma A7, N(u )\ End(S ) 6= Æ. Then Q,t (iii) or (iv) occurs. By Lemma A5(2), the set of vertices of the subgraph Q [ Q is 12 ? equal to V(Q[ P[ C)[fwg or V(Q[ P[ C[ C )[fwg, which contradicts Lemma 4 or (I). Axioms 2023, 12, 411 22 of 25 This contradiction shows that x v (P) 2 / E(G). 0 + + 0 Lemma A14. Let P 2 S , x 2 T (P). If x v (P) 2 / E(G), then N(x )\ End(S ) = Æ. 2 R P C + 0 Proof of Lemma A14. Assuming a contradiction, let us suppose that N(x )\ End(S ) 6= Æ. + 0 By Lemma 2(ii), x 2 / fv (P), v (P)g. Note that N(x)\ End(S ) 6= Æ. We assume that L R 0 + 0 + v 2 N(x) \ End(S ). By Lemma 6(1), N(x ) \ End(S ) = fv g. If x v 2 E(G), C C C C C ! ! then the set of vertices of the subgraph v (P) P xCx P v (P) in G is equal to V(P[ C), L R which contradicts Lemma 4. Therefore, jV(C)j 2 and x v 2 / E(G). By Lemma A9, + + + + N(v )\ X = fv g. By Lemma A12(3), N(x )\ X = fv g. Then (Xnfv g)[fx , v g C C C C C forms an independent set of size k + k + 1; this would contradict the fact that a(G) = k + k. This contradiction demonstrates that Lemma A14 is true. 0 + Lemma A15. Let P 2 S with V(P) \ U 6= Æ. Then u 6= v (P) and N(u ) \ P P,1 P,1 (End(S )[fv (P)g) = Æ. Proof of Lemma A15. Denote X nfv g, if U\ V(S ) 6= Æ, < C C X = X nfu g, if U V(S ) and U \ End(S) = Æ, Q,1 X , if U V(S ) and U \ End(S) 6= Æ. By symmetry and Lemmas A1, A4 and A8, X is an independent set of G with size k + k. Then, u 6= v (P); otherwise, X is an independent set of G with size k + k 1, contradict- P,1 0 0 ing a(G) = k + k. Moreover, N(u )\ (End(S )[fv (P)g) = Æ. Otherwise, X is not an P,1 C independent set of G. Lemma A16. S = Æ. Proof of Lemma A16. By contradiction, suppose that S 6= Æ. Claim A1. V(P)\ U = Æ for any P 2 S . Proof of Claim A1. By contradiction, suppose that V(P)\ U 6= Æ for some P 2 S . Now, we consider the section P[v (P), u ]. By Lemma A15, jV(P[v (P), u ])j 3. L L P,1 P,1 Suppose that v (P)u 2 E(G). Then, by Lemmas A12(2), A13 and A15, we have P,1 N(u )\ X = Æ, contradicting (A2). This contradiction shows that P,1 v (P)u 2 / E(G). (A5) P,1 i+ Hence, jV(P[v (P), u ])j 4. Then there exists a vertex v (P) 2 V(P[v (P), u )) L L L P,1 P,1 i+ (i+1)+ such that v (P)v (P) 2 E(G) for i 1. By Lemmas A12(2) and A13, N(v (P) )\ L L L 0 (i+1)+ (X n (End(S ) [ fv (P)g)) = Æ. By (A2), N(v (P) ) \ X 6= Æ. Then there ex- L L (i+1)+ 0 ists at least one vertex v 2 N(v (P) )\ (End(S )[fv (P)g). Suppose that v 2 L L 0 (i+2)+ End(S ). We know N(v (P) )\ X = Æ by Lemmas A12(3), A13 and A14, contra- dicting (A2). This contradiction shows that v 2 / End(S ). Combining this with (A2) and (i+1)+ 2 Lemmas A12(2), A13, we obtain that N(v (P) )\ X = fv (P)g. Thus, v (P)u 2 L L L P,1 E(G), contradicting (A5). Claim A1 is proved. According to Lemma 4, for any path P 2 S , v (P)v (P) 2 / E(G). We can select the P L R vertex x from V(P) such that V(P[v (P), x ]) N(v (P)) and x 2 / N(v (P)). Denote P L L P L x , if x 6= v (P) and x v (P) 2 / E(G); P R P R xb = x , if x = v (P) or x v (P) 2 E(G). P P R P R Axioms 2023, 12, 411 23 of 25 If x 6= v (P) and x v (P) 2 / E(G), then, by the deﬁnition of x and Lemma A12(2), (A2), P R P R P N(x )\ End(S ) 6= Æ. Claim A2. For any path P 2 S , the following two characterizations are true. (1) f (G[V(P[v (P), x ])]) = 1 and f (G[V(P[xb , v (P)])]) = 1; L P R (2) Either x or x is a cut vertex of G. Proof of Claim A2. First, we will prove Claim A2(1). If x = v (P) or x v (P) 2 E(G), P R P R then Claim A2(1) holds. Therefore, x 6= v (P) and x v (P) 2 / E(G). Note that N(x )\ P R P R P + + End(S ) 6= Æ. Suppose ﬁrst that x = v (P). Claim A2(1) holds. Suppose now that x 6= C P P + + v (P). Then we consider the neighbourhood of the vertex x . If x v (P) 2 / E(G), then, by R R P P + + Lemmas A12(3) and A14, N(x )\ X = Æ, contradicting (A2). Therefore, x v (P) 2 E(G). P P Claim A2(1) holds. Next, we will prove Claim A2(2). Since G is connected, N(V(P[v (P), x )))\ (V(G)n V(P[v (P), x ))) 6= Æ. For z 2 N(V(P[v (P), x )))\ (V(G)n V(P[v (P), x ))), there ex- L L L P P P 0 0 ists a vertex x 2 V(P[v (P), x )) with x z 2 E(G). By the deﬁnition of x and Claim A2(1), L P v (P)x 2 E(G). According to Claim A1 and Lemma A11(2), G[V(P[v (P), x ))] is a clique. (A6) We will demonstrate that z belongs to V(P). To begin, we assume the opposite, z is not an element of V(P). By Lemma 6(2) and (A6), z 2 / V(S )[ V( H). To arrive at a contradiction, we differentiate between the following two cases: 0 0 0 0 0 • Assume that z 2 V(P ) with P 2 S nfPg. By Lemma 6(2), z 2 / fv (P ), v (P )g. L R 0 0 0 Therefore, z 2 V(P )nfv (P ), v (P )g. By the deﬁnition of x 0 and Claim A2(1), L R 0 0 0 v (P )x 2 E(G) and f (G[V(P [xb 0 , v (P )])]) = 1. Then, by Claim A1 and Lem- L P R 0 0 0 0 mas A11(1)(2), G[V(P [v (P ), x ))] and G[V(P (xb 0 , v (P )])] are cliques. Hence, L 0 R ! ! 0 0 0 0 + 0 0 + 0 there exists a Q 2 fv (P ) P z v (P ), z P v (P )z g with f (Q ) = 1. By L L R 0 0 0 Lemma A11(2), the set of vertices of the subgraph G[E(P n Q )]x G[V(P[v (P), x )n fx g)]x 0 0 P v (P) and Q in G is equal to V(P[ P ), which contradicts Lemma 4. • Suppose that z 2 V(Q). To arrive at a contradiction, we differentiate between the following two cases: – Assume that z 2 V(Q[u , u ]). Then, by Lemmas A8(1)(2), A11(1)(2) and (A6), Q,1 Q,t the set of vertices of the subgraph ! ! v (P) P x G[V(P[v (P), x ))]z Q u L wL u Q z , if z 2 > R L Q,t Q,t Q,1 Q,1 P P Q Q > V(Q(u , u )), Q,1 Q,t > Q v (P) P x G[V(P[v (P), x ))]z Q u L w, if z = R L Q,t Q,t P P Q Q Q = u , Q,1 v (P) P x G[V(P[v (P), x ))]z Q u L w, if z = > R L Q,1 Q,1 P P u , Q,t Q and Q are equal to V(P [ Q [ C [ C ) [ fwg or V(P [ Q [ C) [ fwg or 6 7 V(P[ Q[ C )[fwg or V(P[ Q)[fwg, which contradicts (I). – Suppose that z 2 V(P[v (Q), u ]) or V(P[u , v (Q)]). By virtue of symme- L R Q,1 Q,t try, we may restrict our consideration to z 2 V(P[v (Q), u ]). By Lemma A8(1), Q,1 N(u )\ End(S ) 6= Æ or f (Q(u , Q)) = 1. Combining this with Lemma A6, Q,1 Q,1 we obtain that u v (Q) 2 E(G). By Claim A1 and Lemma A11(2), G[V(P[v (Q), L L Q,1 u ))] is a clique. Then, by (A6), either v (P) P x G[ V(P[v (P), x ))]z Q v (Q) R L R Q,1 P P Axioms 2023, 12, 411 24 of 25 and Q(z , Q) in G cover V(P)[ V(Q), or v (P) P x G[V(P[ v (P), x ))]v (Q) R L L P P Q v (Q) in G covers V(P[ Q), which contradicts Lemma 4. This contradiction demonstrates that z 2 V(P). (A7) To prove Claim A2(2), we differentiate between the following two cases: • Suppose that xb = x . We will show that there is no pair of edges x x and x x with P P P 1 x 2 V(P[v (P), x )) and x 2 V(P(x , v (P)]). Suppose otherwise that x x 2 E(G) 1 L 2 P R P 1 and x x 2 E(G). Note that G[V(P(x , v (P)])] and G[V(P[v (P), x ))] form cliques. 2 P R L P P Then the set of vertices of the subgraph x G[V(P(x , v (P)])]x G[V(P[v (P), x ))]x P R P L P P P in G is equal to V(P), which contradicts Lemma 4. If either x x 2 E(G) and x x 2 / E(G) or x x 2 / E(G) and x x 2 / E(G), then, by Lemma 6(2) and (A7), P 1 P 1 2 N(x ) V(P[v (P), x ]). Therefore, x is a cut vertex of G. If x x 2 E(G) and L P 1 P P x x 2 / E(G), then, by Lemma 6(2) and (A7), N(x ) V(P[v (P), x ]). Therefore, x 2 L P P is a cut vertex of G. + 0 • Suppose that xb = x . Then x v (P) 2 / E(G) and N(x )\ End(S ) 6= Æ; say v 2 P P R P P C 0 0 N(x )\ End(S ). Suppose, ﬁrst, that N(x )\ V(P[v (P), x )) = Æ. Then x x 2 / P P L C P P E(G); otherwise, the set of vertices of the subgraph Cx G[V(P[v (P), x ])]x P v (P) P L R P P in G is equal to V(P[ C), which contradicts Lemma 4. Combining this with Lemma 6(2) and (A7), we obtain that N(x ) V(P[v (P), x ]). Then, x is a cut vertex of G. Sup- P P pose, now, that N(x )\ V(P[v (P), x )) 6= Æ; say x 2 N(x )\ V(P[v (P), x )). By P L 0 P L P P (A6), G[P[v (P), x ]] has a cycle C = x P v (P)x P x x with V(P[v (P), x ]) = L P x 0 L P 0 L P 0 0 ! ! 0 0 V(C ). By (A6), we structure a new path P such that P = x P v (P)x P x P v (P) x 0 L P R by rearranging the order of the vertices in P. Then v (P ) = x . It is easy to verify that L 0 0 0 0 G[V(P)] G[V(P )]. We will prove that there is no pair of edges x x , x x such that P 1 2 0 0 0 0 0 + 0 + 0 x 2 V(P [v (P ), x )) and x 2 V(P (x , v (P )]). Suppose otherwise that x x 2 L P R 1 2 P P 1 + + + 0 0 0 0 0 E(G) and x x 2 E(G). Then x G[V(P (x , v (P )])]x G[V(P [v (P ), x ])]x in G P R P L 2 P P P P 0 00 00 cover V(P ), contradicting Lemma 4. Let x 2 V(P[v (P), x ]). By (A7), N(x ) 0 + 0 00 V(P). If x x 2 E(G), then x x 2 / E(G). By Lemma 6(2) and (A7), N(x ) 2 P 1 0 0 V(P[v (P), x ]). Therefore, x is a cut vertex of G. If x x 2 E(G), then, x x 2 / E(G). L P P P 1 2 00 + + By Lemma 6(2) and (A7), N(x ) V(P[v (P), x ]). Therefore, x is a cut vertex P P 0 0 of G. If x x 2 / E(G) and x x 2 / E(G), then, according to Lemma 6(2) and (A7), P 1 N(x ) V(P[v (P), x ]). Therefore, x is a cut vertex of G. L P P Claim A2(2) is proved. Claim A2(2) contradicts k 2. Hence, Lemma A16 is proved. Now, let us prove Lemmas 8 and 9 which are mentioned in Section 2. Proof of Lemma 8. By contradiction, suppose that U \ V(S ) 6= Æ. According to Lemma A16, jS j = 0. As G is connected and k 2, there are at least two elements of S P C connected by a path whose inner vertices are in V(G)n V(S), contradicting Lemma 2(i). Therefore, U V(S ). By Lemma A16, jS j = 1 . P P Proof of Lemma 9. By Lemma A8(1)(2), Lemma 9(1)(2) holds. Suppose ﬁrst that End(S)\ U is not empty. Then, by Lemma A1(2), X forms an independent set of G with size + + k + k. Suppose now that End(S)\ U = Æ. By Lemma 9(1), N(u )\ End(S ) 6= Æ or f (P (u , P)) = 1. By Lemmas A1(3), A2(1) and A4, X forms an independent set of G with size k + k. Therefore, Lemma 9(3) holds. Furthermore, by Lemma A6, Lemma 9(4) holds. By Lemma A11(1), Lemma 9(5) holds. By Lemmas A9 and A10, Lemma 9(6) holds. Axioms 2023, 12, 411 25 of 25 References 1. 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Axioms – Multidisciplinary Digital Publishing Institute

**Published: ** Apr 23, 2023

**Keywords: **connectivity; independence number; k-ended tree; maximum independent set

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