EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS 2025, Vol. 18, Issue 4, Article Number 6849 ISSN 1307-5543 – ejpam.com Published by New York Business Global Vertex-Edge Dominating Sets of Some Graphs under Binary Operations Jerry Tayab1,2,∗, Ferdinand P. Jamil1,2, Imelda S. Aniversario1,2 1 Department of Mathematics and Statistics, College of Science and Mathematics, Mindanao State University-Iligan Institute of Technology, 9200 Iligan City, Philippines 2 Center for Mathematical and Theoretical Physical Sciences, Premier Research Institute of Science and Mathematics, Mindanao State University-Iligan Institute of Technology, 9200 Iligan City, Philippines Abstract. Given a simple undirected graph G = (V (G), E(G)), a vertex u ∈ V (G) vertex-edge dominates the edge xy ∈ E(G) if one of the following holds: (1) u = x or u = y, (2) ux ∈ E(G) or uy ∈ E(G). A subset S ⊆ V (G) is a vertex-edge dominating set of G if for each xy ∈ E(G), there exists u ∈ S such that u vertex-edge dominates xy. A vertex-edge dominating set S ⊆ V (G) is a total vertex-edge dominating set if for each u ∈ S, there exists v ∈ S for which uv ∈ E(G). The minimum cardinality of a vertex-edge (resp. total vertex-edge) dominating set of G is the vertex-edge domination number (resp. total vertex-edge domination number) of G. This paper investigates the vertex-edge domination and total vertex-edge domination in the join, corona, lexicographic product, complementary prism and edge corona of graphs. It provides complete characterizations of both the vertex-edge dominating sets and total vertex-edge dominating sets in these families of graphs, and establishes sharp bounds, if not the exact values, for their respective vertex-edge domination and total vertex-edge domination numbers. 2020 Mathematics Subject Classifications: 05C69 Key Words and Phrases: Domination, vertex-edge domination, total vertex-edge domination 1. Introduction Vertex-edge domination in graphs was first introduced by Peters [1] in 1986, and is a graph protection strategy which basically came from the marriage of the two concepts, namely the domination (about static positioning of guards which protect the vertices) and vertex covering (a static positioning of guards which protect the edges) of graph. The importance of vertex-edge domination is best illustrated by the so called “searchlight problem” (see [2], [3], [4]) which, inspired by the famous art gallery problem, attempts to ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v18i4.6849 Email addresses: jerry.tayab@g.msuiit.edu.ph (J. Tayab), ferdinand.jamil@g.msuiit.edu.ph (F. Jamil), imelda.aniversario@g.msuiit.edu.ph (I. Aniversario) https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0) J. Tayab, F. P. Jamil, I. Aniversario / Eur. J. Pure Appl. Math, 18 (4) (2025), 6849 2 of 19 use searchlights to find an intruder in a graph. In this case, the guards, each of whom holds a searchlight, must shine a searchlight down some edge where they think there might be an intruder. Vertex-edge domination as well as its variant, the total vertex-edge domination, is very-well studied in trees (see [5], [6], [7], [8], [9]), in some special graphs (see [8], [1]), in cubic graphs and grids (see [10], [7]), in connected C5-free graphs and connected K1,k - free graphs (see [11]). Peters also dealt with the complexity problems of the parameter in [1]. In the present paper, we investigate the verter-edge and total vertex-edge domination in the join, corona, lexicographic product and edge corona of graphs. All throughout this paper, we consider only graphs which are simple, finite and undirected. Given a graph G = (V (G), E(G)), we call V (G) the vertex set of G and E(G) its edge set. The cardinality |V (G)| of V (G) is the order of G. If E(G) = ∅, then G is an empty graph. All terminologies used here which are not being defined are adapted from [12]. Given a graph G, G − v refers to the resulting graph after removing vertex v and all incident edges from G. If the removal of vertex v from G increases the number of components, i.e., G − v has more components than G, then v is called a cut vertex of G. For S ⊆ V (G), ⟨S⟩ is the induced subgraph of G with vertex set S and edge set {xy ∈ E(G) : x, y ∈ S}. Let G and H be disjoint graphs. The join of G and H is the graph G + H with vertex set V (G) ∪ V (H) and edge set E(G) ∪ E(H) ∪ {uv : u ∈ V (G), v ∈ V (H)}. The corona of G and H is the graph G ◦ H obtained by taking one copy of G and |V (G)| copies of H, and then joining the ith vertex of G to every vertex in the ith copy of H. In particular, we call G ◦ K1 the corona of G, and write cor(G) = G ◦ K1. The edge corona of G and H is the graph G ⋄ H obtained by taking one copy of G and |E(G)| copies of H and joining each of the end vertices u and v of each edge uv of G to every vertex of the copy Huv of H. The composition (or lexicographic product) of G and H is the graph G[H] with V (G[H]) = V (G) × V (H) and (u, v)(u′, v′) ∈ E(G[H]) if and only if either uu′ ∈ E(G) or u = u′ and vv′ ∈ E(H). In any of these graphs, G and H are referred to as their basic component graphs. The complementary prism GG is formed from G and its complement G by adding a perfect matching between corresponding vertices of G and G. If for each v ∈ V (G), v is the vertex in G corresponding to v, then GG is formed by adding the edge vv for every v ∈ V (G). For vertices u and v of a graph G, a u-v geodesic is any shortest path in G joining u and v. The length of a u-v geodesic is the distance between u and v, and is denoted by dG(u, v). The eccentricity of v refers to the quantity e(v) = max{dG(u, v) : v ∈ V (G)}. Customarily, diam(G) = max{e(v) : v ∈ V (G)}. In this paper, we write e(G) = min{e(v) : v ∈ V (G)}. Vertices u and v of a graph G are neighbors if uv ∈ E(G). The open neighborhood of v refers to the set NG(v) consisting of all neighbors of v. If NG(v) = ∅, then v is an isolated vertex. The degree of v refers to the cardinality |NG(v)| of the open neighborhood of v, and δ(G) is the minimum degree of a vertex of G. The closed neighborhood of v is the set NG[v] = NG(v) ∪ {v}. Customarily, for S ⊆ V (G), NG(S) = ∪v∈SNG(v) and J. Tayab, F. P. Jamil, I. Aniversario / Eur. J. Pure Appl. Math, 18 (4) (2025), 6849 3 of 19 NG[S] = ∪v∈SNG[v]. A subset S ⊆ V (G) is a dominating set of G if NG[S] = V (G). A dominating set S of G is a total dominating set if S ⊆ NG(S). The minimum cardinality γ(G) of a dominating set of G is the domination number of G. The minimum cardinality γt(G) of a total dominating set is the total domination number of G. A dominating (resp. total dominating) set of cardinality γ(G) (resp. γt(G)) is called a γ-set (resp. γt-set of G. The reader is referred to [13], [14], [15], [16], [17], and [18] for the history, fundamental concepts and recent developments of domination in graphs as well as its various applications. Vertex u of G is said to vertex-edge dominate (or ve-dominate) edge xy ∈ E(G) if one of the following holds: • u = x or u = y; • ux ∈ E(G) or uy ∈ E(G). A subset S ⊆ V (G) is a ve-dominating set of G if for every xy ∈ E(G), there exists u ∈ S for which u ve-dominates xy. A ve-dominating set is a total ve-dominating set if, in addition, ⟨S⟩ has no isolated vertex. The minimum cardinality of a ve-dominating set is the ve-domination number of G. Similarly, the minimum cardinality of a total ve-dominating set is the total ve-domination number of G. We use the symbols γve(G) and γt ve(G) to refer to the ve-domination number and total ve-domination number of G, respectively. We also use the terms γve-set (resp. γt ve-set) to refer to any ve-dominating (resp. total ve-dominating) set of cardinality γve(G) (resp. γt ve(G)). For every nontrivial connected graph G, γve(G) ≤ γt ve(G) ≤ 2γve(G)[11]. 2. Preliminary results The following formalizes the alternative definition given by W. Klostermeyer et al. in [10]. Proposition 1. S ⊆ V (G) is a ve-dominating set of G if and only if V (G) \ NG(S) is either empty or a nonempty independent subset of V (G). Proof. First, let S be a ve-dominating set of G. If V (G) \ NG(S) = ∅, then we are done. Suppose that V (G) \ NG(S) ̸= ∅ and let x, y ∈ V (G) \ NG(S). Suppose that xy ∈ E(G). Then there exists u ∈ S for which u ve-dominates xy. If u = x (resp. u = y), then y ∈ NG(S) (resp. x ∈ NG(S)), a contradiction. However, if ux ∈ E(G) (resp. uy ∈ E(G)), then x ∈ NG(S) (resp. y ∈ NG(S)), a contradiction. Since x and y are arbitrary, V (G) \ NG(S) is an independent set. Conversely, first suppose that V (G) \ NG(S) = ∅. Let xy ∈ E(G). In particular, since V (G) = NG(S), there exists u ∈ S for which ux ∈ E(G). Observe that u, and thus S, ve-dominates xy. Next, suppose that V (G) \ NG(S) is a nonempty independent set, and J. Tayab, F. P. Jamil, I. Aniversario / Eur. J. Pure Appl. Math, 18 (4) (2025), 6849 4 of 19 let xy ∈ E(G). Suppose that S does not ve-dominate xy. Then x /∈ NG(S) and y /∈ NG(S). Since V (G) \ NG(S) is an independent set, xy /∈ E(G), a contradiction. ■ If G = Kn, then every S ⊆ V (G) is a ve-dominating set of G. Hence, γve(G) = 0 if and only if G = Kn. If G = Kn or G is the complete multipartite Kn1,n2,...,nk , then S is a ve-dominating set of G for every nonempty S ⊆ V (G). Proposition 2. [19] For the complete graph Kn, the complete bipartite Km,n, complete r-partite Kn1,n2,...,nr , path Pn and cycle Cn, we have: (i) γve(Kn) = γve(Km,n) = γve(Kn1,n2,...,nr ) = 1; (ii) γve(Pn) = ⌊n+2 4 ⌋; (iii) γve(Cn) = ⌊n+3 4 ⌋. Proposition 3. [1] For any graph of order n, γve(G) ≤ n 2 . Observation 1. Let G be a disconnected graph with nontrivial components G1, G2, . . ., Gn. Then S ⊆ V (G) is a (total) ve-dominating set of G if and only if S ∩ V (Gk) is a (total) ve-dominating set of Gk for each k = 1, 2, . . . , n. In particular, γve(G) = n∑ k=1 γve(Gk) and γt ve(G) = n∑ k=1 γt ve(Gk). Proposition 4. Let G be any graph of order n. Then (i) γve(G) = n − 1 if and only if G = {K1, P2}. (ii) γve(G) = 1 if and only if G has exactly one nontrivial component G′, where there exists a vertex v ∈ V (G′) such that dG(v, x) = 1 or dG(v, y) = 1 for every xy ∈ E(G). Proof. The case where G is nonempty follows directly from Proposition 3. Suppose G is an empty graph. Then γve(G) = 0. If γve(G) = n − 1, then n = 1. Conversely, if n = 1, then γve(G) = n − 1. Thus, (i) holds. To prove (ii), first, suppose that γve(G) = 1. In view of Observation 1, G has exactly one nontrivial component G′ and γve(G′) = γve(G) = 1. Let S = {v}, where v ∈ V (G′) for which v ve-dominates every xy ∈ E(G). Let xy ∈ E(G). Then xy ∈ E(G′). If x = v (resp. y = v), then dG(v, y) = dG′(v, y) = 1 (resp. dG(v, x) = dG′(v, x) = 1). Suppose that x ̸= v and y ≠ v. Since v ve-dominates xy, dG(v, x) = dG′(v, x) = 1 or dG(v, y) = dG′(v, y) = 1. Next, for the converse, let G have exactly one nontrivial component G′, where there exists a vertex v ∈ V (G′) such that dG(v, x) = dG′(v, x) = 1 or dG(v, y) = dG′(v, y) = 1 for every xy ∈ E(G). It is worth noting that xy ∈ E(G) if and only if xy ∈ E(G′). Let x, y ∈ V (G) \ NG(v) with x ≠ y. Suppose that xy ∈ E(G). Then, by the assumption, dG(v, x) = dG′(v, x) = 1 or dG(v, y) = dG′(v, y) = 1. That is, x ∈ NG(v) or y ∈ NG(v), a contradiction. Therefore, V (G) \ NG(v) is an independent set. By Proposition 1, S = {v} is a ve-dominating set of G. Consequently, γve(G) = |S| = 1. ■ J. Tayab, F. P. Jamil, I. Aniversario / Eur. J. Pure Appl. Math, 18 (4) (2025), 6849 5 of 19 Observation 2. A graph G admits a total ve-dominating set if and only if G is not an empty graph. Observation 3. For the complete graph Kn (n ≥ 2), the complete bipartite Km,n, complete r-partite Kn1,n2,...,nr , path Pn and cycle Cn, we have: (i) γt ve(Kn) = γt ve(Km,n) = γt ve(Kn1,n2,...,nr ) = 2; (ii) For n ≥ 2, γt ve(Pn) =  2, if n = 2, 3 + 2⌊n−3 5 ⌋, if n ̸= 2 but n ≡ 2 mod 5, 2 + 2⌊n−3 5 ⌋, if n ≡ 0, 1, 3, 4 mod 5. (iii) For n ≥ 3, γt ve(Cn) =  2, if n ∈ {3, 4, 5}, 3, if n = 6, 5 + 2⌊n−7 5 ⌋, if n ≥ 7 and n ≡ 1 mod 5, 4 + 2⌊n−7 5 ⌋, if n ≥ 7 and n ≡ 0, 2, 3, 4 mod 5. Proposition 5. Let G be a nonempty graph of order n. Then (i) γt ve(G) = n if and only if G is the (disjoint) union of copies of K2 ; (ii) γt ve(G) = n − 1 if and only if G is one of the following: P3, K3, the (disjoint) union of K1 and copies of K2, the (disjoint) union of P3 and copies of K2, the (disjoint) union of K3 and copies of K2; (iii) γt ve(G) = 2 if and only if there exists uv ∈ E(G) such that for every xy ∈ E(G)\{uv}, E(G) ∩ {ux, vx, uy, vy} ̸= ∅. Proof. Suppose that γt ve(G) = n, and let K be a nontrivial component of G. Suppose that K ̸= K2. Pick a non-cut vertex x ∈ V (K). Since S = V (K) \ {x} is a total ve-set of K, γt ve(K) ≤ |V (K)| − 1. By Observation 1, γt ve(G) ≤ n − 1, a contradiction. Thus, G is the union of copies of K2. The converse is clear. Thus, (i) holds. Suppose that γt ve(G) = n− 1, and let x ∈ V (G) such that S = V (G) \{x} is a γt ve-set of G. Let C be the component of G for which x ∈ V (C). Put SC = S ∩ V (C) = V (C) \ {x}. In view of Observation 1, SC is a γt ve-set of C. Suppose that |V (C)| ≥ 4, and let y ∈ V (C) ∩ NC(x). Since SC is a total ve-dominating set, there exists q ∈ SC ∩ NG(y) (hence, |NC(y) \ {x}| ≥ 1). Suppose ⟨SC \ {y}⟩ has no isolated vertex. Let w ∈ SC \ {y} and let p ∈ V (C) ∩ NC(w). Then w ve-dominates pw. Since w was arbitrarily chosen, it follows that SC \ {y} is a ve-dominating set of C. Moreover, since ⟨SC \ {y}⟩ has no isolated vertex, SC \ {y} is a total ve-dominating set of C, a contradiction. Next, suppose ⟨SC \ {y}⟩ has an isolated vertex, say p. Since ⟨SC⟩ has no isolated vertex, it follows that yp ∈ E(G). Note that ⟨SC \{p}⟩ has no isolated vertex because ⟨SC⟩ has no isolated vertex and p ∈ End(G). It is easy to verify (following n earlier argument) that ⟨SC \ {p}⟩ is a ve-dominating set of C. Again, this gives a contradiction. Therefore, |V (C)| ≤ 3. J. Tayab, F. P. Jamil, I. Aniversario / Eur. J. Pure Appl. Math, 18 (4) (2025), 6849 6 of 19 Moreover, the definition of C implies that C ̸= K2. If G = C (connected), then either C = P3 or C = K3. Otherwise, (i) implies that G is the disjoint union of copies of K2 and exactly one of the following: K1, P3 and K3. The converse of (ii) is clear. Finally, suppose that γt ve(G) = 2, and let S = {u, v} be a γt ve-set of G. Let xy ∈ E(G) \ {uv} and put T = {ux, vx, uy, vy}. If uv and xy have a common vertex, then T ∩ E(G) ̸= ∅. Suppose otherwise. Assume WLOG that u ve-dominates xy. Then either ux ∈ E(G) or uy ∈ E(G). In any case, T ∩ E(G) ̸= ∅. The converse is straightforward.■ Proposition 6. For every positive integers a and b with 2 ≤ a ≤ b ≤ 2a, there exists a connected graph G for which γve(G) = a and γt ve(G) = b. Proof. Suppose that a = b. For each k ∈ {1, 2, . . . , a}, let [x = xk 1, xk 2, xk 3, xk 4] denote the kth copy of P4. Take G = G1, where G1 is the graph provided in Figure 1 obtained by connecting these a copies of P4 by having x1 1 = x2 1 = · · · = xa 1 and through the path [x1 3, x2 3, . . . , xa 3].The set S = {x1 3, x2 3, . . . , xa 3} is both a γve-set and a γt ve-set of G. Thus, .................................... .................................... .................................... .................................... • • • • .................................... .................................... .................................... .................................... .................................... ................... .................. .................. ....... ............ ........... ........... ........... ........... ........... ........ .............................................................. ........................................................................... ............................................................................ ............................................................................ ............................................................................ ............................................................................ .................................... .................................... .................................... .................................... ................... .................. .................. ....... .............................................................. ............ ........... ........... ........... ........... ........... ........... ......... ....................................................................................... ... ... ... ......... ........ ........ ........ ........ ........ ..... ......... ........ ........ ........ ........ ........ ..... ......... ........ ........ .... ......... ........ ........ .... G1 x x1 3 x2 3 xa−1 3 xa 3 .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .............................................................. .................................... .............................................................. .................................... ............................................................................ ................................................................................................................ ........................................................................ ................... .................. .................. ....... ... ................................. ................... .................. .................. ....... ... ................................. ............................................................................ ................................................................................................................ ........................................................................ ............ ........... ........... ........... ........... ........... ........... ......... .................................... ............ ........... ........... ........... ........... ........... ........ .................................... ............................................................................ ................................................................................................................ ........................................................................ .................................... .................................... .................................... .................................... • • • • .................................... .................................... .................................... .................................... .................................... ................... .................. .................. ....... ............ ........... ........... ........... ........... ........... ........ .............................................................. ........................................................................... ............................................................................ ............................................................................ ............................................................................ ............................................................................ .................................... .................................... .................................... .................................... ................... .................. .................. ....... .............................................................. ............ ........... ........... ........... ........... ........... ........... ......... ....................................................................................... ... ... ... ......... ........ ........ ........ ........ ........ ..... ......... ........ ........ ........ ........ ........ ..... ......... ........ ........ .... ......... ........ ........ .... G2 x x1 3 x2 3 xa−k−1 3 xa−k 3 ... ... ... •• •• •• y1z1 y2z2 ykzk Figure 1: Graphs satisfying the conditions in Proposition 6 γve(G) = γt ve(G) = a. Now, let b = a + k, where 1 ≤ k ≤ a. Construct a graph G1 as above but using only a − k copies of P4. Obtain G as the graph G2 in Figure 1 obtained by connecting to G1 k copies of P5 using vertex x. Then S = {x1 3, x2 3, . . . , xa−k 3 } ∪ {y1, y2, . . . , yk} is a γve-set of G. Also, S ∪ {z1, z2, . . . , zk} is a γt ve-set of G. For this G, γve(G) = (a − k) + k = a and γt ve(G) = a + k = b. ■ Corollary 1. The difference γt ve(G) − γve(G) can be made arbitrarily large. Proof. Let k be any positive integer. Choose an integer a ≥ k and put b = a + k. By Proposition 6, there exists a connected graph G for which γt ve(G) − γve(G) = b − a = k. Since k is arbitrary, the conclusion follows. ■ 3. On families of graphs under binary operations Proposition 7. Let G and H be any graphs, and let S ⊆ (G + H) with S ̸= ∅. Then S is a (total) ve-dominating set of G + H if and only if one of the following holds: J. Tayab, F. P. Jamil, I. Aniversario / Eur. J. Pure Appl. Math, 18 (4) (2025), 6849 7 of 19 (i) S ⊆ V (G) and S is a (total) ve-dominating set of G; (ii) S ⊆ V (H) and S is a (total) ve-dominating set of H; (iii) S ∩ V (G) ̸= ∅ and S ∩ V (H) ̸= ∅. Proof. Assume S is a (total) ve-dominating set of G + H. Suppose that S ⊆ V (G). If S = V (G), then we are done. Suppose that S ̸= V (G). Since V (G) \ NG(S) = V (G + H) \ NG+H(S), (1) V (G) \ NG(S) is an independent set. By Proposition 1, S is a (total) ve-dominating set of G, and (i) holds. Similarly, if S ⊆ V (H), then (ii) holds. If both (i) and (ii) do not hold, then (iii) holds. Conversely, if (i) holds, then Equation (1) implies that V (G + H) \ NG+H(S) is an independent set. Consequently, S is a (total) ve-dominating set of G + H. The same conclusion is attained if (ii) holds. Now, suppose that (iii) holds for S. Since V (G + H) \ NG+H(S) = ∅, S is a (total) ve-dominating set of G + H by Proposition 1.■ Corollary 2. Let G and H be any graphs. Then γt ve(G + H) = 2, and γve(G + H) = { 1, if G or H is empty; min{γve(G), γve(H), 2}, else. In particular, γve(Km,n) = 1 and γt ve(Km,n) = 2 for all m, n ≥ 1. Proposition 8. Let G be a connected graph and H be a nonempty graph. Then S ⊆ V (G ◦ H) is a ve-dominating set of G ◦ H if and only if S = A ∪ ( ∪v∈V (G)Sv ) , (2) where A ⊆ V (G) and Sv ⊆ V (Hv) for each v ∈ V (G) such that Sv is a ve-dominating set of Hv for all v ∈ V (G) \ A. Proof. First, assume that S is a ve-dominating set of G ◦ H. Put A = S ∩ V (G) and Sv = S ∩ V (Hv) for each v ∈ V (G). Then S satisfies Equation 1. Let v ∈ V (G) \ A, and xy ∈ E(Hv). Since S is a ve-dominating set, there exists u ∈ S such that u ve-dominates xy. Since v /∈ S, u ∈ Sv. Accordingly, Sv is a ve-dominating set of Hv. Conversely, suppose S is as given in Equation 1 such that Sv is a ve-dominating set of Hv for all v ∈ V (G) \ A. Let xy ∈ E(G ◦ H). We consider the following cases: Case 1: x ∈ V (G) or y ∈ V (G) WLOG assume x ∈ V (G). If x ∈ A, then S ve-dominates xy. Suppose that x /∈ A. Since H is nonempty, Hx contains an edge ab. Because x /∈ S, there exists u ∈ Sx such that u ve-dominates ab. Since ux ∈ E(G ◦ H), u and therefore, S ve-dominates xy. J. Tayab, F. P. Jamil, I. Aniversario / Eur. J. Pure Appl. Math, 18 (4) (2025), 6849 8 of 19 Case 2: x /∈ V (G) and y /∈ V (G) There exists v ∈ V (G) for which xy ∈ E(Hv). If v ∈ A, then v, and therefore S, ve-dominates xy. Suppose that v /∈ A. Then Sv ve-dominates xy by the assumption. Thus, S ve-dominates xy. Since xy is arbitrary, S is a ve-dominating set of G ◦ H. ■ Corollary 3. Let G be a connected graph of order n ≥ 2 and H any graph. (i) If H is an empty graph, then γve(G ◦ H) = γ(G) and γt ve(G ◦ H) = γt(G). (ii) If H is a nonempty graph, then γve(G ◦ H) = γt ve(G ◦ H) = n. Proof. Let H be an empty graph. Let S ⊆ V (G) be a (total) dominating set of G. We claim that S is a (total) ve-dominating set of G ◦ H. Let xy ∈ E(G ◦ H). Assume WLOG that x ∈ V (G). If x ∈ S, then S ve-dominates xy. Suppose that x /∈ S. Since S is a dominating set of G, there exists u ∈ S for which ux ∈ E(G). This means u, hence S, ve-dominates xy. Therefore, S is a (total) ve-dominating set of G ◦ H. Since S is arbitrary, γve(G ◦ H) ≤ γ(G) and γt ve(G ◦ H) ≤ γt(G). To get the other inequalities, first let S ⊆ V (G◦H) be a γve-set of G◦H. Put A = S ∩V (G) and B = {v ∈ V (G) : S ∩ V (Hv) ̸= ∅}. Define S∗ = A ∪ B. Then |S∗| ≤ |S| = γve(G ◦ H). We claim that S∗ is a dominating set of G. Let v ∈ V (G)\S∗. Pick u ∈ V (Hv). There exists w ∈ S such that w ve-dominates uv. Since S ∩ V (Hv) = ∅, w ∈ A. Thus, w ∈ S∗ ∩ NG(v). This means that S∗ is a dominating set of G. Therefore, γ(G) ≤ γve(G ◦ H). Next, suppose that S is a γt ve-set of G ◦ H. Let A = S ∩ V (G) and B = {v ∈ A : S ∩ V (Hv) ̸= ∅}. For each v ∈ B, choose uv ∈ NG(v). Define S∗ = A ∪ {uv : v ∈ B}. Let v ∈ V (G) \ S∗ and let u ∈ V (Hv). There exists w ∈ S such that w ve-dominates uv. Since S is a total ve-dominating set and v /∈ S, u /∈ S. Hence, w ∈ A showing that S∗ is a dominating set of G. Let w ∈ S∗. If w /∈ A, then w = uv for some v ∈ B ⊆ A. Note here that v ∈ S∗ and wv ∈ E(G). Suppose that w ∈ A. If S ∩ V (Hw) = ∅, then since S is a total ve-dominating set, there exists z ∈ A ∩ NG(w). If S ∩ V (Hw) ̸= ∅, then w ∈ B and uw ∈ S∗ ∩ NG(w). This completely shows that S∗ is a total dominating set of G. Thus, γt(G) ≤ |S∗| ≤ |S| = γt ve(G ◦ H). This proves (i). Now, we prove (ii). By Proposition 8, V (G) is a ve-dominating set, and therefore a total ve-dominating set, of G ◦ H. Thus, γve(G ◦ H) ≤ n and γt ve(G ◦ H) ≤ n. J. Tayab, F. P. Jamil, I. Aniversario / Eur. J. Pure Appl. Math, 18 (4) (2025), 6849 9 of 19 Let S ⊆ V (G ◦ H) be a γve-set of G ◦ H. By Proposition 8, S = A ∪ ( ∪v∈V (G)Sv ) , where A ⊆ V (G) and Sv ⊆ V (Hv) for each v ∈ V (G) such that Sv is a ve-dominating set of Hv for all v ∈ V (G) \ A. Thus, γt ve(G ◦ H) ≥ γve(G ◦ H) = |S| ≥ |A| + ∑ v∈V (G)\A |Sv| ≥ |V (G)| = n. ■ Proposition 9. Let G be a graph of order n. Then (i) γve(GG) = 1 if and only if G ∈ {Kn, Kn}; (ii) γt ve(GG) = 2 if and only if one of the following holds: (a) There exists v ∈ V (G) for which {v} and {v} are ve-dominating sets of G and G, respectively. (b) G has a γt-set {u, v} such that xy ∈ E(G) for all x, y ∈ V (G) \ {u, v}. (c) G has a γt-set {u, v} such that xy ∈ E(G) for all x, y ∈ V (G) \ {u, v}; (iii) γt ve(GG) ̸= 2n − 1; and (iv) γt ve(GG) = 2n if and only if G = K1. Proof. If G = Kn and v ∈ V (G), then dGG(x, v) = 1 or dGG(y, v) = 1 for each xy ∈ E(GG). By Proposition 4, γve(GG) = 1. In case G = Kn, we use the same argument on G = Kn. Conversely, suppose that γve(GG) = 1. Assume G /∈ {Kn, Kn}. Let v ∈ V (GG) such that {v} is a ve-dominating set of GG. WLOG assume that v ∈ V (G). We consider the following cases: Case 1: Suppose that dG(u, v) = 1 for all u ∈ V (G) \ {v}. Since G ̸= Kn, there exist x, y ∈ V (G) for which dG(x, y) = 2. Observe that x y ∈ E(GG) and v does not ve-dominate x y, a contradiction. Case 2: Suppose that dG(u, v) = 2 for some u ∈ V (G). In this case, it is easy to see that v does not ve-dominate uu ∈ E(GG), a contradiction. The above contradictions imply that G ∈ {Kn, Kn}. This proves (i). Suppose that γt ve(GG) = 2, and let S = {u, v} be a γt ve-set of GG. Suppose that u = v. Assume v ∈ V (G). Let xy ∈ E(G). If x = v or y = v, then v ve-dominates xy. Suppose that x ̸= v and y ≠ v. Since S ve-dominates xy, v ve-dominates xy. Similarly, if xy ∈ E(G), then v ve-dominates xy. This proves (ii)(a). Suppose that S ⊆ V (G). First, we claim that S is a total dominating set of G. Let x ∈ V (G)\S. Since u or v ve-dominates xx, ux ∈ E(G) or vx ∈ E(G). Thus, S is a dominating set, hence a total dominating set, of G. Next, let x, y ∈ V (G) \ S. Suppose that x y ∈ E(G). Then u or v ve-dominates x y. This is impossible since min{dGG(u, x), dGG(u, y), dGG(v, x), dGG(v, y)} ≥ 2. Thus, x y /∈ E(G) and (ii)(b) holds. Similarly, if S ⊆ V (G), then (ii)(c) holds. J. Tayab, F. P. Jamil, I. Aniversario / Eur. J. Pure Appl. Math, 18 (4) (2025), 6849 10 of 19 Assume that (ii)(a) holds for G. Let xy ∈ E(GG) \ {vv}. If x ∈ {v, v} or y ∈ {v, v}, then E(GG) ∩ {xv, xv, yv, yv} ≠ ∅. Suppose that {x, y} ∩ {v, v} = ∅. If xy ∈ E(G), then v ve-dominates xy. Thus, E(GG) ∩ {vx, vy} ̸= ∅. Similarly, if xy ∈ E(G), then E(GG) ∩ {xv, yv} ̸= ∅. Now, suppose that y = x and assume x ∈ V (G). Then either vx ∈ E(G) or v x ∈ E(G). Thus, E(GG) ∩ {xv, yv} ̸= ∅. By Proposition 5, γt ve(GG) = 2. Next, assume that (ii)(b) holds. Let xy ∈ E(GG) \ {uv}. If x ∈ {u, v} or y ∈ {u, v}, then E(GG) ∩ {ux, uy, vx, vy} ≠ ∅. Suppose that {x, y} ∩ {u, v} = ∅. If x ∈ V (G), then since {u, v} is a dominating set of G, E(GG) ∩ {ux, vx} ̸= ∅. Suppose that x, y ∈ V (G). Then (exactly) one of the following is true: x = v, y = v, x = u, y = u. In any case, E(GG) ∩ {ux, uy, vx, vy} ≠ ∅. By Proposition 5, γt ve(GG) = 2. Similarly, if (ii)(c) holds, then γt ve(GG) = 2. Statements (iii) and (iv) immediately follow from Proposition 5(ii) and Proposition 5(i), respectively. ■ Graphs G in Proposition 9(ii) need not have to be complete. Graphs G1 = K1,3 and G2 = C4 in Figure 2 are examples for statements (ii)(a) and (ii)(b), respectively. In each case, γt ve(GG) = 2. .................................... .................................... .................................... .................................... .................................... .................................... .................................... .............................................................................................................................................................................................................................................................................................................................................................................................................................. ............................................................................................................................................................................................................................................................ .......................... ......................... ......................... ......................... ......................... .......................... ......................... ......................... ......................... ......................... .................................... ................................................................................................................................................. .................................... ............................................................................................................................................................................................................................................................ .................................... .................................... • • ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ...... ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ...... ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ...... ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ...... v u G1 : G1 : .................................... .................................... .................................... .................................... .................................... .................................... .................................... .............................................................. ......................... ......................... ......................... ......................... .................................... ............................................................................................................................................................................................................................................................ .................................... .......................... ......................... ......................... ......................... ......................... ................................................................................................................................................................................................................................................................................................ .................................... .................................... .......................................................................................................................................................................................................................................................................................................................................................................................... ................................................................................................................................................. ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ...... ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ...... ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ...... ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ...... • • G2 : G2 : u v Figure 2: Examples of a noncomplete graph G for which γt ve(GG) = 2. Proposition 10. For any graph G, (i) γve(GG) ≤ min{γ(G) + γve(G), γ(G) + γve(G)}; (ii) provided G and G have no isolated vertices, γt ve(GG) ≤ min{γt(G) + γt ve(G), γt(G) + γt ve(G)}. Proof. Let S ⊆ V (G) be a γ-set of G and S∗ ⊆ V (G) a γve-set of G. Then S ∪ S∗ ve-dominates V (G) ∪ V (G). Let u ∈ V (G). There exists v ∈ S such that u ∈ NG[u]. If u = v, then v ve-dominates uu. If v ≠ u, then uv ∈ E(G) and v ve-dominates uu. In any case, S ∪ S∗ ve-dominates uu. Since u is arbitrary, S ∪ S∗ is a ve-dominating set of GG. Hence, γve(GG) ≤ |S| + |S∗| = γ(G) + γve(G). Similarly, γve(GG) ≤ γ(G) + γve(G). This proves (i). The inequality in (ii) is proved similarly. ■ J. Tayab, F. P. Jamil, I. Aniversario / Eur. J. Pure Appl. Math, 18 (4) (2025), 6849 11 of 19 Proposition 11. Let G be a graph with one isolated vertex v. Then γve(GG) ≤ 1 + γve(G − v) (3) and γt ve(GG) ≤ 2 + γt ve(G − v). (4) Equality in Equation 3 is attained if GG has a γve-set which contains v. Under the same condition 1 + γve(G − v) ≤ γt ve(GG), (5) and this lower bound is sharp. Proof. Equation 3 follows from Proposition 10. Equation 4 follows from the fact that if S ⊆ V (G − v) is a γt ve-set of G − v, then S ∪ {v, v} is a total ve-dominating set of GG. Let S ⊆ V (GG) be a γve-set of GG with v ∈ S. Let S∗ = {w ∈ V (G − v) : w ∈ S}, and put T = S∗ ∪ (S ∩ V (G − v)). Let xy ∈ E(G − v). There exists w ∈ S for which w ve-dominates xy. If w ∈ V (G − v), then T ve-dominates xy. Suppose that w ∈ V (G). Then w ∈ S∗ and either w = x or w = y. In either case, T ve-dominates xy. Since xy is arbitrary, T is a ve-dominating set of G − v. Hence, γve(GG) = |S| ≥ 1 + |T | ≥ 1 + γve(G − v). If S were taken as a γt ve-set of GG, then the above argument implies that γt ve(GG) = |S| ≥ 1 + |T | ≥ 1 + γve(G − v). If, in particular, G is the graph G1 = K1 ∪ P3 in Figure 2(a), then γt ve(GG) = 2 = 1 + γve(G), showing that the lower bound in Equation 5 is sharp. ■ Strict inequality can be attained in Proposition 11. Consider the graph G2 = K1 ∪ (K2 ∪ K2). Then G2 is the wheel K1 + C4, and G2G2 is the graph in Figure 3(b).Observe that γve(GG) = 2 and is determined by the ve-dominating set S = {x, y} of GG. Note, on the other hand, that 1 + γve(G − v) = 3, where v is the isolated vertex of G. J. Tayab, F. P. Jamil, I. Aniversario / Eur. J. Pure Appl. Math, 18 (4) (2025), 6849 12 of 19 .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... ................... .................. .................. .................. ............. .................................... ................................................................................................................................................................................................................................................. .................................... ....................................................................................................................... ....................................................................................................................................................................... ........................................ ........................................ ........................................ ........................................ ........ .................................................................... ........................................................................................................................................................................................................................................................................... ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ...... ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ...... ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ...... ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ...... v w • •G1 : G1 : (a) .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .......................... ......................... ......................... ......................... ......................... .................................... ............................................................................................................................................................................................................................................................ .................................... .......................... ......................... ......................... ......................... ......................... ................................................................................................................................................................................................................................................................................................ .................................... .................................... ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ...... ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ...... ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ...... ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ...... .......................................................................................................................................................................................................................................................................................................................................................................................... ................................................................................................................................................. .................................... .................................... ............. ............. .............. .............. ............... ................ ................. ................... ...................... .............................. ...................................................................................................................................................................................................................................................................................... .................. ................... .................... ..................... ...................... ....................... ......................... ............................ ................................. ............................................. ..................................................................................................................................................................................................................................................................................................................................................... ................................................ ................................................ ................................................ ................................................ ........... ....................................................................................................................... ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ......G2 : G2 : • • x y v (b) Figure 3: Examples of graphs illustrating Proposition 11. Proposition 12. Let G and H be connected graphs with H nontrivial, and C ⊆ V (G[H]). Then C is a ve-dominating set of G[H] if and only if C = ∪x∈S ({x} × Tx) , (6) where S ⊆ V (G) and Tx ⊆ V (H) for each x ∈ S such that each of the following holds: (i) S is a dominating set of G; (ii) Tx is a ve-dominating set of H for each x ∈ S \ NG(S). Proof. First, assume that C is a ve-dominating set of G[H]. Let S be the G-projection CG of C, i.e., S = CG = {x ∈ V (G) : (x, y) ∈ C for some y ∈ V (H)}. Then C = ∪x∈S ({x} × Tx), where Tx = {y ∈ V (H) : (x, y) ∈ C} for each x ∈ S. We claim that S is a dominating set of G. Suppose not, and let x ∈ V (G) \ NG[S]. Let wz ∈ E(H). Then (x, w)(x, z) ∈ E(G[H]) and there exists (u, v) ∈ C such that (u, v) ve-dominates (x, w)(x, z). Since x /∈ S, u ̸= x. Thus, ux ∈ E(G), a contradiction. This means that V (G) = NG[S], and the claim is done. Therefore, (i) holds. Now, to prove (ii), let x ∈ S \ NG(S). Let wz ∈ E(H). Then (x, w)(x, z) ∈ E(G[H]) and there exists (u, v) ∈ C for which (u, v) ve-dominates (x, w)(x, z) in G[H]. If (u, v) = (x, w), then u = x so that v = w ∈ Tx and Tx ve-dominates wz. Similarly, if (u, v) = (x, z), then Tx ve-dominates wz. Suppose that (u, v)(x, w) ∈ E(G[H]). Since x /∈ NG(S), u = x and vw ∈ E(H). This means that v, and hence Tx, ve-dominates wz. Similarly, if (u, v)(x, z) ∈ E(G[H]), then Tx ve-dominates wz. Accordingly, Tx is a ve-dominating set of H. Next, assume that C satisfies Equation 6 where S ⊆ V (G) and Tx ⊆ V (H) satisfying conditions (i) and (ii), respectively. Note first that since S is a dominating set, S is a ve-dominating set of G. Let (x, y)(w, z) ∈ E(G[H]). We consider the following cases: Case 1: Suppose that xw ∈ E(G). By a ve-dominating set, there exists u ∈ S such that u ve-dominates xw. Pick v ∈ Tu. If u = x (resp. u = w), then (u, v)(w, z) ∈ E(G[H]) (resp. (u, v)(x, y) ∈ E(G[H])). This means that (u, v), and hence C, ve-dominates (x, y)(w, z). On the other hand, if ux ∈ E(G) (resp. uw ∈ E(G)), then (u, v)(x, y) ∈ E(G[H]) (resp. (u, v)(w, z) ∈ E(G[H])). Thus, (u, v), and hence, C ve-dominates (x, y)(w, z). J. Tayab, F. P. Jamil, I. Aniversario / Eur. J. Pure Appl. Math, 18 (4) (2025), 6849 13 of 19 Case 2: Now suppose that x = w and yz ∈ E(H). If x ∈ NG(S) and u ∈ S ∩ NG(x), pick v ∈ Tu. Then (u, v) ∈ C and (u, v)(x, y) ∈ E(G[H]). This means that (u, v), and hence C, ve-dominates (x, y)(w, z). Suppose, on the other hand, that x /∈ NG(S). Since S is a dominating set, x ∈ S \ NG(S). By condition (ii), there exists v ∈ Tx such that v ve-dominates yz. This means that (x, v) ∈ C and (x, v), and hence C, ve-dominates (x, y)(w, z). Accordingly, C is a ve-dominating set of G[H]. ■ Lemma 1. [13, 14] If S is a dominating set of a graph G without isolated vertices, then γt(G) ≤ |S ∩ NG(S)| + 2|S \ NG(S)|. Consequently, γt(G) ≤ 2γ(G). Corollary 4. Let G and H be connected graphs with H nontrivial. (i) If γve(H) = 1, then γve(G[H]) = γ(G). (ii) If γve(H) ≥ 2, then γve(G[H]) = γt(G). Proof. Let S ⊆ V (G) be a γ-set and v ∈ V (H) for which {v} is a ve-dominating set of H. Put Tx = {v} for all x ∈ S. By Proposition 12, S × {v} = ∪x∈S ({x} × Tx) is a ve-dominating set of G[H] yielding γve(G[H]) ≤ |S| = γ(G). For the other inequality, note from Proposition 12 that every ve-dominating set of G[H] is of the form C = ∪x∈S ({x} × Tx), where S ⊆ V (G) is a dominating set of G and Tx ⊆ V (H) is a ve-dominating set of H for each x ∈ S \ NG(S). For any such C, we have |C| = ∑ x∈S |Tx| ≥ |S| = γ(G). Therefore, γve(G[H]) ≥ γ(G). This proves (i). Now we prove (ii) using Proposition 12. For all total dominating sets S of G and v ∈ V (H), since S \ NG(S) = ∅, C = S × {v} = ∪x∈S ({x} × {v}) is a ve-dominating set of G[H] so that γve(G[H]) ≤ |S| = γt(G). Now, let C = ∪x∈S ({x} × Tx), where S ⊆ V (G) and Tx ⊆ V (H), be a ve-dominating set of G[H]. Necessarily, S is a dominating set of G and Tx is a ve-dominating set of H for each x ∈ S \ NG(S). By Lemma 1, |C| = ∑ x∈S |Tx| ≥ 2|S| ≥ 2γ(G) ≥ γt(G). Since C is arbitrary, γve(G[H]) ≥ γt(G). ■ The following immediately follows from Proposition 12 J. Tayab, F. P. Jamil, I. Aniversario / Eur. J. Pure Appl. Math, 18 (4) (2025), 6849 14 of 19 Proposition 13. Let G and H be connected nontrivial graphs, and let C ⊆ V (G[H]). Then C is a total ve-dominating set of G[H] if and only if C = ∪x∈S ({x} × Tx) , (7) where S ⊆ V (G) and Tx ⊆ V (H) for each x ∈ S such that one of the following holds: (i) S is a total dominating set of G; (ii) Each of the following holds: (a) S is a dominating set of G; and (b) Tx is a total ve-dominating set of H for each x ∈ S \ NG(S). Corollary 5. If G and H are connected nontrivial graphs, then γt ve(G[H]) = γt(G). Proof. By Proposition 13, S × {y} is a total ve-dominating set of G[H] for all total dominating sets S ⊆ V (G) of G and all y ∈ V (H). Therefore, γt ve(G[H]) ≤ γt(G). Now, let C = ∪x∈S ({x} × Tx) ⊆ V (G[H]) be a γt ve-set of G[H]. If S is a total dominating set of G, then |C| ≥ |S| ≥ γt(G). Otherwise, Lemma 1 and Proposition 13 imply that |C| = ∑ x∈S∩NG(S) |Tx| + ∑ x∈S\NG(S) |Tx| ≥ |S ∩ NG(S)| + 2|S \ NG(S)| ≥ γt(G). In any case, γt ve(G[H]) ≥ γt(G). ■ For each uv ∈ E(G), Huv denotes that copy of H being joined to G through edge uv. We also write Huv + uv to denote that subgraph of G ⋄ H induced by V (Huv) ∪ {u, v}. Proposition 14. Let G be a nontrivial connected graph and H be any nonempty graph. Then S ⊆ V (G ⋄ H) is a ve-dominating set of G ⋄ H if and only if S = A ∪ ( ∪uv∈E(G)Suv ) , (8) where A ⊆ V (G) and Suv ⊆ V (Huv) for each uv ∈ E(G) such that Suv is a ve-dominating set of Huv whenever {u, v} ∩ A = ∅. J. Tayab, F. P. Jamil, I. Aniversario / Eur. J. Pure Appl. Math, 18 (4) (2025), 6849 15 of 19 Proof. Let S ⊆ V (G ⋄ H) be a ve-dominating set of G ⋄ H. Put A = S ∩ V (G) and Suv = S ∩ V (Huv) for each uv ∈ E(G). Then S satisfies Equation 8. Let uv ∈ E(G) such that {u, v} ∩ A = ∅, and let xy ∈ E(Huv). Since S is a ve-dominating set of G ⋄ H, there exists w ∈ S such that w ve-dominates xy. Since {u, v} ∩ S = ∅, w ∈ Suv. Thus, Suv is a ve-dominating set of Huv. Conversely, suppose that the set S in Equation 8 has the property that Suv is a ve- dominating set of Huv for each uv ∈ E(G) with {u, v} ∩ A = ∅. Let xy ∈ E(G ⋄ H) and let uv ∈ E(G) such that xy ∈ E(Huv + uv). First, suppose that {u, v} ∩ A = ∅. Then Suv is a ve-dominating set of Huv. If {x, y} ∩ {u, v} ̸= ∅, then since {zu, zv} ⊆ E(G ⋄ H), z ve-dominates xy for all z ∈ Suv. If, on the other hand, {x, y} ∩ {u, v} = ∅, then xy ∈ E(Huv) and there exists z ∈ Suv such that z ve-dominates xy. In this case, S ve-dominates xy. Next, suppose that {u, v} ∩ A ≠ ∅. WLOG, assume that u ∈ A. If x = u, then u, and hence S, ve-dominates xy. If x ≠ u, then ux ∈ E(G ⋄ H) and so u, and hence S, ve-dominates xy. Accordingly, S is a ve-dominating set of G ⋄ H. ■ Corollary 6. Let G be a nontrivial connected graph. (i) If H is an empty graph, then γve(G ⋄ H) = γ(G). (ii) If H is a nonempty graph, then γve(G ⋄ H) = min{|S| : S ⊆ V (G) and NG(V (G) \ S) ⊆ S}. Proof. Suppose that H is an empty graph. Let S ⊆ V (G) be a γ-set of G. Let xy ∈ E(G ⋄ H), and let uv ∈ E(G) such that xy ∈ E(Huv + uv). If u ∈ S (resp. v ∈ S), then u (resp. v) ve-dominates xy. Suppose that u /∈ S and v /∈ S. Since S is a dominating set, there exist w, z ∈ S for which uw ∈ E(G) and yz ∈ E(G). In this case, w ve-dominates xy or z ve-dominates xy. Thus, S is a ve-dominating set of G ⋄ H. Consequently, γve(G ⋄ H) ≤ |S| = γ(G). To get the other inequality, let S ⊆ V (G ⋄ H) be a ve-dominating set of G ⋄ H. Put S1 = S ∩V (G). For each uv ∈ E(G), put Suv = S ∩V (Huv). For each uv ∈ E(G) for which Suv ̸= ∅, put either xuv = u or xuv = v. Define S2 = {xuv : uv ∈ E(G) with Suv ≠ ∅} and define S∗ = S1 ∪ S2. We claim that S∗ is a dominating set of G. Let z ∈ V (G) \ S∗ and choose w ∈ V (G) such that zw ∈ E(G). If Szw ̸= ∅, then since z /∈ S2, w ∈ S2. In this case, z ∈ NG(S∗). Suppose that Szw = ∅. Pick t ∈ V (Hzw). There exists u ∈ S for which u ve-dominates zt. If u ∈ V (G), then u ∈ S1 and uz ∈ E(G) so that z ∈ NG(S∗). If u /∈ V (G), then u ∈ Svz for some v ∈ NG(z). Then xvz = v ∈ S2 so that z ∈ NG(S∗). Since z is arbitrary, S∗ is a dominating set of G. Hence, γ(G) ≤ |S∗| ≤ |S|. Since S is arbitrary, γ(G) ≤ γve(G ⋄ H). This completely proves (i). Now we prove (ii). Note first that if S = V (G)\{x}, where x ∈ V (G), then NG(V (G)\ S) = NG(x) ⊆ S. Put α = min{|S| : S ⊆ V (G) and NG(V (G) \ S) ⊆ S}. Let S ⊆ V (G) J. Tayab, F. P. Jamil, I. Aniversario / Eur. J. Pure Appl. Math, 18 (4) (2025), 6849 16 of 19 such that NG(x) ⊆ S for each x ∈ V (G) \ S. Write S = A ∪ ( ∪uv∈E(G)Suv ) , where A = S and Suv = ∅ for all uv ∈ E(G). Note that by the definition of S, {u, v} ∩ A ̸= ∅ for all uv ∈ E(G). By Proposition 14, S is a ve-dominating set of G ⋄ H. Consequently, γve(G ⋄ H) ≤ |S|. Since S is arbitrary, γve(G ⋄ H) ≤ α. For the other inequality, let S ⊆ V (G ⋄ H) be a γve-set of G ⋄ H. By Proposition 14, S = A ∪ ( ∪uv∈E(G)Suv ) , where A = S ∩ V (G) and Suv ⊆ V (Huv) with Suv a ve-dominating set of Huv, hence nonempty, whenever {u, v}∩A = ∅. For each uv ∈ E(G) with {u, v}∩A = ∅, put either xuv = u or xuv = v. Define B = {xuv : uv ∈ E(G) with {u, v} ∩ A = ∅} and put S∗ = A ∪ B. It is worth noting that A ∩ B = ∅ and |S∗| = |A| + |B| ≤ |A| + ∑ uv∈E(G) |Suv| = |S| = γve(G ⋄ H). Let v ∈ V (G) \ S∗, and let u ∈ NG(v). If u /∈ S∗, then {u, v} ∩ A = ∅. This means that xuv = u or xuv = v, which is impossible since xuv ∈ B ⊆ S∗. Therefore, u ∈ S∗. Since u is arbitrary, NG(v) ⊆ S∗. Hence, α ≤ |S∗| ≤ γve(G ⋄ H). ■ It is clear from Corollary 6(ii) that if G is a connected nontrivial graph and H is a nonempty graph, then γ(G) ≤ γve(G ⋄ H) ≤ |V (G)| − 1. In view of Proposition 14, the following is clear. Proposition 15. Let G be a nontrivial connected graph and H be any nonempty graph. Then S ⊆ V (G ⋄ H) is a total ve-dominating set of G ⋄ H if and only if S = A ∪ ( ∪uv∈E(G)Suv ) , (9) where A ⊆ V (G) and Suv ⊆ V (Huv) for each uv ∈ E(G) satisfying the following: (i) Suv is a total ve-dominating set of Huv for each uv ∈ E(G) for which {u, v}∩A = ∅. (ii) A ∩ NG(u) ̸= ∅ for all u ∈ A for which Suv = ∅ for all v ∈ NG(u). Corollary 7. Let G be a nontrivial connected graph. (i) If H is an empty graph, then γt ve(G ⋄ H) = γt(G). (ii) If H is a nonempty graph, then γt ve(G ⋄ H) = min{|S| : S ⊆ V (G) with S ⊆ NG(S) and NG(V (G) \ S) ⊆ S}. J. Tayab, F. P. Jamil, I. Aniversario / Eur. J. Pure Appl. Math, 18 (4) (2025), 6849 17 of 19 Proof. Assume H is an empty graph. Let S ⊆ V (G) be a γt-set of G. Since S is a dominating set of G, following the necessity proof of Corollary 6(i) will show that S is a ve-dominating set of G ⋄ H. Consequently, S is a total ve-dominating set of G ⋄ H. Thus, γt ve(G ⋄ H) ≤ |S| = γt(G). To get the other inequality, let S ⊆ V (G ⋄ H) be a γt ve-set of G ⋄ H. Define S1 = S ∩ V (G) and Suv = S ∩ V (Huv) for each uv ∈ E(G). Note that since S is a total ve-dominating set, if Suv ̸= ∅, then u ∈ S1 or v ∈ S1. For each uv ∈ E(G) for which Suv ̸= ∅, write u = xuv whenever u /∈ S1 and write v = xuv whenever v /∈ S1. Put S2 = {xuv : uv ∈ E(G) with Suv ̸= ∅}. Let S∗ = S1 ∪ S2. Then u, v ∈ S∗ for all uv ∈ E(G) for which Suv ̸= ∅. First, let z ∈ V (G) \ S∗. Choose w ∈ V (G) such that zw ∈ E(G). If w ∈ S∗, then z ∈ NG(S∗). Suppose that w /∈ S∗. Since S is a ve-dominating set, there exists x ∈ S such that x ve-dominates zw. Since z /∈ S∗ and w /∈ S∗, x ∈ S1. This means that z ∈ NG(S∗). Next, let z ∈ S∗. If z ∈ S2, then z ∈ NG(S∗). Suppose that z ∈ S1. Since S is a total ve-dominating set, there exists x ∈ S ∩ NG⋄H(z). If x ∈ S1, then z ∈ NG(S∗). If, on the other hand, x /∈ S1 and w ∈ V (G) for which x ∈ Szw, then w = xzw ∈ S∗. Hence z ∈ NG(S∗). The above implies that S∗ is a total dominating set of G. Hence, γt ve(G ⋄ H) = |S| ≥ |S∗| ≥ γt(G). This proves (i). Now we prove (ii). Put α = min{|S| : S ⊆ V (G) with S ⊆ NG(S) and NG(V (G)\S) ⊆ S}. First, note that if S ⊆ V (G) such that S ⊆ NG(S) and NG(V (G) \ S) ⊆ S, then following the argument in the necessity proof of Corollary (ii) will show that S is a total ve-dominating set of G ⋄ H. Hence, γt ve(G ⋄ H) ≤ α. Next, let S ⊆ V (G ⋄ H) be a γt ve-set of G ⋄ H. By Proposition 15, S = A ∪ ( ∪uv∈E(G)Suv ) , where A = S ∩ V (G) and Suv ⊆ V (Huv) with Suv a total ve-dominating set of Huv, hence |Suv| ≥ 2, whenever {u, v} ∩ A = ∅. Moreover, A ∩ NG(u) ̸= ∅ for all u ∈ A for which Suv = ∅ for all v ∈ NG(u). For each u ∈ A, we say u has the property PA and write u ∈ PA if there exists v ∈ NG(u) such that Suv ̸= ∅. For each u ∈ PA, we pick one such v and write v = vu. Put C = {vu : u ∈ PA} and let B = {u ∈ V (G) \ A : ∃v ∈ V (G) \ A for which uv ∈ E(G)}. Define S∗ = A ∪ B ∪ C, and let X = {uv ∈ E(G) : {u, v} ∩ A = ∅} and Y = {uv ∈ E(G) : {u, v} ∩ A ̸= ∅}. Then γt ve(G ⋄ H) = |S| = |A| + ∑ uv∈E(G) |Suv| = |A| + ∑ uv∈Y |Suv| + ∑ uv∈X |Suv| ≥ |A ∪ C| + |B| = |S∗|. Let u ∈ B. Then there exists v ∈ B such that uv ∈ E(G) and {u, v}∩A = ∅. Consequently, v ∈ S∗ so that u ∈ NG(S∗). Let u ∈ A. If u ∈ PA, then there exists vu ∈ C ∩ NG(u). J. Tayab, F. P. Jamil, I. Aniversario / Eur. J. Pure Appl. Math, 18 (4) (2025), 6849 18 of 19 Thus, u ∈ NG(S∗). On the other hand, if u /∈ PA, then A ∩ NG(u) ̸= ∅. This means u ∈ NG(S∗). Now, let u ∈ C. Then u = ux for some x ∈ PA and so u ∈ NG(S∗). All these imply that S∗ ⊆ NG(S∗). Finally, let v ∈ V (G) \ S∗, and let u ∈ NG(v). If u /∈ S∗, then {u, v} ∩ A = ∅. This means that u, v ∈ B, which is impossible. Therefore, u ∈ S∗. Since u is arbitrary, NG(v) ⊆ S∗. Hence, α ≤ |S∗| ≤ γt ve(G ⋄ H). ■ It also follows from Corollary7(ii) that if G is a connected graph of order n ≥ 3 and H is a nonempty graph, then γt(G) ≤ γt ve(G ⋄ H) ≤ n − 1. 4. Conclusion This study has established several characterizations concerning the vertex-edge domina- tion and total vertex-edge domination in graphs under various binary operations, namely the join, corona, complementary prism, lexicographic product, and edge corona. Each result provided necessary and sufficient conditions for a subset of vertices to be a (total) vertex-edge dominating set within the resulting graph. The results demonstrate that the (total) vertex-edge domination behavior of these resulting graphs is fundamentally determined by the vertex-edge domination properties of their component graphs. In the join of graphs, vertex-edge domination depends on the subset relationships among the component vertex sets; in the corona and edge corona, the vertex-edge domination proper- ties are inherited through the copies of the secondary graph attached to each vertex or edge of the primary graph. Moreover, in the complementary prism, specific bounds for the vertex-edge domination and total vertex-edge domination numbers were established, while in the lexicographic product, the (total) vertex-edge dominating sets were expressed through unions of product sets derived from the component graphs. Acknowledgements This research is fully supported by the Department of Science and Technology - Accelerated Science and Technology Human Resource Development Program (DOST- ASTHRDP), Philippines, and the MSU-Iligan Institute of Technology through the Office of the Vice Chancellor for Research and Enterprise (OVCRE). 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