EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS 2025, Vol. 18, Issue 4, Article Number 6851 ISSN 1307-5543 – ejpam.com Published by New York Business Global Strong and Weak Dominating Sets of Graphs under Some Binary Operations Jerra Mae C. Molles1, Ferdinand P. Jamil1,2, Sergio R. Canoy, Jr.1,2,∗ 1 Department of Mathematics and Statistics, College of Science and Mathematics, MSU-Iligan Institute of Technology, Iligan City, Philippines 2 Center of Mathematical and Theoretical Physical Sciences - PRISM, MSU-Iligan Institute of Technology, Iligan City, Philippines Abstract. A set S of vertices of a graph G is a strong (resp. weak) dominating set of G if for every vertex v of G outside of S, there is a vertex u inside of S such that u and v are adjacent and degG(v) ≤ degG(u) (resp. degG(v) ≥ degG(u)). The minimum cardinality of a strong (resp. weak) dominating set is called the strong (resp. weak) domination number of G, and is denoted by γs(G) (resp. γw(G)). In this paper, we characterize the strong and weak dominating sets of graphs under some binary operations. As a result, we also determine the exact values of or sharp bounds for the corresponding strong and weak domination numbers. 2020 Mathematics Subject Classifications: 05C69 Key Words and Phrases: Strong dominating, weak dominating, shadow, join, corona, edge corona, lexicographic product 1. Introduction All throughout this paper, we consider only graphs which are simple, finite and undi- rected. 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. All terminologies used here which are not being defined are adapted from [1]. Let G and H be disjoint graphs. By G ∪ H, we mean the graph with V (G ∪ H) = V (G)∪V (H) and E(G∪H) = E(G)∪E(H). 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). 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)}. ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v18i4.6851 Email addresses: Jerramae.molles@g.msuiit.edu.ph (J. M. Molles), ferdinand.jamil@g.msuiit.edu.ph (F. P. Jamil), sergio.canoy@g.msuiit.edu.ph (S. R. Canoy Jr.) https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0) J. M. Molles, F. P. Jamil, S. R. Canoy / Eur. J. Pure Appl. Math, 18 (4) (2025), 6851 2 of 18 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. 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 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. 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. The degree of v refers to the cardinality |NG(v)| of the open neighborhood of v, ∆(G) is the maximum degree of a vertex of G and δ(G) is the minimum degree of a vertex of G. If |NG(v)| = 1, then v is an endvertex, and, in this case, if u ∈ NG(v), then u is the support vertex of v. The symbols End(G) and Supp(G) denote the set of all endvertices and the set of all support vertices of G, respectively. The closed neighborhood of v is the set NG[v] = NG(v) ∪ {v}. Customarily, for S ⊆ V (G), NG(S) = ∪v∈SNG(v) and NG[S] = ∪v∈SNG[v]. A subset S ⊆ V (G) is a dominating set of G if NG[S] = V (G). In case NG(S) = V (G), then S is a total dominating set of G. The minimum cardinality γ(G) of a dominating set of G is the domination number of G, and the minimum cardinality γt(G) of a total dominating set is the total domination number of G. A dominating set of cardinality γ(G) is called a γ-set of G. Similarly, a γt-set is a total dominating set of cardinality γt(G). The reader is referred to [2–7] for the history, fundamental concepts and recent developments of domination in graphs as well as its various applications. For two vertices u, v ∈ V (G), v is said to strongly dominate u in G if uv ∈ E(G) and degG(v) ≥ degG(u). In this case, we also say that u weakly dominates v. We write v ≽G u or u ≼G v to mean that v strongly dominates u or, equivalently, u weakly dominates v. A subset S ⊆ V (G) is said to strongly dominate (resp. weakly dominate) u ∈ V (G) \ S in G if there exists v ∈ S for which v ≽G u (resp. u ≽G v) in G. In this case we write u ≼G S (resp. u ≽G S). For S,D ⊆ V (G), S is said to strongly dominate (resp. weakly dominate) D if S strongly dominates (resp. weakly dominates) every vertex v ∈ D \ S. We say S is a strong dominating set (resp. weak dominating set) of G if S strongly dominates (resp. weakly dominates) V (G), i.e., for each u ∈ V (G) \ S, u ≼G S (resp. u ≽G S) in G. The minimum cardinality of a strong dominating set (resp. weak dominating set) of G is the strong domination number (resp. weak domination number) of G, which is denoted by γs(G) (resp. γw(G)). Any strong dominating set (resp. weak dominating set) of G of cardinality γs(G) (resp. γw(G)) is called a γs-set (resp. γw-set) of G. The concepts of strong and weak domination were first introduced by E. Sampathku- mar and L. Pushpa Latha [8] in 1996. Thereafter, several further studies have been done on these two concepts (see [9, 10], [11]- [12], [13, 14], [15]-[16]). In particular, properties and characteristics of strong and weak dominating sets are explored in [12, 17]. Bounds on J. M. Molles, F. P. Jamil, S. R. Canoy / Eur. J. Pure Appl. Math, 18 (4) (2025), 6851 3 of 18 γs(G) and γw(G) are studied in [18–21], and investigation of strong and weak domination in families of graphs are done in [22–25]. In this present study, we continue the investigation of these two concepts, particularly on characterizing the strong and weak dominating sets in families of graphs involving the complementary prism of graphs, join, corona, edge corona and lexicographic product of graphs. For v ∈ V (G), write NG(v ≽) = {u ∈ V (G) : u ≼ v}, NG(v ≼) = {u ∈ V (G) : u ≽ v}, NG[v ≽] = NG(v ≽) ∪ {v} and NG[v ≼] = NG(v ≼) ∪ {v}. For S ⊆ V (G), NG(S ≽) = ∪v∈SNG(v ≽) and NG(S ≼) = ∪v∈SNG(v ≼). We also write NG[S ≽] = NG(S ≽)∪ S and NG[S ≼] = NG(S ≼)∪ S. Hence, S is a strong (resp. weak) dominating set of G if and only if NG[S ≽] = V (G) (resp. NG[S ≼] = V (G)). The symbol Γs(G) (resp. Γw(G)) denotes the family of all strong (resp. weak) dominat- ing sets of G. Thus, γs(G) = min{|S| : S ∈ Γs(G)} and γw(G) = min{|S| : S ∈ Γw(G)}. 2. Preliminary results It is worth noting that strong and weak dominating sets are necessarily dominating sets. Hence, for a graph of order n, γ(G) ≤ γs(G) ≤ n−∆(G) and γ(G) ≤ γw(G) ≤ n− δ(G) [26]. The following are immediate observations. Remark 1. Let G be any graph of order n. Then (i) γs(G) = 1 if and only if γ(G) = 1; (ii) γw(G) = 1 if and only if G = Kn; (iii) γs(G) = n (resp. γw(G) = n) if and only if G = Kn; (iv) γs(G) = n− 1 if and only if G = K2 or G = K2 ∪Kn−2; and (v) γw(G) = n− 1 if and only if G = K1,n−1 or G = K1,(n−k−1) ∪Kk. Remark 2. Let G be a connected graph. Then γs(G) = 2 (resp. γw(G) = 2) if and only if γ(G) = 2 and G has a γ-set S = {u, v} for which x ≼ u (resp. x ≽ u) or x ≼ v (resp. x ≽ v) for each x ∈ V (G) \ S. J. M. Molles, F. P. Jamil, S. R. Canoy / Eur. J. Pure Appl. Math, 18 (4) (2025), 6851 4 of 18 Lemma 1. If G is connected of order n ≥ 3, then for each S ∈ Γs(G) there exists S∗ ∈ Γs(G) for which S∗ ∩ End(G) = ∅ and |S∗| ≤ |S|. Consequently, G has a γs-set S for which S ∩ End(G) = ∅. Proof. Let S ∈ Γs(G). If S∩End(G) = ∅, then let S∗ = S. Suppose that S∩End(G) ̸= ∅. For each v ∈ S ∩ End(G), let xv be the support vertex v. Then |{xv /∈ S : v ∈ S ∩ End(G)}| ≤ |S ∩ End(G)|. Put S∗ = (S \ End(G)) ∪ {xv /∈ S : v ∈ S ∩ End(G)}. Then S∗ ∈ Γs(G) and |S∗| = |S \ End(G)|+ |{xv /∈ S : v ∈ S ∩ End(G)}| ≤ |S|. Lemma 2. Let G be a connected graph. Then for each S ∈ Γs(G) (resp. S ∈ Γw(G)), S contains a vertex v for which degG(v) = ∆(G) (resp. degG(v) = δ(G)). Proof. Let S ⊆ V (G) be a strong dominating set of G. Let v ∈ V (G) for which degG(v) = ∆(G). If v ∈ S, then we are done. Suppose v /∈ S. Since S ∈ Γs(G), there exists u ∈ S for which v ≼ u. Necessarily, degG(u) = ∆(G). Parallel arguments will prove the case of the weak domination. Remark 3. [27] (i) For a cycle Cn, γs(Cn) = γw(Cn) = ⌈n3 ⌉. (ii) For a path Pn, γs(Pn) = ⌈n 3 ⌉ and γw(Pn) = { ⌈n3 ⌉, if n = 1(mod 3), 1 + ⌈n3 ⌉, else. In what follows, for the purpose of emphasis, we write x ≼G y to mean x ≼ y in G. 3. In the complementary prism of graphs Proposition 1. Let G be any graph. Then (i) γs(GG) = 1 (resp. γw(GG) = 1) if and only if G = K1; (ii) γs(GG) = 2 if and only if exactly one of the following holds: (a) G has an isolated vertex x for which γ(G− x) = 1. (b) G has an isolated vertex x for which γ(G− x) = 1. (iii) γw(GG) = 2 if and only if G ∈ {K2,K2}. J. M. Molles, F. P. Jamil, S. R. Canoy / Eur. J. Pure Appl. Math, 18 (4) (2025), 6851 5 of 18 Proof. Statement (i) immediately follows from Observation 1(i). We prove (ii). Sup- pose that γs(GG) = 2. By (i), G ̸= K1. By Observation 2, there exists a γ-set S = {u, v} of GG for which x ≼GG u or x ≼GG v for each x ∈ V (GG) \ S. If u, v ∈ V (G) (resp. u, v ∈ V (G)), then G = K2 (resp. G = K2) and (b) holds (resp. (a) holds). Assume WLOG that u ∈ V (G) and v ∈ V (G). Suppose further that uv ∈ E(GG) and G ̸= K2. Necessarily, u ≽G V (G) \ {u} and v ≽G V (G) \ {u, v} ̸= ∅. The former implies that NG[u] = V (G) and consequently, u is a (unique) isolated vertex of G. The latter implies that NG−u[v] = V (G−u). This shows that (b) holds. Similarly, if uv ∈ E(G) and G ̸= K2, then u is an isolated vertex of G and γ(G− u) = 1, showing that (a) holds. Conversely, suppose that (a) holds for G. Let x be an isolated vertex of G and let z ∈ V (G − x) such that NG−x[z] = V (G − x). Clearly, G ̸= K1 so that γs(GG) ≥ 2. Put S = {x, z}. Since NG[x] = V (G), x ≽GG V (G) ∪ {x} \ {x}. On the other hand, z ≽G V (G) \ {x, z}. By Observation 2, S ∈ Γs(GG). Thus, γs(GG) ≤ 2. Similarly, if (b) holds, then γs(GG) = 2. This proves (ii). Now suppose that γw(GG) = 2, and let S = {u, v} be a γ-set of GG such that u ≼GG x or v ≼GG x for each x ∈ V (GG) \ S. Suppose that u ∈ V (G) and v ∈ V (G). Assume uv ∈ E(G). Since u ≼GG u, there exists w ∈ V (G) such that u w ∈ E(G). Because uw /∈ E(G), wv ∈ E(GG), which is impossible. Thus, u, v ∈ V (G) or u, v ∈ V (G). This implies that G = K2 or G = K2. The converse is easy. This proves (iii). Remark 4. Let G be any graph. Then S ⊆ V (GG) is a strong dominating set of GG if and only if S = SG ∪SG with SG ⊆ V (G) and SG such that for each x ∈ V (G) \SG (resp. x ∈ V (G) \ SG), x ≼G SG (resp. x ≼G SG) or x ∈ SG (resp. x ∈ SG) and x ≼GG x. Let G be of order n. Put SG = {x ∈ V (G) : x ≼GG x} and SG = {x : x ∈ V (G) \ SG}. By Observation 4, S = SG ∪ SG ∈ Γs(GG), showing γs(GG) ≤ |S| = n. If γ(G) = 1 or G has an isolated vertex, this bound coincides with the bound given in Equation (1) for γs(GG). Replacing “≼GG” with “≽GG” in the definition of SG will show that γw(GG) ≤ n. In particular, if G ∈ {Kn,Kn}, then γs(GG) = n = γw(GG). Proposition 2. (i) For all n ≥ 3, γs(PnPn) = { 1 + ⌈n3 ⌉, if 3 ≤ n ≤ 5; 2 + ⌈n−2 3 ⌉, if n ≥ 6, and γw(PnPn) =  3, if n = 3, 4, if 4 ≤ n ≤ 6; 2 + ⌈n3 ⌉, if n ≥ 7, (ii) For n ≥ 4, γs(CnCn) = γw(CnCn) = { 3, if n = 5; 2 + ⌈n3 ⌉, else. J. M. Molles, F. P. Jamil, S. R. Canoy / Eur. J. Pure Appl. Math, 18 (4) (2025), 6851 6 of 18 Proof. For (i), put G = Pn = [x1, x2, . . . , xn] and for any S ⊆ V (GG), define SG = S∩V (G) and SG = S∩V (G). The value of γs(GG) can readily be checked when 3 ≤ n ≤ 5. Let n ≥ 6. Let A ⊆ {x2, x4, . . . , xn−1} be a γ-set of the path P = [x2, x4, . . . , xn−1]. Then by Observation 4, S = A ∪ {x1, xn} ∈ Γs(GG). Hence, γs(GG) ≤ |S| = 2 + γ(Pn−2) = 2 + ⌈n−2 3 ⌉. To get the other inequality, let S ⊆ V (GG) be a γs-set of GG. Since degG(x) < degG(x) for all x ∈ V (G), Observation 4 implies that SG ∈ Γs(G). In view of Lemma 2, we may assume x1 ∈ SG. Since V (G) \ {x2} ≼G x1, |SG| ≥ 2. In case x2 ∈ SG, choose x = x2. Otherwise, choose x ∈ V (G) such that x ∈ SG \ {x1} and x2 ≼ x. If x /∈ SG, then S∗ = S \ {x, x1} strongly dominates V (G) \ {x, x1}. In this case, |S| ≥ 2 + γs(Pn−2). If x ∈ SG, then S \ {x, x1} strongly dominates V (G) \ {x1} so that |S| ≥ 2 + γs(Pn−1) ≥ 2 + γs(Pn−2). In any case, γs(GG) ≥ 2 + γs(Pn−2) = 2 + ⌈n−2 3 ⌉. For γw(GG), the case where 3 ≤ n ≤ 6 can be readily verified. Let n ≥ 7. Put S = {x1, x2, x3, x6, . . . , x3k} whenever n = 3k; otherwise write S = {x1, x2, x3, x6, . . . , x3⌊n 3 ⌋, xn}. By Observation 4, S is a weak dominating set of GG. Thus, γw(GG) ≤ |S| = 2 + ⌈n3 ⌉. Now, let S ⊆ V (GG) be a γw-set of GG. Because degG(x) < degG(x) for all x ∈ V (G), SG ∈ Γw(G). Hence, γw(GG) = |S| ≥ |SG| + γw(G). Moreover, since n ≥ 7, |S| ≤ 2 + ⌈n3 ⌉ ≤ n − 2, and consequently, SG ̸= ∅. If n = 1(mod 3), then |SG| ≥ 2. In view of Observation 3, in any case, γw(GG) = |S| ≥ 2 + ⌈n3 ⌉. For (ii), the case where 4 ≤ n ≤ 5 can be readily verified. Similar arguments used in the proof of statement (i) will prove the case where n ≥ 6. 4. In the join of graphs Remark 5. Let G and H be connected graphs of orders m and n, respectively. Then (i) For u, v ∈ V (G), u ≼G+H v if and only if u ≼G v. (ii) For v ∈ V (G) and u ∈ V (H), u ≼G+H v if and only if degG(v) + n ≥ degH(u) +m. Theorem 1. Let G,H be connected graphs of orders m and n, respectively. Let S ⊆ V (G+H). Then S ∈ Γs(G+H) if and only if one of the following holds: (i) S ⊆ V (G) such that S ∈ Γs(G) and S contains a vertex v for which degG(v) ≥ ∆(H) +m− n. (ii) S ⊆ V (H) such that S ∈ Γs(H) and S contains a vertex v for which degH(v) ≥ ∆(G) + n−m. (iii) SG = S ∩ V (G) ̸= ∅ and SH = S ∩ V (H) ̸= ∅ and one of the following holds for each u ∈ V (G+H) \ S : (a) u ∈ V (G) and u ≼G SG or there exists v ∈ SH for which degH(v) ≥ degG(u) + n−m; J. M. Molles, F. P. Jamil, S. R. Canoy / Eur. J. Pure Appl. Math, 18 (4) (2025), 6851 7 of 18 (b) u ∈ V (H) and u ≼H SH or there exists v ∈ SG for which degG(v) ≥ degH(u)+ m− n. Proof. Assume that S ∈ Γs(G +H). Suppose that S ⊆ V (G). By Observation 5(i), S ∈ Γs(G). Pick u ∈ V (H) for which degH(u) = ∆(H). Since S ∈ Γs(G + H) there exists v ∈ S for which u ≼G+H v. By Observation 5(ii), degG(v) + n ≥ ∆(H) + m or, equivalently, degG(v) ≥ ∆(H) + m − n. Similarly, if S ⊆ V (H), then (ii) holds. Next, suppose that S intersects both V (G) and V (H). Let u ∈ V (G+H) \S, and suppose that u ∈ V (G). Then there exists v ∈ S for which u ≼G+H v. If v ∈ V (G), then u ≼G v. This means that u ≼G SG. If v ∈ V (H), then by Observation 5(ii), degH(v) ≥ degG(u)+n−m, and (a) holds. Similarly, if u ∈ V (H), then (b) holds. Conversely, suppose that (i) holds for S. Let u ∈ V (G + H) \ S. If u ∈ V (G), then u ≼G S, and hence u ≼G+H S. Suppose that u ∈ V (H). There exists v ∈ S for which degG(v) ≥ ∆(H) +m− n. This means degG+H(u) = degH(u) +m ≤ ∆(H) +m ≤ degG(v) + n = degG+H(v). Thus, u ≼G+H v, and consequently, u ≼G+H S. Accordingly, S ∈ Γs(G+H). Similarly, if (ii) holds, then S ∈ Γs(G+H). Finally, suppose that (iii) holds for S. Let u ∈ V (G) \S. If u ≼G SG, then u ≼G+H S. Suppose that SG does not strongly dominate u. By condition (a), there exists v ∈ SH for which degH(v)+m ≥ degG(u)+n. This means that degG+H(v) ≥ degG+H(u) and u ≼G+H v. Thus, u ≼G+H S. Similarly, if u ∈ V (H) \ S and SH does not strongly dominate u, then there exists v ∈ SG for which u ≼G+H v, and therefore u ≼G+H S. Therefore, S ∈ Γs(G+H). Corollary 1. Let G and H be connected graphs of orders m and n, respectively. (i) γs(G+H) = 1 if and only if γ(G) = 1 or γ(H) = 1. (ii) Assume γ(G) ≥ 2 and γ(H) ≥ 2. (a) If ∆(G) + n = ∆(H) +m, then γs(G+H) = 2. (b) If ∆(G) + n > ∆(H) +m, then γs(G+H) = min{γs(G), 1 + γs(K)}, where K = ⟨V (G) \NG+H [u ≽]⟩ and u ∈ V (H) for which degH(u) = ∆(H). Proof. Statement (i) follows immediately from Theorem 1(i). Assume that γ(G) ≥ 2 and γ(H) ≥ 2. Suppose that ∆(G) + n = ∆(H) +m. Pick u ∈ V (G) and v ∈ V (H) such that degG(u) = ∆(G) and degH(v) = ∆(H). Since S = {u, v} satisfies condition (iii) of Theorem 1, S ∈ Γs(G+H). In this case, γs(G+H) = |S| = 2 and (ii)(a) holds. To prove (ii)(b), suppose that ∆(G) + n > ∆(H) + m. First, let S ⊆ V (G) be a γs-set of G. By J. M. Molles, F. P. Jamil, S. R. Canoy / Eur. J. Pure Appl. Math, 18 (4) (2025), 6851 8 of 18 Lemma 2, there exists v ∈ S such that degG(v) = ∆(G). Then degG(v) + n > ∆(H) +m. By Theorem 1, S ∈ Γs(G+H). Thus, γs(G+H) ≤ |S| = γs(G). Next, pick u ∈ V (H) for which degH(u) = ∆(H). Since ∆(G) + n > ∆(H) +m, V (G) \ NG+H [u ≽] ̸= ∅. Let K = ⟨V (G) \ NG+H [u ≽]⟩. Choose a γs-set S∗ of K. Put S = S∗ ∪ {u}. Since S satisfies Theorem 1(iii), S ∈ Γs(G+H). Thus, γs(G+H) ≤ |S| = 1 + γs(K). Therefore, γs(G + H) ≤ min{γs(G), 1 + γs(K)}. Now, let S ⊆ V (G + H) be a γs-set of G+H. By Lemma 2, S ⊈ V (H). If S ⊆ V (G), then by Theorem 1, S ∈ Γs(G), showing γs(G) ≤ |S| = γs(G +H). Suppose that SG = S ∩ V (G) ̸= ∅ and SH = S ∩ V (H) ̸= ∅. Since ∆(G)+n > ∆(H)+m, V (G) \NG+H [SH ≽] ̸= ∅. Put K = ⟨V (G) \NG+H [SH ≽]⟩. We claim that SG is a strong dominating set of K. Let u ∈ V (K)\SG. Since u ∈ V (G)\S, there exist v ∈ S for which uv ∈ E(G + K) and u ≼G+K v. Because u is not strongly dominated by SH in G +H, v ∈ SG. Since u is arbitrary, SG is a strong dominating set of K. Thus, γs(K) ≤ |SG|. Since |SH | ≥ 1, γs(G +H) = |SG| + |SH | ≥ 1 + γs(K). The above results imply that γs(G+H) = min{γs(G), 1 + γs(K)}. Verify that γs(P4 + P6) = γs(P4) = 2. On the other hand, if G is as in Figure 1, then γs(G + P8) = 1 + γs(K) = 2, where K = ⟨V (G) \ NG+P8 [u ≽]⟩ = ⟨{v}⟩ and u ∈ V (P8) with degP8(u) = 2. .................................................................................................. .................................................................................................. .................................................................................................. .................................................................................................. .................................................................................................. .................................... .................................... .................................... .................................... .................................... ............ ........... ........... ....... ......................................... ......................................... ............ ........... ........... ....... G : v • Figure 1: Graph G The following versions for weak domination follow similar proofs. Theorem 2. Let G,H be connected graphs of orders m and n, respectively. Let S ⊆ V (G+H). Then S ∈ Γw(G+H) if and only if one of the following holds: (i) S ⊆ V (G) such that S ∈ Γw(G) and S contains a vertex v for which degG(v) ≤ δ(H) +m− n. (ii) S ⊆ V (H) such that S ∈ Γw(H) and S contains a vertex v for which degH(v) ≤ δ(G) + n−m. (iii) SG = S ∩ V (G) ̸= ∅ and SH = S ∩ V (H) ̸= ∅ and one of the following holds for each u ∈ V (G+H) \ S : (a) u ∈ V (G) and u ≽G SG or there exists v ∈ SH for which degH(v) ≤ degG(u) + n−m; J. M. Molles, F. P. Jamil, S. R. Canoy / Eur. J. Pure Appl. Math, 18 (4) (2025), 6851 9 of 18 (b) u ∈ V (H) and u ≽H SH or there exists v ∈ SG for which degG(v) ≤ degH(u)+ m− n. Corollary 2. Let G and H be connected graphs of orders m and n, respectively. Assume γ(G) ≥ 2 and γ(H) ≥ 2. Then the following holds: (a) If δ(G) + n = δ(H) +m, then γw(G+H) = 2. (b) If δ(G) + n < δ(H) +m, then γw(G+H) = min{γw(G), 1 + γw(K)}, where K = ⟨V (G) \NG+H [u ≼]⟩ and u ∈ V (H) for which degH(u) = δ(H). Example 1. For m,n ≥ 4, (1) γs(Pm + Pn) = { 2, if m = n; ⌈m3 ⌉, if m < n, (2) γw(Pm + Pn) =  2, if m = n; 3, if m = n+ 1; ⌈m3 ⌉, if m ≥ n+ 2;m ≡ 1 (mod 3); ⌈m3 ⌉+ 1, if m ≥ n+ 2;m ≡ 0, 2 (mod 3), (3) γs(Cm + Cn) = { 2, if m = n; ⌈m3 ⌉, if m < n, γw(Cm + Cn) = { 2, if m = n; ⌈n3 ⌉, if m < n; 5. In the corona of graphs Proposition 3. Let G be a nontrivial connected graph and H any graph, and let S ⊆ V (G ◦H). Then S ∈ Γs(G ◦H) if and only if S = A ∪ ( ∪v∈V (G)Sv ) , (1) where A ⊆ V (G) and Sv ⊆ V (Hv) such that the following hold: (i) A ∈ Γs(G); and (ii) For each v ∈ V (G) \A, Sv ∈ Γs(H v). Proof. Assume that S ∈ Γs(G ◦H). Put A = S ∩ V (G) and Sv = A ∩ V (Hv) for all v ∈ V (G). Then Equation 1 holds. To prove (i), let v ∈ V (G) \A. There exists u ∈ S for which v ≼G◦H u. Since degG◦H(v) > degG◦H(w) for all w ∈ V (Hv), u /∈ Sv. Hence, u ∈ A. This means that v ≼G A. Thus, A ∈ Γs(G), and (i) holds. To prove (ii), let v ∈ V (G) \A J. M. Molles, F. P. Jamil, S. R. Canoy / Eur. J. Pure Appl. Math, 18 (4) (2025), 6851 10 of 18 and let u ∈ V (Hv) \ Sv. There exists w ∈ S for which u ≼G◦H w. Since w ̸= v, w ∈ Sv. Thus, u ≼Hv Sv. Therefore, Sv ∈ Γs(H v), and (ii) holds. Conversely, suppose that Equation 1 holds for S together with conditions (i) and (ii). Let w ∈ V (G ◦ H) \ S, and let v ∈ V (G) for which w ∈ V (Hv + v). If w ∈ V (Hv) and v ∈ A, then w ≼G◦H S. On the other hand, if w ∈ V (Hv) and v /∈ A, then w ≼Hv Sv by (ii) so that w ≼G◦H S. Suppose that w = v. By (i), w ≼G A and, consequently, w ≼G◦H A. Thus, w ≼G◦H S. Accordingly, S ∈ Γs(G ◦H). Proposition 4. Let G be a nontrivial connected graph and H any graph, and let S ⊆ V (G ◦H). Then S ∈ Γw(G ◦H) if and only if S = A ∪ ( ∪v∈V (G)Sv ) , where A ⊆ V (G) and Sv ∈ Γw(H v) for each v ∈ V (G). Proof. Suppose that S ∈ Γw(G◦H). Then S = A∪ ( ∪v∈V (G)Sv ) , where A = S ∩V (G) and Sv = S ∩ V (Hv) for each v ∈ V (G). Let v ∈ V (G), and let x ∈ V (Hv) \ Sv. There exists y ∈ S for which x ≽G◦H y. Since degG◦H(x) < degG◦H(v), y ∈ Sv and x ≽Hv y. Thus, Sv ∈ Γw(H v). Conversely, suppose that S = A ∪ ( ∪v∈V (G)Sv ) , where A ⊆ V (G) and Sv ∈ Γw(H v) for each v ∈ V (G). Let x ∈ V (G ◦ H) \ S, and let v ∈ V (G) such that x ∈ V (Hv + v). Note that Sv ̸= ∅. If x = v, then pick any w ∈ Sv. Then x ≽G◦H w. If x ̸= v, then x ∈ V (Hv) \ Sv, and there exists y ∈ Sv such that x ≽Hv y. This means x ≽G◦H y. Therefore, S ∈ Γw(G ◦H). Corollary 3. If G is a connected graph of order n ≥ 2. Then (i) [25] γs(G ◦H) = n for any graph H. (ii) γw(G ◦H) = nγw(H) for any graph H. Proof. By Proposition 3, V (G) is a strong dominating set of G◦H. Thus, γs(G◦H) ≤ n. Now, let S ⊆ V (G ◦ H) be a strong dominating set of G ◦ H, and let A ⊆ V (G) and Sv ⊆ V (Hv) be as provided in Proposition 3 such that S = A ∪ ( ∪v∈V (G)Sv ) . By Proposition 3(ii), |Sv| ≥ 1 for all v ∈ V (G) \ A. Thus, |S| ≥ |A|+ |V (G) \ A| = n. Since S is arbitrary, γs(G ◦H) ≥ n. Statement (ii) is easy. 5.1. In the edge corona of graphs Given graphs G and H, we write Huv to denote that copy of H that is being joined with the endvertices of the edge uv ∈ E(G) in the edge corona G ⋄ H. For uv ∈ E(G), we write Huv + uv = Huv + ⟨{u, v}⟩. If x ∈ V (H), then we write xuv to denote the corresponding vertex in Huv. J. M. Molles, F. P. Jamil, S. R. Canoy / Eur. J. Pure Appl. Math, 18 (4) (2025), 6851 11 of 18 Proposition 5. Let G and H be a nontrivial connected graphs with δ(G) ≥ 2 or γ(H) ≥ 2, and let S ⊆ V (G ⋄H). Then S ∈ Γs(G ⋄H) if and only if S = A ∪ ( ∪uv∈E(G)Suv ) , (2) where A ⊆ V (G) and Suv ⊆ V (Huv satisfying the following: (i) A ∈ Γs(G); and (ii) For each uv ∈ E(G) for which {u, v} ∩A = ∅, Suv ∈ Γs(H uv). Proof. Assume that S is strong dominating set of G ◦ H. Put A = S ∩ V (G) and Suv = S ∩ V (Huv) for each uv ∈ E(G). Then Equation 2 holds. Let v ∈ V (G) \A. There exists u ∈ S for which v ≼G⋄H u. If δ(G) ≥ 2 or γ(H) ≥ 2, then degG⋄H(w) < degG⋄H(v) for all w ∈ V (Hxv), for all x ∈ NG(v). Thus, u ∈ A. Hence, u ≼G A and A ∈ Γs(G). This proves (i). Now, let uv ∈ E(G) with u /∈ A and v /∈ A, and let w ∈ V (Huv) \ Suv. There exists z ∈ S for which w ≼G⋄H z. Clearly, z ∈ Suv so that w ≼ Suv. Therefore, Suv is a strong dominating set of Huv. Conversely, assume that (i) and (ii) hold for S. Let v ∈ V (G ⋄H) \ S. There exists xy ∈ E(G) such that v ∈ V (Hxy + xy). First, suppose that v ∈ V (Hxy). If x ∈ A, then v ≼ x. Similarly, if y ∈ A, then v ≼ y. In any case, v ≼ S. Suppose that x, y /∈ A. By (ii), v ≼ Sxy. Thus v ≼ S. Next, suppose that v = x or v = y. By (i), there exists w ∈ A for which v ≼ w. Therefore, v ≼ S. Accordingly, S is a strong dominating set of G ⋄H. The set A in Proposition 5 need not be a strong dominating set of G whenever δ(G) = 1 = γ(H). Consider the edge corona P5 ⋄ P3 in Figure 2. The set S of darkened vertices is a strong dominating set of P5 ⋄ P3. However A = S ∩ V (P5) is not a strong dominating set of P5. ................................................................................................................................................................................................................................................. ................................................................................................................................................................................................................................................. ................................................................................................................................................................................................................................................. 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...................................................................................................................................................................................................... ......................................................................................................................................................................... ...................................................................................................................................................... • • • Figure 2: Graph P5 ⋄ P3 Proposition 6. Let G and H be a nontrivial connected graphs where γ(H) ̸= 1, and let S ⊆ V (G ⋄H). Then S ∈ Γw(G ⋄H) if and only if S = A ∪ ( ∪uv∈E(G)Suv ) , where A ⊆ V (G) and Suv ∈ Γw(H uv) for each uv ∈ E(G). Proof. Assume that S ∈ Γw(G ⋄ H). Let A = S ∩ V (G) and Suv = S ∩ V (Huv) for each uv ∈ E(G). Then S = A∪ ( ∪uv∈E(G)Suv ) . Let uv ∈ E(G) and let x ∈ V (Huv) \Suv. J. M. Molles, F. P. Jamil, S. R. Canoy / Eur. J. Pure Appl. Math, 18 (4) (2025), 6851 12 of 18 There exists y ∈ S such that x ≽G⋄H y. Since min{degG⋄H(u), degG⋄H(v)} > degG⋄H(x), y /∈ {u, v}. Thus, y ∈ Suv and y ≽Huv Suv. This shows that Suv ∈ Γw(H uv). Conversely, Let x ∈ V (G ⋄H) \ S. Let uv ∈ E(G) such that x ∈ V (Huv + uv). Since Suv ∈ Γw(H uv), Suv ̸= ∅. If x ∈ {u, v}, then x ≽G⋄H w for each w ∈ Suv. Also, if x ∈ V (Huv) \ Suv, then there exists w ∈ Suv for which x ≽Huv w. This means x ≽G⋄H w. Therefore, S ∈ Γw(G ⋄H). For a nonempty A ⊆ V (G), define Ae = {uv ∈ E(G) : u /∈ A and v /∈ A}. Corollary 4. Let G and H be connected graphs where G is nontrivial. Then (i) γs(G ⋄H) = min{|A|+ |Ae| : A ∈ Γs(G)}. (ii) γw(G ⋄H) = |E(G)|γw(H). Proof. To prove (i), put α = min{|A|+ |Ae| : A ∈ Γs(G)}. Let n = |V (G)|. If n = 2, then γs(G ⋄H) = 1 = α. Assume that n ≥ 3. We consider the following cases: Case 1: Suppose that δ(G) ≥ 2 or γ(H) ≥ 2. Let A ∈ Γs(G). For each uv ∈ Ae, denote by wuv exactly one of u and v. Then wuv strongly dominates V (Huv). Define A∗ = A ∪ {wuv : uv ∈ Ae}. Then A∗ ∈ Γs(G) with (A∗)e = ∅ and |{wuv : uv ∈ Ae}| ≤ |Ae|. By Proposition 5, A∗ ∈ Γs(G ⋄ H) so that γs(G ⋄H) ≤ |A∗| ≤ |A|+ |Ae|. Since A is arbitrary, γs(G ⋄H) ≤ α. To get the other inequality, let S ⊆ V (G ⋄H) be a γs-set of G ⋄H. By Proposition 5, there exists A ∈ Γs(G) such that S = A ∪ ( ∪uv∈E(G)Suv ) , where Suv ∈ Γs(H uv) for each uv ∈ Ae. We have γs(G ⋄H) = |S| ≥ |A|+ ∑ uv∈Ae |Suv| ≥ |A|+ |Ae|γs(H) ≥ α. Case 2: Suppose that δ(G) = 1 = γ(H). For each uv ∈ E(G), let xuv ∈ V (Huv) such that NHuv [xuv] = V (Huv). It is worth noting that if u ∈ End(G) or v ∈ End(G), say u ∈ End(G), then degG⋄H(xuv) = degG⋄H(u). Also, degG⋄H(xuv) < degG⋄H(y) for all y ∈ {u, v} \ End(G) and that {xuv} ∈ Γs(H uv). Choose A ∈ Γs(G) such that |A|+ |Ae| = α. Construct an A∗ ∈ Γs(G) as in the proof of Lemma 1 such that |A∗| ≤ |A| and A∗∩End(G) = ∅. More precisely, A∗ = (A\End(G))∪ {xv /∈ A : v ∈ A ∩ End(G)}. Define S∗ = A∗ ∪ (∪uv∈Ae{xuv}). Then S∗ ∈ Γs(G ⋄ H). Thus, γs(G ⋄H) ≤ |S∗| ≤ |A|+ |Ae| = α. J. M. Molles, F. P. Jamil, S. R. Canoy / Eur. J. Pure Appl. Math, 18 (4) (2025), 6851 13 of 18 To get the other inequality, let S ⊆ V (G ⋄H) be a γs-set of G ⋄H. Put A = S ∩V (G) and Suv = S ∩ V (Huv) for all uv ∈ E(G). Let Ae = {uv ∈ E(G) : u, v /∈ A}. If uv ∈ Ae, then Suv is a γs-set of H uv. Hence, Suv = {xuv} where xuv ∈ V (Huv) such that NHuv [xuv] = V (Huv). Thus, γs(G ⋄H) = |S| = |A|+ ∑ uv∈Ae |Suv|+ ∑ uv∈E(G)\Ae |Suv| = |A|+ |Ae|+ ∑ uv∈E(G)\Ae |Suv| ≥ |A|+ |Ae| = α. To prove (ii), note that if γ(H) ̸= 1, then it follows from Proposition 6 that S ⊆ V (G ⋄ H) is a γw-set of G ⋄ H if and only if S = ∪uv∈E(GSuv, where Suv is a γw-set of Huv for each uv ∈ E(G). In this case, γw(G ⋄ H) = |E(G)|γw(H). Now suppose that γ(H) = 1. If H is complete, then γw(G ⋄H) = |E(G)| = |E(G)|γw(H). Assume H is not complete. In view of Lemma 2, S contains xuv for which δ(Huv) = degHuv(xuv) for each uv ∈ E(G). Since S is a γw-set and u ≽G⋄H xuv and v ≽G⋄H xuv, {u, v} ∩ Suv = ∅ for all uv ∈ E(G). Consequently, S = ∪uv∈E(G)Suv where Suv is a γs-set of H uv. Therefore, γs(G ⋄H) = |S| = |E(G)|γw(H). The value of γs(G ⋄H) in Corollary 4 is not necessarily determined by a γs-set A of G. Observe that γs(C6 ⋄ H) = 3 for any connected graph H, and is not determined by any γs-set of C6. Example 2. Let G be any graph. For positive integers n ≥ 2 and m ≥ 3, (i) [25] γs(Pn ⋄G) = ⌊n2 ⌋ and γs(Cm ⋄G) = ⌈m2 ⌉; (ii) γw(Pn ⋄G) = (n+ 1)γs(G) and γw(Cm ⋄G) = mγw(G). 6. In the lexicographic product of graphs Here we note that for (u, v) ∈ V (G[H]), degG[H]((u, v)) = |V (H)|degG(u) + degH(v). For S ⊆ V (G[H]), the projection of S with respect to G refers to the set SG = {x ∈ V (G) : ∃y ∈ V (H) for which (x, y) ∈ S}. If S1 ∈ Γs(G) (resp. S1 ∈ Γw(G)) and S2 ∈ Γs(H) (resp. S2 ∈ Γw(H)), then S1 × S2 ∈ Γs(G[H]) (resp. S1 × S2 ∈ Γw(G[H])). Consequently, γs(G[H]) ≤ γs(G)γs(H) (resp. γw(G[H]) ≥ γw(G)γw(H)). J. M. Molles, F. P. Jamil, S. R. Canoy / Eur. J. Pure Appl. Math, 18 (4) (2025), 6851 14 of 18 Proposition 7. Let G and H be nontrivial connected graphs, and let S ∈ Γs(G[H]). Then S = ∪x∈A ({x} × Tx), where A ⊆ V (G) and Tx ⊆ V (H) satisfying the following. (i) A ∈ Γs(G; and (ii) For each x ∈ A \NG(A ≽), Tx ∈ Γs(H). Proof. Put A = SG, the projection of S under G. For each x ∈ A, define Tx = {y ∈ V (H) : (x, y) ∈ S}. Then S = ∪x∈A ({x} × Tx). Let x ∈ V (G) \ A, and pick z ∈ V (H) such that degH(z) = ∆(H). Since (x, z) /∈ S, there exists (u, v) ∈ S for which (x, z) ≼G[H] (u, v). Since x ̸= u, u ∈ A ∩ NG(x). Moreover, since (x, z) ≼G[H] (u, v) and degH(z) ≥ degH(v), degG(x) ≤ degG(u), i.e., x ≼G u. Thus, A ∈ Γs(G), and (i) holds. To show (ii), put n = |V (H)| and let x ∈ A \ NG(A ≽). We claim that Tx ∈ Γs(H). To this end, let y ∈ V (H) \ Tx. Since (x, y) /∈ S, there exists (u, v) ∈ S for which (x, y) ≼G[H] (u, v). If x ̸= u, then since x /∈ NG(A ≽), degG(u) < degG(x). Thus, n ≤ n[degG(x)− degG(u)] ≤ degH(v)− degH(y), which is impossible. Thus, x = u so that v ∈ Tx ∩NH(y), and necessarily, y ≼H v. Accordingly, Tx ∈ Γs(H). Proposition 8. Let G and H be connected nontrivial graphs, and S = ∪x∈A ({x} × Tx), where A ⊆ V (G) and Tx = {y ∈ V (H) : (x, y) ∈ S}. Suppose that each of the following holds for A: (i) A ∈ Γs(G); (ii) For each x ∈ A ∩NG(A ≽), there exists y ∈ Tx such that degH(y) = ∆(H); and (iii) For each x ∈ A \NG(A ≽), Tx ∈ Γs(H). Then S ∈ Γs(G[H]). Proof. Let (x, y) ∈ V (G[H]) \ S. We consider the following cases: Case 1: x /∈ A If x /∈ A, then by (i), there exists u ∈ A such that x ≼G u. If u /∈ NG(A ≽), then by (iii), Tu ∈ Γs(H) so that, by Lemma 2, Tu contains a vertex w for which degH(w) = ∆(H). Here we have (u,w) ∈ S ∩ NG[H]((x, y)) and (x, y) ≼G[H] (u,w). Suppose that u ∈ NG(A ≽). Then by (ii), there exists v ∈ Tu for which degH(v) = ∆(H). It means (u, v) ∈ S ∩NG[H]((x, y)) and (x, y) ≼G[H] (u, v). Case 2: x ∈ A ∩NG(A ≽) If x ∈ A ∩ NG(A ≽) and u ∈ A such that x ≼G u, then by (ii), there exists v ∈ Tu such that degH(v) = ∆(H). Thus, (u, v) ∈ S ∩NG[H]((x, y)) and (x, y) ≼G[H] (u, v). Case 3: x ∈ A \NG(A ≽) Suppose that x ∈ A \ NG(A ≽). Then Tx ∈ Γs(H). Thus, there exists w ∈ Tx for which y ≼H w. Here we have (x,w) ∈ S ∩NG[H]((x, y)) and (x, y) ≼G[H] (x,w). All 3 cases above imply that S ∈ Γs(G[H]). Similar arguments will also prove the following two propositions for weak domination: J. M. Molles, F. P. Jamil, S. R. Canoy / Eur. J. Pure Appl. Math, 18 (4) (2025), 6851 15 of 18 Proposition 9. Let G and H be nontrivial connected graphs, and let S ∈ Γw(G[H])). Then S = ∪x∈A ({x} × Tx), where A ⊆ V (G) and Tx ⊆ V (H) satisfying the following. (i) A ∈ Γw(G); and (ii) For each x ∈ A \NG(A ≼), Tx ∈ Γw(H). Proposition 10. Let G and H be connected nontrivial graphs, and S = ∪x∈A ({x} × Tx), where A ⊆ V (G) and Tx = {y ∈ V (H) : (x, y) ∈ S}. Suppose that each of the following holds for A: (i) A ∈ Γw(G); (ii) For each x ∈ A ∩NG(A ≼), there exists y ∈ Tx such that degH(y) = δ(H); and (iii) For each x ∈ A \NG(A ≼), Tx ∈ Γw(H). Then S ∈ Γw(G[H]). Proposition 11. Let G and H be nontrivial connected graphs. Then for each S∗ ∈ Γs(G[H]), there exists S = ∪x∈A ({x} × Tx) ∈ Γs(G[H]) satisfying the following: (i) |S| = |S∗|; (ii) For each x ∈ A ∩NG(A ≽), there exists y ∈ Tx for which degH(y) = ∆(H). Proof. Let v ∈ V (H) such that degH(v) = ∆(H). Write S∗ = ∪x∈A∗ ({x} × T ∗ x ), where A∗ ∈ Γs(G) and T ∗ x ∈ Γs(H) for each x ∈ A∗ \NG(A ∗ ≽). For each x ∈ A∗ ∩NG(A ∗ ≽), let yx ∈ Tx. Define the following: • A = A∗; • Tx = T ∗ x for each x ∈ A∗ \NG(A ∗ ≽); and • Tx = (T ∗ x \ {yx}) ∪ {v} for each x ∈ A∗ ∩NG(A ∗ ≽). By Proposition 8, S = ∪x∈A ({x} × Tx) ∈ Γs(G[H]). Moreover, by the construction of S, |S| = |S∗|. Corollary 5. Let G and H be nontrivial connected graphs. Then γs(G[H]) = min{|A ∩NG(A ≽)|+ γs(H)|A \NG(A ≽)| : A ∈ Γs(G)}, and γw(G[H]) = min{|A ∩NG(A ≼)|+ γw(H)|A \NG(A ≼)| : A ∈ Γw(G)}. J. M. Molles, F. P. Jamil, S. R. Canoy / Eur. J. Pure Appl. Math, 18 (4) (2025), 6851 16 of 18 Proof. Put α = min{|A∩NG(A ≽)|+γs(H)|A\NG(A ≽)| : A ∈ Γs(G)}. Let v ∈ V (H) such that degH(v) = ∆(H), and let A ∈ Γs(G) such that α = |A|+ γs(H)|A \NG(A ≽)|. For each x ∈ A ∩NG(A ≽), let Tx = {v} and for each x ∈ A \NG(A ≽), let Tx ⊆ V (H) be a γs-set of H. By Proposition 8, S = ∪x∈A ({x} × Tx) ∈ Γs(G[H]). Thus, γs(G[H]) ≤ |S| = |A ∩NG(A ≽)|+ γs(H)|A \NG(A ≽) = α. Let S = ∪x∈A ({x} × Tx) be a γs-set of G[H]. In view of Proposition 11, S = ∪x∈A ({x} × Tx) with A ⊆ V (G) and Tx ⊆ V (H), where A ∈ Γs(G), Tx ̸= ∅ for each x ∈ A ∩NG(A ≽), and Tx ∈ Γs(H) for each x ∈ A \NG(A ≽). Thus, γs(G[H]) = |S| ≥ |A ∩NG(A ≽)|+ γs(H)|A \NG(A ≽)| ≥ α. Proof for the weak domination case is similar. Corollary 6. The following hold for nontrivial connected graphs G and H: (i) If γ(H) = 1, then γs(G[H]) = γs(G). (ii) If H is a regular graph, then γs(G[H]) = min{|A ∩NG(A ≽)|+ γ(H)|A \NG(A ≽)| : A ∈ Γs(G)} and γw(G[H]) = min{|A ∩NG(A ≼)|+ γ(H)|A \NG(A ≼)| : A ∈ Γw(G)}. Acknowledgements This project is fully supported by the DOST-ASTHRD of the Philippines and the OVCRE of MSU-Iligan Institute of Technology. References [1] F. Buckley and F. Harary. Distance in Graphs. Addison-Wesley, Redwood City, CA, 1990. [2] Claude Berge. Théorie des graphes et ses applications. Dunod, Paris, 1958. English translation: The Theory of Graphs and its Applications, Methuen (London) andWiley (New York), 1962. [3] E. Cockayne and S. 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