Product vague graphs and its applications

Abstract

A vague graph is a generalized structure of a fuzzy graph that gives more precision, flexibility and compatibility to a system when compared with systems that are designed using fuzzy graphs. In this paper, we defined the notions of product and complete product vague graphs. We investigated some properties of union, join and complement on product vague graphs and introduced the multiplication of the product vague graphs. Finally we give an application of product vague graphs. Product and complete product vague graphs are highly utilized by computer science, geometry, algebra, number theory and operation research. Likewise, these properties will also be helpful to study large vague graph as a combination of small, vague graphs and to derive its properties from those of the smaller ones.

Keywords

Vague graphs product vague graphs multiplication of vague graphs

1 Introduction

Presently, science and technology is featured with complex processes and phenomena for which complete information is not always available. For such cases, mathematical models are developed to handle various types of systems containing elements of uncertainty. A large number of these models is based on an extension of the ordinary set theory, namely, fuzzy sets. Graph theory has numerous application to problem in computer science, electrical engineering, system analysis, operations research, economics, networking routing, and transportation. Also, the major role of graph theory in computer applications is the development of graph algorithms. A number of algorithms are used to solve problems that are modeled in the form of graphs and the corresponding computer science application problems. In the classical set theory introduced by Cantor, values of elements in a set are either 0 or 1. That is, for any element, there are only two possibilities: the element is either in the set or it is not. Therefore, Cantor’s set theory cannot handle data with ambiguity and uncertainty. In 1965 Zadeh [15] first proposed the theory of fuzzy sets. The most important feature of a fuzzy set is that a fuzzy set A is a class of object that satisfy a certain (or several) property. Gau and Buehrer [3] proposed the concept of vague set in 1993, by replacing the value of an element in a set with a subinterval of [0, 1]. Namely, a true-membership function t_v (x) and a false-membership function f_v (x) are used to describe the boundaries of the membership degree. The initial definition given by Kaufmann [4] of a fuzzy graph was based on the fuzzy relation proposed by Zadeh [15]. Later Rosenfeld [8] introduced the fuzzy analogue of several basic graph-theoretic concepts. Mordeson and Nair [5] defined the concept of complement of fuzzy graph and studied some operations on fuzzy graphs. Akram et al. [1, 2] introduced bipolar fuzzy graphs and vague hypergraphs. Ramakrishna [7] introduced the concept of vague graphs and studied some of their properties. Pal and Rashmanlou [6] studied irregular interval valued fuzzy graphs. Also, they defined antipodal interval valued fuzzy graphs [9], balanced interval valued fuzzy graphs [10], some properties of highly irregular interval valued fuzzy graphs [11] and a study on bipolar fuzzy graphs [12]. Rashmanlou and Yong Bae Jun investigated complete interval valued fuzzy graphs [13]. Sunitha and Vijayakumar [14] studied some properties of complement on fuzzy graphs. In this paper, we defined the notions of product vague graphs, complete product vague graphs and the complement of a product vague graphs. We investigated some properties of union, join and complement on product vague graphs and introduced the multiplication of the product vague graphs. Finally we give an application of product vague graphs.

2 Preliminaries

By a graph G^* = (V, E), we mean a non-trivial, finite, connected and undirected graph without loops or multiple edges. Formally, given a graph G^* = (V, E), two vertices x, y ∈ V are said to be neighbors, or adjacent nodes, if xy ∈ E.

Definition 2.1. [15] A fuzzy subset μ on a set X is a map μ : X → [0, 1]. A fuzzy binary relation on X is a fuzzy subset μ on X × X.

Definition 2.2. [3] A vague set on an ordinary finite non-empty set X is a pair (t_A, f_A), where t_A : X → [0, 1], f_A : X → [0, 1] are true and false membership functions, respectively such that 0 ≤ t_A (x) + f_A (x) ≤1, for all x ∈ X.

In the above definition, t_A (x) is considered as the lower bound for degree of membership of x in A (based on evidence), and f_A (x) is the lower bound for negation of membership of x in A (based on evidence against). So the degree of membership of x in the vague set A is characterized by the interval [t_A (x) , 1 - f_A (x)]. Therefore, a vague set is a special case of interval valued sets studied by many mathematicians. The interval [t_A (x) , 1 - f_A (x)] is called the vague value of x in A, and is denoted by V_A (x).

Definition 2.3. [3] Let X and Y be ordinary finite non-empty sets. We call a vague relation to be a vague subset of X × Y, that is, an expression R defined by: $R = {〈 (x, y), t_{R} (x, y), f_{R} (x, y) 〉 | x \in X, y \in Y}$ where t_R : X × Y → [0, 1], f_R : X × Y → [0, 1], which satisfies the condition 0 ≤ t_R (x, y) + f_R (x, y) ≤1 for all (x, y) ∈ X × Y. A vague relation R on X is called reflexive if t_R (x, x) =1 and f_R (x, x) =0, for all x ∈ X. A vague relation R is symmetric if t_R (x, y) = t_R (y, x) and f_R (x, y) = f_R (y, x) , for all x, y ∈ X. Since an edge xy ∈ E is identified with an ordered pair (x, y) ∈ V × V, a vague relation on E can be identified with a vague set on E.

Definition 2.4. [7] Let G^* = (V, E) be a crisp graph. A pair G = (A, B) is called a vague graph on a crisp graph G^*, where A = (t_A, f_A) is a vague set on V and B = (t_B, f_B) is a vague set on E ⊆ V × V such that t_B (xy) ≤ min(t_A (x) , t_A (y)) and f_B (xy) ≥ max(f_A (x) , f_A (y)), for each edge xy ∈ E.

A vague graph G is called complete if t_B (xy) = min(t_A (x) , t_A (y)) and f_B (xy) = max(f_A (x) , f_A (y)) , for each edge xy ∈ E.

3 Complement of product vague graphs

In this section we investigate some interesting properties of join, union and complement on product vague graphs and introduce the multiplication of the product vague graphs.

Definition 3.1. Let G = (A, B) be a vague graph of a graph G^* = (V, E). If t_B (xy) ≤ t_A (x) × t_A (y) and f_B (xy) ≥ f_A (x) × f_A (y) , for all x, y ∈ V, then the vague graph G is called product vague graph of G^*. A product vague graph G = (A, B) is said to be complete if t_B (xy) = t_A (x) × t_A (y) and f_B (xy) = f_A (x) × f_A (y) , for all x, y ∈ V.

Example 3.2. (Fig. 1) Consider a vague graph G such that V = {x, y, z}, E = {xy, yz, xz}.

By routine computation, it is easy to show that G is a product vague graph.

Remark 3.3. If G = (A, B) is a product vague graph, since t_A (x) and t_A (y) are less than or equal to 1, then it follows that t_B (xy) ≤ t_A (x) × t_A (y) ≤ t_A (x) ∧ t_A (y), for all x, y ∈ V. Similarly, f_B (xy) ≥ f_A (x) × f_A (y) ≥ f_A (x) ∨ f_A (y) , for all x, y ∈ V. Thus, every product vague graph is a vague graph.

Definition 3.4. The complement of product vague graph G = (A, B) is a vague graph G^c = (A^c, B^c), where A^c = A = (t_A, f_A) and $B^{c} = (t_{B}^{c}, f_{B}^{c})$ is defined by: $\begin{matrix} t_{B}^{c} (xy) & = & t_{A} (x) \times t_{A} (y) - t_{B} (xy), f_{B}^{c} (xy) \\ = & f_{B} (xy) - f_{A} (x) \times f_{A} (y) . \end{matrix}$

Example 3.5. (Figs. 2, and 3) Consider a product vague graph G such that V = {x, y, z} and E = {xy, yz, xz}.

The complement of G is as follows.

Remark 3.6. It follows that G^c is a product vague graph and (G^c) ^c = G.

Note 3.7. Throughout this paper suppose that G^* is a crisp graph, G₁ = (A₁, B₁) and G₂ = (A₂, B₂) are product vague graphs of $G_{1}^{*} = (V_{1}, E_{1})$ and $G_{2}^{*} = (V_{2}, G_{2})$ , respectively.

Definition 3.8. We denote the union of G₁ and G₂ by G₁ ∪ G₂ = (A₁ ∪ A₂, B₁ ∪ B₂) and define as follow: $\begin{matrix} (i) {\begin{matrix} (t_{A_{1}} \cup t_{A_{2}}) (x) = t_{A_{1}} (x), if x \in V_{1} and x \notin V_{2}, \\ (t_{A_{1}} \cup t_{A_{2}}) (x) = t_{A_{2}} (x), if x \in V_{2} and x \notin V_{1}, \\ (t_{A_{1}} \cup t_{A_{2}}) (x) = max (t_{A_{1}} (x), t_{A_{2}} (x)), \\ if x \in V_{1} \cap V_{2} . \end{matrix} \end{matrix}$

$\begin{matrix} (ii) {\begin{matrix} (f_{A_{1}} \cup f_{A_{2}}) (x) = f_{A_{1}} (x), if x \in V_{1} and x \notin V_{2}, \\ (f_{A_{1}} \cup f_{A_{2}}) (x) = f_{A_{2}} (x), if x \in V_{2} and x \notin V_{1}, \\ (f_{A_{1}} \cup f_{A_{2}}) (x) = min (f_{A_{1}} (x), f_{A_{2}} (x)), \\ if x \in V_{1} \cap V_{2} . \end{matrix} \end{matrix}$ $\begin{matrix} (iii) {\begin{matrix} (t_{B_{1}} \cup t_{B_{2}}) (xy) = t_{B_{1}} (xy), if xy \in E_{1} and \\ xy \notin E_{2}, \\ (t_{B_{1}} \cup t_{B_{2}}) (xy) = t_{B_{2}} (xy), if xy \in E_{2} and \\ xy \notin E_{1}, \\ (t_{B_{1}} \cup t_{B_{2}}) (xy) = max (t_{B_{1}} (xy), t_{B_{2}} (xy)), \\ if xy \in E_{1} \cap E_{2} . \end{matrix} \end{matrix}$ $\begin{matrix} (iv) {\begin{matrix} (f_{B_{1}} \cup f_{B_{2}}) (xy) = f_{B_{1}} (xy), if xy \in E_{1} and \\ xy \notin E_{2}, \\ (f_{B_{1}} \cup f_{B_{2}}) (xy) = f_{B_{2}} (xy), if xy \in E_{2} and \\ xy \notin E_{1}, \\ (f_{B_{1}} \cup f_{B_{2}}) (xy) = min (f_{B_{1}} (xy), f_{B_{2}} (xy)), \\ if xy \in E_{1} \cap E_{2} . \end{matrix} \end{matrix}$

Example 3.9. (Figs. 4 and 5) Let G₁ = (A₁, B₁) and G₂ = (A₂, B₂) be two product vague graphs asfollows.

The union of G₁ and G₂, that is G₁ ∪ G₂ is asfollows.

By routine computation, it is easy to show that G₁ ∪ G₂ is a product vague graph.

Proposition 3.10. The union of G₁ and G₂ is a product vague graph also.

Proof. We prove that G₁ ∪ G₂ is a product vague graph of the graph $G_{1}^{*} \cup G_{2}^{*} = (V_{1} \cup V_{2}, E_{1} \cup E_{2})$ . Let xy ∈ E₁ ∩ E₂. Then $\begin{matrix} (t_{B_{1}} \cup t_{B_{2}}) (xy) = max (t_{B_{1}} (xy), t_{B_{2}} (xy)) \\ \leq max (t_{A_{1}} (x) \times t_{A_{1}} (y), t_{A_{2}} (x) \times t_{A_{2}} (y)) \\ \leq max {t_{A_{1}} (x), t_{A_{2}} (x)} \times max {t_{A_{1}} (y), t_{A_{2}} (y)} \\ = (t_{A_{1}} \cup t_{A_{2}}) (x) \times (t_{A_{1}} \cup t_{A_{2}}) (y), \end{matrix}$ $\begin{matrix} (f_{B_{1}} \cup f_{B_{2}}) (xy) = min (f_{B_{1}} (xy), f_{B_{2}} (xy)) \\ \geq min (f_{A_{1}} (x) \times f_{A_{1}} (y), f_{A_{2}} (x) \times f_{A_{2}} (y)) \\ \geq min {f_{A_{1}} (x), f_{A_{2}} (x)} \times min {f_{A_{1}} (y), f_{A_{2}} (y)} \\ = (f_{A_{1}} \cup f_{A_{2}}) (x) \times (f_{A_{1}} \cup f_{A_{2}}) (y) . \end{matrix}$ If xy ∈ E₁ and xy ∉ E₂, then $\begin{matrix} (t_{B_{1}} \cup t_{B_{2}}) (xy) & = & t_{B_{1}} (xy) \leq t_{A_{1}} (x) \times t_{A_{1}} (y) \\ = & (t_{A_{1}} \cup t_{A_{2}}) (x) \times (t_{A_{1}} \cup t_{A_{2}}) (y), \end{matrix}$ $\begin{matrix} (f_{B_{1}} \cup f_{B_{2}}) (xy) & = & f_{B_{1}} (xy) \geq f_{A_{1}} (x) \times f_{A_{1}} (y) \\ = & (f_{A_{1}} \cup f_{A_{2}}) (x) \times (f_{A_{1}} \cup f_{A_{2}}) (y) . \end{matrix}$ Similarly, if xy ∈ E₂ and xy ∉ E₁, then we get $\begin{matrix} (t_{B_{1}} \cup t_{B_{2}}) (xy)) & \leq & (t_{A_{1}} \cup t_{A_{2}}) (x) \times (t_{A_{1}} \cup t_{A_{2}}) (y), \\ (f_{B_{1}} \cup f_{B_{2}}) (xy) & \geq & (f_{A_{1}} \cup f_{A_{2}}) (x) \times (f_{A_{1}} \cup f_{A_{2}}) (y) . \end{matrix}$ □

Theorem 3.11. G₁ ∪ G₂ = (A₁ ∪ A₂, B₁ ∪ B₂) is a product vague graph of $G_{1}^{*} \cup G_{2}^{*}$ if and only if G₁ and G₂ are product vague graphs of $G_{1}^{*}$ and $G_{2}^{*}$ respectively.

Proof. Let G₁ ∪ G₂ be a product vague graph of $G_{1}^{*} \cup G_{2}^{*}$ and xy ∈ E₁. Then, xy ∉ E₂ and x, y ∈ V₁ - V₂. Hence, $\begin{matrix} t_{B_{1}} (xy) & = & (t_{B_{1}} \cup t_{B_{2}}) (xy) \\ \leq & (t_{A_{1}} \cup t_{A_{2}}) (x) \times (t_{A_{1}} \cup t_{A_{2}}) (y) \\ = & t_{A_{1}} (x) \times t_{A_{1}} (y), \end{matrix}$

$\begin{matrix} f_{B_{1}} (xy) & = & (f_{B_{1}} \cup f_{B_{2}}) (xy) \\ \geq & (f_{A_{1}} \cup f_{A_{2}}) (x) \times (f_{A_{1}} \cup f_{A_{2}}) (y) \\ = & f_{A_{1}} (x) \times f_{A_{1}} (y) . \end{matrix}$ Therefore, G₁ is a product vague graph. Similarly we can prove that G₂ is a product vague graph also. By Proposition 3.10, we get the converse. □

Definition 3.12. We denote the join of G₁ and G₂ by G₁ + G₂ = (A₁ + A₂, B₁ + B₂) and define as follow $\begin{matrix} (i) {\begin{matrix} (t_{A_{1}} + t_{A_{2}}) (x) = (t_{A_{1}} \cup t_{A_{2}}) (x) \\ (f_{A_{1}} + f_{A_{2}}) (x) = (f_{A_{1}} \cap f_{A_{2}}) (x), \\ if x \in V_{1} \cup V_{2} . \end{matrix} \end{matrix}$ $\begin{matrix} (ii) {\begin{matrix} (t_{B_{1}} + t_{B_{2}}) (xy) = (t_{B_{1}} \cup t_{B_{2}}) (xy) \\ (f_{B_{1}} + f_{B_{2}}) (xy) = (f_{B_{1}} \cap f_{B_{2}}) (xy), \\ if xy \in E_{1} \cup E_{2} . \end{matrix} \end{matrix}$ $\begin{matrix} (iii) {\begin{matrix} (t_{B_{1}} + t_{B_{2}}) (xy) = (t_{A_{1}} (x) \times t_{A_{2}} (y)) \\ (f_{B_{1}} + f_{B_{2}}) (xy) = (f_{A_{1}} (x) \times f_{A_{2}} (y)), \\ if xy \in E^{'} . \end{matrix} \end{matrix}$ where E′ is the set of all edges joining the nodes of V₁ and V₂.

Example 3.13. (Fig. 6) In Example 3.9 the join of G₁ and G₂ is as follows.

Proposition 3.14. The join of G₁ and G₂ is a product vague graph also.

Proof. In the view of Proposition 3.10, it is sufficient to verify when xy ∈ E′. In this case we have: $\begin{matrix} (t_{B_{1}} + t_{B_{2}}) (xy) = t_{A_{1}} (x) \times t_{A_{2}} (y) \\ \leq (t_{A_{1}} + t_{A_{2}}) (x) \times (t_{A_{1}} + t_{A_{2}}) (y), \\ (f_{B_{1}} + f_{B_{2}}) (xy) = f_{A_{1}} (x) \times f_{A_{2}} (y) \\ \geq (f_{A_{1}} + f_{A_{2}}) (x) \times (f_{A_{1}} + f_{A_{2}}) (y) . \end{matrix}$ □

The converse of Proposition 3.14 is true when G₁ and G₂ be two complete product vague graphs.

Proposition 3.15. If G₁ + G₂ is complete product vague graph then, G₁ and G₂ are both complete product vague graphs.

Proof. Let G₁ + G₂ be complete product vague graph. First we show that G₁ is complete. Let x, y ∈ V₁. Then we have $\begin{matrix} (t_{B_{1}} + t_{B_{2}}) (xy) & = & t_{B_{1}} (xy) and (f_{B_{1}} + f_{B_{2}}) (xy) \\ = & f_{B_{1}} (xy) . \end{matrix}$ Since G₁ + G₂ is complete, $\begin{matrix} (t_{B_{1}} + t_{B_{2}}) (xy) & = & (t_{A_{1}} + t_{A_{2}}) (x) \times (t_{A_{1}} + t_{A_{2}}) (y) \\ = & t_{A_{1}} (x) \times t_{A_{1}} (y), \\ (f_{B_{1}} + f_{B_{2}}) (xy) & = & (f_{A_{1}} + f_{A_{2}}) (x) \times (f_{A_{1}} + f_{A_{2}}) (y) \\ = & f_{A_{1}} (x) \times f_{A_{1}} (y) . \end{matrix}$ From above equalities we get t_{B
₁} (xy) = t_{A
₁} (x) × t_{A
₁} (y) and f_{B
₁} (xy) = f_{A
₁} (x) × f_{A
₁} (y). Therefore G₁ is complete product vague graph. Similarly we can prove that G₂ is complete product vague graph. □

Proposition 3.16. Let G₁ and G₂ be two product vague graphs of $G_{1}^{*} = (V_{1}, E_{1})$ and $G_{2}^{*} = (V_{2}, E_{2})$ , respectively such that V₁∩ V₂ = ∅. Then $(G_{1} + G_{2})^{c} = G_{1}^{c} \cup G_{2}^{c}$ .

Proof. Let u ∈ V₁, then (t_{A
₁} + t_{A
₂}) ^c (u) = (t_{A
₁} + t_{A
₂}) (u) = t_{A
₁} (u), and $max (t_{A_{1}}^{c} (u), t_{A_{2}}^{c} (u)) = max (t_{A_{1}} (u), t_{A_{2}} (u)) = t_{A_{1}} (u)$ . Hence, $(t_{A_{1}} + t_{A_{2}})^{c} (u) = (t_{A_{1}}^{c} \cup t_{A_{2}}^{c}) (u)$ . Also, (f_{A
₁} + f_{A
₂}) ^c (u) = (f_{A
₁} + f_{A
₂}) (u) = f_{A
₁} (u) ∧ f_{A
₂} (u) = f_{A
₁} (u) and $min (f_{A_{1}}^{c} (u) + f_{A_{2}}^{c} (u)) = min (f_{A_{1}} (u), f_{A_{2}} (u)) = f_{A_{1}} (u)$ . So, (f_{A
₁}+ $f_{A_{2}})^{c} (u) = (f_{A_{1}}^{c} \cup f_{A_{2}}^{c}) (u)$ . Similarly, we can prove that $(t_{A_{1}} + t_{A_{2}})^{c} (u) = (t_{A_{1}}^{c} \cup t_{A_{2}}^{c}) (u)$ and $(f_{A_{1}} + f_{A_{2}})^{c} (u) = (f_{A_{1}}^{c} \cup f_{A_{2}}^{c}) (u)$ for all u ∈ V₂. Suppose that uv ∈ E₁ then, u, v ∈ V₁ and we have $\begin{matrix} (t_{B_{1}} + t_{B_{2}})^{c} (uv) = (t_{A_{1}} + t_{A_{2}}) (u) \\ \times (t_{A_{1}} + t_{A_{2}}) (v) - (t_{B_{1}} + t_{B_{2}}) (uv) \\ = t_{A_{1}} (u) \times t_{A_{1}} (v) - t_{B_{1}} (uv) = t_{B_{1}}^{c} (uv) \end{matrix}$ and $max (t_{B_{1}}^{c} (uv), t_{B_{2}}^{c} (uv)) = t_{B_{1}}^{c} (uv)$ . Hence $(t_{B_{1}} + t_{B_{2}})^{c} (uv) = (t_{B_{1}}^{c} \cup t_{B_{2}}^{c}) (uv)$ . Also, $\begin{matrix} (f_{B_{1}} + f_{B_{2}})^{c} (uv) = (f_{B_{1}} + f_{B_{2}}) (uv) \\ - (f_{A_{1}} + f_{A_{2}}) (u) \times (f_{A_{1}} + f_{A_{2}}) (v) \\ = f_{B_{1}} (uv) - f_{A_{1}} (u) \times f_{A_{1}} (v) = f_{B_{1}}^{c} (uv), \end{matrix}$ and $min (f_{B_{1}}^{c} (uv), f_{B_{2}}^{c} (uv)) = f_{B_{1}}^{c} (uv)$ . Thus, $(f_{B_{1}} + f_{B_{2}})^{c} (uv) = (f_{B_{1}}^{c} \cup f_{B_{2}}^{c}) (uv)$ . Similarly, we can prove that $(t_{B_{1}} + t_{B_{2}})^{c} (uv) = (t_{B_{1}}^{c} \cup t_{B_{2}}^{c}) (uv)$ and $(f_{B_{1}} + f_{B_{2}})^{c} (uv) = (f_{B_{1}}^{c} \cup f_{B_{2}}^{c}) (uv)$ for all uv ∈ E₂. Now suppose that (u, v) ∈ E′ then, u ∈ V₁, v ∈ V₂ and we have $\begin{matrix} (t_{B_{1}} + t_{B_{2}})^{c} (uv) = (t_{A_{1}} + t_{A_{2}}) (u) \\ \times (t_{A_{1}} + t_{A_{2}}) (v) - (t_{B_{1}} + t_{B_{2}}) (uv) \\ = t_{A_{1}} (u) \times t_{A_{2}} (v) - (t_{A_{1}} (u) \times t_{A_{2}} (v)) = 0 . \end{matrix}$ Also, $max (t_{B_{1}}^{c} (uv), t_{B_{2}}^{c} (uv)) = 0$ [Since u ∈ V₁ and v ∈ V₂]. In the other hand $\begin{matrix} (f_{B_{1}} + f_{B_{2}})^{c} (uv) = (f_{B_{1}} + f_{B_{2}}) (uv) \\ - (f_{A_{1}} + f_{A_{2}}) (u) \times (f_{A_{1}} + f_{A_{2}}) (v) \\ = f_{A_{1}} (u) \times f_{A_{2}} (v) - (f_{A_{1}} (u) \times f_{A_{2}} (v)) = 0 \end{matrix}$ and $min (f_{B_{1}}^{c} (uv), f_{B_{2}}^{c} (uv)) = 0$ [Since u ∈ V₁ and v ∈ V₂]. Therefore, $(G_{1} + G_{2})^{c} = G_{1}^{c} \cup G_{2}^{c}$ . □

Proposition 3.17. Let G₁ and G₂ be two product vague graphs with V₁∩ V₂ = ∅. Then $(G_{1} \cup G_{2})^{c} = G_{1}^{c} + G_{2}^{c}$ .

Proof. (i): Let u ∈ V₁. Then, (t_{A
₁} ∪ t_{A
₂}) ^c (u) = (t_{A
₁} ∪ t_{A
₂}) (u) = t_{A
₁} (u) and $(t_{A_{1}}^{c} + t_{A_{2}}^{c}) (u) = max (t_{A_{1}}^{c} (u),$

$t_{A_{2}}^{c} (u)) = max (t_{A_{1}} (u), t_{A_{2}} (u)) = t_{A_{1}} (u)$ . Hence, $(t_{A_{1}} \cup t_{A_{2}})^{c} (u) = (t_{A_{1}}^{c} + t_{A_{2}}^{c}) (u)$ for all u ∈ V₁. Also, (f_{A
₁} ∪ f_{A
₂}) ^c (u) = (f_{A
₁} ∪ f_{A
₂}) (u) = f_{A
₁} (u) and $(f_{A_{1}}^{c} + f_{A_{2}}^{c}) (u) = min (f_{A_{1}}^{c} (u), f_{A_{2}}^{c} (u)) = min (f_{A_{1}} (u), f_{A_{2}} (u)) = f_{A_{1}} (u)$ . So, $(f_{A_{1}} \cup f_{A_{2}})^{c} (u) = (f_{A_{1}}^{c} + f_{A_{2}}^{c}) (u)$ for all u ∈ V₁. Similarly we can prove when u ∈ V₂.

(ii): If uv ∈ E₁, then u, v ∈ V₁. Hence, $\begin{matrix} (t_{B_{1}} \cup t_{B_{2}})^{c} (uv) & = & (t_{A_{1}} \cup t_{A_{2}}) (u) \times (t_{A_{1}} \cup t_{A_{2}}) (v) \\ - (t_{B_{1}} \cup t_{B_{2}}) (uv) \\ = & t_{A_{1}} (u) \times t_{A_{1}} (v) - t_{B_{1}} (uv) \\ = & t_{B_{1}}^{c} (uv) = (t_{B_{1}}^{c} + t_{B_{2}}^{c}) (uv) . \end{matrix}$ Also, $\begin{matrix} (f_{B_{1}} \cup f_{B_{2}})^{c} (uv) = (f_{B_{1}} \cup f_{B_{2}}) (uv) \\ - (f_{A_{1}} \cup f_{A_{2}}) (u) \times (f_{A_{1}} \cup f_{A_{2}}) (v) \\ = f_{B_{1}} (uv) - f_{A_{1}} (u) \times f_{A_{1}} (v) \\ = f_{B_{1}}^{c} (uv) = (f_{B_{1}}^{c} + f_{B_{2}}^{c}) (uv) . \end{matrix}$

(iii) If uv ∈ E₂, then, u, v ∈ V₂ and we have $\begin{matrix} (t_{B_{1}} \cup t_{B_{2}})^{c} (uv) = (t_{A_{1}} \cup t_{A_{2}}) (u) \\ \times (t_{A_{1}} \cup t_{A_{2}}) (v) - (t_{B_{1}} \cup t_{B_{2}}) (uv) \\ = t_{A_{2}} (u) \times t_{A_{2}} (v) - t_{B_{2}} (uv) \\ = t_{B_{2}}^{c} (uv) = (t_{B_{1}}^{c} + t_{B_{2}}^{c}) (uv) . \end{matrix}$ Also, $\begin{matrix} (f_{B_{1}} \cup f_{B_{2}})^{c} (uv) = (f_{B_{1}} \cup f_{B_{2}}) (uv) \\ - (f_{A_{1}} \cup f_{A_{2}}) (u) \times (f_{A_{1}} \cup f_{A_{2}}) (v) \\ = f_{B_{2}} (uv) - f_{A_{2}} (u) \times f_{A_{2}} (v) \\ = f_{B_{2}}^{c} (uv) = (f_{B_{1}}^{c} + f_{B_{2}}^{c}) (uv) . \end{matrix}$

(iv) If uv ∈ E′, then u ∈ V₁ and v ∈ V₂. Hence $\begin{matrix} (t_{B_{1}} \cup t_{B_{2}})^{c} (uv) = (t_{A_{1}} \cup t_{A_{2}}) (u) \\ \times (t_{A_{1}} \cup t_{A_{2}}) (v) - (t_{B_{1}} \cup t_{B_{2}}) (uv) \\ = t_{A_{1}} (u) \times t_{A_{2}} (v) \\ = t_{A_{1}}^{c} (u) \times t_{A_{2}}^{c} (v) = t_{B_{2}}^{c} (uv) . \end{matrix}$ Therefore, $(t_{B_{1}} \cup t_{B_{2}})^{c} = t_{B_{1}}^{c} + t_{B_{2}}^{c}$ . Similarly we can show that $(f_{B_{1}} \cup f_{B_{2}})^{c} = f_{B_{1}}^{c} + f_{B_{2}}^{c}$ . □

Now we introduce the multiplication of two vague graphs and conclude by giving a necessary and sufficient condition for a product vague graph to be the multiplication of two product vague graphs.

Let $G_{1}^{*}$ and $G_{2}^{*}$ be two graphs whose vertex sets are V₁ and V₂ respectively. Consider a new graph $G^{*} = G_{1}^{*} \times G_{2}^{*}$ whose vertex set is V₁ × V₂ and edge set is a subset of (V₁ × V₂) × (V₁ × V₂). Let G₁ and G₂ be two product vague graphs of $G_{1}^{*}$ and $G_{2}^{*}$ , respectively. If v₁ ∈ V₁ and v₂ ∈ V₂, then we define: $(t_{A_{1}} \times t_{A_{2}}) (v_{1}, v_{2}) = t_{A_{1}} (v_{1}) \times t_{A_{2}} (v_{2}),$ $(f_{A_{1}} \times f_{A_{2}}) (v_{1}, v_{2}) = f_{A_{1}} (v_{1}) \times f_{A_{2}} (v_{2}) .$ Also, if u₁, v₁ ∈ V₁ and u₂, v₂ ∈ V₂, then we define: $(t_{B_{1}} \times t_{B_{2}}) ((u_{1}, u_{2}) (v_{1}, v_{2})) = t_{B_{1}} (u_{1} v_{1}) \times t_{B_{2}} (u_{2} v_{2})$ and (f_{B
₁} × f_{B
₂}) ((u₁, u₂) (v₁, v₂)) = f_{B
₁} (u₁v₁) × f_{B
₂} (u₂v₂).

Then, A = (t_{A
₁} × t_{A
₂}, f_{A
₁} × f_{A
₂}) is a vague subset on V = V₁ × V₂ and B = (t_{B
₁} × t_{B
₂}, f_{B
₁} × f_{B
₂}) is a vague subset on (V₁ × V₂) × (V₁ × V₂). In fact G = (A, B) is a vague graph of $G_{1}^{*} \times G_{2}^{*}$ that is denoted by G₁ × G₂.

Proposition 3.18. Let G₁ and G₂ be two product vague graphs of $G_{1}^{*}$ and $G_{2}^{*}$ , respectively. Then G₁ × G₂ is a product vague graph of $G_{1}^{*} \times G_{2}^{*}$ .

Proof. Let u₁, v₁ ∈ V₁ and u₂, v₂ ∈ V₂. Then we have $\begin{matrix} (t_{B_{1}} \times t_{B_{2}}) ((u_{1}, u_{2}) (v_{1}, v_{2})) \\ = t_{B_{1}} (u_{1} v_{1}) \times t_{B_{2}} (u_{2} v_{2}) \\ \leq (t_{A_{1}} (u_{1}) \times t_{A_{1}} (v_{1})) \times (t_{A_{2}} (u_{2}) \times t_{A_{2}} (v_{2})) \\ = (t_{A_{1}} (u_{1}) \times t_{A_{2}} (u_{2})) \times (t_{A_{1}} (v_{1}) \times t_{A_{2}} (v_{2})) \\ = (t_{A_{1}} \times t_{A_{2}}) (u_{1}, u_{2}) \times (t_{A_{1}} \times t_{A_{2}}) (v_{1}, v_{2}) . \end{matrix}$ Also, $\begin{matrix} (f_{B_{1}} \times f_{B_{2}}) ((u_{1}, u_{2}) (v_{1}, v_{2})) \\ = f_{B_{1}} (u_{1} v_{1}) \times f_{B_{2}} (u_{2} v_{2}) \\ \geq (f_{A_{1}} (u_{1}) \times f_{A_{1}} (v_{1})) \times (f_{A_{2}} (u_{2}) \times f_{A_{2}} (v_{2})) \\ = (f_{A_{1}} (u_{1}) \times f_{A_{2}} (u_{2})) \times (f_{A_{1}} (v_{1}) \times f_{A_{2}} (v_{2})) \\ = (f_{A_{1}} \times f_{A_{2}}) (u_{1}, u_{2}) \times (f_{A_{1}} \times f_{A_{2}}) (v_{1}, v_{2}) . \end{matrix}$ □

Theorem 3.19. Let G₁ and G₂ be two product vague graphs. Then G₁ × G₂ is complete if and only if both G₁ and G₂ are complete.

Proof. Let G₁ and G₂ be complete and u₁, v₁ ∈ V₁, u₂, v₂ ∈ V₂. Then $\begin{matrix} (t_{B_{1}} \times t_{B_{2}}) ((u_{1}, u_{2}) (v_{1}, v_{2})) \\ = t_{B_{1}} (u_{1} v_{1}) \times t_{B_{2}} (u_{2} v_{2}) \\ = (t_{A_{1}} (u_{1}) \times t_{A_{1}} (v_{1})) \times (t_{A_{2}} (u_{2}) \times t_{A_{2}} (v_{2})) \\ = (t_{A_{1}} (u_{1}) \times t_{A_{2}} (u_{2})) \times (t_{A_{1}} (v_{1}) \times t_{A_{2}} (v_{2})) \\ = (t_{A_{1}} \times t_{A_{2}}) (u_{1}, u_{2}) \times (t_{A_{1}} \times t_{A_{2}}) (v_{1}, v_{2}) . \end{matrix}$ Similarly we can prove that $\begin{matrix} (f_{B_{1}} \times f_{B_{2}}) ((u_{1}, u_{2}) (v_{1}, v_{2})) & = & (f_{A_{1}} \times f_{A_{2}}) (u_{1}, u_{2}) \\ \times & (f_{A_{1}} \times f_{A_{2}}) (v_{1}, v_{2}) . \end{matrix}$ Therefore, G₁ × G₂ is complete. Conversely, let G₁ × G₂ be complete. We will prove that G₁ and G₂ both are complete. Suppose that G₁ is not complete then there exists u₁, v₁ ∈ V₁ for which one of the following inequalities hold.

t_B
₁ (u₁v₁) < t_A
₁ (u₁) × t_A
₁ (v₁),

f_B
₁ (u₁v₁) > f_A
₁ (u₁) × f_A
₁ (v₁).

Assume that t_{B
₁} (u₁v₁) < t_{A
₁} (u₁) × t_{A
₁} (v₁). Now by considering ((u₁, u₂) , (v₁, v₂)) ∈ (V₁ × V₂) × (V₁ × V₂) we have $\begin{matrix} (t_{B_{1}} \times t_{B_{2}}) ((u_{1}, u_{2}) (v_{1}, v_{2})) \\ = t_{B_{1}} (u_{1} v_{1}) \times t_{B_{2}} (u_{2} v_{2}) \\ < (t_{A_{1}} (u_{1}) \times t_{A_{1}} (v_{1})) \times (t_{A_{2}} (u_{2}) \times t_{A_{2}} (v_{2})) \\ = (t_{A_{1}} (u_{1}) \times t_{A_{2}} (u_{2})) \times (t_{A_{1}} (v_{1}) \times t_{A_{2}} (v_{2})) \\ = (t_{A_{1}} \times t_{A_{2}}) (u_{1}, u_{2}) \times (t_{A_{1}} \times t_{A_{2}}) (v_{1}, v_{2}), \end{matrix}$ which is a contradiction, since that G₁ × G₂ is complete. Similarly if f_{B
₁} (u₁v₁) > f_{A
₁} (u₁) × f_{A
₁} (v₁), a contradiction can be obtained. Hence, G₁ is complete. By the same argument as above we can prove that G₂ is complete. □

Theorem 3.20. Let G = (A, B) be a complete product vague graph where t_A and f_A are normal. Then $t_{B}^{n} (xy) = t_{B} (xy)$ and $f_{B}^{n} (xy) = f_{B} (xy)$ for all x, y ∈ V, in which for all positive integer n ≥ 2,

$t_{B}^{n} (xy) = ⋁_{z \in V} {t_{B}^{n - 1} (xz) \times t_{B} (zy)}$ , $f_{B}^{n} (xy) = ⋁_{z \in V} {f_{B}^{n - 1} (xz) \times f_{B} (zy)}$ .

Proof. We prove by induction on n. Let n = 2. Then for all x, y ∈ V, we have: $\begin{matrix} t_{B}^{2} (xy) & = & ⋁_{z \in V} {t_{B} (xz) \times t_{B} (zy)} \\ = & ⋁_{z \in V} {(t_{A} (x) \times t_{A} (z)) \times (t_{A} (z) \times t_{A} (y))} \\ = & ⋁_{z \in V} {t_{A} (x) \times t_{A} (y) \times t_{A} (z)^{2}} \end{matrix}$ Since t_A (z) ² ≤ 1 for all z (t_A (z) ≤1), $t_{B}^{2} (xy) \leq ⋁_{z \in V} {t_{A} (x) \times t_{A} (y)} = t_{A} (x) \times t_{A} (y)$ .

Hence $t_{B}^{2} (xy) \leq t_{B} (xy) (1)$ .

Since t_A is normal, t_A (e) =1 for some e. Then $\begin{matrix} t_{B}^{2} (xy) & = & ⋁_{z \in V} {t_{A} (x) \times t_{A} (y) \times t_{A} (z)^{2}} \\ \geq & t_{A} (x) \times t_{A} (y) \times t_{A} (e)^{2} \\ = & t_{A} (x) \times t_{A} (y) (t_{A} (e)^{2} = 1) . \end{matrix}$ Therefore, $t_{B}^{2} (xy) \geq t_{B} (xy), (2) [Since t_{B} (xy) = t_{A} (x) \times t_{A} (y)]$ .

From (1) and (2) we get $t_{B}^{2} (xy) = t_{B} (xy)$ . Also, $\begin{matrix} f_{B}^{2} (xy) & = & ⋁_{z \in V} {f_{B} (xz) \times f_{B} (zy)} \\ = & ⋁_{z \in V} {(f_{A} (x) \times f_{A} (z)) \times (f_{A} (z) \times f_{A} (y))} \\ = & ⋁_{z \in V} {f_{A} (x) \times f_{A} (y) \times f_{A} (z)^{2}} \end{matrix}$ Since f_A (z) ² ≤ 1 for all z, $f_{B}^{2} (xy) \geq ⋁_{z \in V} {f_{A} (x) \times f_{A} (y)} = f_{A} (x) \times f_{A} (y) .$ Hence, $f_{B}^{2} (xy) \geq f_{B} (xy) (3)$ .

Since f_A is normal, f_A (e) =1 for some e. Then $f_{B}^{2} (xy) = ⋁_{z \in V} {f_{A} (x) \times f_{A} (y) \times f_{A} (z)^{2}} \leq f_{A} (x) \times f_{A} (y) \times f_{A} (e)^{2} = f_{A} (x) \times f_{A} (y) .$ Hence, $f_{B}^{2} (xy) \leq f_{B} (xy) (4)$ .

From (3) and (4) we get $f_{B}^{2} (xy) = f_{B} (xy)$ . Let $t_{B}^{k} (xy) = t_{B} (xy)$ and $f_{B}^{k} (xy) = f_{B} (xy)$ . We will prove that $t_{B}^{k + 1} (xy) = t_{B} (xy)$ and $f_{B}^{k + 1} (xy) = f_{B} (xy)$ . We have $\begin{matrix} t_{B}^{k + 1} (xy) & = & ⋁_{z \in V} {t_{B}^{k} (xz) \times t_{B} (zy)} \\ = & ⋁_{z \in V} {t_{B} (xz) \times t_{B} (zy)} \\ = & t_{B}^{2} (xy) = t_{B} (xy), \end{matrix}$ $\begin{matrix} f_{B}^{k + 1} (xy) & = & ⋁_{z \in V} {f_{B}^{k} (xz) \times f_{B} (zy)} \\ = & ⋁_{z \in V} {f_{B} (xz) \times f_{B} (zy)} \\ = & f_{B}^{2} (xy) = f_{B} (xy) . □ \end{matrix}$

Theorem 3.21. Let V₁ = {v₁₁, v₁₂, ⋯ , v_1n} and V₂ = {v₂₁, v₂₂, ⋯ , v_2m} be the vertex sets of product vague graphs G₁ and G₂, respectively and G = (A, B) be the multiplication of G₁ and G₂. Then, the following equations have solutions in [0, 1].

x_i × y_j = t_A (v_1i, v_2j) (i = 1, 2, ⋯ , n, j = 1, 2, ⋯ , m)

z_ik × w_jl = t_B ((v_1i, v_2j) (v_1k, v_2l)) (i, k = 1, 2, ⋯ , n, j, l = 1, 2, ⋯ , m)

x_i × y_j = f_A (v_1i, v_2j) (i = 1, 2, ⋯ , n, j = 1, 2, ⋯ , m)

z_ik × w_jl = f_B ((v_1i, v_2j) (v_1k, v_2l)) (i, k = 1, 2, ⋯ , n, j, l = 1, 2, ⋯ , m).

Proof. Let G = (A, B) be the multiplication of product vague graphs G₁ and G₂. Then, (t_A, t_B) = (t_{A
₁} × t_{A
₂}, t_{B
₁} × t_{B
₂}) and (f_A, f_B) = (f_{A
₁} × f_{A
₂}, f_{B
₁} × f_{B
₂}). Now we have $\begin{matrix} t_{A} (v_{1 i}, v_{2 j}) & = & (t_{A_{1}} \times t_{A_{2}}) (v_{1 i}, v_{2 j}) \\ = & t_{A_{1}} (v_{1 i}) \times t_{A_{2}} (v_{2 j}) = x_{i} \times y_{j} \end{matrix}$ where x_i = t_{A
₁} (v_1i) ∈ [0, 1] and y_j = t_{A
₂} (v_2j) ∈ [0, 1]. Also, $\begin{matrix} f_{A} (v_{1 i}, v_{2 j}) & = & (f_{A_{1}} \times f_{A_{2}}) (v_{1 i}, v_{2 j}) \\ = & f_{A_{1}} (v_{1 i}) \times f_{A_{2}} (v_{2 j}) = x_{i} \times y_{j} \end{matrix}$ where x_i = f_{A
₁} (v_1i) ∈ [0, 1] and y_j = f_{A
₂} (v_2j) ∈ [0, 1] this follows that the equations (i) and (iii) have solutions in [0, 1]. If v_1i, v_1k ∈ V₁ and v_2j, v_2l ∈ V₂, then $\begin{matrix} t_{B} ((v_{1 i}, v_{2 j}) (v_{1 k}, v_{2 l})) \\ = (t_{B_{1}} \times t_{B_{2}}) ((v_{1 i}, v_{2 j}) (v_{1 k}, v_{2 l})) \\ = t_{B_{1}} (v_{1 i} v_{1 k}) \times t_{B_{2}} (v_{2 j} v_{2 l}) \\ = z_{ik} \times w_{jl}, \end{matrix}$ where z_ik = t_{B
₁} (v_1iv_1k) ∈ [0, 1] and w_jl = t_{B
₂} (v_2jv_2l) ∈ [0, 1]. Also, we have $\begin{matrix} f_{B_{1}} ((v_{1 i}, v_{2 j}) (v_{1 k}, v_{2 l})) \\ = (f_{B_{1}} \times f_{B_{2}}) ((v_{1 i}, v_{2 j}) (v_{1 k}, v_{2 l})) \\ = f_{B_{1}} (v_{1 i} v_{1 k}) \times f_{B_{2}} (v_{2 j} v_{2 l}) \\ = z_{ik} \times w_{jl}, \end{matrix}$ where z_ik = f_{B
₁} (v_1iv_1k) and w_jl = f_{B
₂} (v_2jv_2l). Therefore, the equations (ii) and (iv) have solutions in [0, 1].

□

Theorem 3.22. Let G^* be a product of two graphs $G_{1}^{*}$ and $G_{2}^{*}$ and G = (A, B) be a product vague graph of G^* where t_B and f_B are normal. Moreover, suppose that the following equations have solutions in [0, 1],

x_i × y_j = t_A (v_1i, v_2j) (i = 1, 2, ⋯ , n, j = 1, 2, ⋯ , m) ,

z_ik × w_jl = t_B ((v_1i, v_2j) (v_1k, v_2l)) (i, k = 1, 2, ⋯ , n, j, l = 1, 2, ⋯ , m) ,

s_i × t_j = f_A (v_1i, v_2j) (i = 1, 2, ⋯ , n, j = 1, 2, ⋯ , m) ,

p_ik × w_jl = f_B ((v_1i, v_2j) (v_1k, v_2l)) (i, k = 1, 2, ⋯ , n, j, l = 1, 2, ⋯ , m).

Then G is the multiplication of a product vague graph of $G_{1}^{*}$ and a product vague graph of $G_{2}^{*}$ .

Proof. Define $\begin{matrix} t_{A_{1}} : V_{1} \to [0, 1], t_{A_{1}} (v_{1 i}) = x_{i}, \\ t_{A_{2}} : V_{2} \to [0, 1], t_{A_{2}} (v_{2 j}) = y_{j}, \\ f_{B_{1}} : V_{1} \times V_{1} \to [0, 1], f_{B_{1}} (v_{1 i} v_{1 k}) = z_{ik}, \\ t_{B_{2}} : V_{2} \times V_{2} \to [0, 1], t_{B_{2}} (v_{2 j} v_{2 l}) = w_{jl}, \\ f_{A_{1}} : V_{1} \to [0, 1], f_{A_{1}} (v_{1 i}) = s_{i}, \\ f_{A_{2}} : V_{2} \to [0, 1], f_{A_{2}} (v_{2 j}) = t_{j}, \\ t_{B_{1}} : V_{1} \times V_{1} \to [0, 1], t_{B_{1}} (v_{1 i} v_{1 k}) = p_{k}, \\ f_{B_{2}} : V_{2} \times V_{2} \to [0, 1], f_{B_{2}} (v_{2 j} v_{2 l}) = p_{jl} . \end{matrix}$ We will prove the following.

G₁ = (A₁, B₁) is a product vague graph of $G_{1}^{*}$ .

G₂ = (A₂, B₂) is a product vague graph of $G_{2}^{*}$ .

t_A = t_A
₁ × t_A
₂, t_B = t_B
₁ × t_B
₂.

f_A = f_A
₁ × f_A
₂, f_B = f_B
₁ × f_B
₂.

If v_1i, v_1k ∈ V₁, then for all v_2j, v_2l ∈ V₂, we have $\begin{matrix} t_{B} ((v_{1 i}, v_{2 j}) (v_{1 k}, v_{2 l})) \leq t_{A} (v_{1 i}, v_{2 j}) \times t_{A} (v_{1 k}, v_{2 l}) \\ = (x_{i} \times y_{j}) \times (x_{k} \times y_{l}) = (x_{i} \times x_{k}) \times (y_{j} \times y_{l}) \\ \leq x_{i} \times x_{k} (Since y_{j}, y_{l} \leq 1) = t_{A_{1}} (v_{1 i}) \times t_{A_{1}} (v_{1 k}) . \end{matrix}$ Hence z_ik × w_jl ≤ t_{A
₁} (v_1i) × t_{A
₁} (v_1k) for all j, l. Since t_B is normal, t_B ((v_1p, v_2a) (v_1q, v_2b)) =1 for some p, q, a and b. Thus z_pq × w_ab = 1 and z_pq = w_ab = 1 (Since z_pq, w_pq ∈ [0, 1]). Replacing j by a and l by b, we get t_{B
₁} (v_1iv_1k) = z_ik = z_ik × w_ab ≤ t_{A
₁} (v_1i) × t_{A
₁} (v_1k). Similarly we can get f_{B
₁} (v_1iv_1k) ≥ f_{A
₁} (v_1i) × f_{A
₁} (v_1k). Therefore, G₁ = (A₁, B₁) is a product vague graph of $G_{1}^{*}$ . Similarly, we can prove that G₂ = (A₂, B₂) is a product vague graph of $G_{2}^{*}$ . Now if v_1i ∈ V₁ and v_2j ∈ V₂, then we have $\begin{matrix} (t_{A_{1}} \times t_{A_{2}}) (v_{1 i}, v_{2 j}) & = & t_{A_{1}} (v_{1 i}) \times t_{A_{2}} (v_{2 j}) = x_{i} \times y_{j} \\ = & t_{A} (v_{1 i}, v_{2 j}) \end{matrix}$ and $\begin{matrix} (f_{A_{1}} \times f_{A_{2}}) (v_{1 i}, v_{2 j}) & = & f_{A_{1}} (v_{1 i}) \times f_{A_{2}} (v_{2 j}) = s_{i} \times t_{j} \\ = & f_{A} (v_{1 i}, v_{2 j}) . \end{matrix}$ It follows that t_A = t_{A
₁} × t_{A
₂} and f_A = f_{A
₁} × f_{A
₂}. If v_1i, v_1k ∈ V₁ and v_2j, v_2l ∈ V₂ then $\begin{matrix} (t_{B_{1}} \times t_{B_{2}}) ((v_{1 i}, v_{2 j}) (v_{1 k}, v_{2 l})) \\ = t_{B_{1}} (v_{1 i} v_{1 k}) \times t_{B_{2}} (v_{2 j} v_{2 l}) \\ = z_{ik} \times w_{jl} = t_{B} ((v_{1 i}, v_{2 j}) (v_{1 k}, v_{2 l})) . \end{matrix}$ Thus t_B = t_{B
₁} × t_{B
₂}. Similarly, we can prove that f_B = f_{B
₁} × f_{B
₂}. □

Definition 3.23. The ring sum of two product vague graphs G₁ and G₂ is denoted by G₁ ⊕ G₂ = (A₁ ⊕ A₂, B₁ ⊕ B₂) where $(t_{A_{1}} \oplus t_{A_{2}}) (u) = (t_{A_{1}} \cup t_{A_{2}}) (u),$ $(f_{A_{1}} \oplus f_{A_{2}}) (u) = (f_{A_{1}} \cup f_{A_{2}}) (u),$ for all u ∈ V₁ ∪ V₂ and $\begin{matrix} {\begin{matrix} (t_{B_{1}} \oplus t_{B_{2}}) (uv) = t_{B_{1}} (uv) if uv \in E_{1} - E_{2} \\ (t_{B_{1}} \oplus t_{B_{2}}) (uv) = t_{B_{2}} (uv) if uv \in E_{2} - E_{1} \\ 0 otherwise . \end{matrix} \end{matrix}$ $\begin{matrix} {\begin{matrix} (f_{B_{1}} \oplus f_{B_{2}}) (uv) = f_{B_{1}} (uv) if uv \in E_{1} - E_{2} \\ (f_{B_{1}} \oplus f_{B_{2}}) (uv) = f_{B_{2}} (uv) if uv \in E_{2} - E_{1} \\ 0 otherwise . \end{matrix} \end{matrix}$

Proposition 3.24. The ring sum of G₁ and G₂ is a product vague graph.

Proof. We have to show that the ring sum of two product vague graphs G = G₁ ⊕ G₂ = (A₁ ⊕ A₂, B₁ ⊕ B₂) is a product vague graph. To prove this we should show that $\begin{matrix} (t_{B_{1}} \oplus t_{B_{2}}) (uv) \leq (t_{A_{1}} \oplus t_{A_{2}}) (u) \times (t_{A_{1}} \oplus t_{A_{2}}) (v), \\ (f_{B_{1}} \oplus f_{B_{2}}) (uv) \geq (f_{A_{1}} \oplus f_{A_{2}}) (u) \times (f_{A_{1}} \oplus f_{A_{2}}) (v) . \end{matrix}$

Case 1. If uv ∈ E₁ - E₂ and u, v ∈ V₁ - V₂, then $\begin{matrix} (t_{B_{1}} \oplus t_{B_{2}}) (uv) & = & t_{B_{1}} (uv) \leq (t_{A_{1}} (u) \times t_{A_{1}} (v)) \\ = & (t_{A_{1}} \cup t_{A_{2}}) (u) \times (t_{A_{1}} \cup t_{A_{2}}) (v) \\ = & (t_{A_{1}} \oplus t_{A_{2}}) (u) \times (t_{A_{1}} \oplus t_{A_{2}}) (v) \end{matrix}$ $\begin{matrix} (f_{B_{1}} \oplus f_{B_{2}}) (uv) & = & f_{B_{1}} (uv) \geq (f_{A_{1}} (u) \times f_{A_{1}} (v)) \\ = & (f_{A_{1}} \cup f_{A_{2}}) (u) \times (f_{A_{1}} \cup f_{A_{2}}) (v) \\ = & (f_{A_{1}} \oplus f_{A_{2}}) (u) \times (f_{A_{1}} \oplus f_{A_{2}}) (v) . \end{matrix}$

Case 2. If uv ∈ E₁ - E₂ and u ∈ V₁ - V₂, v ∈ V₁ ∩ V₂, then $\begin{matrix} (t_{B_{1}} \oplus t_{B_{2}}) (uv) = t_{B_{1}} (uv) \\ \leq (t_{A_{1}} \cup t_{A_{2}}) (u) \times max (t_{A_{1}} (v), t_{A_{2}} (v)) \\ = (t_{A_{1}} \cup t_{A_{2}}) (u) \times (t_{A_{1}} \cup t_{A_{2}}) (v) \\ = (t_{A_{1}} \oplus t_{A_{2}}) (u) \times (t_{A_{1}} \oplus t_{A_{2}}) (v) \end{matrix}$ i.e. (t_{B
₁} ⊕ t_{B
₂}) (uv) ≤ (t_{A
₁} ⊕ t_{A
₂}) (u) × (t_{A
₁} ⊕ t_{A
₂}) (v). Similarly, we can show that

$(f_{B_{1}} \oplus f_{B_{2}}) (uv) \geq (f_{A_{1}} \oplus f_{A_{2}}) (u) \times (f_{A_{1}} \oplus f_{A_{2}}) (v) .$

Case 3. If uv ∈ E₁ - E₂ and u, v ∈ V₁ ∩ V₂, then $\begin{matrix} (t_{B_{1}} \oplus t_{B_{2}}) (uv) = t_{B_{1}} (uv) \\ \leq max (t_{A_{1}} (u), t_{A_{2}} (u)) \times max (t_{A_{1}} (v), t_{A_{2}} (v)) \\ = (t_{A_{1}} \cup t_{A_{2}}) (u) \times (t_{A_{1}} \cup t_{A_{2}}) (v) \\ = (t_{A_{1}} \oplus t_{A_{2}}) (u) \times (t_{A_{1}} \oplus t_{A_{2}}) (v) \end{matrix}$ i.e. (t_{B
₁} ⊕ t_{B
₂}) (uv) ≤ (t_{A
₁} ⊕ t_{A
₂}) (u) × (t_{A
₁} ⊕ t_{A
₂}) (v). It is easy to show that

(f_{B
₁} ⊕ f_{B
₂}) (uv) ≥ (f_{A
₁} ⊕ f_{A
₂}) (u) × (f_{A
₁} ⊕ f_{A
₂}) (v) if u, v ∈ E₁ - E₂ and u, v ∈ V₁ ∩ V₂. Similarly, we can show that if u, v ∈ E₂ - E₁ then (t_{B
₁} ⊕ t_{B
₂}) (uv) ≤ (t_{A
₁} ⊕ t_{A
₂}) (u) × (t_{A
₁} ⊕ t_{A
₂}) (v) and (f_{B
₁} ⊕ f_{B
₂}) (uv) ≥ (f_{A
₁} ⊕ f_{A
₂}) (u) × (f_{A
₁} ⊕ f_{A
₂}) (v).

□

Corollary 3.25. The converse of Proposition 3.24 is not true.

Proof. If xy ∈ E₁ - E₂, x ∈ V₁ - V₂, y ∈ V₁ ∩ V₂ then, the following inequalities are not hold.

t_{B
₁} (xy) ≤ t_{A
₁} (x) × t_{A
₁} (y) and f_{B
₁} (xy) ≥ f_{A
₁} (x) × f_{A
₁} (y).

Note that in this case max (t_{A
₁} (x) , t_{A
₂} (x)) and min

(f_{A
₁} (x) , f_{A
₂} (x)) are not clear. Hence, G₁ is not product vague graph. Similarly we can show that G₂ is not product vague graph too. In the same way we can prove that G₁ and G₂ are not product vague graphs when xy ∈ E₁ - E₂ and x, y ∈ V₁ ∩ V₂. □

Example 3.26. (Figs. 7 and 8) Let G₁ = (A₁, B₁) and G₂ = (A₂, B₂) be two product vague graphs as follows.

The ring sum of G₁ and G₂, that is G₁ ⊕ G₂ is as follows

Remark 3.27. If G = G₁ ⊕ G₂ = (A₁ ⊕ A₂, B₁ ⊕ B₂) is a product vague graph, then we have (A₁ ⊕ A₂) ^c (u) = (A₁ ⊕ A₂) (u) and (B₁ ⊕ B₂) ^c (uv) = (A₁ ⊕ A₂) ^c (u) × (A₁ ⊕ A₂) ^c (v) - (B₁ ⊕ B₂) ^c (uv) = (A₁ ⊕ A₂) (u) × (A₁ ⊕ A₂) (v) - [(A₁ ⊕ A₂) (u) × (A₁ ⊕ A₂) (v) - (B₁ ⊕ B₂) (uv)] = (B₁ ⊕ B₂) (uv) i.e. (B₁ ⊕ B₂) ^c (uv) = (B₁ ⊕ B₂) (uv).

Theorem 3.28. Let G₁ and G₂ be two product vague graphs. Then $(G_{1} + G_{2})^{c} = G_{1}^{c} \oplus G_{2}^{c}$ .

Proof. It is sufficient to show that $(t_{A_{1}} + t_{A_{2}})^{c} (u) = (t_{A_{1}}^{c} \oplus t_{A_{2}}^{c}) (u)$ and $(f_{A_{1}}^{c} + f_{A_{2}}^{c}) (u) = (f_{A_{1}}^{c} \oplus f_{A_{2}}^{c}) (u)$ and $(t_{B_{1}} + t_{B_{2}})^{c} (uv) = (t_{B_{1}}^{c} \oplus t_{B_{2}}^{c}) (uv)$ and $(f_{B_{1}} + f_{B_{2}})^{c} (uv) = (f_{B_{1}}^{c} \oplus f_{B_{2}}^{c}) (uv)$ . We just prove that $(t_{B_{1}} + t_{B_{2}})^{c} (uv) = (t_{B_{1}}^{c} \oplus t_{B_{2}}^{c}) (uv)$ , $(t_{A_{1}} + t_{A_{2}})^{c} (u) = (t_{A_{1}}^{c} \oplus t_{A_{2}}^{c}) (u)$ . Proofs are similar for $(f_{B_{1}} + f_{B_{2}})^{c} (uv) = (f_{B_{1}}^{c} \oplus f_{B_{2}}^{c}) (uv)$ and $(f_{A_{1}} + f_{A_{2}})^{c} (u) = (f_{A_{1}}^{c} \oplus f_{A_{2}}^{c}) (u)$ . Now, $\begin{matrix} (t_{A_{1}} + t_{A_{2}})^{c} (u) & = & (t_{A_{1}} + t_{A_{2}}) (u) \\ = & (t_{A_{1}} \cup t_{A_{2}}) (u) \\ = & (t_{A_{1}}^{c} \oplus t_{A_{2}}^{c}) (u) \end{matrix}$ i.e. $(t_{A_{1}} + t_{A_{2}})^{c} (u) = (t_{A_{1}}^{c} \oplus t_{A_{2}}^{c}) (u)$ . (i)

Next $\begin{matrix} (t_{B_{1}} + t_{B_{2}})^{c} (uv) = (t_{A_{1}} + t_{A_{2}})^{c} (u) \\ \times (t_{A_{1}} + t_{A_{2}})^{c} (v) - (t_{B_{1}} + t_{B_{2}}) (uv) . \end{matrix}$ $\begin{matrix} (t_{B_{1}} + t_{B_{2}})^{c} (uv) = \\ {\begin{matrix} (t_{A_{1}} \cup t_{A_{2}}) (u) \times (t_{A_{1}} \cup t_{A_{2}}) (v) - (t_{B_{1}} \cup t_{B_{2}}) (uv), \\ if uv \in E_{1} \cup E_{2} \\ (t_{A_{1}} (u) \times t_{A_{2}} (v)) - (t_{A_{1}} (u) \times t_{A_{2}} (v)), \\ if u \in V_{1}, v \in V_{2} . \end{matrix} \end{matrix}$ Now, we discuss different cases.

Case 1. If uv ∈ E′ i.e. u ∈ V₁ and v ∈ V₂ then (t_{B
₁} + t_{B
₂}) ^c (uv) = (t_{A
₁} (u) × t_{A
₂} (v)) - (t_{A
₁} (u) × t_{A
₂} (v)) =0 i.e. (t_{B
₁} + t_{B
₂}) ^c (uv) =0.

Case 2. If uv ∈ E₁ - E₂ and u, v ∈ V₁ - V₂ $\begin{matrix} (t_{B_{1}} + t_{B_{2}})^{c} (uv) & = & (t_{A_{1}} (u) \times t_{A_{2}} (v)) - t_{B_{1}} (uv) \\ = & t_{B_{1}}^{c} (uv) \end{matrix}$ i.e. $(t_{B_{1}} + t_{B_{2}})^{c} (uv) = t_{B_{1}}^{c} (uv)$ for all uv ∈ E₁ - E₂.

Case 3. If uv ∈ E₁ - E₂ and u, v ∈ V₁ ∩ V₂ $\begin{matrix} (t_{B_{1}} + t_{B_{2}})^{c} (uv) & = & (t_{A_{1}} \cup t_{A_{2}}) (u) \times (t_{A_{1}} \cup t_{A_{2}}) (v) \\ - & t_{B_{1}} (uv) = t_{B_{1}}^{c} (uv) . \end{matrix}$ Hence, (t_{A
₁} ∪ t_{A
₂}, t_{B
₁}) is product vague graphs i.e. $(t_{B_{1}} + t_{B_{2}})^{c} (uv) = t_{B_{1}}^{c} (uv)$ .

Case 4. If uv ∈ E₁ - E₂ and u ∈ V₁ - V₂, v ∈ V₁ ∩ V₂ $\begin{matrix} (t_{B_{1}} + t_{B_{2}})^{c} (uv) = t_{A_{1}} (u) \times (t_{A_{1}} (v) \cup t_{A_{2}} (v)) \\ = t_{B_{1}} (uv) = (t_{A_{1}} (u) \cup t_{A_{2}} (u)) \times (t_{A_{1}} (v) \\ \times t_{A_{2}} (v)) - t_{B_{1}} (uv) \\ = (t_{A_{1}} \cup t_{A_{2}}) (u) \times (t_{A_{1}} \cup t_{A_{2}}) (v) - t_{B_{1}} (uv) \\ = t_{B_{1}}^{c} (uv) . \end{matrix}$ So, (t_{A
₁} ∪ t_{A
₂}, t_{B
₁}) is a product vague graph i.e. $(t_{B_{1}} + t_{B_{2}})^{c} (uv) = t_{B_{1}}^{c} (uv)$ .

Case 5. If uv ∈ E₂ - E₁ and u, v ∈ V₂ - V₁ then it is obvious that $(t_{B_{1}} + t_{B_{2}})^{c} (uv) = t_{B_{2}}^{c} (uv)$ .

Case 6. If uv ∈ E₂ - E₁ and u, v ∈ V₁ ∩ V₂ then it is obvious that $(t_{B_{1}} + t_{B_{2}})^{c} (uv) = t_{B_{2}}^{c} (uv)$ .

Case 7. If uv ∈ E₂ - E₁ and u ∈ V₂ - V₁, v ∈ V₁ ∩ V₂ then it is obvious that $(t_{B_{1}} + t_{B_{2}})^{c} (uv) = t_{B_{2}}^{c} (uv)$ . Thus from case 1 to 7 we have $\begin{matrix} (t_{B_{1}} + t_{B_{2}})^{c} (uv) & = & {\begin{matrix} t_{B_{1}}^{c} (uv) if uv \in E_{1} - E_{2} \\ t_{B_{2}}^{c} (uv) if uv \in E_{2} - E_{1} \\ 0 otherwise . \end{matrix} \\ = & (t_{B_{1}}^{c} \oplus t_{B_{2}}^{c}) (uv) . \end{matrix}$ i.e. $(t_{B_{1}} + t_{B_{2}})^{c} (uv) = (t_{B_{1}}^{c} \oplus t_{B_{2}}^{c}) (uv)$ . (ii)

Similarly, we can show that $(f_{B_{1}} + f_{B_{2}})^{c} (uv) = (f_{B_{1}}^{c} \oplus f_{B_{2}}^{c}) (uv)$ . (iii)

From (i), (ii) and (iii) it follows that $(G_{1} + G_{2})^{c} ≅ G_{1}^{c} \oplus G_{2}^{c}$ . □

Theorem 3.29. If G is a ring sum of two subgraphs G₁ and G₂ with E₁∩ E₂ = ∅, then every complete product vague subgraph (A, B) of G is ring sum of complete product vague subgraph of G₁ and complete product vague subgraph of G₂.

Proof. We define the vague subsets A₁, A₂, B₁ and B₂ of V₁, V₂, E₁ and E₂ as follow

t_{A
₁} (u) = t_A (u), f_{A
₁} (u) = f_A (u) if u ∈ V₁ - V₂ and t_{A
₂} (u) = t_A (u), f_{A
₂} (u) = f_A (u) if u ∈ V₂ - V₁,t_{B
₁} (uv) = t_B (uv), f_{B
₁} (uv) = f_B (uv) if uv ∈ E₁ - E₂ and t_{B
₂} (uv) = t_B (u), f_{B
₂} (uv) = f_B (uv) if uv ∈ E₂ - E₁.

So (A₁, B₁) is a product vague graph of G₁ and (A₂, B₂) is a product vague graph of G₂ and A = (A₁ ⊕ A₂) is a product vague graphs G₁ and G₂, too. If uv ∈ E₁ ∪ E₂ then $t_{B} (uv) = (t_{B_{1}} \oplus t_{B_{2}}) (uv), f_{B} (uv) = (f_{B_{1}} \oplus f_{B_{2}}) (uv) .$ If uv ∈ E₁ - E₂ then $t_{B} (uv) = (t_{B_{1}} \oplus t_{B_{2}}) (uv), f_{B} (uv) = (f_{B_{1}} \oplus f_{B_{2}}) (uv) .$ (By definition of ring sum of product vague graph).

If uv ∈ E₂ - E₁ then $\begin{matrix} t_{B} (uv) & = & (t_{B_{1}} \oplus t_{B_{2}}) (uv), f_{B} (uv) \\ = & (f_{B_{1}} \oplus f_{B_{2}}) (uv) . \end{matrix}$ If uv ∈ E′ i.e. u ∈ V₁ and v ∈ V₂ then $\begin{matrix} (t_{B_{1}} \oplus t_{B_{2}}) (uv) & = & 0 = t_{B} (uv), (f_{B_{1}} \oplus f_{B_{2}}) (uv) \\ = & 0 = f_{B} (uv) . \end{matrix}$ Other equalities are also holds because (A, B) is a complete product vague graph. □

4 Application in social networks

The social networks are the suitable examples of product vague graphs. In such networks an account of individual or organization or a group of people is taken as node. If there is some relationship between the nodes then they are connected by an edge. In such networks we assume that a node (i.e. a person, organization, etc.) has both membership (good) and negation of membership (bad) activities. It may be observed that, two person have membership attitude for some types of activities (such as teaching method, student evaluation, etc.) and they also have negation of membership mind in some other types of activities (say political view, food habit, etc.). Thus, there are two types of edge membership values, viz. membership and negation of membership. This type of network is an ideal example of product vague graph. Now, we consider a network consisting of four important cities (vertices) in a country. They are interconnected by roads (edges). The network is shown in Fig. 9. The number adjacent to an edge represents the distance between the cities (vertices). The above network can be represented with the help of a classical matrix A = [a_ij], i, j = 1, 2, ⋯ , n, where n is the total number of nodes. The ij th element a_ij of A is defined as $a_{ij} = {\begin{matrix} 0, if i = j \\ \infty, the vertices i and j are not directly \\ connected by an edge \\ w_{ij}, w_{ij} is the distance of the road \\ connecting i and j . \end{matrix}$

Thus the adjacent matrix of the network of Fig. 9 is $[\begin{matrix} 0 & 20 & 30 & 60 \\ 20 & 0 & 50 & 35 \\ 30 & 50 & 0 & 40 \\ 60 & 25 & 40 & 0 \end{matrix}]$ Since the distance between the two vertices are known, precisely, so the matrix A is obviously a classical matrix. Generally, the distance between two cities are crisp value, so the corresponding matrix is crisp matrix. Now we consider the crowdness of the roads connecting cities. It is clear that the crowdness of the roads connecting cities. It is clear that the crowdness of a road obviously, is a fuzzy quantity. The amount of crowdness depends on the decision makers mentality, habits, natures, etc, i.e., completely depends on the decision maker. The measurement of crowdness as a point is a difficult task for the decision maker. So, here we consider amount of crowdness as an interval instead of a point (Table 1).

As the crossing road increases, the possibility of crowdness increases and kill time, so the traveling cost increases. Thus, construct the roads in such a way that the number of crossing degreases. In the network of Fig. 9, there is only one crossing between the roads (1, 4) and (2, 3). To model up the given network as a product vague graph we consider the membership values of each vertices as interval number [0, 1]. Since that the strength of an edge (a, b) is defined by an interval number $I_{(a, b)} = [I_{(a, b)}^{t}, I_{(a, b)}^{f}]$ , where $I_{(a, b)}^{t} = \frac{t_{B} (a, b)}{min (f_{A} (a), f_{A} (b))}$ and $I_{(a, b)}^{f} = \frac{f_{B} (a, b)}{min (f_{A} (a), f_{A} (b))}$ , we have $I_{(1, 4)}^{t} = \frac{t_{B} (1, 4)}{min (f_{A} (1), f_{A} (4))} = \frac{0.4}{1} = 0.4$ . Similarly $I_{(1, 4)}^{f} = \frac{f_{B} (1, 4)}{min (f_{A} (1), f_{A} (4))} = \frac{0.7}{1} = 0.7$ .

$I_{(3, 4)}^{t} = 0.6$ , $I_{(3, 4)}^{f} = 0.7$ , $I_{(2, 3)}^{t} = 0.1$ , $I_{(2, 3)}^{f} = 0.1$ .

5 Conclusion

Graph theory has several interesting applications in system analysis, operations research, computer applications, and economics. Since most of the time the aspects of graph problems are uncertain, it is nice to deal with these aspects via the methods of fuzzy systems. It is known that fuzzy graph theory has numerous applications in modern sciences and technology, especially in the fields of neural networks, artificial intelligence and decision making. In this paper, we defined the notions of product vague graphs, complete product vague graphs and the complement of a product vague graph. We investigated some properties of join, union and complement on product vague graphs and introduced the multiplication of the product vague graphs. In our future work we will focus on categorical properties on vague graphs.

Acknowledgments

The authors would like to thank the referees for their valuable comments and suggestions.

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