Two computational formulae for similarity measures on intuitionistic fuzzy sets based on intuitionistic fuzzy equivalencies

Abstract

The measure of the similarity between intuitionistic fuzzy sets (IFSs) is an important topic in IFSs theory. In this paper, we propose two computational formulae for similarity measures on IFSs based on a quaternary function called intuitionistic fuzzy equivalence. We first propose the concept of intuitionistic fuzzy equivalence. Then we give a computational formula for intuitionistic fuzzy equivalencies (i.e., Eq. (1)), which is obtained from combining dissimilarity functions and fuzzy equivalencies. Based on Eq. (1), we obtain two computational formulae for similarity measures on IFSs. The first one is obtained by aggregating Eq. (1). The second one is obtained by respectively aggregating the numerator and the denominator of Eq. (1). We also examine some properties of the proposed similarity measures on IFSs. Finally, we make a comparison between the proposed similarity measures on IFSs and those existing ones in the literature through several counter-intuitive cases.

Keywords

Intuitionistic fuzzy sets Similarity measures Dissimilarity functions Fuzzy equivalencies Intuitionistic fuzzy equivalencies

1 Introduction

The theory of fuzzy sets (FSs), proposed by Zadeh [36], has achieved a great success in many fields due to its capability of coping with uncertainty. In order to handle the uncertainty that may exist due to information impression, Atanassov [1, 2] proposed the concept of intuitionistic fuzzy set (IFS), which is characterized both by a membership degree and a nonmembership degree. Thus it is a more effective way to deal with vagueness than FS. Although Gau and Buehrer [14] presented the concept of vague set, it was pointed out by Bustince and Burillo [6] that the notion of vague set coincides with that of IFS.

The measure of the similarity between IFSs is an important topic in IFSs theory. In [8], Chen presented two similarity measures for measuring the degree of similarity between vague sets. In [21], Lee-kwang et al. studied similarity measures between fuzzy sets and between elements and considered an example of behavior analysis in an organization. In [9], Chen extended the work of Lee-kwang et al. [21] to apply the vague sets theory in behavior analysis of an organization. In [22], Li and Cheng defined several similarity measures on IFSs and showed how it may be used in pattern recognition. Mitchell [31] showed that the similarity measure of Li and Cheng [22] may give counter-intuitive results and thus presented a modified formula for Li and Cheng’s similarity measure from a statistical viewpoint. Meanwhile, Liang and Shi [29] gave some examples to show that the Li and Cheng’s similarity measure was not always effective in some cases and then proposed several new similarity measures on IFSs. Hung and Yang [16] introduced a new similarity measure on IFSs based on the Hausdorff distance and showed that the proposed similarity measure was much simpler than the existing methods. Li et al. [23] analyzed and compared some existing similarity measures between IFSs through pointing out some weaknesses of them and examining the conditions or reasons they did not work. Xu [33] developed some similarity measures on IFSs and defined the notions of positive ideal IFS and negative ideal IFS and applied the similarity measures to multiple attribute decision making based on intuitionistic fuzzy information. Xu and Chen [34] defined some continuous distance and similarity measures on IFSs based on the weighted Hamming distance, the weighted Euclidean distance, and the weighted Hausdorff distance, respectively. Hung and Yang [18] reviewed several popular similarity measures between FSs and then extended them to IFSs. Ye [35] proposed a cosine similarity measure on IFSs based on the concept of cosine similarity measure for FSs. Hwang et al. [19] proposed a new similarity measure on IFSs induced by the Sugeno integral. Farhadinia [12] presented a new similarity measure on IFSs by making use of the convex combination of endpoints of the membership-degree interval and focusing on the properties of min and max operators. Chen et al. [10, 11] proposed a novel similarity measure between IFSs based on the centroid points of the transformed right-angled triangular fuzzy numbers for handling pattern recognition problems. Beg and Rashid [3] proposed a new measure of similarity between IFSs based on the intuitionistic fuzzy inclusion measure.

Although similarity measures on IFSs have been widely studied in previous studies, there are few studies about unified forms of similarity measures on IFSs. In this paper, we propose two computational formulae for similarity measures on IFSs based on a quaternary function called intuitionistic fuzzy equivalence. We propose the concept of intuitionistic fuzzy equivalence. We give a computational formula for intuitionistic fuzzy equivalencies (i.e., Eq. (1)), which are obtained from combining dissimilarity functions and fuzzy equivalencies. By taking different fuzzy equivalencies and dissimilarity functions in Eq. (1), we obtain different intuitionistic fuzzy equivalencies. Based on Eq. (1), we obtain two computational formulae for similarity measures on IFSs. The first one is obtained by aggregating Eq. (1). The second one is obtained by respectively aggregating the numerator and the denominator of Eq. (1). We examine some properties of the proposed similarity measures on IFSs. We also make a comparison between the proposed similarity measures on IFSs and those existing ones through several counter-intuitive cases.

The rest of the paper is organized as follows. In Section 2, we briefly describe some basic definitions and notions related to FSs and IFSs. In Section 3, we define a quaternary function called intuitionistic fuzzy equivalence to construct similarity measures on IFSs. We present a computational formula for intuitionistic fuzzy equivalencies. Based on this computational formula, we obtain two computational formulae for similarity measures on IFSs. We also examine and compare some properties of the proposed similarity measures on IFSs. In Section 4, we make a comparison between the proposed similarity measures on IFSs and those existing ones through several counter-intuitive cases. In Section 4, we conclude the work along with some comments on possible future works.

2 Preliminaries

In this section, we briefly describe some basic definitions and notions related to FSs and IFSs. Throughout this paper, we use X = {x₁, x₂, …, x_n} to denote the discourse set. In order to avoid the denominators of some functions being zero, we set $\frac{0}{0} = 1$ .

Definition 1. [36] A fuzzy set $\tilde{A}$ in X is defined as $\begin{matrix} \tilde{A} = {〈 x_{i}, \tilde{A} (x_{i}) 〉 : x_{i} \in X}, \end{matrix}$ where $\tilde{A} : X ⟶ [0, 1]$ is the membership function of $\tilde{A}$ and $\tilde{A} (x_{i})$ is the degree of membership of x_i ∈ X in $\tilde{A}$ . The complement set of $\tilde{A}$ is ${\tilde{A}}^{c}$ and defined as ${\tilde{A}}^{c} = {〈 x_{i}, 1 - \tilde{A} (x_{i}) 〉 : x_{i} \in X}$ .

Definition 2. [1] An intuitionistic fuzzy set A in X is defined as $\begin{matrix} A = {〈 x_{i}, \hat{A} (x_{i}), \overset{ˇ}{A} (x_{i}) 〉 : x_{i} \in X}, \end{matrix}$ where $\hat{A}, \overset{ˇ}{A} : X ⟶ [0, 1]$ are such that $0 \leq \hat{A} (x_{i}) + \overset{ˇ}{A} (x_{i}) \leq 1$ for all x_i ∈ X. The numbers $\hat{A} (x_{i})$ and $\overset{ˇ}{A} (x_{i})$ respectively represent the degrees of membership and nonmembership of x_i ∈ X in A.

The complement set of A is A^c and defined as $A^{c} = {〈 x_{i}, \overset{ˇ}{A} (x_{i}), \hat{A} (x_{i}) 〉 : x_{i} \in X}$ . We denote all the FSs and IFSs in X by FS (X) and IFS (X), respectively. The intuitionistic fuzzy index (or hesitancy degree) [1, 2] of an element x_i ∈ X in A is defined as $\bar{A} (x_{i}) = 1 - \hat{A} (x_{i}) - \overset{ˇ}{A} (x_{i})$ with $\bar{A} (x_{i}) \in [0, 1]$ .

Definition 3. [1] If A, B ∈ IFS (X) defined by $A = {〈 x_{i}, \hat{A} (x_{i}), \overset{ˇ}{A} (x_{i}) 〉 : x_{i} \in X}$ and $B = {〈 x_{i}, \hat{B} (x_{i}), \overset{ˇ}{B} (x_{i}) 〉 : x_{i} \in X}$ , then for each x_i ∈ Xčň

$A \subseteq B \Leftrightarrow \hat{A} (x_{i}) \leq \hat{B} (x_{i})$ and $\overset{ˇ}{A} (x_{i}) \geq \overset{ˇ}{B} (x_{i})$ .

$A = B \Leftrightarrow \hat{A} (x_{i}) = \hat{B} (x_{i})$ and $\overset{ˇ}{A} (x_{i}) = \overset{ˇ}{B} (x_{i})$ .

$A = \emptyset \Leftrightarrow \hat{A} (x_{i}) = 0$ and $\overset{ˇ}{A} (x_{i}) = 1$ .

$A \cup B = {〈 x_{i}, max (\hat{A} (x_{i}), \hat{B} (x_{i})), min (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i})) 〉}$ .

$A \cap B = {〈 x_{i}, min (\hat{A} (x_{i}), \hat{B} (x_{i})), max (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i})) 〉}$ .

Definition 4. [7] A continuous, strictly increasing function φ : [0, 1] ⟶ [0, 1] with φ (0) =0 and φ (1) =1 is called an automorphism of [0, 1].

Definition 5. [7] If a decreasing function n : [0, 1] ⟶ [0, 1] satisfies the boundary conditions n (0) =1 and n (1) =0, then n is called a fuzzy negation.

Definition 6. [13] A binary function E : [0, 1] ² ⟶ [0, 1] is called a fuzzy equivalence if it satisfies the following properties:

E (α, β) = E (β, α) for all α, β ∈ [0, 1].

E (α, α) =1 for each α ∈ [0, 1].

E (1, 0) =0.

For all α, β, γ ∈ [0, 1], if α ≤ β ≤ γ, then min(E (α, β) , E (β, γ)) ≥ E (α, γ).

The following properties can be considered for fuzzy equivalencies [24 , 27].

E (α, β) =1 ⇔ α = β for all α, β ∈ [0, 1].

E (α, β) =0 ⇔ min(α, β) =0 and max(α, β) ≠0 for all α, β ∈ [0, 1].

E (α, 1) = α for each α ∈ [0, 1].

E (α, β) = E (1 - α, 1 - β) for all α, β ∈ [0, 1].

Example 1. [13] As examples of E, we can take

E₁ (α, β) =1 - |α - β| for all α, β ∈ [0, 1].

$E_{2} (α, β) = \frac{min (α, β)}{max (α, β)}$ for all α, β ∈ [0, 1].

Lemma 1. [30] Given a fuzzy equivalence E, for all α, β, γ ∈ [0, 1] we have

E (max(α, γ) , max(β, γ)) ≥ E (α, β).

E (min(α, γ) , min(β, γ)) ≥ E (α, β).

Definition 7. [32] A function N : FS (X) × FS (X) ⟶ [0, 1] is called a similarity measure on FSs if N satisfies the following properties:

N (X, ∅) =0 and $N (\tilde{A}, \tilde{A}) = 1$ for all $\tilde{A} \in FS (X)$ .

$N (\tilde{A}, \tilde{B}) = N (\tilde{B}, \tilde{A})$ whenever $\tilde{A}, \tilde{B} \in FS (X)$ .

For all $\tilde{A}, \tilde{B}, \tilde{C} \in FS (X)$ , $N (\tilde{A}, \tilde{C}) \leq min (N (\tilde{A}, \tilde{B}), N (\tilde{B}, \tilde{C}))$ whenever $\tilde{A} \subseteq \tilde{B} \subseteq \tilde{C}$ .

Definition 8. [20] A function M : [0, 1] ⁿ ⟶ [0, 1] is called an aggregation function if it satisfies the following properties:

$M (\underset{n times}{\underset{︸}{0, \dots, 0}}) = 0$ .

M (1, …, 1) =1.

M (α, …, α) = α.

M is monotonically increasing in all of its arguments.

Example 2. [20] As examples of M, we can take

$M_{1} (α_{1}, \dots, α_{n}) = \frac{1}{n} \sum_{i = 1}^{n} α_{i}$ .

$M_{2} (α_{1}, \dots, α_{n}) = \frac{1}{\sqrt{n}} \sqrt{\sum_{i = 1}^{n} α_{i}^{2}}$ .

$M_{3} (α_{1}, \dots, α_{n}) = \frac{1}{\sqrt[p]{n}} \sqrt[p]{\sum_{i = 1}^{n} α_{i}^{p}}$ with p ∈ [1, + ∞).

$M_{4} (α_{1}, \dots, α_{n}) = \sum_{i = 1}^{n} ω_{i} α_{i}$ with ω_i ∈ [0, 1].

$M_{5} (α_{1}, \dots, α_{n}) = \sqrt[p]{\sum_{i = 1}^{n} ω_{i} α_{i}^{p}}$ with p ∈ [1, + ∞) , ω_i ∈ [0, 1].

Lemma 2. [25] Let E be a fuzzy equivalence and M be an aggregation function. Suppose N : FS (X) × FS (X) ⟶ [0, 1] is a function defined for all $\tilde{A}, \tilde{B} \in FS (X)$ by $\begin{matrix} N (\tilde{A}, \tilde{B}) = M_{i = 1}^{n} E (\tilde{A} (x_{i}), \tilde{B} (x_{i})), \end{matrix}$ then N is a similarity measure on FSs.

Definition 9. [22] A function $N : IFS (X) \times IFS (X) ⟶ [0, 1]$ is called a similarity measure on IFSs if $N$ satisfies the following properties:

$N (A, A) = 1$ whenever A ∈ IFS (X).

$N (A, B) = N (B, A)$ whenever A, B ∈ IFS (X).

For all A, B, C ∈ IFS (X), $N (A, C) \leq min (N (A, B), N (B, C))$ whenever A ⊆ B ⊆ C.

Definition 10. [26] A function d : [0, 1] ² ⟶ [0, 1] is called a dissimilarity function if it satisfies the following properties:

d (α, β) = d (β, α) for all α, β ∈ [0, 1].

d (α, α) =0 for all α ∈ [0, 1].

d (1, 0) =1.

For all α, β, γ ∈ [0, 1], if α ≤ β ≤ γ, then max(d (α, β) , d (β, γ)) ≤ d (α, γ).

The following properties can be considered for dissimilarity functions [26, 28].

d (α, β) =0 ⇔ α = β for all α, β ∈ [0, 1].

d (α, 0) = α for each α ∈ [0, 1].

d (α, 1) =1 - α for each α ∈ [0, 1].

d (α, β) = d (1 - α, 1 - β) for all α, β ∈ [0, 1].

Lemma 3. [26] Given a dissimilarity function d, for all α, β, γ ∈ [0, 1] we have

d (max(α, γ) , max(β, γ)) ≤ d (α, β).

d (min(α, γ) , min(β, γ)) ≤ d (α, β).

Example 3. [26] As examples of d, we can take

d₁ (α, β) = |α - β| for all α, β ∈ [0, 1].

$d_{2} (α, β) = \frac{| α - β |}{max (α, β)}$ for all α, β ∈ [0, 1].

3 Two computational formulae for similarity measures on IFSs based on intuitionistic fuzzy equivalencies

Definition 11. A quaternary function $E : [0, 1]^{4} ⟶ [0, 1]$ is called an intuitionistic fuzzy equivalence if $E$ satisfies the following properties:

$E (α, β, γ, δ) = E (γ, δ, α, β)$ .

$E (α, β, α, β) = 1$ .

If α ≤ γ₁ ≤ γ₂ and β ≥ δ₁ ≥ δ₂, then $E (α, β, γ_{1}, δ_{1}) \geq E (α, β, γ_{2}, δ_{2})$ . If α₂ ≤ α₁ ≤ γ and β₂ ≥ β₁ ≥ δ, then $E (α_{1}, β_{1}, γ, δ) \geq E (α_{2}, β_{2}, γ, δ)$ .

Besides above-mentioned properties in its concept, we can consider other properties for an intuitionistic fuzzy equivalence. For all α, β, γ, δ, χ, ζ ∈ [0, 1],

$E (α, β, γ, δ) = 1 \Leftrightarrow α = γ$ and β = δ.

$E (1, 0, 0, 1) = 0$ .

$E (α, β, γ, δ) = E (max (α, γ), min (β, δ), min (α, γ), max (β, δ))$ .

$E (α, β, γ, δ) = E (β, α, δ, γ)$ .

$min (E (α, β, α, δ), E (α, β, γ, β)) \geq E (α, β, γ, δ)$ .

$E (α, β, α, δ) = E (χ, β, χ, δ)$ and $E (α, β, γ, β) = E (α, ζ, γ, ζ)$ for all α, β, γ, δ, χ, ζ ∈ [0, 1].

$E (max (α, χ), min (β, ζ), max (γ, χ), min (δ, ζ)) \geq E (α, β, γ, δ)$ .

$E (min (α, χ), max (β, ζ), min (γ, χ), max (δ, ζ)) \geq E (α, β, γ, δ)$ .

Proposition 1 shows that properties $E 10$ and $E 11$ are direct consequences of properties $E 8$ and $E 9$ .

Proposition 1. Let $E$ be an intuitionistic fuzzy equivalence. If $E$ satisfies $E 8$ and $E 9$ , then it satisfies $E 10$ and $E 11$ .

Proof. For arbitrary α, γ, χ, there exist six permutations α ≤ γ ≤ χ, α ≤ χ ≤ γ, χ ≤ α ≤ γ, γ ≤ α ≤ χ, γ ≤ χ ≤ α and χ ≤ γ ≤ α. For arbitrary β, δ, ζ, there exist six permutations β ≤ δ ≤ ζ, β ≤ ζ ≤ δ, ζ ≤ β ≤ δ, δ ≤ β ≤ ζ, δ ≤ ζ ≤ β and ζ ≤ δ ≤ β. Since $E (α, β, γ, δ) = E (γ, δ, α, β)$ , it is enough to consider the preceding three cases for α, γ, χ and those for β, δ, ζ. Let $E (max (α, χ), min (β, ζ), max (γ, χ), min (δ, ζ)) = Δ_{1}$ and $E (min (α, χ), max (β, ζ), min (γ, χ), max (δ, ζ)) = Δ_{2}$ . Table 1 lists values of Δ₁ and Δ₂ under different conditions.

Table 1
The values of Δ₁ and Δ₂ under different conditions

α, γ, χ β, δ, ζ Δ ₁ Δ ₂

α ≤ γ ≤ χ β ≤ δ ≤ ζ $E (χ, β, χ, δ)$ $E (α, ζ, γ, ζ)$

α ≤ γ ≤ χ β ≤ ζ ≤ δ $E (χ, β, χ, ζ)$ $E (α, ζ, γ, δ)$

α ≤ γ ≤ χ ζ ≤ β ≤ δ $E (χ, ζ, χ, ζ)$ $E (α, β, γ, δ)$

α ≤ χ ≤ γ β ≤ δ ≤ ζ $E (χ, β, γ, δ)$ $E (α, ζ, χ, ζ)$

α ≤ χ ≤ γ β ≤ ζ ≤ δ $E (χ, β, γ, ζ)$ $E (α, ζ, χ, δ)$

α ≤ χ ≤ γ ζ ≤ β ≤ δ $E (χ, ζ, γ, ζ)$ $E (α, β, χ, δ)$

χ ≤ α ≤ γ β ≤ δ ≤ ζ $E (α, β, γ, δ)$ $E (χ, ζ, χ, ζ)$

χ ≤ α ≤ γ β ≤ ζ ≤ δ $E (α, β, γ, ζ)$ $E (χ, ζ, χ, δ)$

χ ≤ α ≤ γ ζ ≤ β ≤ δ $E (α, ζ, γ, ζ)$ $E (χ, β, χ, δ)$

α, γ, χ	β, δ, ζ	Δ ₁	Δ ₂
α ≤ γ ≤ χ	β ≤ δ ≤ ζ	$E (χ, β, χ, δ)$	$E (α, ζ, γ, ζ)$
α ≤ γ ≤ χ	β ≤ ζ ≤ δ	$E (χ, β, χ, ζ)$	$E (α, ζ, γ, δ)$
α ≤ γ ≤ χ	ζ ≤ β ≤ δ	$E (χ, ζ, χ, ζ)$	$E (α, β, γ, δ)$
α ≤ χ ≤ γ	β ≤ δ ≤ ζ	$E (χ, β, γ, δ)$	$E (α, ζ, χ, ζ)$
α ≤ χ ≤ γ	β ≤ ζ ≤ δ	$E (χ, β, γ, ζ)$	$E (α, ζ, χ, δ)$
α ≤ χ ≤ γ	ζ ≤ β ≤ δ	$E (χ, ζ, γ, ζ)$	$E (α, β, χ, δ)$
χ ≤ α ≤ γ	β ≤ δ ≤ ζ	$E (α, β, γ, δ)$	$E (χ, ζ, χ, ζ)$
χ ≤ α ≤ γ	β ≤ ζ ≤ δ	$E (α, β, γ, ζ)$	$E (χ, ζ, χ, δ)$
χ ≤ α ≤ γ	ζ ≤ β ≤ δ	$E (α, ζ, γ, ζ)$	$E (χ, β, χ, δ)$

We only prove the first three cases. If α ≤ γ ≤ χ and β ≤ δ ≤ ζ, then $Δ_{1} = E (χ, β, χ, δ)$ and $Δ_{2} = E (α, ζ, γ, ζ)$ . Since $E$ satisfies $E 9$ , we have $Δ_{1} = E (α, β, α, δ)$ and $Δ_{2} = E (α, β, γ, β)$ . According to $E 8$ , we have $min (E (α, β, α, δ), E (α, β, γ, β)) \geq E (α, β, γ, δ)$ for all α, β, γ, δ ∈ [0, 1] and thus $min (Δ_{1}, Δ_{2}) \geq E (α, β, γ, δ)$ . If α ≤ γ ≤ χ and β ≤ ζ ≤ δ, then $Δ_{1} = E (χ, β, χ, ζ)$ and $Δ_{2} = E (α, ζ, γ, δ) \geq E (α, β, γ, δ)$ . Since $E$ satisfies $E 8$ , we have $Δ_{1} = E (α, β, α, ζ) \geq E (α, β, α, δ) \geq E (α, β, γ, δ)$ . Thus $min (Δ_{1}, Δ_{2}) \geq E (α, β, γ, δ)$ . If α ≤ γ ≤ χ and ζ ≤ β ≤ δ, then $Δ_{1} = E (χ, ζ, χ, ζ) = 1$ and $Δ_{2} = E (α, β, γ, δ)$ . Thus $min (Δ_{1}, Δ_{2}) = E (α, β, γ, δ)$ . The other cases can be proved similarly. □

Now we give a computational formula for intuitionistic fuzzy equivalencies.

Theorem 1. Let E and d be a fuzzy equivalence and a dissimilarity function, respectively. Suppose that $E$ is a function defined for all α, β, γ, δ ∈ [0, 1] by

$\begin{matrix} E (α, β, γ, δ) = \\ \frac{a (E (α, γ) + E (β, δ)) + b (min (α, γ) + min (β, δ))}{a (E (α, γ) + E (β, δ)) + b (min (α, γ) + min (β, δ)) + c (d (α, γ) + d (β, δ))} \end{matrix}$ (1) with a ≥ 0, b ≥ 0, c > 0 and max(a, b) ≠0, then $E$ is an intuitionistic fuzzy equivalence.

Proof. Since 0 ≤ a (E (α, γ) + E (β, δ)) + b (min(α, γ) + min(β, δ)) ≤ a (E (α, γ) + E (β, δ)) + b (min(α, γ) + min(β, δ)) + c (d (α, γ) + d (β, δ)), we have $0 \leq E (α, β, γ, δ) \leq 1$ . Note that $E (α, β, γ, δ) = E (γ, δ, α, β)$ and $E (α, β, α, β) = 1$ . If α ≤ γ₁ ≤ γ₂ and β ≥ δ₁ ≥ δ₂, then E (α, γ₁) ≥ E (α, γ₂) and E (β, δ₁) ≥ E (β, δ₂), d (α, γ₁) ≤ d (α, γ₂) and d (β, δ₁) ≤ d (β, δ₂). Note that

$\begin{matrix} E (α, β, γ_{1}, δ_{1}) = \\ \frac{a (E (α, γ_{1}) + E (β, δ_{1})) + b (α + δ_{1})}{a (E (α, γ_{1}) + E (β, δ_{1})) + b (α + δ_{1}) + c (d (α, γ_{1}) + d (β, δ_{1}))}, \\ E (α, β, γ_{2}, δ_{2}) = \\ \frac{a (E (α, γ_{2}) + E (β, δ_{2})) + b (α + δ_{2})}{a (E (α, γ_{2}) + E (β, δ_{2})) + b (α + δ_{2}) + c (d (α, γ_{2}) + d (β, δ_{2}))}, \end{matrix}$

and a (E (α, γ₁) + E (β, δ₁)) + b (α + δ₁) ≥ a (E (α, γ₂) + E (β, δ₂)) + b (α + δ₂). If $E (α, β, γ_{2}, δ_{2}) = 0$ , then $E (α, β, γ_{1}, δ_{1}) \geq E (α, β, γ_{2}, δ_{2})$ . If $E (α, β, γ_{2}, δ_{2}) \neq 0$ , then $E (α, β, γ_{1}, δ_{1}) \neq 0$ . In this case,

$\begin{matrix} \frac{1}{E (α, β, γ_{1}, δ_{1})} & = & 1 + \frac{c (d (α, γ_{1}) + d (β, δ_{1}))}{a (E (α, γ_{1}) + E (β, δ_{1})) + b (α + δ_{1})}, \\ \frac{1}{E (α, β, γ_{2}, δ_{2})} & = & 1 + \frac{c (d (α, γ_{2}) + d (β, δ_{2}))}{a (E (α, γ_{2}) + E (β, δ_{2})) + b (α + δ_{2})} . \end{matrix}$

Since c (d (α, γ₁) + d (β, δ₁)) ≤ c (d (α, γ₂) + d (β, δ₂)), we have $\frac{1}{E (α, β, γ_{1}, δ_{1})} \leq \frac{1}{E (α, β, γ_{2}, δ_{2})}$ and thus $E (α, β, γ_{1}, δ_{1}) \geq E (α, β, γ_{2}, δ_{2})$ . If α₂ ≤ α₁ ≤ γ and β₂ ≥ β₁ ≥ δ, then E (α₁, γ) ≥ E (α₂, γ) and E (β₁, δ) ≥ E (β₂, δ), d (α₁, γ) ≤ d (α₂, γ) and d (β₁, δ) ≤ d (β₂, δ). Note that

$\begin{matrix} E (α_{1}, β_{1}, γ, δ) = \\ \frac{a (E (α_{1}, γ) + E (β_{1}, δ)) + b (α_{1} + δ)}{a (E (α_{1}, γ) + E (β_{1}, δ)) + b (α_{1} + δ) + c (d (α_{1}, γ) + d (β_{1}, δ))}, \\ E (α_{2}, β_{2}, γ, δ) = \\ \frac{a (E (α_{2}, γ) + E (β_{2}, δ)) + b (α_{2} + δ)}{a (E (α_{2}, γ) + E (β_{2}, δ)) + b (α_{2} + δ) + c (d (α_{2}, γ) + d (β_{2}, δ))}, \end{matrix}$

and a (E (α₁, γ) + E (β₁, δ)) + b (α₁ + δ) ≥ a (E (α₂, γ) + E (β₂, δ)) + b (α₂ + δ). If $E (α_{2}, β_{2}, γ, δ) = 0$ , then $E (α_{1}, β_{1}, γ, δ) \geq E (α_{2}, β_{2}, γ, δ)$ . If $E (α_{2}, β_{2}, γ, δ) \neq 0$ , then $E (α_{1}, β_{1}, γ, δ) \neq 0$ . In this case, $\begin{matrix} \frac{1}{E (α_{1}, β_{1}, γ, δ)} & = & 1 + \frac{c (d (α_{1}, γ) + d (β_{1}, δ))}{a (E (α_{1}, γ) + E (β_{1}, δ)) + b (α_{1} + δ)}, \\ \frac{1}{E (α_{2}, β_{2}, γ, δ)} & = & 1 + \frac{c (d (α_{2}, γ) + d (β_{2}, δ))}{a (E (α_{2}, γ) + E (β_{2}, δ)) + b (α_{2} + δ)} . \end{matrix}$ Since c (d (α₁, γ) + d (β₁, δ)) ≤ c (d (α₂, γ) + d (β₂, δ)), we have $\frac{1}{E (α_{1}, β_{1}, γ, δ)} \leq \frac{1}{E (α_{2}, β_{2}, γ, δ)}$ and thus $E (α_{1}, β_{1}, γ, δ) \geq E (α_{2}, β_{2}, γ, δ)$ . □

Remark 1. For the intuitionistic fuzzy equivalence defined by Eq. (1), we have $E (0, 0, 0, 0) = \frac{a}{a}$ . If a ≠ 0, then $E (0, 0, 0, 0) = 1$ . If a = 0, then $E (0, 0, 0, 0) = \frac{0}{0} = 1$ since we have set $\frac{0}{0} = 1$ to avoid the denominators of some functions being zero in Section 2.

By taking different fuzzy equivalencies and dissimilarity functions in Eq. (1), we can obtain different intuitionistic fuzzy equivalencies. For the fixed fuzzy equivalence and dissimilarity function, there exist monotonic relationships between the values of intuitionistic fuzzy equivalencies and the values of their parameters.

Proposition 2. Let E and d be a fuzzy equivalence and a dissimilarity function. Suppose that α, β, γ and δ are constants in [0, 1]. If we define a function f as $f (a, b, c) = \frac{a (E (α, γ) + E (β, δ)) + b (min (α, γ) + min (β, δ))}{a (E (α, γ) + E (β, δ)) + b (min (α, γ) + min (β, δ)) + c (d (α, γ) + d (β, δ))}$ with a ≥ 0, b ≥ 0, c > 0 and max(a, b) ≠0, then f is increasing with respect to a, b and decreasing with respect to c.

Proof. Let us denote E (α, γ) + E (β, δ) , min(α, γ) + min(β, δ) and d (α, γ) + d (β, δ) by three constants C₁, C₂ and C₃, respectively, then 0 ≤ C₁ ≤ 2, 0 ≤ C₂ ≤ 2, 0 ≤ C₃ ≤ 2 and thus $f (a, b, c) = \frac{{aC}_{1} + {bC}_{2}}{{aC}_{1} + {bC}_{2} + {cC}_{3}}$ . Then we have $\frac{\partial f}{\partial a} = \frac{{cC}_{1} C_{3}}{({aC}_{1} + {bC}_{2} + {cC}_{3})^{2}} \geq 0$ , $\frac{\partial f}{\partial b} = \frac{{cC}_{2} C_{3}}{({aC}_{1} + {bC}_{2} + {cC}_{3})^{2}} \geq 0$ and $\frac{\partial f}{\partial c} = \frac{- C_{3} ({aC}_{1} + {bC}_{2})}{({aC}_{1} + {bC}_{2} + {cC}_{3})^{2}} \leq 0$ . Thus f is increasing with respect to a, b and decreasing with respect to c. □

Let us examine some properties of intuitionistic fuzzy equivalencies defined by Eq. (1).

Proposition 3. Let $E$ be an intuitionistic fuzzy equivalence defined by Eq. (1). If d satisfies d5, then $E$ satisfies $E 4$ .

Proof. $\frac{a (E (α, γ) + E (β, δ)) + b (min (α, γ) + min (β, δ))}{a (E (α, γ) + E (β, δ)) + b (min (α, γ) + min (β, δ)) + c (d (α, γ) + d (β, δ))} = 1 \Leftrightarrow c (d (α, γ) + d (β, δ)) = 0$ . Since d satisfies d5, we have c (d (α, γ) + d (β, δ)) =0 ⇔ α = γ and β = δ for all α, β, γ, δ ∈ [0, 1].

Proposition 4. Let $E$ be an intuitionistic fuzzy equivalence defined by Eq. (1). Then $E$ satisfies $E 5, E 6, E 7$ and $E 8$ .

Proof.

$\frac{a (E (1, 0) + E (0, 1)) + b (min (1, 0) + min (0, 1))}{a (E (1, 0) + E (0, 1)) + b (min (1, 0) + min (0, 1)) + c (d (1, 0) + d (0, 1))} = 0$ .

Since E (max(α, γ) , min(α, γ)) = E (α, γ) , E (min(β, δ) , max(β, δ)) = E (β, δ), min(min(α, γ) , max(α, γ)) = min(α, γ) , min(min(β, δ) , max(β, δ)) = min(β, δ) , d (max(α, γ) , min(α, γ)) = d (α, γ) and d (min(β, δ) , max(β, δ)) = d (β, δ), we have $E (α, β, γ, δ) = E (max (α, γ), min (β, δ), min (α, γ), max (β, δ))$ for all α, β, γ, δ ∈ [0, 1].

$\frac{a (E (α, γ) + E (β, δ)) + b (min (α, γ) + min (β, δ))}{a (E (α, γ) + E (β, δ)) + b (min (α, γ) + min (β, δ)) + c (d (α, γ) + d (β, δ))} = \frac{a (E (β, δ) + E (α, γ)) + b (min (β, δ) + min (α, γ))}{a (E (β, δ) + E (α, γ)) + b (min (β, δ) + min (α, γ)) + c (d (β, δ) + d (α, γ))}$ for all α, β, γ, δ ∈ [0, 1].

If $E (α, β, γ, δ) = 0$ , then $min (E (α, β, α, δ), E (α, β, γ, β)) \geq E (α, β, γ, δ)$ . Note that $E (α, β, α, δ) = \frac{a (1 + E (β, δ)) + b (α + min (β, δ))}{a (1 + E (β, δ)) + b (α + min (β, δ)) + cd (β, δ)}$ and $E (α, β, γ, β) = \frac{a (E (α, γ) + 1) + b (min (α, γ) + β)}{a (E (α, γ) + 1) + b (min (α, γ) + β) + cd (α, γ)}$ . If $E (α, β, γ, δ) \neq 0$ , then we have $min (E (α, β, α, δ), E (α, β, γ, β)) \neq 0$ . Thus $\frac{1}{E (α, β, α, δ)} = 1 + \frac{cd (β, δ)}{a (1 + E (β, δ)) + b (α + min (β, δ))}$ , $\frac{1}{E (α, β, γ, β)} = 1 + \frac{cd (α, γ)}{a (E (α, γ) + 1) + b (min (α, γ) + β)}$ and $\frac{1}{E (α, β, γ, δ)} = 1 + \frac{c (d (α, γ) + d (β, δ))}{a (E (α, γ) + E (β, δ)) + b (min (α, γ) + min (β, δ))}$ . It is obtained that $\frac{1}{E (α, β, γ, δ)} \geq max (\frac{1}{E (α, β, α, δ)}, \frac{1}{E (α, β, γ, β)})$ . Therefore, we have $min (E (α, β, α, δ), E (α, β, γ, β)) \geq E (α, β, γ, δ)$ for all α, β, γ, δ ∈ [0, 1]. □

Proposition 5. Let $E$ be an intuitionistic fuzzy equivalence defined by Eq. (1). If b = 0, then $E$ satisfies $E 9, E 10$ and $E 11$ .

Proof.

If b = 0, then we have $E (α, β, γ, δ) = \frac{a (E (α, γ) + E (β, δ))}{a (E (α, γ) + E (β, δ)) + c (d (α, γ) + d (β, δ))}$ .

$E (α, β, α, δ) = \frac{a (1 + E (β, δ))}{a (1 + E (β, δ)) + cd (β, δ)}$ and $E (χ, β, χ, δ) = \frac{a (1 + E (β, δ))}{a (1 + E (β, δ)) + cd (β, δ)}$ . Thus $E (α, β, α, δ) = E (χ, β, χ, δ)$ and $E (α, β, γ, β) = E (α, ζ, γ, ζ)$ can be proved similarly.

Since $E$ satisfies $E 8$ and $E 9$ , according to Proposition 3, we have $E$ satisfies $E 10$ .

Since $E$ satisfies $E 8$ and $E 9$ , according to Proposition 3, we have $E$ satisfies $E 11$ . □

In the following, we introduces two computational formulae for similarity measures on IFSs.

Lemma 4. Let M be an aggregation function and let $E$ be an intuitionistic fuzzy equivalence. Suppose $N$ is a function defined for all A, B ∈ IFS (X) by $\begin{matrix} N (A, B) = M_{i = 1}^{n} E (\hat{A} (x_{i}), \overset{ˇ}{A} (x_{i}), \hat{B} (x_{i}), \overset{ˇ}{B} (x_{i})), \end{matrix}$ then $N$ is a similarity measure on IFSs.

Proof. Note that $N$ is a function from IFS (X) × IFS (X) to [0, 1], $N (A, B) = M_{i = 1}^{n} E (\hat{A} (x_{i}), \overset{ˇ}{A} (x_{i}), \hat{B} (x_{i}), \overset{ˇ}{B} (x_{i})) = M_{i = 1}^{n} E (\hat{B} (x_{i}), \overset{ˇ}{B} (x_{i}), \hat{A} (x_{i}), \overset{ˇ}{A} (x_{i})) = N (B, A)$ and $N (A, A) = M_{i = 1}^{n} E (\hat{A} (x_{i}), \overset{ˇ}{A} (x_{i}), \hat{A} (x_{i}), \overset{ˇ}{A} (x_{i})) = 1$ . Thus $N$ satisfies $N 1$ and $N 2$ . To prove the property $N 3$ , we proceed as follows. Let A ⊆ B ⊆ C, then $\hat{A} (x_{i}) \leq \hat{B} (x_{i}) \leq \hat{C} (x_{i})$ and $\overset{ˇ}{A} (x_{i}) \geq \overset{ˇ}{B} (x_{i}) \geq \overset{ˇ}{C} (x_{i})$ for each x_i ∈ X. According to the properties of $E$ , we have $E (\hat{A} (x_{i}), \overset{ˇ}{A} (x_{i}), \hat{B} (x_{i}), \overset{ˇ}{B} (x_{i})) \geq E (\hat{A} (x_{i}), \overset{ˇ}{A} (x_{i}), \hat{C} (x_{i}), \overset{ˇ}{C} (x_{i}))$ and $E (\hat{B} (x_{i}), \overset{ˇ}{B} (x_{i}), \hat{C} (x_{i}), \overset{ˇ}{C} (x_{i})) \geq E (\hat{A} (x_{i}), \overset{ˇ}{A} (x_{i}), \hat{C} (x_{i}), \overset{ˇ}{C} (x_{i}))$ . Thus $min (N (A, B), N (B, C)) \geq N (A, C)$ . □

Theorem 2. Let M, E and d be an aggregation function, a fuzzy equivalence and a dissimilarity function, respectively. Suppose $N$ is a function defined for all A, B ∈ IFS (X) by $N (A, B) = M_{i = 1}^{n} \frac{Φ (\hat{A} (x_{i}), \overset{ˇ}{A} (x_{i}), \hat{B} (x_{i}), \overset{ˇ}{B} (x_{i}))}{Ψ (\hat{A} (x_{i}), \overset{ˇ}{A} (x_{i}), \hat{B} (x_{i}), \overset{ˇ}{B} (x_{i}))},$ where $Φ (\hat{A} (x_{i}), \overset{ˇ}{A} (x_{i}), \hat{B} (x_{i}), \overset{ˇ}{B} (x_{i})) = a (E (\hat{A} (x_{i}), \hat{B} (x_{i})) + E (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i}))) + b (min (\hat{A} (x_{i}), \hat{B} (x_{i})) + min (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i})))$ and $Ψ (\hat{A} (x_{i}), \overset{ˇ}{A} (x_{i}), \hat{B} (x_{i}), \overset{ˇ}{B} (x_{i})) = a (E (\hat{A} (x_{i}), \hat{B} (x_{i})) + E (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i}))) + b (min (\hat{A} (x_{i}), \hat{B} (x_{i})) + min (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i}))) + c (d (\hat{A} (x_{i}), \hat{B} (x_{i})) + d (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i})))$ with a ≥ 0, b ≥ 0, c > 0 and max(a, b) ≠0, then $N$ is a similarity measure on IFSs.

Proof. It is straightforward from Theorem 1 and Lemma 4. □

Theorem 3. Let E and d be a fuzzy equivalence and a dissimilarity function, respectively. Suppose $N$ is a function defined for all A, B ∈ IFS (X) by $N (A, B) = \frac{\sum_{i = 1}^{n} Φ (\hat{A} (x_{i}), \overset{ˇ}{A} (x_{i}), \hat{B} (x_{i}), \overset{ˇ}{B} (x_{i}))}{\sum_{i = 1}^{n} Ψ (\hat{A} (x_{i}), \overset{ˇ}{A} (x_{i}), \hat{B} (x_{i}), \overset{ˇ}{B} (x_{i}))},$ where $Φ (\hat{A} (x_{i}), \overset{ˇ}{A} (x_{i}), \hat{B} (x_{i}), \overset{ˇ}{B} (x_{i})) = a (E (\hat{A} (x_{i}), \hat{B} (x_{i})) + E (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i}))) + b (min (\hat{A} (x_{i}), \hat{B} (x_{i})) + min (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i})))$ and $Ψ (\hat{A} (x_{i}), \overset{ˇ}{A} (x_{i}), \hat{B} (x_{i}), \overset{ˇ}{B} (x_{i})) = a (E (\hat{A} (x_{i}), \hat{B} (x_{i})) + E (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i}))) + b (min (\hat{A} (x_{i}), \hat{B} (x_{i})) + min (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i}))) + c (d (\hat{A} (x_{i}), \hat{B} (x_{i})) + d (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i})))$ with a ≥ 0, b ≥ 0, c > 0 and max(a, b) ≠0, then $N$ is a similarity measure on IFSs.

Proof. Note that $N$ is a function from IFS (X) × IFS (X) to [0, 1] satisfying $N 1$ and $N 2$ . To prove the property $N 3$ , we proceed as follows. Let A ⊆ B ⊆ C, then $\hat{A} (x_{i}) \leq \hat{B} (x_{i}) \leq \hat{C} (x_{i})$ and $\overset{ˇ}{A} (x_{i}) \geq \overset{ˇ}{B} (x_{i}) \geq \overset{ˇ}{C} (x_{i})$ for each x_i ∈ X. According to the properties of E and d, we obtain that $E (\hat{A} (x_{i}), \hat{B} (x_{i})) + E (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i})) \geq E (\hat{A} (x_{i}), \hat{C} (x_{i})) + E (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{C} (x_{i}))$ , $d (\hat{A} (x_{i}), \hat{B} (x_{i})) + d (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i})) \leq d (\hat{A} (x_{i}), \hat{C} (x_{i})) + d (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{C} (x_{i}))$ , $E (\hat{B} (x_{i}), \hat{C} (x_{i})) + E (\overset{ˇ}{B} (x_{i}), \overset{ˇ}{C} (x_{i})) \geq E (\hat{A} (x_{i}), \hat{C} (x_{i})) + E (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{C} (x_{i}))$ , $d (\hat{B} (x_{i}), \hat{C} (x_{i})) + d (\overset{ˇ}{B} (x_{i}), \overset{ˇ}{C} (x_{i})) \leq d (\hat{A} (x_{i}), \hat{C} (x_{i})) + d (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{C} (x_{i}))$ for each x_i ∈ X. Therefore, we obtain that $a (E (\hat{A} (x_{i}), \hat{B} (x_{i})) + E (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i}))) + b (min (\hat{A} (x_{i}), \hat{B} (x_{i})) + min (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i}))) \geq a (E (\hat{A} (x_{i}), \hat{C} (x_{i})) + E (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{C} (x_{i}))) + b (min (\hat{A} (x_{i}), \hat{C} (x_{i})) + min (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{C} (x_{i})))$ and $a (E (\hat{B} (x_{i}), \hat{C} (x_{i})) + E (\overset{ˇ}{B} (x_{i}), \overset{ˇ}{C} (x_{i}))) + b (min (\hat{B} (x_{i}), \hat{C} (x_{i})) + min (\overset{ˇ}{B} (x_{i}), \overset{ˇ}{C} (x_{i}))) \geq a (E (\hat{A} (x_{i}), \hat{C} (x_{i})) + E (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{C} (x_{i}))) + b (min (\hat{A} (x_{i}), \hat{C} (x_{i})) + min (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{C} (x_{i})))$ . If $N (A, C) = 0$ , then $min (N (A, B), N (B, C)) \geq N (A, C)$ . If $N (A, C) \neq 0$ , then $N (A, B) \neq 0$ and $N (B, C) \neq 0$ . In this case,

$\frac{1}{N (A, B)} - 1 =$

$\frac{\sum_{i = 1}^{n} c (d (\hat{A} (x_{i}), \hat{B} (x_{i})) + d (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i})))}{\sum_{i = 1}^{n} a (E (\hat{A} (x_{i}), \hat{B} (x_{i})) + E (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i}))) + b (min (\hat{A} (x_{i}), \hat{B} (x_{i})) + min (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i})))},$

$\frac{1}{N (B, C)} - 1 =$

$\frac{\sum_{i = 1}^{n} c (d (\hat{B} (x_{i}), \hat{C} (x_{i})) + d (\overset{ˇ}{B} (x_{i}), \overset{ˇ}{C} (x_{i})))}{\sum_{i = 1}^{n} a (E (\hat{B} (x_{i}), \hat{C} (x_{i})) + E (\overset{ˇ}{B} (x_{i}), \overset{ˇ}{C} (x_{i}))) + b (min (\hat{B} (x_{i}), \hat{C} (x_{i})) + min (\overset{ˇ}{B} (x_{i}), \overset{ˇ}{C} (x_{i})))},$

$\frac{1}{N (A, C)} - 1 =$

$\frac{\sum_{i = 1}^{n} c (d (\hat{A} (x_{i}), \hat{C} (x_{i})) + d (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{C} (x_{i})))}{\sum_{i = 1}^{n} a (E (\hat{A} (x_{i}), \hat{C} (x_{i})) + E (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{C} (x_{i}))) + b (min (\hat{A} (x_{i}), \hat{C} (x_{i})) + min (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{C} (x_{i})))} .$

Then we have $max (\frac{1}{N (A, B)}, \frac{1}{N (B, C)}) \leq \frac{1}{N (A, C)}$ and thus $min (N (A, B), N (B, C)) \geq N (A, C)$ . □

Remark 2. If there is only an element in the discourse set, then the similarity measures on IFSs constructed by the above-mentioned two computational formulae are equivalent.

For similarity measures on IFSs, there exist some special properties, which can express natural characteristics of the meaning of them. For all A, B, C ∈ IFS (X),

$N (A, B) = 1 \Leftrightarrow A = B$ .

If A is a crisp set, then $N (A, A^{c}) = 0$ .

$N (A \cap B, A \cup B) = N (A, B)$ .

$N (A, B) = N (A^{c}, B^{c})$ .

$min (N (A, A \cup B), N (A \cap B, B)) \geq N (A, B)$ .

$N (A \cup C, B \cup C) \geq N (A, B)$ .

$N (A \cap C, B \cap C) \geq N (A, B)$

Let us study some properties of similarity measures on IFSs built by the above-mentioned two computational formulae.

Proposition 6. Let $E$ be an intuitionistic fuzzy equivalence and let $N$ be a similarity measure on IFSs defined by $N (A, B) = M_{i = 1}^{n} E (\hat{A} (x_{i}), \overset{ˇ}{A} (x_{i}), \hat{B} (x_{i}), \overset{ˇ}{B} (x_{i}))$ for all A, B ∈ IFS (X). If $E$ respectively satisfies $E 4, E 5, E 6, E 7, E 8, E 10, E 11$ , then $N$ respectively satisfies $N 4, N 5, N 6, N 7, N 8, N 9, N 10$ .

Proof. We only prove that if $E$ satisfies $E 8$ , then $N$ satisfies $N 8$ . The other situation can be proved similarly. It is shown that $N (A, A \cup B) = M_{i = 1}^{n} E (\hat{A} (x_{i}), \overset{ˇ}{A} (x_{i}), max (\hat{A} (x_{i}), \hat{B} (x_{i})), min (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i})))$ and $N (A \cap B, B) = M_{i = 1}^{n} E (min (\hat{A} (x_{i}), \hat{B} (x_{i})), max (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i})), \hat{B} (x_{i}), \overset{ˇ}{B} (x_{i}))$ . Let us denote $E (\hat{A} (x_{i}), \overset{ˇ}{A} (x_{i}), max (\hat{A} (x_{i}), \hat{B} (x_{i})), min (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i})))$ and $E (min (\hat{A} (x_{i}), \hat{B} (x_{i})), max (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i})), \hat{B} (x_{i}), \overset{ˇ}{B} (x_{i}))$ by Δ₁ and Δ₂. If $\hat{A} (x_{i}) \geq \hat{B} (x_{i}), \overset{ˇ}{A} (x_{i}) \geq \overset{ˇ}{B} (x_{i}),$ then $Δ_{1} = E (\hat{A} (x_{i}), \overset{ˇ}{A} (x_{i}), \hat{A} (x_{i}), \overset{ˇ}{B} (x_{i}))$ and $Δ_{2} = E (\hat{B} (x_{i}), \overset{ˇ}{A} (x_{i}), \hat{B} (x_{i}), \overset{ˇ}{B} (x_{i}))$ . If $\hat{A} (x_{i}) \geq \hat{B} (x_{i}), \overset{ˇ}{A} (x_{i}) \leq \overset{ˇ}{B} (x_{i}),$ then $Δ_{1} = E (\hat{A} (x_{i}), \overset{ˇ}{A} (x_{i}), \hat{A} (x_{i}), \overset{ˇ}{A} (x_{i})) = 1$ and $Δ_{2} = E (\hat{B} (x_{i}), \overset{ˇ}{B} (x_{i}), \hat{B} (x_{i}), \overset{ˇ}{B} (x_{i})) = 1$ . If $\hat{A} (x_{i}) \leq \hat{B} (x_{i}), \overset{ˇ}{A} (x_{i}) \geq \overset{ˇ}{B} (x_{i}),$ then $Δ_{1} = E (\hat{A} (x_{i}), \overset{ˇ}{A} (x_{i}), \hat{B} (x_{i}), \overset{ˇ}{B} (x_{i}))$ and $Δ_{2} = E (\hat{A} (x_{i}), \overset{ˇ}{A} (x_{i}), \hat{B} (x_{i}), \overset{ˇ}{B} (x_{i}))$ . If $\hat{A} (x_{i}) \leq \hat{B} (x_{i}), \overset{ˇ}{A} (x_{i}) \leq \overset{ˇ}{B} (x_{i}),$ then $Δ_{1} = E (\hat{A} (x_{i}), \overset{ˇ}{A} (x_{i}), \hat{B} (x_{i}), \overset{ˇ}{A} (x_{i}))$ and $Δ_{2} = E (\hat{A} (x_{i}), \overset{ˇ}{B} (x_{i}), \hat{B} (x_{i}), \overset{ˇ}{B} (x_{i}))$ . Therefore, we have $min (E (\hat{A} (x_{i}), \overset{ˇ}{A} (x_{i}), max (\hat{A} (x_{i}), \hat{B} (x_{i})), min (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i}))), E (min (\hat{A} (x_{i}), \hat{B} (x_{i})), max (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i})), \hat{B} (x_{i}), \overset{ˇ}{B} (x_{i}))) \geq E (\hat{A} (x_{i}), \overset{ˇ}{A} (x_{i}), \hat{B} (x_{i}), \overset{ˇ}{B} (x_{i}))$ for each x_i ∈ X. Thus $min (N (A, A \cup B), N (A \cap B, B)) \geq N (A, B)$ for all A, B ∈ IFS (X). □

Proposition 7. Let $N$ be a similarity measure on IFSs defined by Theorem 3. If d satisfies d5, then $N$ satisfies $N 4$ .

Proof. It is obvious from Propositions 3 and 6. □

Proposition 8. Let $N$ be a similarity measure on IFSs defined by Theorem 2. Then $N$ satisfies $N 5, N 6, N 7$ and $N 8$ .

Proof. It is obvious from Propositions 4 and 6. □

Proposition 9. Let $N$ be a similarity measure on IFSs defined by Theorem 3. If b = 0, then $N$ satisfies $N 9$ and $N 10$ .

Proof. It is obvious from Propositions 5 and 6. □

Proposition 10. Let $N$ be a similarity measure on IFSs defined by Theorem 3. If d satisfies d5, then $N$ satisfies $N 4$ .

Proof. Note that $N (A, B) = 1 \Leftrightarrow c (d (\hat{A} (x_{i}), \hat{B} (x_{i})) + d (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i}))) = 0$ . Since d satisfies d5, we obtain that $c (d (\hat{A} (x_{i}), \hat{B} (x_{i})) + d (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i}))) = 0 \Leftrightarrow \hat{A} (x_{i}) = \hat{B} (x_{i})$ and $\overset{ˇ}{A} (x_{i}) = \overset{ˇ}{B} (x_{i})$ . □

Proposition 11. Let $N$ be a similarity measure on IFSs defined by Theorem 3. Then $N$ satisfies $N 5, N 6, N 7$ and $N 8$ .

Proof.

If A is a crisp set, then $\hat{A} (x_{i}) = 1$ or 0 and $\overset{ˇ}{A} (x_{i}) = 0$ or 1 and thus $E (\hat{A} (x_{i}), \overset{ˇ}{A} (x_{i})) = min (\overset{ˇ}{A} (x_{i}), \hat{A} (x_{i})) = 0$ . Therefore, $N (A, A^{c}) = 0$ .

Note that $E (\hat{A} (x_{i}), \hat{B} (x_{i})) + E (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i})) = E (min (\hat{A} (x_{i}), \hat{B} (x_{i})), max (\hat{A} (x_{i}), \hat{B} (x_{i}))) + E (max (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i})), min (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i})))$ , $min (\hat{A} (x_{i}), \hat{B} (x_{i})) + min (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i})) = min (min (\hat{A} (x_{i}), \hat{B} (x_{i})), max (\hat{A} (x_{i}), \hat{B} (x_{i}))) + min (max (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i})), min (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i})))$ and $d (\hat{A} (x_{i}), \hat{B} (x_{i})) + d (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i})) = d (min (\hat{A} (x_{i}), \hat{B} (x_{i})), max (\hat{A} (x_{i}), \hat{B} (x_{i}))) + d (max (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i})), min (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i})))$ . Thus we have $N (A \cap B, A \cup B) = N (A, B)$ for all A, B ∈ IFS (X).

Since $A^{c} = {〈 x_{i}, \overset{ˇ}{A} (x_{i}), \hat{A} (x_{i}) 〉}$ , $B^{c} = {〈 x_{i}, \overset{ˇ}{B} (x_{i}), \hat{B} (x_{i}) 〉}$ , by the commutativity of min , E and d we have $N (A^{c}, B^{c}) = N (A, B)$ .

Note that $N (A, A \cup B) = \frac{\sum_{i = 1}^{n} Δ_{1} (\hat{A} (x_{i}), \overset{ˇ}{A} (x_{i}), \hat{B} (x_{i}), \overset{ˇ}{B} (x_{i}))}{\sum_{i = 1}^{n} Δ_{2} (\hat{A} (x_{i}), \overset{ˇ}{A} (x_{i}), \hat{B} (x_{i}), \overset{ˇ}{B} (x_{i}))}$ with $Δ_{1} (\hat{A} (x_{i}), \overset{ˇ}{A} (x_{i}), \hat{B} (x_{i}), \overset{ˇ}{B} (x_{i})) = a (E (\hat{A} (x_{i}), max (\hat{A} (x_{i}), \hat{B} (x_{i}))) + E (\overset{ˇ}{A} (x_{i}), min (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i})))) + b (\hat{A} (x_{i}) + min (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i})))$ and $Δ_{2} (\hat{A} (x_{i}), \overset{ˇ}{A} (x_{i}), \hat{B} (x_{i}), \overset{ˇ}{B} (x_{i})) = a (E (\hat{A} (x_{i}), max (\hat{A} (x_{i}), \hat{B} (x_{i}))) + E (\overset{ˇ}{A} (x_{i}), min (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i})))) + b (\hat{A} (x_{i}) + min (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i}))) + c (d (\hat{A} (x_{i}), max (\hat{A} (x_{i}), \hat{B} (x_{i}))) + d (\overset{ˇ}{A} (x_{i}), min (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i}))))$ . Meanwhile, $N (A \cap B, B) = \frac{\sum_{i = 1}^{n} Δ_{3} (\hat{A} (x_{i}), \overset{ˇ}{A} (x_{i}), \hat{B} (x_{i}), \overset{ˇ}{B} (x_{i}))}{\sum_{i = 1}^{n} Δ_{4} (\hat{A} (x_{i}), \overset{ˇ}{A} (x_{i}), \hat{B} (x_{i}), \overset{ˇ}{B} (x_{i}))}$ with $Δ_{3} (\hat{A} (x_{i}), \overset{ˇ}{A} (x_{i}), \hat{B} (x_{i}), \overset{ˇ}{B} (x_{i})) = a (E (min (\hat{A} (x_{i}), \hat{B} (x_{i})), \hat{B} (x_{i})) + E (max (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i})), \overset{ˇ}{B} (x_{i}))) + b (min (\hat{A} (x_{i}), \hat{B} (x_{i})) + \overset{ˇ}{B} (x_{i}))$ and $Δ_{4} (\hat{A} (x_{i}), \overset{ˇ}{A} (x_{i}), \hat{B} (x_{i}), \overset{ˇ}{B} (x_{i})) = a (E (min (\hat{A} (x_{i}), \hat{B} (x_{i})), \hat{B} (x_{i})) + E (max (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i})), \overset{ˇ}{B} (x_{i}))) + b (min (\hat{A} (x_{i}), \hat{B} (x_{i})) + \overset{ˇ}{B} (x_{i})) + c (d (min (\hat{A} (x_{i}), \hat{B} (x_{i})), \hat{B} (x_{i})) + d (max (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i})), \overset{ˇ}{B} (x_{i})))$ . Note that $\sum_{i = 1}^{n} Δ_{1} (\hat{A} (x_{i}), \overset{ˇ}{A} (x_{i}), \hat{B} (x_{i}), \overset{ˇ}{B} (x_{i})) \geq \sum_{i = 1}^{n} Φ (\hat{A} (x_{i}), \overset{ˇ}{A} (x_{i}), \hat{B} (x_{i}), \overset{ˇ}{B} (x_{i}))$ and $\sum_{i = 1}^{n} Δ_{3} (\hat{A} (x_{i}), \overset{ˇ}{A} (x_{i}), \hat{B} (x_{i}), \overset{ˇ}{B} (x_{i})) \geq \sum_{i = 1}^{n} Φ (\hat{A} (x_{i}), \overset{ˇ}{A} (x_{i}), \hat{B} (x_{i}), \overset{ˇ}{B} (x_{i}))$ . If $N (A, B) = 0$ , then we have $min (N (A, A \cup B), N (A \cap B, B)) \geq N (A, B)$ . If $N (A, B) \neq 0$ , then $N (A, A \cup B) \neq 0$ and $N (A \cap B, B) \neq 0$ and thus

$\frac{1}{N (A, A \cup B)} - 1$

$\leq \frac{\sum_{i = 1}^{n} c (d (\hat{A} (x_{i}), \hat{B} (x_{i})) + d (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i})))}{\sum_{i = 1}^{n} a (E (\hat{A} (x_{i}), \hat{B} (x_{i})) + E (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i}))) + b (min (\hat{A} (x_{i}), \hat{B} (x_{i})) + min (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i})))}$

$= \frac{1}{N (A, B)}$ , $\frac{1}{N (A \cap B, B)} - 1$

$= \frac{1}{N (A, B)}$ .

Therefore, we have $min (N (A, A \cup B), N (A \cap B, B)) \geq N (A, B)$ for all A, B ∈ IFS (X).

□

Proposition 12. Let $N$ be a similarity measure on IFSs defined by Theorem 3. If b = 0, then $N$ satisfies $N 9$ and $N 10$ .

Proof. Suppose that b = 0, then we have $N (A, B) =$ $\frac{\sum_{i = 1}^{n} a (E (\hat{A} (x_{i}), \hat{B} (x_{i})) + E (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i})))}{\sum_{i = 1}^{n} a (E (\hat{A} (x_{i}), \hat{B} (x_{i})) + E (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i}))) + c (d (\hat{A} (x_{i}), \hat{B} (x_{i})) + d (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i})))}$ . According to Lemmas 2 and 2, we have $E (max (\hat{A} (x_{i}), \hat{C} (x_{i})), max (\hat{B} (x_{i}), \hat{C} (x_{i}))) \geq E (\hat{A} (x_{i}), \hat{B} (x_{i}))$ , $E (min (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{C} (x_{i})), min (\overset{ˇ}{B} (x_{i}), \overset{ˇ}{C} (x_{i}))) \geq E (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i}))$ , $d (max (\hat{A} (x_{i}), \hat{C} (x_{i})), max (\hat{B} (x_{i}), \hat{C} (x_{i}))) \leq d (\hat{A} (x_{i}), \hat{B} (x_{i}))$ , $d (min (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{C} (x_{i})), min (\overset{ˇ}{B} (x_{i}), \overset{ˇ}{C} (x_{i}))) \leq d (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i}))$ . If $N (A, B) = 0$ , then $N (A \cup C, B \cup C) \geq N (A, B)$ . If $N (A, B) \neq 0$ , then $N (A \cup C, B \cup C) \neq 0$ and thus $\begin{matrix} \frac{1}{N (A \cup C, B \cup C)} - 1 \\ \leq \frac{\sum_{i = 1}^{n} c (d (\hat{A} (x_{i}), \hat{B} (x_{i})) + d (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i})))}{\sum_{i = 1}^{n} a (E (\hat{A} (x_{i}), \hat{B} (x_{i})) + E (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i}))) + b (min (\hat{A} (x_{i}), \hat{B} (x_{i})) + min (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i})))} \\ = \frac{1}{N (A, B)} . \end{matrix}$ Therefore, we have $N (A \cup C, B \cup C) \geq N (A, B)$ for all A, B, C ∈ IFS (X). The case of $N (A \cap C, B \cap C) \geq N (A, B)$ can be proved in the same manner.

□

Example 4. For a similarity measure on IFSs $N$ defined by Theorem 2 or Theorem 3, we cannot always conclude that $N$ satisfies $N 9$ and $N 10$ in case that b ≠ 0. Let us consider M = M₁, E = E₁ and d = d₁ in Theorem 3 and let a = 0, b = c. Then we have $N_{1} (A, B) = \frac{1}{n} \sum_{i = 1}^{n} \frac{min (\hat{A} (x_{i}), \hat{B} (x_{i})) + min (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i})))}{max (\hat{A} (x_{i}), \hat{B} (x_{i})) + max (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i}))}$ . Let us consider E = E₁ and d = d₁ in Theorem 3 and let a = 0, b = c. Then we have $N_{2} (A, B) =$ $\frac{\sum_{i = 1}^{n} (min (\hat{A} (x_{i}), \hat{B} (x_{i})) + min (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i})))}{\sum_{i = 1}^{n} (max (\hat{A} (x_{i}), \hat{B} (x_{i})) + max (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i})))}$ . Suppose that A = {〈x, 0.2, 0.7〉}, B = {〈x, 0.3, 0.8〉}, C = {〈x, 0.1, 0.2〉}, then A ∪ C = {〈x, 0.2, 0.2〉}, B ∪ C = {〈x, 0.3, 0.2〉} and thus $N_{1} (A \cup C, B \cup C) = N_{2} (A \cup C, B \cup C) = 0.8 < N_{1} (A, B) = N_{2} (A, B) = 0.818$ . Suppose that A = {〈x, 0.7, 0.2〉}, B = {〈x, 0.8, 0.3〉}, C = {〈x, 0.2, 0.1〉}, then A ∩ C = {〈x, 0.2, 0.2〉}, B ∩ C = {〈x, 0.2, 0.3〉} and thus $N_{1} (A \cap C, B \cap C) = N_{2} (A \cap C, B \cap C) = 0.8 < N_{1} (A, B) = N_{2} (A, B) = 0.818$ .

By studying the properties of the proposed two computational formulae of similarity measure on IFSs, we know that both of the two computational formulae satisfy $N 5, N 6, N 7$ and $N 8$ in any case. Both of them satisfy $N 4$ in case that d satisfies d5 and satisfy $N 9$ and $N 10$ in case that b = 0. The properties of the first computational formula are straightforward from the properties of the corresponding intuitionistic fuzzy equivalencies. Thus the properties of the first computational formula are easier to obtain than those of the second computational formula. However, the second computational formula may be more effective than the first computational formula in some case. Let us first take E = E₁, d = d₁ and M = M₁ in Theorems 3 and 3 and obtain $N^{'} (A, B)$ and $N^{″} (A, B)$ shown in Eqs. (2) and (3). Then we consider the following example. $\begin{matrix} N^{'} (A, B) = \\ \frac{1}{n} \sum_{i = 1}^{n} \frac{a (2 - | \hat{A} (x_{i}) - \hat{B} (x_{i}) | - | \overset{ˇ}{A} (x_{i}) - \overset{ˇ}{B} (x_{i}) |) + b (min (\hat{A} (x_{i}), \hat{B} (x_{i})) + min (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i})))}{a (2 - | \hat{A} (x_{i}) - \hat{B} (x_{i}) | - | \overset{ˇ}{A} (x_{i}) - \overset{ˇ}{B} (x_{i}) |) + b (min (\hat{A} (x_{i}), \hat{B} (x_{i})) + min (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i}))) + c (| \hat{A} (x_{i}) - \hat{B} (x_{i}) | + | \overset{ˇ}{A} (x_{i}) - \overset{ˇ}{B} (x_{i}) |)}, \end{matrix}$ (2) $\begin{matrix} N^{″} (A, B) = \\ \frac{\sum_{i = 1}^{n} a (2 - | \hat{A} (x_{i}) - \hat{B} (x_{i}) | - | \overset{ˇ}{A} (x_{i}) - \overset{ˇ}{B} (x_{i}) |) + b (min (\hat{A} (x_{i}), \hat{B} (x_{i})) + min (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i})))}{\sum_{i = 1}^{n} a (2 - | \hat{A} (x_{i}) - \hat{B} (x_{i}) | - | \overset{ˇ}{A} (x_{i}) - \overset{ˇ}{B} (x_{i}) |) + b (min (\hat{A} (x_{i}), \hat{B} (x_{i})) + min (\overset{ˇ}{A} (x_{i}), \overset{ˇ}{B} (x_{i}))) + c (| \hat{A} (x_{i}) - \hat{B} (x_{i}) | + | \overset{ˇ}{A} (x_{i}) - \overset{ˇ}{B} (x_{i}) |)} . \end{matrix}$ (3)

Example 5. Let us consider three known patterns P₁, P₂ and P₃ represented by the following IFSs in X = {x₁, x₂, x₃}:

$\begin{matrix} P_{1} = {〈 x_{1}, 0.5, 0.4 〉, 〈 x_{2}, 0.8, 0 〉, 〈 x_{3}, 0.3, 0.7 〉}, \\ P_{2} = {〈 x_{1}, 0.6, 0.3 〉, 〈 x_{2}, 0.9, 0.1 〉, 〈 x_{3}, 0.6, 0.4 〉}, \\ P_{3} = {〈 x_{1}, 0.6, 0.3 〉, 〈 x_{2}, 0.9, 0.1 〉, 〈 x_{3}, 0.5, 0.5 〉} . \end{matrix}$

It is aimed to classify an unknown pattern, represented by an IFS Q in X into one of the patterns P₁, P₂ and P₃, where

$\begin{matrix} Q = {〈 x_{1}, 0, 0 〉, 〈 x_{2}, 0, 0 〉, 〈 x_{3}, 0, 0.1 〉} . \end{matrix}$

In order to do that, the similarity degrees between Q and P₁, P₂ and P₃ should be calculated.

If we take the similarity measure $N^{'}$ , then

$\begin{matrix} N^{'} (P_{1}, Q) = \frac{1}{3} (\frac{11 a}{11 a + 9 c} + \frac{3 a}{3 a + 2 c} + \frac{11 a + b}{11 a + b + 9 c}), \\ N^{'} (P_{2}, Q) = \frac{1}{3} (\frac{11 a}{11 a + 9 c} + \frac{a}{a + c} + \frac{11 a + b}{11 a + b + 9 c}), \\ N^{'} (P_{3}, Q) = \frac{1}{3} (\frac{11 a}{11 a + 9 c} + \frac{a}{a + c} + \frac{11 a + b}{11 a + b + 9 c}) . \end{matrix}$

If we take $N^{'}$ with a = 0, then $N^{'} (P_{2}, Q) = N^{'} (P_{2}, Q) = N^{'} (P_{3}, Q)$ and the classification result cannot be determined. If we take $N^{'}$ with a ≠ 0, then $N^{'} (P_{2}, Q) = N^{'} (P_{3}, Q) < N^{'} (P_{1}, Q)$ . Therefore, only the similarity measure $N^{'}$ with a ≠ 0 can classify the unknown pattern Q into the pattern P₁.

If we take the similarity measure $N^{″}$ , then

$\begin{matrix} N^{″} (P_{1}, Q) = \frac{34 a + b}{34 a + b + 26 c}, \\ N^{″} (P_{2}, Q) = \frac{32 a + b}{32 a + b + 28 c}, \\ N^{″} (P_{3}, Q) = \frac{32 a + b}{32 a + b + 28 c} . \end{matrix}$

Therefore, we have $N^{″} (P_{2}, Q) = N^{″} (P_{3}, Q) < N^{″} (P_{1}, Q)$ for any parameters. Thus the similarity measure $N^{″}$ with arbitrary parameters a, b and c can classify the unknown pattern Q into the pattern P₁. The result coincides with the one obtained in [11].

4 A comparison between the proposed similarity measures on IFSs and some existing ones

Some similarity measures on IFSs have been proposed by researchers in order to measure the degree of similarity between two IFSs. We list some of them as follows and demonstrate their counter-intuitive cases.

(1) Chen [8, 9]

$N_{C} (A, B) = 1 - \frac{1}{n} \sum_{i = 1}^{n} | \frac{\hat{A} (x_{i}) - \overset{ˇ}{A} (x_{i})}{2} - \frac{\hat{B} (x_{i}) - \overset{ˇ}{B} (x_{i})}{2} | .$

(2) Li and Cheng [22]

$N_{LC} (A, B) = 1 - \frac{1}{\sqrt[p]{n}} \sqrt[p]{\sum_{i = 1}^{n} {| \frac{\hat{A} (x_{i}) - \overset{ˇ}{A} (x_{i})}{2} - \frac{\hat{B} (x_{i}) - \overset{ˇ}{B} (x_{i})}{2} |}^{p}}$

with p ∈ [1, + ∞). (3) Hong and Kim [15]

$N_{HK} (A, B) = 1 - \frac{1}{n} \sum_{i = 1}^{n} \frac{| \hat{A} (x_{i}) - \hat{B} (x_{i}) | + | \overset{ˇ}{A} (x_{i}) - \overset{ˇ}{B} (x_{i}) |}{2} .$

(4) Burillo and Bustince [5]

$N_{B} (A, B) = 1 - \frac{1}{\sqrt{n}} \sqrt{\sum_{i = 1}^{n} \frac{| \hat{A} (x_{i}) - \hat{B} (x_{i}) |^{2} + | \overset{ˇ}{A} (x_{i}) - \overset{ˇ}{B} (x_{i}) |^{2}}{2}} .$

(5) Hung and Yang [17]

$N_{HY} (A, B) = 1 - \frac{1}{n} \sum_{i = 1}^{n} \sqrt[p]{\frac{| \hat{A} (x_{i}) - \hat{B} (x_{i}) |^{p} + | \overset{ˇ}{A} (x_{i}) - \overset{ˇ}{B} (x_{i}) |^{p}}{2}}$

with p ∈ [1, + ∞). (6) Xu and Chen [34]

$N_{XC} (A, B) = 1 - \frac{1}{\sqrt{n}} \sqrt{\sum_{i = 1}^{n} max (| \hat{A} (x_{i}) - \hat{B} (x_{i}) |^{2}, | \overset{ˇ}{A} (x_{i}) - \overset{ˇ}{B} (x_{i}) |^{2})} .$

(7) Liang and Shi [29] $\begin{matrix} N_{LS} (A, B) = 1 - \\ \frac{1}{\sqrt[p]{n}} \sqrt[p]{\sum_{i = 1}^{n} \frac{(| 3 (\hat{A} (x_{i}) - \hat{B} (x_{i})) + \overset{ˇ}{B} (x_{i}) - \overset{ˇ}{A} (x_{i}) | + | 3 (\overset{ˇ}{B} (x_{i}) - \overset{ˇ}{A} (x_{i})) + \hat{A} (x_{i}) - \hat{B} (x_{i}) |)^{p}}{8^{p}}} \end{matrix}$ with p ∈ [1, + ∞). (8) Boran and Akay [4] $\begin{matrix} N_{BA} (A, B) = 1 - \\ \frac{1}{\sqrt[p]{n}} \sqrt[p]{\sum_{i = 1}^{n} \frac{| t (\hat{A} (x_{i}) - \hat{B} (x_{i})) + \overset{ˇ}{B} (x_{i}) - \overset{ˇ}{A} (x_{i}) |^{p} + | t (\overset{ˇ}{B} (x_{i}) - \overset{ˇ}{A} (x_{i})) + \hat{A} (x_{i}) - \hat{B} (x_{i}) |^{p}}{2 (t + 1)^{p}}} \end{matrix}$ with p ∈ [1, + ∞) , t ∈ [2, + ∞).

Example 6. Let A = {〈x, 0.1, 0.1〉} , B = {〈x, 0.9, 0.9〉}, which are indeed not equal to each other. If we take the similarity measures $N_{C}$ and $N_{LC}$ to calculate the similarity between A and B, then we have $N_{C} (A, B) = N_{LC} (A, B) = 1$ . Thus $N_{C}$ and $N_{LC}$ get an unreasonable result. If we take the similarity measures $N^{'}$ and $N^{″}$ , then we have $N^{'} (A, B) = N^{″} (A, B) = \frac{2 a + b}{2 a + b + 8 c}$ and the final results of $N^{'}$ and $N^{″}$ depend on the values of their parameters.

Example 7. Let us consider three known patterns P₁, P₂ and P₃ represented by the following IFSs in X = {x₁, x₂, x₃, x₄}:

$\begin{matrix} P_{1} = {〈 x_{1}, 0.5, 0.3 〉, 〈 x_{2}, 0.7, 0 〉, 〈 x_{3}, 0.4, 0.5 〉, 〈 x_{4}, 0.7, 0.3 〉}, \\ P_{2} = {〈 x_{1}, 0.5, 0.2 〉, 〈 x_{2}, 0.6, 0.1 〉, 〈 x_{3}, 0.2, 0.7 〉, 〈 x_{4}, 0.7, 0.3 〉}, \\ P_{3} = {〈 x_{1}, 0.5, 0.4 〉, 〈 x_{2}, 0.7, 0.1 〉, 〈 x_{3}, 0.4, 0.6 〉, 〈 x_{4}, 0.7, 0.2 〉} . \end{matrix}$

It is aimed to classify an unknown pattern, represented by an IFS Q in X into one of the patterns P₁, P₂ and P₃, where

$\begin{matrix} Q = {〈 x_{1}, 0.4, 0.3 〉, 〈 x_{2}, 0.7, 0.1 〉, 〈 x_{3}, 0.3, 0.6 〉, 〈 x_{4}, 0.7, 0.3 〉} . \end{matrix}$

In order to do that, the similarity degrees between Q and P₁, P₂, P₃ should be calculated. If we take the similarity measure $N_{HK}$ to calculate the similarity between the above-mentioned IFSs, then we have $N_{HK} (P_{1}, Q) = 0.95, N_{HK} (P_{2}, Q) = 0.9375$ and $N_{HK} (P_{3}, Q) = 0.95$ and thus $N_{HK} (P_{2}, Q) < N_{HK} (P_{1}, Q) = N_{HK} (P_{3}, Q)$ . If we take the similarity measure $N_{B}$ to calculate the similarity between the above-mentioned IFSs, then we obtain that $N_{B} (P_{1}, Q) = 0.929, N_{B} (P_{2}, Q) = 0.921, N_{B} (P_{3}, Q) = 0.929$ and thus $N_{B} (P_{2}, Q) < N_{B} (P_{1}, Q) = N_{B} (P_{3}, Q)$ . If we take the similarity measure $N_{HY}$ to calculate the similarity between the above-mentioned IFSs, then we have $N_{HY} (P_{1}, Q) = 1 - 0.25 \times (0.1 + \frac{0.2}{\sqrt[p]{2}}), N_{HY} (P_{2}, Q) = 1 - 0.25 \times (0.2 + \frac{0.1}{\sqrt[p]{2}})$ and $N_{HY} (P_{3}, Q) = 1 - 0.25 \times (0.1 + \frac{0.2}{\sqrt[p]{2}})$ and thus $N_{HY} (P_{2}, Q) < N_{HY} (P_{1}, Q) = N_{HY} (P_{3}, Q)$ . If we take the similarity measure $N_{XC}$ to calculate the similarity between the above-mentioned IFSs, then we have $N_{XC} (P_{1}, Q) = 0.913, N_{XC} (P_{2}, Q) = 0.913, N_{XC} (P_{3}, Q) = 0.913$ and thus $N_{XC} (P_{2}, Q) = N_{XC} (P_{1}, Q) = N_{XC} (P_{3}, Q)$ . In these cases, the classification result cannot be determined.

If we take the similarity measure $N^{'}$ , then we have

$\begin{matrix} N^{'} (P_{1}, Q) = \frac{1}{4} (\frac{19 a + 7 b}{19 a + 7 b + c} + \frac{19 a + 7 b}{19 a + 7 b + c} + \frac{9 a + 4 b}{9 a + 4 b + c} + 1), \\ N^{'} (P_{2}, Q) = \frac{1}{4} (\frac{9 a + 3 b}{9 a + 3 b + c} + \frac{19 a + 7 b}{19 a + 7 b + c} + \frac{9 a + 4 b}{9 a + 4 b + c} + 1), \\ N^{'} (P_{3}, Q) = \frac{1}{4} (\frac{18 a + 7 b}{18 a + 7 b + 2 c} + \frac{19 a + 9 b}{19 a + 9 b + c} + \frac{19 a + 9 b}{19 a + 9 b + c} + 1) . \end{matrix}$

It is obtained that $N^{'} (P_{3}, Q) > N^{'} (P_{1}, Q) > N^{'} (P_{2}, Q)$ in case that b ≠ 0.

If we take the similarity measure $N^{″}$ , then we have

$\begin{matrix} N^{″} (P_{1}, Q) = \frac{76 a + 32 b}{76 a + 32 b + 4 c}, \\ N^{″} (P_{2}, Q) = \frac{75 a + 31 b}{75 a + 31 b + 5 c}, \\ N^{″} (P_{3}, Q) = \frac{76 a + 35 b}{76 a + 35 b + 4 c} . \end{matrix}$

It is obtained that $N^{″} (P_{3}, Q) > N^{″} (P_{1}, Q) > N^{″} (P_{2}, Q)$ in case that b ≠ 0. Therefore, if we take $N^{'}$ or $N^{″}$ with b ≠ 0, then we can classify the unknown pattern Q into the pattern P₃. The result coincides with the one obtained in [4, 11].

Example 8. Let us consider three known patterns P₁, P₂ and P₃ represented by the following IFSs in X = {x₁, x₂, x₃, x₄}:

$\begin{matrix} P_{1} = {〈 x_{1}, 0.3, 0.3 〉, 〈 x_{2}, 0.4, 0.4 〉, 〈 x_{3}, 0.4, 0.4 〉, 〈 x_{4}, 0.4, 0.4 〉}, \\ P_{2} = {〈 x_{1}, 0.5, 0.5 〉, 〈 x_{2}, 0.1, 0.1 〉, 〈 x_{3}, 0.5, 0.5 〉, 〈 x_{4}, 0.1, 0.1 〉}, \\ P_{3} = {〈 x_{1}, 1, 0 〉, 〈 x_{2}, 0.3, 0.3 〉, 〈 x_{3}, 0.5, 0.5 〉, 〈 x_{4}, 0.6, 0 〉} . \end{matrix}$

It is aimed to classify an unknown pattern, represented by an IFS Q in X into one of the patterns P₁, P₂ and P₃, where

$\begin{matrix} Q = {〈 x_{1}, 0.4, 0.4 〉, 〈 x_{2}, 0.5, 0.5 〉, 〈 x_{3}, 0.2, 0.2 〉, 〈 x_{4}, 0.3, 0.3 〉} . \end{matrix}$

In order to do that, the similarity degrees between Q and P₁, P₂, P₃ should be calculated. If we take the similarity measures $N_{C}$ and $N_{LC}$ to calculate the similarity between the above-mentioned IFSs, then we have $N_{C} (P_{1}, Q) = N_{C} (P_{2}, Q) = 1 > N_{C} (P_{3}, Q)$ and $N_{LC} (P_{1}, Q) = N_{LC} (P_{2}, Q) = 1 > N_{LC} (P_{3}, Q)$ . In this case, the classification result cannot be determined.

If we take the similarity measure $N^{'}$ , then we have $\begin{matrix} N^{'} (P_{1}, Q) = \\ \frac{1}{4} (\frac{9 a + 3 b}{9 a + 3 b + c} + \frac{9 a + 4 b}{9 a + 4 b + c} + \frac{4 a + b}{4 a + b + c} + \frac{9 a + 3 b}{9 a + 3 b + c}), \\ N^{'} (P_{2}, Q) = \\ \frac{1}{4} (\frac{9 a + 4 b}{9 a + 4 b + c} + \frac{6 a + b}{6 a + b + 4 c} + \frac{7 a + 2 b}{7 a + 2 b + 3 c} + \frac{8 a + b}{8 a + b + 2 c}), \\ N^{'} (P_{3}, Q) = \\ \frac{1}{4} (\frac{5 a + 2 b}{5 a + 2 b + 5 c} + \frac{8 a + 3 b}{8 a + 3 b + 2 c} + \frac{7 a + 2 b}{7 a + 2 b + 3 c} + \frac{14 a + 3 b}{14 a + 3 b + 6 c}) . \end{matrix}$ Thus $N^{'} (P_{1}, Q) > max (N^{'} (P_{2}, Q), N^{'} (P_{3}, Q))$ .

If we take the similarity measure $N^{″}$ , then we have $\begin{matrix} N^{″} (P_{1}, Q) = \frac{70 a + 24 b}{70 a + 24 b + 10 c}, \\ N^{″} (P_{2}, Q) = \frac{60 a + 16 b}{60 a + 16 b + 20 c}, \\ N^{″} (P_{3}, Q) = \frac{54 a + 17 b}{54 a + 17 b + 26 c} . \end{matrix}$ Thus $N^{'} (P_{1}, Q) > N^{'} (P_{2}, Q) > N^{'} (P_{3}, Q)$ .

Therefore, the similarity measures $N^{'}$ and $N^{″}$ with arbitrary parameters can classify the unknown pattern Q into the pattern P₁. The result coincides with the one obtained in [4].

Example 9. Let us consider three known patterns P₁, P₂ and P₃ represented by the following IFSs in X = {x₁, x₂, x₃}:

It is aimed to classify an unknown pattern, represented by an IFS Q in X into one of the patterns P₁, P₂ and P₃, where

$\begin{matrix} Q = {〈 x_{1}, 0, 0 〉, 〈 x_{2}, 0, 0 〉, 〈 x_{3}, 0, 0.1 〉} . \end{matrix}$

In order to do that, the similarity degrees between Q and P₁, P₂, P₃ should be calculated.

If we take the similarity measures $N_{C}$ and $N_{LC}$ to calculate the similarity between the above-mentioned IFSs, then we have $N_{C} (P_{1}, Q) = N_{C} (P_{3}, Q) > N_{C} (P_{2}, Q)$ and $N_{LC} (P_{1}, Q) = N_{LC} (P_{3}, Q) > N_{LC} (P_{2}, Q)$ and the classification result cannot be determined. If we take the similarity measure $N_{LS}$ , then we have $N_{LS} (P_{1}, Q) = N_{LS} (P_{2}, Q) = N_{LS} (P_{3}, Q)$ and thus classification result cannot be determined. If we take the similarity measure $N_{BA}$ , then we have

$1 - N_{BA} (P_{1}, Q) =$

$\frac{1}{\sqrt[p]{3}} \sqrt[p]{\frac{| 0.5 t - 0.4 |^{p} + | 0.4 t - 0.5 |^{p}}{2 (t + 1)^{p}} + \frac{| 0.8 t |^{p} + | 0.8 |^{p}}{2 (t + 1)^{p}} + \frac{| 0.6 t - 0.3 |^{p} + | 0.3 t - 0.6 |^{p}}{2 (t + 1)^{p}}},$

$1 - N_{BA} (P_{2}, Q) =$

$\frac{1}{\sqrt[p]{3}} \sqrt[p]{\frac{| 0.6 t - 0.3 |^{p} + | 0.3 t - 0.6 |^{p}}{2 (t + 1)^{p}} + \frac{| 0.9 t - 0.1 |^{p} + | 0.1 t - 0.9 |^{p}}{2 (t + 1)^{p}} + \frac{| 0.6 t - 0.3 |^{p} + | 0.3 t - 0.6 |^{p}}{2 (t + 1)^{p}}},$

$1 - N_{BA} (P_{3}, Q) =$

$\frac{1}{\sqrt[p]{3}} \sqrt[p]{\frac{| 0.6 t - 0.3 |^{p} + | 0.3 t - 0.6 |^{p}}{2 (t + 1)^{p}} + \frac{| 0.9 t - 0.1 |^{p} + | 0.1 t - 0.9 |^{p}}{2 (t + 1)^{p}} + \frac{| 0.5 t - 0.4 |^{p} + | 0.4 t - 0.5 |^{p}}{2 (t + 1)^{p}}} .$

If 2 ≤ t ≤ 9, then $N_{BA} (P_{1}, Q) = N_{BA} (P_{2}, Q) = N_{BA} (P_{3}, Q)$ and the classification result cannot be determined.

According to Example 3, we know that the similarity measure $N^{'}$ with a ≠ 0 and the similarity measure $N^{″}$ with arbitrary parameters can classify the unknown pattern Q into the pattern P₁. The result coincides with the one obtained in [11].

5 Conclusions

In this paper, two computational formulae for similarity measures on IFSs were proposed based on a quaternary function called intuitionistic fuzzy equivalence. First, the concept of intuitionistic fuzzy equivalence was proposed. A computational formula for intuitionistic fuzzy equivalencies (i.e., Eq. (1)) was obtained from combining dissimilarity functions and fuzzy equivalencies. By taking different fuzzy equivalencies and dissimilarity functions in Eq. (1), different intuitionistic fuzzy equivalencies can be obtained. Some properties of intuitionistic fuzzy equivalencies were examined and compared. Based on Eq. (1), two computational formulae for similarity measures on IFSs were introduced. The first one was obtained by aggregating Eq. (1). The second one was obtained by respectively aggregating the numerator and the denominator of Eq. (1). Some properties of the proposed similarity measures on IFSs were also proved. The comparison between the proposed similarity measures on IFSs and some existing ones in the literature was made through several counter-intuitive cases.

In the future, we will apply the proposed similarity measures to larger data sets and use the proposed intuitionistic fuzzy equivalencies to construct other intuitionistic fuzzy measures such as entropy and subsethood measures on IFSs.

Footnotes

Acknowledgments

This research is supported by the National Natural Science Foundation of China (Grant Nos. 61603307, 61673320 and 61473239) and the Grant from MOE (Ministry of Education in China) Project of Humanities and Social Sciences (Grant No. 19YJCZH048).

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