Comment on “Improvement of the distance between intuitionistic fuzzy sets and its applications”

Abstract

Here, necessary corrections on the proof the Theorem 1 of Xu (J Intell Fuzzy Syst 33(3): 1563-1575, 2017) are stated in brief. Throughout, we use the same notations and equation numbers as in Xu.

Intuitionistic fuzzy sets(IFSs) were proposed by Atanassov [1] as a generalization of the fuzzy sets. As the most interesting topics in IFSs theory, distance measures are involved in fuzzy decision making, patter recognition, fuzzy reasoning, etc, [2–6].

In 2017, a measuring distance between intuitionistic fuzzy sets, proposed by Xu [5], was successfully applied into pattern recognition problems and medical diagnosis. However, there is a small mistake about the proof of the Theorem 1 in Xu [5]. In order to show the detailed correction instructions, the definitions involved in the paper [5] are as follows.

Definition 1. A metric distance D in a non-empty set X is a real value function D : X × X → [0, + ∞), which satisfies the following conditions, ∀x, y, z  ∈  X:

(MD1) D (x, y) =0 if and only if x = y;

(MD2) D (x, y) = D (y, x);

(MD3) D (x, y) + D (y, z) ≥ D (x, z).

Definition 2. [7] Let D be a mapping: IFSs (X) × IFSs (X) → [0, 1]. For ∀A, B, C∈ IFSs(X), D (A, B) is a distance measure between IFSs A and B, if D satisfies the following properties:

(DP1) 0 ≤ D (A, B) ≤1;

(DP2) D (A, B) =0 if and only if A = B;

(DP3) D (A, B) = D (B, A);

(DP4) If A ⊆ B ⊆ C, then D (A, C) ≥ D (A, B), D (A, C) ≥D (B, C).

Definition 3. [5] Let A={〈x, μ _A  (x) , v _A  (x) 〉|x ∈ X} be a IFSs in X={x}, then the assignments of the hesitancy degree π _A  (x) to membership degree μ _A  (x) and non-membership degree ν _A  (x) are defined as $\begin{array}{cc}\hfill {\mathit{Assign}}_A^{\pi \mu }\left(x\right)& =\left[{\pi }_A\right(x)+2{\mu }_A(x\left)\right]/2,\hfill \\ \hfill {\mathit{Assign}}_A^{\pi \nu }\left(x\right)& =\left[{\pi }_A\right(x)+2{\nu }_A(x\left)\right]/2.\hfill \end{array}$(1)

We take the four parts μ _A  (x), ν _A  (x), ${\mathit{Assign}}_A^{\pi \mu }\left(x\right)$ and ${\mathit{Assign}}_A^{\pi \nu }\left(x\right)$ into account the distances between IFSs, thereby a new distance measure, denoted as D_IFSs, is defined.

Definition 4. [5] Let A = {〈x, μ _A  (x) , v _A  (x) 〉|x ∈ X} , B = {〈x, μ _B  (x) , v _B  (x) 〉|x ∈ X} be two IFSs in X={x}, then the distance measure between A and B is defined as $\begin{array}{cc}\hfill D_{\mathrm{IFSs}}& (A,B)=\hfill \\ \hfill & \frac{1}{2}\sqrt{{\left({\Delta }_{\mu }^{\mathit{AB}}\right)}^2+{\left({\Delta }_{\nu }^{\mathit{AB}}\right)}^2+{\left({\Delta }_{\pi \mu }^{\mathit{AB}}\right)}^2+{\left({\Delta }_{\pi \nu }^{\mathit{AB}}\right)}^2},\hfill \end{array}$(2) where, ${\Delta }_{\mu }^{\mathit{AB}}$ =μ _A -μ _B , ${\Delta }_{\nu }^{\mathit{AB}}$ =ν _A -ν _B , ${\Delta }_{\pi \mu }^{\mathit{AB}}={\mathit{Assign}}_A^{\pi \mu }$ - ${\mathit{Assign}}_B^{\pi \mu }$ and ${\Delta }_{\pi \nu }^{\mathit{AB}}$ = ${\mathit{Assign}}_A^{\pi \nu }$ - ${\mathit{Assign}}_B^{\pi \nu }$ .

Theorem 1. [5] Let A={〈x, μ _A  (x) , v _A  (x) 〉|x ∈ X}, B= {〈x, μ _B  (x) , v _B  (x) 〉|x ∈ X} be two IFSs in X={x}, then D_IFSs (A, B) is a distance measure satisfying the Definition 1 and Definition 2.

In the paper [5], the proof of the third step is as follows.

3) For ∀A, B, C ∈ IFSs (X), we have

$\begin{array}{cc}\hfill ({\Delta }_{\mu }^{\mathit{AC}}{)}^2& =({\Delta }_{\mu }^{\mathit{AB}}+{\Delta }_{\mu }^{\mathit{BC}}{)}^2\le ({\Delta }_{\mu }^{\mathit{AB}}{)}^2+({\Delta }_{\mu }^{\mathit{BC}}{)}^2;\hfill \\ \hfill ({\Delta }_{\nu }^{\mathit{AC}}{)}^2& =({\Delta }_{\nu }^{\mathit{AB}}+{\Delta }_{\nu }^{\mathit{BC}}{)}^2\le ({\Delta }_{\nu }^{\mathit{AB}}{)}^2+({\Delta }_{\nu }^{\mathit{BC}}{)}^2;\hfill \\ \hfill ({\Delta }_{\pi \mu }^{\mathit{AC}}{)}^2& =({\Delta }_{\pi \mu }^{\mathit{AB}}+{\Delta }_{\pi \mu }^{\mathit{BC}}{)}^2\le ({\Delta }_{\pi \mu }^{\mathit{AB}}{)}^2+({\Delta }_{\pi \mu }^{\mathit{BC}}{)}^2;\hfill \\ \hfill ({\Delta }_{\pi \nu }^{\mathit{AC}}{)}^2& =({\Delta }_{\pi \nu }^{\mathit{AB}}+{\Delta }_{\pi \nu }^{\mathit{BC}}{)}^2\le ({\Delta }_{\pi \nu }^{\mathit{AB}}{)}^2+({\Delta }_{\pi \nu }^{\mathit{BC}}{)}^2.\hfill \end{array}$(3)

Thus, D_IFSs (A, C) ≤ D_IFSs (A, B) + D_IFSs (B, C), which indicates that D_IFSs satisfies (MD3).

However, the conclusion D_IFSs (A, C) ≤ D_IFSs (A, B) + D_IFSs (B, C) is derived from the formula (3), which is a wrong logical reasoning. Where, it should be noted that the formula (3) is correct. In fact, from the formula (3) and the property of inequality, we can obtained

$\begin{array}{cc}\hfill \left[\right({\Delta }_{\mu }^{\mathit{AC}}{)}^2+({\Delta }_{\nu }^{\mathit{AC}}{)}^2+({\Delta }_{\pi \mu }^{\mathit{AC}}{)}^2+\left({\Delta }_{\pi \nu }^{\mathit{AC}}{)}^2\right]\le & \hfill \\ \hfill \left\{\right[({\Delta }_{\mu }^{\mathit{AB}}{)}^2+({\Delta }_{\nu }^{\mathit{AB}}{)}^2+({\Delta }_{\pi \mu }^{\mathit{AB}}{)}^2+({\Delta }_{\pi \nu }^{\mathit{AB}}{)}^2]& \hfill \\ \hfill +\left[\right({\Delta }_{\mu }^{\mathit{BC}}{)}^2+({\Delta }_{\nu }^{\mathit{BC}}{)}^2+({\Delta }_{\pi \mu }^{\mathit{BC}}{)}^2+\left({\Delta }_{\pi \nu }^{\mathit{BC}}{)}^2\right]\}.& \hfill \end{array}$(4)

While, from the Definition 4, we have $\begin{array}{ccc}\hfill & \hfill \hfill & 4\left[D_{\mathrm{IFSs}}\right(A,C){]}^2=({\Delta }_{\mu }^{\mathit{AC}}{)}^2+({\Delta }_{\nu }^{\mathit{AC}}{)}^2+({\Delta }_{\pi \mu }^{\mathit{AC}}{)}^2+({\Delta }_{\pi \nu }^{\mathit{AC}}{)}^2,\hfill \\ \hfill & \hfill \hfill & 4\left[D_{\mathrm{IFSs}}\right(A,B){]}^2=({\Delta }_{\mu }^{\mathit{AB}}{)}^2+({\Delta }_{\nu }^{\mathit{AB}}{)}^2+({\Delta }_{\pi \mu }^{\mathit{AB}}{)}^2+({\Delta }_{\pi \nu }^{\mathit{AB}}{)}^2,\hfill \\ \hfill & \hfill \hfill & 4\left[D_{\mathrm{IFSs}}\right(B,C){]}^2=({\Delta }_{\mu }^{\mathit{BC}}{)}^2+({\Delta }_{\nu }^{\mathit{BC}}{)}^2+({\Delta }_{\pi \mu }^{\mathit{BC}}{)}^2+({\Delta }_{\pi \nu }^{\mathit{BC}}{)}^2.\hfill \end{array}$

Therefore, $\left[D_{\mathrm{IFSs}}\right(A,C){]}^2\le [D_{\mathrm{IFSs}}(A,B){]}^2+\left[D_{\mathrm{IFSs}}\right(B,C){]}^2.$(5)

However, based on the formula (5), it is not obtained $D_{\mathrm{IFSs}}(A,C)\le D_{\mathrm{IFSs}}(A,B)+D_{\mathrm{IFSs}}(B,C).$ This shows that the proof of the paper [5] is incorrect.

The proper proof of the third step is as follows.

3) According to the Definition 4, the distance measure 2D_IFSs (A, B) can be viewed as the Euclidean distance between two points $({\mu }_A,{\nu }_A,{\mathit{Assign}}_A^{\pi \mu },{\mathit{Assign}}_A^{\pi \nu })$ and $({\mu }_B,{\nu }_B,{\mathit{Assign}}_B^{\pi \mu },{\mathit{Assign}}_B^{\pi \nu })$ in a four-dimensional real number space. For ∀A, B, C ∈ IFSs (X), there are three real points. $\begin{array}{cc}\hfill A:({\mu }_A,{\nu }_A,{\mathit{Assign}}_A^{\pi \mu },{\mathit{Assign}}_A^{\pi \nu }),& \hfill \\ \hfill B:({\mu }_B,{\nu }_B,{\mathit{Assign}}_B^{\pi \mu },{\mathit{Assign}}_B^{\pi \nu }),& \hfill \\ \hfill C:({\mu }_C,{\nu }_C,{\mathit{Assign}}_C^{\pi \mu },{\mathit{Assign}}_C^{\pi \nu }).& \hfill \\ \hfill & \hfill \end{array}$

Since the distance measure 2D_IFSs (A, B) is a metric distance, based on the third condition (MD3) of the Definition 1 (The properties of triangular inequalities for Euclidean distance), we have $2D_{\mathrm{IFSs}}(A,C)\le 2D_{\mathrm{IFSs}}(A,B)+2D_{\mathrm{IFSs}}(B,C),$ which yields $D_{\mathrm{IFSs}}(A,C)\le D_{\mathrm{IFSs}}(A,B)+D_{\mathrm{IFSs}}(B,C).$ The result shows that D_IFSs satisfies (MD3).