**4. Similarity Measure of Quadripartitioned Bipolar Neutrosophic Sets**

Here we provide the definition of similarity measure between two QSVBNS *A*, *B* over a universe *X*.

**Definition 14.** *Let QSVBNS(X) indicate the set of all QSVBNS over the universe X. Let S* : *QSVBNS*(*X*) × *QSVBNS*(*X*) → [0, 1] *be a function satisfying the following properties for all A*, *B* ∈ *QSVBNS*(*X*)*:*

*(S1)* 0 ≤ *S*(*A*, *B*) < 1 *and S*(*A*, *B*) = 1 *iff A* = *B, (S2) S*(*A*, *B*) = *S*(*B*, *A*)*, (S3) for any A*, *B*, *C* ∈ *QSVBNS*(*X*) *with A* ⊆ *B* ⊆ *C, S*(*A*, *C*) ≤ *S*(*A*, *B*) ∧ *S*(*B*, *C*)*.*

*then S is said to be a similarity measure.*

Based on the membership functions of two QSVBNS, we prescribe some functions which measures the differences between the membership values of two QSVBNS. Let *A*, *B* ∈ QSVBNS. For each *x<sup>k</sup>* ∈ *X*, *k* = 1, 2, ..., *n* and for each *i* = 1, 2, 3, 4 and *j* = 1, 2 define the functions *δ A*,*B i* , *λ A*,*B j* : QSVBNS(X)×*QSVBNS*(*X*) → [0, 1] respectively,

$$\bullet \,\,\delta\_1^{A,B}(\mathfrak{x}\_k) = \frac{1}{2} (|T\_A^P(\mathfrak{x}\_k) - T\_B^P(\mathfrak{x}\_k)| + |T\_A^N(\mathfrak{x}\_k) - T\_B^N(\mathfrak{x}\_k)|) \,\,\,\omega$$

$$\bullet \ \delta\_2^{A,B}(\mathbf{x}\_k) = \frac{1}{2} (|\boldsymbol{F}\_A^P(\mathbf{x}\_k) - \boldsymbol{F}\_B^P(\mathbf{x}\_k)| + |\boldsymbol{F}\_A^N(\mathbf{x}\_k) - \boldsymbol{F}\_B^N(\mathbf{x}\_k)|),$$

$$\bullet \,\,\delta\_3^{A,B}(\mathbf{x}\_k) = \frac{1}{4} (\delta\_1^{A,B}(\mathbf{x}\_k) + \delta\_2^{A,B}(\mathbf{x}\_k) + |\mathsf{C}\_A^P(\mathbf{x}\_k) - \mathsf{C}\_B^P(\mathbf{x}\_k)| + |\mathsf{C}\_A^N(\mathbf{x}\_k) - \mathsf{C}\_B^N(\mathbf{x}\_k)|),$$

$$\bullet \ \delta\_{4}^{A,B}(\mathbf{x}\_{k}) = \frac{1}{2}(|\mathcal{U}\_{A}^{P}(\mathbf{x}\_{k}) - \mathcal{U}\_{B}^{P}(\mathbf{x}\_{k})| + |\mathcal{U}\_{A}^{N}(\mathbf{x}\_{k}) - \mathcal{U}\_{B}^{N}(\mathbf{x}\_{k})|)\_{\lambda}$$

$$\begin{array}{rcl} \bullet \ \boldsymbol{\lambda}\_{1}^{A,\mathcal{B}}(\mathbf{x}\_{k}) &=& \frac{1}{4} (|\boldsymbol{T}\_{\boldsymbol{A}}^{\mathcal{P}}(\mathbf{x}\_{k})\boldsymbol{\mathsf{C}}\_{\boldsymbol{A}}^{\mathcal{P}}(\mathbf{x}\_{k}) - \boldsymbol{T}\_{\boldsymbol{B}}^{\mathcal{P}}(\mathbf{x}\_{k})\boldsymbol{\mathsf{C}}\_{\boldsymbol{B}}^{\mathcal{P}}(\mathbf{x}\_{k})| + |\boldsymbol{T}\_{\boldsymbol{A}}^{\mathcal{N}}(\mathbf{x}\_{k})\boldsymbol{\mathsf{C}}\_{\boldsymbol{A}}^{\mathcal{N}}(\mathbf{x}\_{k}) - \boldsymbol{T}\_{\boldsymbol{B}}^{\mathcal{N}}(\mathbf{x}\_{k})\boldsymbol{\mathsf{C}}\_{\boldsymbol{B}}^{\mathcal{N}}(\mathbf{x}\_{k})| + |\boldsymbol{\mathsf{C}}\_{\boldsymbol{A}}^{\mathcal{N}}(\mathbf{x}\_{k}) - \boldsymbol{T}\_{\boldsymbol{B}}^{\mathcal{N}}(\mathbf{x}\_{k})|) \\ & & \boldsymbol{\mathsf{C}}\_{\boldsymbol{B}}^{\mathcal{P}}(\mathbf{x}\_{k})| + |\boldsymbol{\mathsf{C}}\_{\boldsymbol{A}}^{\mathcal{N}}(\mathbf{x}\_{k}) - \boldsymbol{\mathsf{C}}\_{\boldsymbol{B}}^{\mathcal{N}}(\mathbf{x}\_{k})|), \end{array}$$

$$\bullet \ \lambda\_2^{A,B}(\mathbf{x}\_k) = \frac{1}{4}(\delta\_1^{A,B}(\mathbf{x}\_k) + \delta\_2^{A,B}(\mathbf{x}\_k) + |\mathcal{U}\_A^P(\mathbf{x}\_k) - \mathcal{U}\_B^P(\mathbf{x}\_k)| + |\mathcal{U}\_A^N(\mathbf{x}\_k) - \mathcal{U}\_B^N(\mathbf{x}\_k)|).$$

**Remark 2.** *The function* <sup>T</sup>1(*A*, *<sup>B</sup>*) *defined by*T1(*A*, *<sup>B</sup>*) = <sup>1</sup> <sup>−</sup> <sup>1</sup> 4*n n* ∑ *k*=1 4 ∑ *i*=1 *δ A*,*B i* (*x<sup>k</sup>* ), *A*, *B* ∈ *QSVBNS*(*X*) *is shown to be a similarity measure in the following theorem.*

**Theorem 2.** T1(*A*, *B*) *is a similarity measure between two quadripartitioned bipolar neutrosophic sets A and B over X.*

**Proof.** Since *T P A* (*x<sup>k</sup>* ), *C P A* (*x<sup>k</sup>* ), *U P A* (*x<sup>k</sup>* ), *F P A* (*x<sup>k</sup>* ) ∈ [0, 1] and *T N A* (*x<sup>k</sup>* ), *C N A* (*x<sup>k</sup>* ), *U N A* (*x<sup>k</sup>* ), *F N A* (*x<sup>k</sup>* ) ∈ [−1, 0] for all *x<sup>k</sup>* ∈ *X* it follows that *δ A*,*B* 1 (*x<sup>k</sup>* ) attains its maximum value 1 whenever one of the positive truth membership values corresponding to *A* and *B* is 1 and the other is 0 and one of the negative truth membership values corresponding to *A* and *B* is −1 and the other is 0. Similarly for each *x<sup>k</sup>* ∈ *X*, *δ A*,*B* 1 (*x<sup>k</sup>* ) attains its minimum value 0 whenever *T P A* (*x<sup>k</sup>* ) = *T P B* (*x<sup>k</sup>* ) and *T N A* (*x<sup>k</sup>* ) = *T N B* (*x<sup>k</sup>* ). Thus for all *x<sup>k</sup>* ∈ *X*, 0 ≤ *δ A*,*B* 1 (*x<sup>k</sup>* ) ≤ 1, ∀*x<sup>k</sup>* ∈ *X*. Likewise it can be shown that 0 ≤ *δ A*,*B* 2 (*x<sup>k</sup>* ), *δ A*,*B* 3 (*x<sup>k</sup>* ), *δ A*,*B* 4 (*x<sup>k</sup>* ) ≤ 1. Therefore for all *x<sup>k</sup>* ∈ *X* we have,

$$\begin{array}{lcl} & 0 \le \delta\_1^{A,B}(\mathbf{x}\_k) + \delta\_2^{A,B}(\mathbf{x}\_k) + \delta\_3^{A,B}(\mathbf{x}\_k) + \delta\_4^{A,B}(\mathbf{x}\_k) \le 4\\ \implies & 0 \le \frac{1}{4} [\delta\_1^{A,B}(\mathbf{x}\_k) + \delta\_2^{A,B}(\mathbf{x}\_k) + \delta\_3^{A,B}(\mathbf{x}\_k) + \delta\_4^{A,B}(\mathbf{x}\_k)] \le 1\\ \implies & 0 \le \frac{1}{4n} \sum\_{k=1}^n \sum\_{i=1}^4 \delta\_i^{A,B}(\mathbf{x}\_k) \le 1\\ \implies & -1 \le -\frac{1}{4n} \sum\_{k=1}^n \sum\_{i=1}^4 \delta\_i^{A,B}(\mathbf{x}\_k) \le 0\\ \implies & 0 \le 1 - \frac{1}{4n} \sum\_{k=1}^n \sum\_{i=1}^4 \delta\_i^{A,B}(\mathbf{x}\_k) \le 1\\ \implies & 0 \le T\_1(A,B) \le 1 \end{array}$$

Again, T1(*A*, *B*) = 1 ⇐⇒ *n* ∑ *k*=1 4 ∑ *i*=1 *δ A*,*B i* (*x<sup>k</sup>* ) = 0 ⇐⇒ *δ A*,*B* 1 (*x<sup>k</sup>* ) = *δ A*,*B* 2 (*x<sup>k</sup>* ) = *δ A*,*B* 3 (*x<sup>k</sup>* ) = *δ A*,*B* 4 (*x<sup>k</sup>* ) = 0 ⇐⇒ *T P A* (*x<sup>k</sup>* ) = *T P B* (*x<sup>k</sup>* ), *C P A* (*x<sup>k</sup>* ) = *C P B* (*x<sup>k</sup>* ), *U P A* (*x<sup>k</sup>* ) = *U P B* (*x<sup>k</sup>* ), *F P A* (*x<sup>k</sup>* ) = *F P B* (*x<sup>k</sup>* ) and *T N A* (*x<sup>k</sup>* ) = *T N B* (*x<sup>k</sup>* ), *C N A* (*x<sup>k</sup>* ) = *C N B* (*x<sup>k</sup>* ), *U N A* (*x<sup>k</sup>* ) = *U N B* (*x<sup>k</sup>* ), *F N A* (*x<sup>k</sup>* ) = *F N B* (*x<sup>k</sup>* ) ⇐⇒ *A* = *B*.

It is easy to prove that T1(*A*, *B*) = T1(*B*, *A*).

Finally let *A* ⊂ *B* ⊂ *C*. Then for all *x<sup>k</sup>* ∈ *X*, we have *T P A* (*x<sup>k</sup>* ) ≤ *T P B* (*x<sup>k</sup>* ) ≤ *T P C* (*x<sup>k</sup>* ), *T N A* (*x<sup>k</sup>* ) ≥ *T N B* (*x<sup>k</sup>* ) ≥ *T N C* (*x<sup>k</sup>* ), *C P A* (*x<sup>k</sup>* ) ≤ *C P B* (*x<sup>k</sup>* ) ≤ *C P C* (*x<sup>k</sup>* ), *C N A* (*x<sup>k</sup>* ) ≥ *C N B* (*x<sup>k</sup>* ) ≥ *C N C* (*x<sup>k</sup>* ), *U P A* (*x<sup>k</sup>* ) ≥ *U P B* (*x<sup>k</sup>* ) ≥ *U P C* (*x<sup>k</sup>* ), *U N A* (*x<sup>k</sup>* ) ≤ *U N B* (*x<sup>k</sup>* ) ≤ *U N C* (*x<sup>k</sup>* ), *F P A* (*x<sup>k</sup>* ) ≥ *F P B* (*x<sup>k</sup>* ) ≥ *F P C* (*x<sup>k</sup>* ), *F N A* (*x<sup>k</sup>* ) ≤ *F N B* (*x<sup>k</sup>* ) ≤ *F N C* (*x<sup>k</sup>* ). Then, |*T P* (*x<sup>k</sup>* ) − *T P* (*x<sup>k</sup>* )| ≤ |*T P* (*x<sup>k</sup>* ) − *T P* (*x<sup>k</sup>* )|, |*T N* (*x<sup>k</sup>* ) − *T N* (*x<sup>k</sup>* )| ≤ |*T N* (*x<sup>k</sup>* ) − *T N* (*x<sup>k</sup>* )|.

*A B A C A B A C* Therefore, <sup>1</sup> 2 (|*T P A* (*x<sup>k</sup>* ) − *T P B* (*x<sup>k</sup>* )| + |*T N A* (*x<sup>k</sup>* ) − *T N B* (*x<sup>k</sup>* )|) <sup>≤</sup> <sup>1</sup> 2 (|*T P A* (*x<sup>k</sup>* ) − *T P C* (*x<sup>k</sup>* )| + |*T N A* (*x<sup>k</sup>* ) − *T N C* (*x<sup>k</sup>* )|) ⇒ *δ A*,*B* 1 (*x<sup>k</sup>* ) ≤ *δ A*,*C* 1 (*x<sup>k</sup>* ), ∀*x<sup>k</sup>* ∈ *X* ...........................(∗)

Considering a similar process, it can be shown that *δ A*,*B* 2 (*x<sup>k</sup>* ) ≤ *δ A*,*C* 2 (*x<sup>k</sup>* ), *δ A*,*B* 3 (*x<sup>k</sup>* ) ≤ *δ A*,*C* 3 (*x<sup>k</sup>* ) and *δ A*,*B* 4 (*x<sup>k</sup>* ) ≤ *δ A*,*C* 4 (*x<sup>k</sup>* ).

Therefore,

$$
\sum\_{i=1}^{4} \delta\_i^{A, \mathsf{C}}(\mathbf{x}\_k) \ge \sum\_{i=1}^{4} \delta\_i^{A, B}(\mathbf{x}\_k)
$$

$$
\Rightarrow \sum\_{k=1}^{n} \sum\_{i=1}^{4} \delta\_i^{A, \mathsf{C}}(\mathbf{x}\_k) \ge \sum\_{k=1}^{n} \sum\_{i=1}^{4} \delta\_i^{A, B}(\mathbf{x}\_k)
$$

$$
\Rightarrow 1 - \frac{1}{4n} \sum\_{k=1}^{n} \sum\_{i=1}^{4} \delta\_i^{A, B}(\mathbf{x}\_k) \ge 1 - \frac{1}{4n} \sum\_{k=1}^{n} \sum\_{i=1}^{4} \delta\_i^{A, \mathsf{C}}(\mathbf{x}\_k)
$$

$$
\Rightarrow \mathcal{T}\_1(A, B) \ge \mathcal{T}\_1(A, \mathsf{C}).
$$

Similarly T1(*B*, *C*) ≥ T1(*A*, *C*). Therefore, T1(*A*, *C*) ≤ T1(*A*, *B*) ∧ T1(*B*, *C*). Hence the proof.

**Remark 3.** *Define* T2(*A*, *B*) = 1 − h 1 *n n* ∑ *k*=1 1 4 *δ A*,*B* 1 (*x<sup>k</sup>* ) + *δ A*,*B* 2 (*x<sup>k</sup>* ) + *λ A*,*B* 1 (*x<sup>k</sup>* ) + *λ A*,*B* 2 (*x<sup>k</sup>* ) *<sup>p</sup>* i 1 *p , where p, a positive integer, is defined to be the order of the similarity.*

**Theorem 3.** T2(*A*, *B*) *is a similarity measure between two quadripartitioned bipolar neutrosophic sets A and B over X.*

**Proof.** T2(*A*, *B*) = T2(*B*, *A*) is quite obvious.

Under any of the the following condition *λ A*,*B* 1 (*x<sup>k</sup>* ) attains its maximum value 1,


Also the minimum value of *λ A*,*B* 1 (*x<sup>k</sup>* ) is 0. Thus 0 ≤ *λ A*,*B* 1 (*x<sup>k</sup>* ) ≤ 1 for all *x<sup>k</sup>* ∈ *X*. Also 0 ≤ *λ A*,*B* 2 (*x<sup>k</sup>* ) ≤ 1 follows from similar arguments. Therefore,

0 ≤ *δ A*,*B* 1 (*x<sup>k</sup>* ) + *δ A*,*B* 2 (*x<sup>k</sup>* ) + *λ A*,*B* 1 (*x<sup>k</sup>* ) + *λ A*,*B* 2 (*x<sup>k</sup>* ) ≤ 4 =⇒ 0 ≤ 1 4 *δ A*,*B* 1 (*x<sup>k</sup>* ) + *δ A*,*B* 2 (*x<sup>k</sup>* ) + *λ A*,*B* 1 (*x<sup>k</sup>* ) + *λ A*,*B* 2 (*x<sup>k</sup>* ) *<sup>p</sup>* <sup>≤</sup> 1, <sup>∀</sup>*<sup>p</sup>* <sup>≥</sup> <sup>1</sup> =⇒ 0 ≤ *n* ∑ *k*=1 1 4 *δ A*,*B* 1 (*x<sup>k</sup>* ) + *δ A*,*B* 2 (*x<sup>k</sup>* ) + *λ A*,*B* 1 (*x<sup>k</sup>* ) + *λ A*,*B* 2 (*x<sup>k</sup>* ) *<sup>p</sup>* <sup>≤</sup> *<sup>n</sup>* <sup>=</sup><sup>⇒</sup> <sup>0</sup> <sup>≤</sup> <sup>1</sup> *n n* ∑ *k*=1 1 4 *δ A*,*B* 1 (*x<sup>k</sup>* ) + *δ A*,*B* 2 (*x<sup>k</sup>* ) + *λ A*,*B* 1 (*x<sup>k</sup>* ) + *λ A*,*B* 2 (*x<sup>k</sup>* ) *<sup>p</sup>* <sup>≤</sup> <sup>1</sup> =⇒ 0 ≤ h 1 *n n* ∑ *k*=1 1 4 *δ A*,*B* 1 (*x<sup>k</sup>* ) + *δ A*,*B* 2 (*x<sup>k</sup>* ) + *λ A*,*B* 1 (*x<sup>k</sup>* ) + *λ A*,*B* 2 (*x<sup>k</sup>* ) *<sup>p</sup>* i 1 *p* ≤ 1 =⇒ 0 ≤ 1 − h 1 *n n* ∑ *k*=1 1 4 *δ A*,*B* 1 (*x<sup>k</sup>* ) + *δ A*,*B* 2 (*x<sup>k</sup>* ) + *λ A*,*B* 1 (*x<sup>k</sup>* ) + *λ A*,*B* 2 (*x<sup>k</sup>* ) *<sup>p</sup>* i 1 *p* ≤ 1 =⇒ 0 ≤ T2(*A*, *B*) ≤ 1

To show the triangular inequality suppose *P* ⊂ *Q* ⊂ *R*. Then for all *x<sup>k</sup>* ∈ *X*, we have, *T P P* (*x<sup>k</sup>* ) ≤ *T P Q* (*x<sup>k</sup>* ) ≤ *T P R* (*x<sup>k</sup>* ), *T N P* (*x<sup>k</sup>* ) ≥ *T N Q* (*x<sup>k</sup>* ) ≥ *T N R* (*x<sup>k</sup>* ), *C P P* (*x<sup>k</sup>* ) ≤ *C P Q* (*x<sup>k</sup>* ) ≤ *C P R* (*x<sup>k</sup>* ), *C N P* (*x<sup>k</sup>* ) ≥ *C N Q* (*x<sup>k</sup>* ) ≥ *C N R* (*x<sup>k</sup>* ), *U P P* (*x<sup>k</sup>* ) ≥ *U P Q* (*x<sup>k</sup>* ) ≥ *U P R* (*x<sup>k</sup>* ), *U N P* (*x<sup>k</sup>* ) ≤ *U N Q* (*x<sup>k</sup>* ) ≤ *U N R* (*x<sup>k</sup>* ), *F P P* (*x<sup>k</sup>* ) ≥ *F P Q* (*x<sup>k</sup>* ) ≥ *F P R* (*x<sup>k</sup>* ), *F N P* (*x<sup>k</sup>* ) ≤ *F N Q* (*x<sup>k</sup>* ) ≤ *F N R* (*x<sup>k</sup>* ).

Then *δ P*,*R* 1 (*x<sup>k</sup>* ) ≥ *δ P*,*Q* 1 (*x<sup>k</sup>* ) and *δ P*,*R* 2 (*x<sup>k</sup>* ) ≥ *δ P*,*Q* 2 (*x<sup>k</sup>* ) follows from (∗) of Theorem 2. Next consider *λ P*,*Q* 1 (*x<sup>k</sup>* ) and *λ P*,*R* 1 (*x<sup>k</sup>* ). From above inequalities we have, *T P P* (*x<sup>k</sup>* )*C P P* (*x<sup>k</sup>* ) ≤ *T P Q* (*x<sup>k</sup>* )*C P Q* (*x<sup>k</sup>* ) ≤ *T P R* (*x<sup>k</sup>* )*C P R* (*x<sup>k</sup>* ) ⇒ |*T P P* (*x<sup>k</sup>* )*C P P* (*x<sup>k</sup>* ) − *T P Q* (*x<sup>k</sup>* )*C P Q* (*x<sup>k</sup>* )| ≤ |*T P P* (*x<sup>k</sup>* )*C P P* (*x<sup>k</sup>* ) − *T P R* (*x<sup>k</sup>* )*C P R* (*x<sup>k</sup>* )|. Similarly, |*T N P* (*x<sup>k</sup>* )*C N P* (*x<sup>k</sup>* ) − *T N Q* (*x<sup>k</sup>* )*C N Q* (*x<sup>k</sup>* )| ≤ |*T N P* (*x<sup>k</sup>* )*C N P* (*x<sup>k</sup>* ) − *T N R* (*x<sup>k</sup>* )*C N R* (*x<sup>k</sup>* )| and |*C P P* (*x<sup>k</sup>* ) − *C P Q* (*x<sup>k</sup>* )| ≤ |*C P P* (*x<sup>k</sup>* ) − *C P Q* (*x<sup>k</sup>* )|, |*C P P* (*x<sup>k</sup>* ) − *C P Q* (*x<sup>k</sup>* )| ≤ |*C P P* (*x<sup>k</sup>* ) − *C P Q* (*x<sup>k</sup>* )|. *P P P P N N N N P*

Then, <sup>1</sup> 4 |*T P* (*x<sup>k</sup>* )*C P* (*x<sup>k</sup>* ) − *T R* (*x<sup>k</sup>* )*C R* (*x<sup>k</sup>* )| + |*T P* (*x<sup>k</sup>* )*C P* (*x<sup>k</sup>* ) − *T R* (*x<sup>k</sup>* )*C R* (*x<sup>k</sup>* )| + |*C P* (*x<sup>k</sup>* ) − *C P R* (*x<sup>k</sup>* )| + |*C N P* (*x<sup>k</sup>* ) − *C N R* (*x<sup>k</sup>* )| <sup>≥</sup> <sup>1</sup> 4 |*T P P* (*x<sup>k</sup>* )*C P P* (*x<sup>k</sup>* ) − *T P Q* (*x<sup>k</sup>* )*C P Q* (*x<sup>k</sup>* )| + |*T N P* (*x<sup>k</sup>* )*C N P* (*x<sup>k</sup>* ) − *T N Q* (*x<sup>k</sup>* )*C N Q* (*x<sup>k</sup>* )| + |*C P P* (*x<sup>k</sup>* ) − *C P Q* (*x<sup>k</sup>* )| + |*C N P* (*x<sup>k</sup>* ) − *C N Q* (*x<sup>k</sup>* )| ⇒ *λ P*,*R* 1 (*x<sup>k</sup>* ) ≥ *λ P*,*Q* 1 (*x<sup>k</sup>* )

Similarly it can shown that *λ P*,*R* 2 (*x<sup>k</sup>* ) ≥ *λ P*,*Q* 2 (*x<sup>k</sup>* ).

1 4 *δ P*,*R* 1 (*x<sup>k</sup>* ) + *δ P*,*R* 2 (*x<sup>k</sup>* ) + *λ P*,*R* 1 (*x<sup>k</sup>* ) + *λ P*,*R* 2 (*x<sup>k</sup>* ) <sup>≥</sup> <sup>1</sup> 4 *δ P*,*Q* 1 (*x<sup>k</sup>* ) + *δ P*,*Q* 2 (*x<sup>k</sup>* ) + *λ P*,*Q* 1 (*x<sup>k</sup>* ) + *λ P*,*Q* 2 (*x<sup>k</sup>* ) ⇒ *n* ∑ *k*=1 1 4 *δ P*,*R* 1 (*x<sup>k</sup>* ) + *δ P*,*R* 2 (*x<sup>k</sup>* ) + *λ P*,*R* 1 (*x<sup>k</sup>* ) + *λ P*,*R* 2 (*x<sup>k</sup>* ) *<sup>p</sup>* <sup>≥</sup> *n* ∑ *k*=1 1 4 *δ P*,*Q* 1 (*x<sup>k</sup>* ) + *δ P*,*Q* 2 (*x<sup>k</sup>* ) + *λ P*,*Q* 1 (*x<sup>k</sup>* ) + *λ P*,*Q* 2 (*x<sup>k</sup>* ) *<sup>p</sup>* <sup>⇒</sup> <sup>1</sup> <sup>−</sup> h 1 *n n* ∑ *k*=1 1 4 *δ P*,*R* 1 (*x<sup>k</sup>* ) + *δ P*,*R* 2 (*x<sup>k</sup>* ) + *λ P*,*R* 1 (*x<sup>k</sup>* ) + *λ P*,*R* 2 (*x<sup>k</sup>* ) *<sup>p</sup>* i 1 *p* ≤ 1 − h 1 *n n* ∑ *k*=1 1 4 *δ P*,*Q* 1 (*x<sup>k</sup>* ) + *δ P*,*Q* 2 (*x<sup>k</sup>* ) + *λ P*,*Q* 1 (*x<sup>k</sup>* ) + *λ P*,*Q* 2 (*x<sup>k</sup>* ) *<sup>p</sup>* i 1 *p* ⇒ T2(*P*, *R*) ≤ T2(*P*, *Q*)

It is identical to show T2(*P*, *R*) ≤ T2(*Q*, *R*). Therefore, T2(*P*, *R*) ≤ T2(*P*, *Q*) ∧ T2(*Q*, *R*). This completes the proof.

We define below the weighted similarity measure between two QSVBNS *A*, *B* over a universe *X*.

**Definition 15.** *The weighted similarity measure of two QSVBNS A*, *B is defined as,* T *w* 2 (*A*, *B*) = 1 − h 1 *n n* ∑ *k*=1 *wk* 1 4 *δ A*,*B* 1 (*x<sup>k</sup>* ) + *δ A*,*B* 2 (*x<sup>k</sup>* ) + *λ A*,*B* 1 (*x<sup>k</sup>* ) + *λ A*,*B* 2 (*x<sup>k</sup>* ) *<sup>p</sup>* i 1 *p , where,* (*w*1, *w*2, ....*wn*) *T is the weight vector assigned to the element x*1, *x*2, .....*x<sup>n</sup> of the universe X, such that* 0 ≤ *w<sup>k</sup>* ≤ 1, *k* = 1, 2, ..., *n and n* ∑ *k*=1 *w<sup>k</sup>* = 1*.*

*It is effortless to find out that* T *w* 2 (*A*, *B*) *satisfies the conditions of similarity measure.*

**Remark 4.** *We define the following functions based on one particular membership function for two QSVBNSs A and B over the universe of discourse X, provided the denominators never vanish*

T *P T* (*A*, *B*) = *n* ∑ *k*=1 *min T P A* (*x<sup>k</sup>* ), *T P B* (*x<sup>k</sup>* ) *n* ∑ *k*=1 *max T P A* (*x<sup>k</sup>* ), *T P B* (*x<sup>k</sup>* ) , <sup>T</sup> *P C* (*A*, *B*) = *n* ∑ *k*=1 *min C P A* (*x<sup>k</sup>* ), *C P B* (*x<sup>k</sup>* ) *n* ∑ *k*=1 *max C P A* (*x<sup>k</sup>* ), *C P B* (*x<sup>k</sup>* ) , T *P U* (*A*, *B*) = *n* ∑ *k*=1 *min U P A* (*x<sup>k</sup>* ), *U P B* (*x<sup>k</sup>* ) *n* ∑ *k*=1 *max U P A* (*x<sup>k</sup>* ), *U P B* (*x<sup>k</sup>* ) , <sup>T</sup> *P F* (*A*, *B*) = *n* ∑ *k*=1 *min F P A* (*x<sup>k</sup>* ), *F P B* (*x<sup>k</sup>* ) *n* ∑ *k*=1 *max F P A* (*x<sup>k</sup>* ), *F P B* (*x<sup>k</sup>* ) , T *N T* (*A*, *B*) = *n* ∑ *k*=1 *max T N A* (*x<sup>k</sup>* ), *T N B* (*x<sup>k</sup>* ) *n* ∑ *k*=1 *min T N A* (*x<sup>k</sup>* ), *T N B* (*x<sup>k</sup>* ) , <sup>T</sup> *N C* (*A*, *B*) = *n* ∑ *k*=1 *max C N A* (*x<sup>k</sup>* ), *C N B* (*x<sup>k</sup>* ) *n* ∑ *k*=1 *min C N A* (*x<sup>k</sup>* ), *C N B* (*x<sup>k</sup>* ) , T *N U* (*A*, *B*) = *n* ∑ *k*=1 *max U N A* (*x<sup>k</sup>* ), *U N B* (*x<sup>k</sup>* ) *n* ∑ *k*=1 *min U N A* (*x<sup>k</sup>* ), *U N B* (*x<sup>k</sup>* ) , <sup>T</sup> *N F* (*A*, *B*) = *n* ∑ *k*=1 *max F N A* (*x<sup>k</sup>* ), *F N B* (*x<sup>k</sup>* ) *n* ∑ *k*=1 *min F N A* (*x<sup>k</sup>* ), *F N B* (*x<sup>k</sup>* ) .

*The following is a definition of generalized similarity measure between two QSVBNS A*, *B whose value set is the set of all* 2 × 4 *matrices over* R*.*

**Definition 16.** *Let X* = {*x*1, *x*2, ....., *xn*} *be a finite universe of discourse. For two QSVBNSs A*, *B define a mapping* L :*QSVBNS*(*X*)× *QSVBNS*(*X*) → M2×4(R) *by*

$$
\mathcal{L}(A,B) = \begin{pmatrix}
\mathcal{T}\_{\scriptscriptstyle{\boldsymbol{\mathcal{T}}}}^{\scriptscriptstyle{\boldsymbol{\mathcal{P}}}}(A,B) & \mathcal{T}\_{\scriptscriptstyle{\boldsymbol{\mathcal{C}}}}^{\scriptscriptstyle{\boldsymbol{\mathcal{P}}}}(A,B) & \mathcal{T}\_{\scriptscriptstyle{\boldsymbol{\mathcal{U}}}}^{\scriptscriptstyle{\boldsymbol{\mathcal{P}}}}(A,B) & \mathcal{T}\_{\scriptscriptstyle{\boldsymbol{\mathcal{E}}}}^{\scriptscriptstyle{\boldsymbol{\mathcal{P}}}}(A,B) \\
\mathcal{T}\_{\scriptscriptstyle{\boldsymbol{\mathcal{T}}}}^{\scriptscriptstyle{\boldsymbol{\mathcal{N}}}}(A,B) & \mathcal{T}\_{\scriptscriptstyle{\boldsymbol{\mathcal{C}}}}^{\scriptscriptstyle{\boldsymbol{\mathcal{N}}}}(A,B) & \mathcal{T}\_{\scriptscriptstyle{\boldsymbol{\mathcal{U}}}}^{\scriptscriptstyle{\boldsymbol{\mathcal{N}}}}(A,B) & \mathcal{T}\_{\scriptscriptstyle{\boldsymbol{\mathcal{T}}}}^{\scriptscriptstyle{\boldsymbol{\mathcal{N}}}}(A,B)
\end{pmatrix},
$$

*and a partial order relation "" on* M2×4(R) *as:*

$$
\begin{pmatrix} a\_1 & a\_2 & a\_3 & a\_4 \\ a\_5 & a\_6 & a\_7 & a\_8 \end{pmatrix} \preceq \begin{pmatrix} b\_1 & b\_2 & b\_3 & b\_4 \\ b\_5 & b\_6 & b\_7 & b\_8 \end{pmatrix}, \text{ if } a\_i \le b\_i \,\,\forall i.
$$

*Also define*

$$
\tilde{0} = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \text{ and } \tilde{1} = \begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \end{pmatrix}.
$$

**Remark 5.** *Then for A*, *B* ∈*QSVBNS*(*X*)

*(1)* 0˜ L(*A*, *<sup>B</sup>*) 1˜*, (2)* L(*A*, *B*) = L(*B*, *A*)*, (3) for A* ⊂ *B* ⊂ *C*, L(*A*, *C*) L(*A*, *B*) ∧ L(*B*, *C*)*.*

*We give an outline of the proof as:*

*Let A* ⊂ *B* ⊂ *C. Then for all x<sup>k</sup>* ∈ *X*, *T N A* (*x<sup>k</sup>* ) ≥ *T N B* (*x<sup>k</sup>* ) ≥ *T N C* (*x<sup>k</sup>* )*. Consequently, min T N A* (*x<sup>k</sup>* ), *T N B* (*x<sup>k</sup>* ) = *T N B* (*x<sup>k</sup>* )*, max T N A* (*x<sup>k</sup>* ), *T N B* (*x<sup>k</sup>* ) = *T N A* (*x<sup>k</sup>* )*, min T N A* (*x<sup>k</sup>* ), *T N C* (*x<sup>k</sup>* ) = *T N C* (*x<sup>k</sup>* )*, max T N A* (*x<sup>k</sup>* ), *T N C* (*x<sup>k</sup>* ) = *T N A* (*x<sup>k</sup>* )*.*

$$\text{Then we have, } \sum\_{k=1}^{n} T\_{\text{B}}^{N}(\mathbf{x}\_{k}) \ge \sum\_{k=1}^{n} T\_{\text{C}}^{N}(\mathbf{x}\_{k}) \Rightarrow \frac{1}{\sum\_{k=1}^{n} T\_{\text{B}}^{N}(\mathbf{x}\_{k})} \le \frac{1}{\sum\_{k=1}^{n} T\_{\text{C}}^{N}(\mathbf{x}\_{k})} \Rightarrow \frac{\sum\_{k=1}^{n} T\_{\text{A}}^{N}(\mathbf{x}\_{k})}{\sum\_{k=1}^{n} T\_{\text{B}}^{N}(\mathbf{x}\_{k})} \ge \frac{\sum\_{k=1}^{n} T\_{\text{A}}^{N}(\mathbf{x}\_{k})}{\sum\_{k=1}^{n} T\_{\text{C}}^{N}(\mathbf{x}\_{k})}$$

⇒ T *<sup>N</sup> T* (*A*, *<sup>B</sup>*) ≥ T *<sup>N</sup> T* (*A*, *C*)*. A similar process follows for* T *N T* (*B*, *<sup>C</sup>*) ≥ T *<sup>N</sup> T* (*A*, *C*)*. Hence* T *N T* (*A*, *<sup>C</sup>*) ≤ T *<sup>N</sup> T* (*A*, *<sup>B</sup>*) ∧ T *<sup>N</sup> T* (*B*, *C*)*. Hence* L *is a generalized similarity measure.*

**Definition 17.** *Let <sup>d</sup>* :*QSVBNS*(*X*)×*QSVBNS*(*X*) <sup>→</sup> <sup>R</sup><sup>+</sup> ∪ {0} *be a mapping satisfying the following conditions:*

*(d1) d*(*A*, *B*) ≥ 0 *and d*(*A*, *B*) = 0 *iff A* = *B, (d2) d*(*A*, *B*) = *d*(*B*, *A*)*, (d3) d*(*A*, *C*) ≤ *d*(*A*, *B*) + *d*(*B*, *C*)*.*

*Then d is said to be a distance based measure between two QSVBNS A and B.*

We now define the Hamming distance, normalized Hamming distance, Euclidean distance, and normalized Euclidean distance between two QSVBNSs *A*, *B* ∈ QSVBNS(*X*),

(1) The Hamming distance:

$$\begin{split} \bullet \quad d\_{H}(A,B) &= \sum\_{k=1}^{n} \left( |\boldsymbol{T}\_{A}^{p}(\mathbf{x}\_{k}) - \boldsymbol{T}\_{\texttt{g}}^{p}(\mathbf{x}\_{k})| + |\boldsymbol{\mathcal{C}}\_{A}^{p}(\mathbf{x}\_{k}) - \boldsymbol{\mathcal{C}}\_{\texttt{g}}^{p}(\mathbf{x}\_{k})| + |\boldsymbol{\mathcal{U}}\_{A}^{p}(\mathbf{x}\_{k}) - \boldsymbol{\mathcal{U}}\_{\texttt{g}}^{p}(\mathbf{x}\_{k})| + |\boldsymbol{\mathcal{F}}\_{A}^{p}(\mathbf{x}\_{k}) - \boldsymbol{\mathcal{U}}\_{\texttt{g}}^{p}(\mathbf{x}\_{k})| \right) \\ \boldsymbol{\mathcal{F}}\_{\texttt{g}}^{p}(\mathbf{x}\_{k})| + |\boldsymbol{T}\_{A}^{N}(\mathbf{x}\_{k}) - \boldsymbol{T}\_{\texttt{g}}^{N}(\mathbf{x}\_{k})| + |\boldsymbol{\mathcal{C}}\_{A}^{N}(\mathbf{x}\_{k}) - \boldsymbol{\mathcal{C}}\_{\texttt{g}}^{N}(\mathbf{x}\_{k})| + |\boldsymbol{\mathcal{U}}\_{A}^{N}(\mathbf{x}\_{k}) - \boldsymbol{\mathcal{U}}\_{\texttt{g}}^{N}(\mathbf{x}\_{k})| + |\boldsymbol{\mathcal{F}}\_{A}^{N}(\mathbf{x}\_{k}) - \boldsymbol{\mathcal{F}}\_{\texttt{g}}^{N}(\mathbf{x}\_{k})| \\ \boldsymbol{\mathcal{D}} \text{s} &= \mathbf{v} \times \mathbf{v} \times \mathbf{v} \end{split}$$


• *<sup>d</sup>E*(*A*, *<sup>B</sup>*) = *n* ∑ *k*=1 |*T P A* (*x<sup>k</sup>* ) − *T P B* (*x<sup>k</sup>* )| <sup>2</sup> <sup>+</sup> <sup>|</sup>*<sup>C</sup> P A* (*x<sup>k</sup>* ) − *C P B* (*x<sup>k</sup>* )| <sup>2</sup> <sup>+</sup> <sup>|</sup>*<sup>U</sup> P A* (*x<sup>k</sup>* ) − *U P B* (*x<sup>k</sup>* )| <sup>2</sup> <sup>+</sup> <sup>|</sup>*<sup>F</sup> P A* (*x<sup>k</sup>* ) − *F P B* (*x<sup>k</sup>* )| <sup>2</sup> <sup>+</sup> <sup>|</sup>*<sup>T</sup> N A* (*x<sup>k</sup>* ) − *T N B* (*x<sup>k</sup>* )| <sup>2</sup> <sup>+</sup> <sup>|</sup>*<sup>C</sup> N A* (*x<sup>k</sup>* ) − *C N B* (*x<sup>k</sup>* )| <sup>2</sup> <sup>+</sup> <sup>|</sup>*<sup>U</sup> N A* (*x<sup>k</sup>* ) − *U N B* (*x<sup>k</sup>* )| <sup>2</sup> <sup>+</sup> <sup>|</sup>*<sup>F</sup> N A* (*x<sup>k</sup>* ) − *F N B* (*x<sup>k</sup>* )| 2 !1 2 ,

(4) Normalized Euclidean distance:

.

• *<sup>d</sup>NE*(*A*, *<sup>B</sup>*) = *<sup>d</sup>E*(*A*,*B*) 2*n* √ 2
