Some new operations on single-valued neutrosophic matrices and their applications in multi-criteria group decision making

Abstract

The single-valued neutrosophic set plays a crucial role to handle indeterminant and inconsistent information during decision making process. In recent research, a development in neutrosophic theory is emerged, called single-valued neutrosophic matrices, are used to address uncertainties. The beauty of single-valued neutrosophic matrices is that the utilizing of several fruitful operations in decision making. In this paper, some novel operations on neutrosophic matrices of are introduced, that is, type-1 product $(\tilde \odot )$, type-2 product $(\tilde \otimes )$ and minus $(\tilde \ominus )$ between two single-valued neutrosophic matrices. Also, we introduced complement, transpose, upper and lower α −level matrices of single-valued neutrosophic matrices and discussed related properties. Furthermore, we propose a multi-criteria group decision making method based on these new operations, and give an application of the proposed method in a real life problem. Finally, we compare proposed method in this paper with proposed methods previously.

Appendix A

1.1 A.1

• 1. Let $\alpha =\mu \tilde \oplus \nu$ and $\beta =\mu \tilde \odot \nu$. Then $\alpha _{pq}^{t}=\mu _{pq}^{t}+\nu _{pq}^{t}-\mu _{pq}^{t}\nu _{pq}^{t}$ and $\beta _{pq}^{t}=\mu _{pq}^{t}\nu _{pq}^{t}$. The proof will be made for membership, indeterminacy and non-membership degrees of elements of SVN-matrices.

  • (a) Suppose that $\mu _{pq}^{t}+\nu _{pq}^{t}-\mu _{pq}^{t}\nu _{pq}^{t}\geq \mu _{pq}^{t}\nu _{pq}^{t}$. Since $0\leq \mu _{pq}^{t}\leq 1$ and $0\leq \nu _{pq}^{t}\leq 1$, $\mu _{pq}^{t}(1-\nu _{pq}^{t})+\nu _{pq}^{t}(1-\mu _{pq}^{t})\geq 0$. Hence $\alpha _{pq}^{t}\geq \beta _{pq}^{t}$.

  • (b) Suppose that $\mu _{pq}^{i}\nu _{pq}^{i}\leq \mu _{pq}^{i}+\nu _{pq}^{i}-\mu _{pq}^{i}\nu _{pq}^{i}$. Since $0\leq \mu _{pq}^{i}\leq 1$ and $0\leq \nu _{pq}^{i}\leq 1$, $0\leq \mu _{pq}^{i}(1-\nu _{pq}^{i})+\nu _{pq}^{i}(1-\mu _{pq}^{i})$. Hence $\alpha _{pq}^{i}\leq \beta _{pq}^{i}$.

  • (c) It can be shown that $\alpha _{pq}^{f}\leq \beta _{pq}^{f}$ with similar way to proof of $\alpha _{pq}^{i}\leq \beta _{pq}^{i}$.

• 2. Let $\alpha =\mu \tilde \oplus \mu$. Then $\alpha =(\mu _{pq}^{t}+\nu _{pq}^{t}-\mu _{pq}^{t}\nu _{pq}^{t}, \mu _{pq}^{i}\nu _{pq}^{i},\mu _{pq}^{f}\nu _{pq}^{f})$. Since each of $\mu _{pq}^{t},\mu _{pq}^{i},\mu _{pq}^{f}$, $\nu _{pq}^{t}, \nu _{pq}^{t}$ and $\nu _{pq}^{f}$ are in interval [0,1], it is clear that $\mu _{pq}^{t}+\nu _{pq}^{t}-\mu _{pq}^{t}\nu _{pq}^{t}\geq \mu _{pq}^{t}$, $\mu _{pq}^{i}\nu _{pq}^{i}\leq \mu _{pq}^{i}$ and $\mu _{pq}^{f}\nu _{pq}^{f}\leq \mu _{pq}^{f}$. Therefore $\mu \tilde \oplus \mu \succeq \mu$.

• 3. Let $\alpha =\mu \tilde \odot \mu$. Then $\alpha =(\mu _{pq}^{t}\mu _{pq}^{t}, \mu _{pq}^{i}+\mu _{pq}^{i}-\mu _{pq}^{i}\mu _{pq}^{i},\mu _{pq}^{f}+\mu _{pq}^{f}-\mu _{pq}^{f}\mu _{pq}^{f})$. Since $\mu _{pq}^{t}\nu _{pq}^{t}\leq \mu _{pq}^{t}$, $\mu _{pq}^{i}+\mu _{pq}^{i}-\mu _{pq}^{i}\mu _{pq}^{i}\geq \mu _{pq}^{i}$ and $\mu _{pq}^{f}+\mu _{pq}^{f}-\mu _{pq}^{f}\mu _{pq}^{f}\geq \mu _{pq}^{f}$, $\mu \tilde \odot \mu \preceq \mu$.

• 4. Let $\alpha _{pq}=\langle \mu _{pq}^{t}+\nu _{pq}^{t}-\mu _{pq}^{t}\nu _{pq}^{t},\mu _{pq}^{i}\nu _{pq}^{i},\mu _{pq}^{f}\nu _{pq}^{f}\rangle$ be an element of SVN-matrix $\mu \tilde \oplus \nu$. From definition of $\mu \tilde \ominus \nu$, we know that for truth-membership values of elements of SVN-matrix $\mu \tilde \ominus \nu$, if $\mu _{pq}^{t}\geq \nu _{pq}^{t}$, $\mu _{pq}^{t}\ominus \nu _{pq}^{t}=\mu _{pq}^{t}$ and if $\mu _{pq}^{t}< \nu _{pq}^{t}$, $\mu _{pq}^{t}\ominus \nu _{pq}^{t}= 0$. Since $\mu _{pq}^{t}+\nu _{pq}^{t}-\mu _{pq}^{t}\nu _{pq}^{t}\geq \mu _{pq}^{t}$ and $\mu _{pq}^{t}+\nu _{pq}^{t}-\mu _{pq}^{t}\nu _{pq}^{t}\geq 0$, truth-membership values of elements of SVN-matrix $\mu \tilde \oplus \nu$ are always greater than or equal to truth-membership values of elements of SVN-matrix $\mu \tilde \ominus \nu$. From definition of $\mu \tilde \ominus \nu$, if $\mu _{pq}^{i}\geq \nu _{pq}^{i}$, $\mu _{pq}^{i}\ominus \nu _{pq}^{i}= 1$, and if $\mu _{pq}^{i}< \nu _{pq}^{i}$, $\mu _{pq}^{i}\ominus \nu _{pq}^{i}=\mu _{pq}$. Since $1\geq \mu _{pq}^{i}\nu _{pq}^{i}$ and $\mu _{pq}^{i}\geq \mu _{pq}^{i}\nu _{pq}^{i}$, indeterminacy-membership values of elements of SVN-matrix $\mu \tilde \oplus \nu$ are always smaller than or equal to indeterminacy-membership values of elements of SVN-matrix $\mu \tilde \ominus \nu$. Also, it can be shown that falsity-membership values of elements of SVN-matrix $\mu \tilde \oplus \nu$ are always smaller than or equal to falsity-membership values of elements of SVN-matrix $\mu \tilde \ominus \nu$. It is concluded that $\mu \tilde \oplus \nu \tilde \succeq \mu \tilde \ominus \nu$.

• 5. Let α_pq and β_pq be the pq th elements of the SVN-matrices $\mu \tilde \oplus (\nu \tilde \ominus \gamma )$ and $(\mu \tilde \oplus \nu )\tilde \ominus (\mu \tilde \oplus \gamma )$, respectively. The proof will be made for three cases.

  ᅟ:

  Case 1: For truth-membership values of the SVN-matrices. From Definition 6,

  $$\alpha_{pq}^{t}=\left\{ \begin{array}{ll} \mu_{pq}^{t}+\nu_{pq}^{t}-\mu_{pq}^{t}\nu_{pq}^{t}, & \nu_{pq}^{t}>\gamma_{pq}^{t}, \\ \mu_{pq}^{t}, & \nu_{pq}^{t}\leq \gamma_{pq}^{t}. \end{array} \right.$$

  Since $\mu _{pq}^{t}+\nu _{pq}^{t}-\mu _{pq}^{t}\nu _{pq}^{t}> \mu _{pq}^{t}+\gamma _{pq}^{t}-\mu _{pq}^{t}\gamma _{pq}^{t}$ for $\nu _{pq}^{t}> \gamma _{pq}^{t}$, and $\mu _{pq}^{t}+\nu _{pq}^{t}-\mu _{pq}^{t}\nu _{pq}^{t}\leq \mu _{pq}^{t}+\gamma _{pq}^{t}-\mu _{pq}^{t}\gamma _{pq}^{t}$ for $\nu _{pq}^{t}\leq \gamma _{pq}^{t}$,

  $$\beta_{pq}^{t}=\left\{ \begin{array}{ll} \mu_{pq}^{t}+\nu_{pq}^{t}-\mu_{pq}^{t}\nu_{pq}^{t}, & \nu_{pq}^{t}> \gamma_{pq}^{t}, \\ 0, & \nu_{pq}^{t}\leq \gamma_{pq}^{t}. \end{array} \right.$$

  Hence $\alpha _{pq}^{t}\geq \beta _{pq}^{t}$.

  ᅟ:

  Case 2: For indeterminacy-membership values of the SVN-matrices. From Definition 6,

  $$\alpha_{pq}^{i}=\left\{ \begin{array}{ll} \mu_{pq}^{i}, & \nu_{pq}^{i}\geq \gamma_{pq}^{i}, \\ \mu_{pq}^{i}\nu_{pq}^{i}, & \nu_{pq}^{i}< \gamma_{pq}^{i}. \end{array} \right.$$

  Since $\mu _{pq}^{i}\nu _{pq}^{i}\geq \mu _{pq}^{i}\gamma _{pq}^{i}$ for $\nu _{pq}^{i}\geq \gamma _{pq}^{i}$, and $\mu _{pq}^{i}\nu _{pq}^{i}<\mu _{pq}^{i}\gamma _{pq}^{i}$ for $\nu _{pq}^{i}< \gamma _{pq}^{i}$,

  $$\beta_{pq}^{i}=\left\{ \begin{array}{ll} 1, & \nu_{pq}^{i}\geq \gamma_{pq}^{i}, \\ \mu_{pq}^{i}\nu_{pq}^{i}, & \nu_{pq}^{i}< \gamma_{pq}^{i}. \end{array} \right.$$

  Thus $\alpha _{pq}^{i}\leq \beta _{pq}^{i}$

  ᅟ:

  Case 3: For falsity-membership values of the SVN-matrices. From Definition 6,

  $$\alpha_{pq}^{f}=\left\{ \begin{array}{ll} \mu_{pq}^{f}, & \nu_{pq}^{f}\geq \gamma_{pq}^{f}, \\ \mu_{pq}^{f}\nu_{pq}^{f}, & \nu_{pq}^{f}< \gamma_{pq}^{f}. \end{array} \right.$$

  Since $\mu _{pq}^{f}\nu _{pq}^{f}\geq \mu _{pq}^{f}\gamma _{pq}^{f}$ for $\nu _{pq}^{f}\geq \gamma _{pq}^{f}$, and $\mu _{pq}^{f}\nu _{pq}^{f}<\mu _{pq}^{f}\gamma _{pq}^{f}$ for $\nu _{pq}^{f}< \gamma _{pq}^{f}$,

  $$\beta_{pq}^{f}=\left\{ \begin{array}{ll} 1, & \nu_{pq}^{f}\geq \gamma_{pq}^{f}, \\ \mu_{pq}^{f}\nu_{pq}^{f}, & \nu_{pq}^{f}< \gamma_{pq}^{f}. \end{array} \right.$$

  Thus $\alpha _{pq}^{f}\leq \beta _{pq}^{f}$ and α_pq ≽ β_pq.

  From Case 1,2 and 3, it is concluded that $\mu \tilde \oplus (\nu \tilde \ominus \gamma )\tilde \succeq (\mu \tilde \oplus \nu )\tilde \ominus (\mu \tilde \oplus \gamma )$.

• 6. The proof can be made with similar way to proof of statement 1.

• 7. Let α_pq, β_pq, 𝜃_pq, σ_pq, δ_pq, τ_pq, ρ_pq and 𝜖_pq be pq th elements of $\mu \tilde \oplus \gamma , \nu \tilde \oplus \gamma , \mu \tilde \odot \gamma , \nu \tilde \odot \gamma , \mu \tilde \otimes \gamma , \nu \tilde \otimes \gamma$ and $\mu \tilde \ominus \gamma , \nu \tilde \ominus \gamma$, respectively. Since $\mu \tilde \preceq \nu$, $\mu _{pq}^{t}\leq \nu _{pq}^{t}, \mu _{pq}^{i}\geq \nu _{pq}^{i}$ and $\mu _{pq}^{f}\geq \nu _{pq}^{f}$. It is clear that $\alpha _{pq}^{t}=\mu _{pq}^{t}+\gamma _{pq}^{t}-\mu _{pq}^{t}\gamma _{pq}^{t}\leq \nu _{pq}^{t}+\gamma _{pq}^{t}-\nu _{pq}^{t}\gamma _{pq}^{t}=\beta _{pq}^{t}$, $\alpha _{pq}^{i}=\mu _{pq}^{i}\gamma _{pq}^{i}\geq \nu _{pq}^{i}\gamma _{pq}^{i}=\beta _{pq}^{t}$ and $\alpha _{pq}^{f}=\mu _{pq}^{f}\gamma _{pq}^{f}\geq \nu _{pq}^{f}\gamma _{pq}^{f}=\beta _{pq}^{f}$. Therefore α_pq ≼ β_pq and $\mu \tilde \oplus \gamma \tilde \preceq \nu \tilde \oplus \gamma$.

  By given condition $\theta _{pq}^{t}=\mu _{pq}^{t}\gamma _{pq}^{t}\leq \nu _{pq}^{t}\gamma _{pq}^{t}=\sigma _{pq}^{t}$, $\theta _{pq}^{i}=\mu _{pq}^{i}+\gamma _{pq}^{i}-\mu _{pq}^{i}\gamma _{pq}^{i}\geq \nu _{pq}^{i}+\gamma _{pq}^{i}-\nu _{pq}^{i}\gamma _{pq}^{i}=\sigma _{pq}^{t}$ and $\theta _{pq}^{f}=\mu _{pq}^{f}+\gamma _{pq}^{f}-\mu _{pq}^{f}\gamma _{pq}^{f}\geq \nu _{pq}^{f}+\gamma _{pq}^{f}-\nu _{pq}^{f}\gamma _{pq}^{f}=\sigma _{pq}^{f}$, 𝜃_pq ≼ σ_pq. Therefore $\mu \tilde \odot \gamma \tilde \preceq \nu \tilde \odot \gamma$.

  By given condition $\delta _{pq}^{t}= 1-{\prod }_{k = 1}^{n}(1-(\mu _{pk}^{t} \gamma _{kq}^{t})) \leq 1-{\prod }_{k = 1}^{n}(1-(\nu _{pk}^{t} \gamma _{kq}^{t}))=\tau _{pq}^{t}$, $\delta _{pq}^{i}={\prod }_{k = 1}^{n}(\mu _{pk}^{i} \gamma _{kq}^{i})\geq {\prod }_{k = 1}^{n}(\nu _{pk}^{i} \nu _{kq}^{i})=\tau _{pq}^{i}$ and $\delta _{pq}^{f}={\prod }_{k = 1}^{n}(\mu _{pk}^{f} \gamma _{kq}^{f})\geq {\prod }_{k = 1}^{n}(\nu _{pk}^{f} \nu _{kq}^{f})=\tau _{pq}^{f}$, δ_pq ≼ τ_pq. Therefore $\mu \tilde \otimes \gamma \tilde \preceq \nu \tilde \otimes \gamma$.

  Since

  $$\rho_{pq}^{t}=\left\{ \begin{array}{ll} \mu_{pq}^{t}, & \mu_{pq}^{t}\geq \gamma_{pq}^{t} \\ 0, & \mu_{pq}^{t}< \gamma_{pq}^{t} \end{array} \right. ,\quad \epsilon_{pq}^{t}=\left\{ \begin{array}{ll} \nu_{pq}^{t}, & \nu_{pq}^{t}\geq \gamma_{pq}^{t} \\ 0, & \nu_{pq}^{t}< \gamma_{pq}^{t} \end{array} \right. $$

  and $\mu _{pq}^{t}\leq \nu _{pq}^{t}$ from hypothesis, $\rho _{pq}^{t}\leq \epsilon _{pq}^{t}$. Also since

  $$\rho_{pq}^{i}=\left\{ \begin{array}{ll} 1, & \mu_{pq}^{i}\geq \gamma_{pq}^{i} \\ \mu_{pq}^{i}, & \mu_{pq}^{i}< \gamma_{pq}^{i} \end{array} \right. ,\quad \epsilon_{pq}^{i}=\left\{ \begin{array}{ll} 1, & \nu_{pq}^{i}\geq \gamma_{pq}^{i} \\ \nu_{pq}^{i}, & \nu_{pq}^{i}< \gamma_{pq}^{i} \end{array} \right. $$

  and $\mu _{pq}^{i}\geq \nu _{pq}^{i}$ from hypothesis, $\rho _{pq}^{i}\geq \epsilon _{pq}^{i}$.

  Similarly, $\rho _{pq}^{f}=\left \{ \begin {array}{ll} 1, & \mu _{pq}^{f}\geq \gamma _{pq}^{f} \\ \mu _{pq}^{f}, & \mu _{pq}^{f}< \gamma _{pq}^{f} \end {array} \right . ,\quad \epsilon _{pq}^{f}=\left \{ \begin {array}{ll} 1, & \nu _{pq}^{f}\geq \gamma _{pq}^{f} \\ \nu _{pq}^{f}, & \nu _{pq}^{f}< \gamma _{pq}^{f} \end {array} \right .$ and $\mu _{pq}^{f}\geq \nu _{pq}^{f}$ from hypothesis, $\rho _{pq}^{f}\geq \epsilon _{pq}^{f}$. Therefore ρ_pq ≼ 𝜖_pq and so $\mu \tilde \ominus \gamma \tilde \preceq \nu \tilde \ominus \gamma$.

1.2 A.2

• 1. Suppose that $\alpha _{pq}=\langle \alpha _{pq}^{t},\alpha _{pq}^{i},\alpha _{pq}^{f}\rangle$ and $\beta _{pq}=\langle \beta _{pq}^{t},\beta _{pq}^{i},\beta _{pq}^{f}\rangle$ are ij th elements of $\mu \tilde \oplus \nu$ and $\mu ^{\prime } \tilde \oplus \nu ^{\prime }$, respectively. Then σ_pq = α_qp is the pq th element of $(\mu \tilde \oplus \nu )'$ and

  $$\begin{array}{@{}rcl@{}} \alpha_{pq}&=&\langle\alpha_{pq}^{t},\alpha_{pq}^{i},\alpha_{pq}^{f}\rangle\\&=& \langle\mu_{pq}^{t}+\nu_{pq}^{t}-\mu_{pq}^{t}\nu_{pq}^{t},\mu_{pq}^{i}\nu_{pq}^{i},\mu_{pq}^{f}\nu_{pq}^{f}\rangle\\ \beta_{pq}&=&\langle\beta_{pq}^{t},\beta_{pq}^{i},\beta_{pq}^{f}\rangle\\&=& \langle\mu_{qp}^{t}+\nu_{qp}^{t}-\mu_{qp}^{t}\nu_{qp}^{t},\mu_{qp}^{i}\nu_{qp}^{i},\mu_{qp}^{f}\nu_{qp}^{f}\rangle\\ \sigma_{pq}&=&\langle\alpha_{qp}^{t},\alpha_{qp}^{i},\alpha_{qp}^{f}\rangle\\&=& \langle\mu_{qp}^{t}+\nu_{qp}^{t}\!-\mu_{qp}^{t}\nu_{qp}^{t},\mu_{qp}^{i}\nu_{qp}^{i},\mu_{qp}^{f}\nu_{qp}^{f}\rangle\\&=&\beta_{pq}. \end{array} $$

  Thus, σ_pq = β_pq for all p, q. Hence $(\mu \tilde \oplus \nu )'=\mu ^{\prime } \tilde \oplus \nu ^{\prime }$.

• 2. Suppose that $\alpha _{pq}=\langle \alpha _{pq}^{t},\alpha _{pq}^{i},\alpha _{pq}^{f}\rangle$ and $\beta _{pq}=\langle \beta _{pq}^{t}, \beta _{pq}^{i},\beta _{pq}^{f}\rangle$ are pq th elements of $\mu \tilde \odot \nu$ and $\mu ^{\prime } \tilde \odot \nu ^{\prime }$, respectively. Suppose that $\alpha _{pq}=\langle \alpha _{pq}^{t},\alpha _{pq}^{i},\alpha _{pq}^{f}\rangle$ and $\beta _{pq}=\langle \beta _{pq}^{t},\beta _{pq}^{i},\beta _{pq}^{f}\rangle$ are pq th elements of $\mu \tilde \odot \nu$ and $\mu ^{\prime } \tilde \odot \nu ^{\prime }$, respectively. Then σ_pq = α_qp is the pq th element of $(\mu \tilde \odot \nu )'$ and

  $$\begin{array}{@{}rcl@{}} \alpha_{pq}&=&\langle\alpha_{pq}^{t},\alpha_{pq}^{i},\alpha_{pq}^{f}\rangle\\&=& \langle\mu_{pq}^{t}\nu_{pq}^{t},\mu_{pq}^{i}+\nu_{pq}^{i}-\mu_{pq}^{i}\nu_{pq}^{i},\mu_{pq}^{f}+\nu_{pq}^{f}-\mu_{pq}^{f}\nu_{pq}^{f}\rangle\\ \beta_{pq}&=&\langle\beta_{pq}^{t},\beta_{pq}^{i},\beta_{pq}^{f}\rangle\\&=& \langle\mu_{qp}^{t}\nu_{qp}^{t},\mu_{qp}^{i}+\nu_{qp}^{i}-\mu_{qp}^{i}\nu_{qp}^{i},\mu_{qp}^{f}+\nu_{qp}^{f}-\mu_{qp}^{f}\nu_{qp}^{f}\rangle\\ \sigma_{pq}&=&\langle\sigma_{pq}^{t},\sigma_{pq}^{i},\sigma_{pq}^{f}\rangle\\&=& \langle\mu_{qp}^{t}\nu_{qp}^{t},\mu_{qp}^{i}+\nu_{qp}^{i}-\mu_{qp}^{i}\nu_{qp}^{i},\mu_{qp}^{f}+\nu_{qp}^{f}-\mu_{qp}^{f}\nu_{qp}^{f}\rangle\\ \end{array} $$

  Thus, σ_pq = β_pq for all p, q. Hence $(\mu \tilde \odot \nu )'=\mu ^{\prime } \tilde \odot \nu ^{\prime }$.

• 3. Suppose that $\alpha _{pq}=\langle \alpha _{pq}^{t},\alpha _{pq}^{i},\alpha _{pq}^{f}\rangle$ and $\beta _{pq}=\langle \beta _{pq}^{t}, \beta _{pq}^{i},\beta _{pq}^{f}\rangle$ are pq th elements of $\mu \tilde \otimes \nu$ and $\nu ^{\prime } \tilde \otimes \mu ^{\prime }$, respectively. Suppose that $\alpha _{pq}=\langle \alpha _{pq}^{t},\alpha _{pq}^{i},\alpha _{pq}^{f}\rangle$ and $\beta _{pq}=\langle \beta _{pq}^{t},\beta _{pq}^{i},\beta _{pq}^{f}\rangle$ are ij th elements of $\mu \tilde \otimes \nu$ and $\nu ^{\prime } \tilde \otimes \mu ^{\prime }$, respectively. Then σ_pq = α_qp is the pq th element of $(\mu \tilde \otimes \nu )'$ and

  $$\begin{array}{@{}rcl@{}} \alpha_{pq}&=&\langle\alpha_{pq}^{t},\alpha_{pq}^{i},\alpha_{pq}^{f}\rangle= \langle1-\prod\limits_{k = 1}^{n}(1-\mu_{pk}^{t}\nu_{kq}^{t}),\\ &&\prod\limits_{k = 1}^{n}(\mu_{pk}^{i}+\nu_{kq}^{i}-\mu_{pq}^{i}\nu_{pq}^{i}),\prod\limits_{k = 1}^{n}(\mu_{pq}^{f}+\nu_{pq}^{f}-\mu_{pq}^{f}\nu_{pq}^{f})\rangle\\ \beta_{pq}&=&\langle\beta_{pq}^{t},\beta_{pq}^{i},\beta_{pq}^{f}\rangle= \langle1-\prod\limits_{k = 1}^{n}(1-\nu_{qk}^{t}\mu_{kp}^{t}),\\ &&\prod\limits_{k = 1}^{n}(\nu_{qk}^{i}+\mu_{kp}^{i}-\nu_{qk}^{i}\mu_{kp}^{i}),\prod\limits_{k = 1}^{n}(\nu_{qk}^{f}+\mu_{kp}^{f}-\nu_{qk}^{f}\mu_{kp}^{f})\rangle\\ \sigma_{pq}&=&\langle\sigma_{pq}^{t},\sigma_{pq}^{i},\sigma_{pq}^{f}\rangle= \langle1-\prod\limits_{k = 1}^{n}(1-\mu_{qk}^{t}\nu_{kp}^{t}),\\ &&\prod\limits_{k = 1}^{n}(\mu_{qk}^{i}+\nu_{kp}^{i}-\mu_{qk}^{i}\nu_{kp}^{i}),\prod\limits_{k = 1}^{n}(\mu_{qk}^{f}+\nu_{kp}^{f}-\mu_{qk}^{f}\nu_{kp}^{f})\rangle\\ &=&\beta_{pq}. \end{array} $$

  Thus, σ_pq = β_pq for all p, q. Hence $(\mu \tilde \otimes \nu )'=\nu ^{\prime } \tilde \otimes \mu ^{\prime }$.

• 4. Suppose that $\alpha _{pq}=\langle \alpha _{pq}^{t},\alpha _{pq}^{i},\alpha _{pq}^{f}\rangle$ and $\beta _{pq}=\langle \beta _{pq}^{t}, \beta _{pq}^{i},\beta _{pq}^{f}\rangle$ are pq th elements of $\mu \tilde \ominus \nu$ and $\mu ^{\prime } \tilde \ominus \nu ^{\prime }$, respectively. Suppose that $\alpha _{pq}=\langle \alpha _{pq}^{t},\alpha _{pq}^{i},\alpha _{pq}^{f}\rangle$ and $\beta _{pq}=\langle \beta _{pq}^{t},\beta _{pq}^{i},\beta _{pq}^{f}\rangle$ are pq th elements of $\mu \tilde \ominus \nu$ and $\mu ^{\prime } \tilde \ominus \nu ^{\prime }$, respectively. Then σ_pq = α_qp is the pq th element of $(\mu \tilde \ominus \nu )'$ and

  $$\begin{array}{@{}rcl@{}} \alpha_{pq}&=&\langle\alpha_{pq}^{t},\alpha_{pq}^{i},\alpha_{pq}^{f}\rangle\,=\, \langle\mu_{pq}^{t}\ominus\nu_{pq}^{t},\mu_{pq}^{i}\ominus\nu_{pq}^{i},\mu_{pq}^{f}\ominus\nu_{pq}^{f}\rangle,\\ \beta_{pq}&=&\langle\beta_{pq}^{t},\beta_{pq}^{i},\beta_{pq}^{f}\rangle= \langle\mu_{qp}^{t}\ominus\nu_{qp}^{t},\mu_{qp}^{i}\ominus\nu_{qp}^{i},\mu_{qp}^{f}\ominus\nu_{qp}^{f}\rangle,\\ \sigma_{pq}&=&\langle\sigma_{pq}^{t},\sigma_{pq}^{i},\sigma_{pq}^{f}\rangle= \langle\mu_{qp}^{t}\ominus\nu_{qp}^{t},\mu_{qp}^{i}\ominus\nu_{qp}^{i},\mu_{qp}^{f}\ominus\nu_{qp}^{f}\rangle,\\ &=&\beta_{pq}. \end{array} $$

  Thus, σ_pq = β_pq for all p, q. Hence $(\mu \tilde \ominus \nu )'=\mu ^{\prime } \tilde \ominus \nu ^{\prime }$.

1.3 A.3

• 1. Let $\alpha _{pq}=\langle \alpha _{pq}^{t}, \alpha _{pq}^{i},\alpha _{pq}^{f}\rangle$ and $\beta _{pq}=\langle \beta _{pq}^{t}, \beta _{pq}^{i},\beta _{pq}^{f}\rangle$ be pq th elements of $\mu \tilde \oplus \nu$ and $\mu ^{c}\tilde \odot \nu ^{c}$, respectively, and let $\sigma _{pq}=\alpha _{pq}^{c}$. Then

  $$\begin{array}{@{}rcl@{}} \alpha_{pq}&=&\left\langle\alpha_{pq}^{t},\alpha_{pq}^{i},\alpha_{pq}^{f}\right\rangle\,=\,\left\langle\mu_{pq}^{t}\,+\,\nu_{pq}^{t}\,-\,\mu_{pq}^{t}\nu_{pq}^{t},\mu_{pq}^{i}\nu_{pq}^{i},\mu_{pq}^{f}\nu_{pq}^{f}\right\rangle\\ \sigma_{pq}&=&\alpha_{pq}^{c}=\left\langle\alpha_{pq}^{f},1\,-\,\alpha_{pq}^{i},\alpha_{pq}^{t}\right\rangle= \left\langle\mu_{pq}^{f}\nu_{pq}^{f},1\,-\,\mu_{pq}^{i}\nu_{pq}^{i},\mu_{pq}^{t}\right.\\ &&\left.+~\nu_{pq}^{t}-\mu_{pq}^{t}\nu_{pq}^{t}\right\rangle\\ \end{array} $$
  $$\begin{array}{@{}rcl@{}} \beta_{pq}&=&\langle\beta_{pq}^{t}, \beta_{pq}^{i},\beta_{pq}^{f}\rangle\,=\,\left\langle\mu_{pq}^{f},1\,-\,\mu_{pq}^{i}, \mu_{pq}^{t}\right\rangle\odot\left\langle\nu_{pq}^{f},1\,-\,\nu_{pq}^{i}, \nu_{pq}^{t}\right\rangle\\ &=&\left\langle\mu_{pq}^{f}\nu_{pq}^{f},1-\mu_{pq}^{i}\nu_{pq}^{i}, \mu_{pq}^{t}+\nu_{pq}^{t}-\mu_{pq}^{t}\nu_{pq}^{t}\right\rangle. \end{array} $$

  Thus, σ_pq = β_pq. Hence $(\mu \tilde \oplus \nu )^{c}=\mu ^{c}\tilde \odot \nu ^{c}$.

• 2. The proof can be proved by similar way to proof of 1.

• 3. Let $\alpha _{pq}=\langle \alpha _{pq}^{t}, \alpha _{pq}^{i},\alpha _{pq}^{f}\rangle$ and $\beta _{pq}=\langle \beta _{pq}^{t}, \beta _{pq}^{i},\beta _{pq}^{f}\rangle$ be pq th elements of $\mu \tilde \ominus \nu$ and $\mu ^{c}\tilde \ominus \nu ^{c}$, respectively, and let $\sigma _{pq}=\alpha _{pq}^{c}$. Then,

  $$\begin{array}{@{}rcl@{}} \alpha_{pq}&=&\langle\alpha_{pq}^{t}, \alpha_{pq}^{i},\alpha_{pq}^{f}\rangle\\&=&\left\langle \left\{ \begin{array}{ll} \mu_{pq}^{t}, & \mu_{pq}^{t}> \nu_{pq}^{t} \\ 0, & \mu_{pq}^{t}\leq \nu_{pq}^{t} \end{array} \right.,\left\{ \begin{array}{ll} 1, & \mu_{pq}^{i}\geq \nu_{pq}^{i} \\ \mu_{pq}^{i}, & \mu_{pq}^{i}< \nu_{pq}^{i} \end{array} \right.,\left\{ \begin{array}{ll} 1, & \mu_{pq}^{f}\geq \nu_{pq}^{f} \\ \mu_{pq}^{t}, & \mu_{pq}^{f}<\nu_{pq}^{f} \end{array} \right.\right\rangle\\ \sigma_{pq}&=&\alpha_{pq}^{c}=\left\langle \left\{\begin{array}{ll} 1, & \mu_{pq}^{f}\geq \nu_{pq}^{f} \\ \mu_{pq}^{t}, & \mu_{pq}^{f}<\nu_{pq}^{f} \end{array} \right.,\left\{ \begin{array}{ll} 0, & \mu_{pq}^{i}\geq \nu_{pq}^{i} \\ 1-\mu_{pq}^{i}, & \mu_{pq}^{i}< \nu_{pq}^{i} \end{array} \right.,\left\{\begin{array}{ll} \mu_{pq}^{t}, & \mu_{pq}^{t}>\nu_{pq}^{t} \\ 0, & \mu_{pq}^{t}\leq \nu_{pq}^{t} \end{array} \right.\right\rangle\\ \beta_{pq}&=&\left\langle \mu_{pq}^{f},1-\mu_{pq}^{i}, \mu_{pq}^{t}\right\rangle \tilde\ominus \left\langle \nu_{pq}^{f},1-\nu_{pq}^{i}, \nu_{pq}^{t}\right\rangle\\ &=&\left\langle \left\{\begin{array}{ll} \mu_{pq}^{f}, & \mu_{pq}^{f}\geq \nu_{pq}^{f} \\ 0, & \mu_{pq}^{f}<\nu_{pq}^{f} \end{array} \right.,\left\{ \begin{array}{ll} 1-\mu_{pq}^{i}, & \mu_{pq}^{i}> \nu_{pq}^{i} \\ 1 & \mu_{pq}^{i}\leq \nu_{pq}^{i} \end{array} \right.,\left\{\begin{array}{ll} 1, & \mu_{pq}^{t}\geq \nu_{pq}^{t} \\ \mu_{pq}^{t}, & \mu_{pq}^{t}< \nu_{pq}^{t} \end{array} \right.\right\rangle. \end{array} $$

  For $\mu _{pq}^{f}\geq \nu _{pq}^{f}$, since $\mu _{pq}^{f}\leq 1$, $1-\mu _{pq}^{i}\geq 0$, $1\geq \mu _{pq}^{t}$ and $0\leq \mu _{pq}^{f}$, $1\geq 1-\mu _{pq}^{i}$, $\mu _{pq}^{i}\geq 0$, σ_pq ≽ β_pq. Thus $(\mu \tilde \ominus \nu )^{c}\tilde \succeq \mu ^{c} \tilde \ominus \nu ^{c}$.

1.4 A.4

• 1. Let α ≼ β. Then α^t ≤ β^t, αⁱ ≥ βⁱ and α^f ≥ β^f. The proof will be made each of $\mu _{ij}^{t},\mu _{pq}^{i}$ and $\mu _{pq}^{f}$. Let $\mu _{pq}^{t}\geq \beta ^{t}$. Then $(\mu _{pq}^{t})^{\beta ^{t}}= 1$. Since β^t ≥ α^t, $\mu _{pq}^{t}\geq \alpha ^{t}$, $(\mu _{pq}^{t})^{\alpha ^{t}}= 1$. Therefore $(\mu _{pq}^{t})^{\alpha ^{t}}=(\mu _{pq}^{t})^{\beta ^{t}}$. If $\mu _{pq}^{t}< \beta ^{t}$, there are two cases.

  ᅟ:

  Case 1: If $\mu _{pq}^{t}<\alpha ^{t}\leq \beta ^{t}$, then $(\mu _{pq}^{t})^{\beta ^{t}}= 0=(\mu _{pq}^{t})^{\alpha ^{t}}$.

  ᅟ:

  Case 2: If $\alpha ^{t}\leq \mu _{pq}^{t}< \beta ^{t}$, then $(\mu _{pq}^{t})^{\beta ^{t}}= 0<1=(\mu _{pq}^{t})^{\alpha ^{t}}$. Thus, $(\mu _{pq}^{t})^{\beta ^{t}}\leq (\mu _{pq}^{t})^{\alpha ^{t}}$.

  Let $\mu _{pq}^{i}\geq \beta ^{i}$. Then, there are two cases;

  ᅟ:

  Case 1: If $\mu _{pq}^{i}\geq \alpha ^{i}\geq \beta ^{i}$, then $(\mu _{pq}^{i})^{\alpha ^{i}}=(\mu _{pq}^{i})^{\beta ^{i}}= 1$.

  ᅟ:

  Case 2: If $\alpha ^{i}\geq \mu _{pq}^{i}\geq \beta ^{i}$, then $(\mu _{pq}^{i})^{\alpha ^{i}}= 0<(\mu _{pq}^{i})^{\beta ^{i}}= 1$. Also if $\mu _{pq}^{i}< \beta ^{i}$, then $(\mu _{pq}^{i})^{\beta ^{i}}= 0$. Since βⁱ ≤ αⁱ, $\mu _{pq}^{i}\leq \alpha ^{i}$, $(\mu _{pq}^{i})^{\alpha ^{i}}= 0$. Thus, $(\mu _{pq}^{i})^{\beta ^{i}}\geq (\mu _{pq}^{i})^{\alpha ^{i}}$.

  Let $\mu _{pq}^{f}\geq \beta ^{f}$. Then, there are two cases;

  ᅟ:

  Case 1: If $\mu _{pq}^{f}\geq \alpha ^{f}\geq \beta ^{f}$, then $(\mu _{pq}^{f})^{\alpha ^{f}}=(\mu _{pq}^{f})^{\beta ^{f}}= 1$.

  ᅟ:

  Case 2: If $\alpha ^{f}\geq \mu _{pq}^{f}\geq \beta ^{f}$, then $(\mu _{pq}^{f})^{\alpha ^{f}}= 0<(\mu _{pq}^{f})^{\beta ^{f}}= 1$. Also if $\mu _{pq}^{f}< \beta ^{f}$, then $(\mu _{pq}^{f})^{\beta ^{f}}= 0$. Since β^f ≤ α^f, $\mu _{pq}^{f}\leq \alpha ^{f}$, $(\mu _{pq}^{f})^{\alpha ^{f}}= 0$. Thus, $(\mu _{pq}^{f})^{\beta ^{f}}\geq (\mu _{pq}^{f})^{\alpha ^{f}}$.

  It is concluded that $\mu ^{(\beta )}\tilde \preceq \mu ^{(\alpha )}$.

• 2. The proof can be made by similar way to proof of statement 1.

1.5 A.5

• 1. The proof will be made for truth, indeterminacy and falsity membership values of elements of SVN-matrices:

  ᅟ:

  Case 1: For truth-membership values:

  ᅟ:

  Subcase 1: Let $\mu _{pq}^{t},\nu _{pq}^{t}\geq \alpha ^{t}$. Then, $\mu _{pq}^{t}\oplus \nu _{pq}^{t}=\mu _{pq}^{t}+\nu _{pq}^{t}-\mu _{pq}^{t}\nu _{pq}^{t}\geq \mu _{pq}^{t}$ and $\nu _{pq}^{t}$. Therefore $(\mu _{pq}^{t}\oplus \nu _{pq}^{t})\geq \alpha ^{t}$ and so $(\mu _{pq}^{t}\oplus \nu _{pq}^{t})= 1$. Also, since $(\mu _{pq}^{t})^{\alpha }= 1$ and $(\nu _{pq}^{t})^{\alpha }= 1$, $(\mu _{pq}^{t})^{\alpha ^{t}}\oplus (\nu _{pq}^{t})^{\alpha ^{t}}= 1$. Thus, $(\mu _{pq}^{t}\oplus \nu _{pq}^{t})^{\alpha ^{t}}=(\mu _{pq}^{t})^{\alpha ^{t}}\oplus (\nu _{pq}^{t})^{\alpha ^{t}}$.

  ᅟ:

  Subcase 2: Let $\mu _{pq}^{t},\nu _{pq}^{t}< \alpha ^{t}$. Then, there are two cases: $\mu _{pq}^{t}\oplus \nu _{pq}^{t}<\alpha ^{t}$ or $\mu _{pq}^{t}\oplus \nu _{pq}^{t}\geq \alpha ^{t}$.

  ᅟ:

  1. If $\mu _{pq}^{t}\oplus \nu _{pq}^{t}<\alpha ^{t}$, $(\mu _{pq}^{t}\oplus \nu _{pq}^{t})^{\alpha ^{t}}= 0$. Since $(\mu _{pq}^{t})^{\alpha ^{t}}= 0$ and $(\nu _{pq}^{t})^{\alpha ^{t}}= 0$, $(\mu _{pq}^{t})^{\alpha ^{t}}\oplus (\nu _{pq}^{t})^{\alpha ^{t}}= 0$. Hence $(\mu _{pq}^{t}\oplus \nu _{pq}^{t})^{\alpha ^{t}}=(\mu _{pq}^{t})^{\alpha ^{t}}\oplus (\nu _{pq}^{t})^{\alpha ^{t}}$.

  ᅟ:

  2. If $\mu _{pq}^{t}\oplus \nu _{pq}^{t}\geq \alpha ^{t}$, $(\mu _{pq}^{t}\oplus \nu _{pq}^{t})^{\alpha ^{t}}= 1$. Since $(\mu _{pq}^{t})^{\alpha ^{t}}= 0$ and $(\nu _{pq}^{t})^{\alpha ^{t}}= 0$, $(\mu _{pq}^{t})^{\alpha ^{t}}\oplus (\nu _{pq}^{t})^{\alpha ^{t}}= 0$. Thus $(\mu _{pq}^{t}\oplus \nu _{pq}^{t})^{\alpha ^{t}}\geq (\mu _{pq}^{t})^{\alpha ^{t}}\oplus (\nu _{pq}^{t})^{\alpha ^{t}}$.

  ᅟ:

  Subcase 3: Let $\mu _{pq}^{t}< \alpha ^{t}$ and $\nu _{pq}^{t}\geq \alpha ^{t}$. Then $\mu _{pq}^{t}\oplus \nu _{pq}^{t}=\mu _{pq}^{t}+\nu _{pq}^{t}-\mu _{pq}^{t}\nu _{pq}^{t}\geq \alpha ^{t}$. Therefore $(\mu _{pq}^{t}\oplus \nu _{pq}^{t})^{\alpha ^{t}}= 1$. Also, since $(\mu _{pq}^{t})^{\alpha ^{t}}= 0$ and $(\nu _{pq}^{t})^{\alpha ^{t}}= 1$, $(\mu _{pq}^{t})^{\alpha ^{t}}\oplus (\nu _{pq}^{t})^{\alpha ^{t}}= 1$. Thus, $(\mu _{pq}^{t}\oplus \nu _{pq}^{t})^{\alpha ^{t}}= (\mu _{pq}^{t})^{\alpha ^{t}}\oplus (\nu _{pq}^{t})^{\alpha ^{t}}$.

  ᅟ:

  Subcase 4: Let $\mu _{pq}^{t}\geq \alpha ^{t}$ and $\nu _{pq}^{t}<\alpha ^{t}$. Then $\mu _{pq}^{t}\oplus \nu _{pq}^{t}=\mu _{pq}^{t}+\nu _{pq}^{t}-\mu _{pq}^{t}\nu _{pq}^{t}\geq \alpha ^{t}$. Therefore $(\mu _{pq}^{t}\oplus \nu _{pq}^{t})^{\alpha ^{t}}= 1$. Also, since $(\mu _{pq}^{t})^{\alpha ^{t}}= 1$ and $(\nu _{pq}^{t})^{\alpha ^{t}}= 0$, $(\mu _{pq}^{t})^{\alpha ^{t}}\oplus (\nu _{pq}^{t})^{\alpha ^{t}}= 1$. Thus, $(\mu _{pq}^{t}\oplus \nu _{pq}^{t})^{\alpha ^{t}}= (\mu _{pq}^{t})^{\alpha ^{t}}\oplus (\nu _{pq}^{t})^{\alpha ^{t}}$.

  ᅟ:

  Case 2: For indeterminacy-membership values:

  ᅟ:

  Subcase 1: Let $\mu _{pq}^{i},\nu _{pq}^{i}> \alpha ^{i}$. Then, there are two cases: $\mu _{pq}^{i}\oplus \nu _{pq}^{i}=\mu _{pq}^{i}\nu _{pq}^{i}\leq \alpha ^{i}$ or $\mu _{pq}^{i}\oplus \nu _{pq}^{i}=\mu _{pq}^{i}\nu _{pq}^{i}>\alpha ^{i}$.

  ᅟ:

  1. If $\mu _{pq}^{i}\oplus \nu _{pq}^{i}=\mu _{pq}^{i}\nu _{pq}^{i}\leq \alpha ^{i}$, then $(\mu _{pq}^{i}\oplus \nu _{pq}^{i})^{\alpha ^{i}}= 0$. Since $(\mu _{pq}^{i})^{\alpha ^{i}}= 1$ and $(\nu _{pq}^{i})^{\alpha ^{i}}= 1$, $(\mu _{pq}^{i})^{\alpha ^{i}}\oplus (\nu _{pq}^{i})^{\alpha ^{i}}= 1$. Hence $(\mu _{pq}^{i}\oplus \nu _{pq}^{i})^{\alpha ^{i}}\leq (\mu _{pq}^{i})^{\alpha ^{i}}\oplus (\nu _{pq}^{i})^{\alpha ^{i}}$.

  ᅟ:

  2. If $\mu _{pq}^{i}\oplus \nu _{pq}^{i}=\mu _{pq}^{i}\nu _{pq}^{i}>\alpha ^{i}$, $(\mu _{pq}^{i}\oplus \nu _{pq}^{i})^{\alpha ^{i}}= 1$. Since $(\mu _{pq}^{i})^{\alpha ^{i}}= 0$ and $(\nu _{pq}^{i})^{\alpha ^{i}}= 0$, $(\mu _{pq}^{i})^{\alpha ^{i}}\oplus (\nu _{pq}^{i})^{\alpha ^{i}}= 1$. Thus $(\mu _{pq}^{i}\oplus \nu _{pq}^{i})^{\alpha ^{i}}=(\mu _{pq}^{i})^{\alpha ^{i}}\oplus (\nu _{pq}^{i})^{\alpha ^{i}}$.

  ᅟ:

  Subcase 2: Let $\mu _{pq}^{i},\nu _{pq}^{i}\leq \alpha ^{i}$. Then, $\mu _{pq}^{i}\oplus \nu _{pq}^{i}=\mu _{pq}^{i}\nu _{pq}^{i}\leq \alpha ^{i}$ and so $(\mu _{pq}^{i}\oplus \nu _{pq}^{i})^{\alpha ^{i}}= 0$. Since $(\mu _{pq}^{i})^{\alpha ^{i}}= 0$ and $(\nu _{pq}^{i})^{\alpha ^{i}}= 0$, $(\mu _{pq}^{i})^{\alpha ^{i}}\oplus (\nu _{pq}^{i})^{\alpha ^{i}}= 0$. Thus $(\mu _{pq}^{i}\oplus \nu _{pq}^{i})^{\alpha ^{i}}=(\mu _{pq}^{i})^{\alpha ^{i}}\oplus (\nu _{pq}^{i})^{\alpha ^{i}}$.

  ᅟ:

  Subcase 3: Let $\mu _{pq}^{i}\leq \alpha ^{i}$ and $\nu _{pq}^{i}>\alpha ^{i}$. If $\mu _{pq}^{i}= 0$, Then $\mu _{pq}^{i}\oplus \nu _{pq}^{i}=\mu _{pq}^{i}\nu _{pq}^{i}= 0\leq \alpha ^{i}$. Therefore $(\mu _{pq}^{i}\oplus \nu _{pq}^{i})^{\alpha ^{i}}= 0$. Since $(\mu _{pq}^{i})^{\alpha ^{i}}= 0$ and $(\nu _{pq}^{i})^{\alpha ^{i}}= 1$, $(\mu _{pq}^{i})^{\alpha ^{i}}\oplus (\nu _{pq}^{i})^{\alpha ^{i}}= 0$. Thus, $(\mu _{pq}^{i}\oplus \nu _{pq}^{i})^{\alpha ^{i}}= (\mu _{pq}^{i})^{\alpha ^{i}}\oplus (\nu _{pq}^{i})^{\alpha ^{i}}$. If $\mu _{pq}^{i}\not = 0$, then $\mu _{pq}^{i}\oplus \nu _{pq}^{i}=\mu _{pq}^{i}\nu _{pq}^{i}\leq \alpha ^{i}$ and so $(\mu _{pq}^{i}\oplus \nu _{pq}^{i})^{\alpha ^{i}}= 0$. Since $(\mu _{pq}^{i})^{\alpha ^{i}}= 0$ and $(\nu _{pq}^{i})^{\alpha ^{i}}= 1$, $(\mu _{pq}^{i})^{\alpha ^{i}}\oplus (\nu _{pq}^{i})^{\alpha ^{i}}= 0$. Thus, $(\mu _{pq}^{i}\oplus \nu _{pq}^{i})^{\alpha ^{i}}= (\mu _{pq}^{i})^{\alpha ^{i}}\oplus (\nu _{pq}^{i})^{\alpha ^{i}}$.

  ᅟ:

  Subcase 4: Let $\mu _{pq}^{i}> \alpha ^{i}$ and $\nu _{pq}^{i}\leq \alpha ^{i}$. If $\nu _{pq}^{i}= 0$, Then $\mu _{pq}^{i}\oplus \nu _{pq}^{i}=\mu _{pq}^{i}\nu _{pq}^{i}= 0\leq \alpha ^{i}$. Therefore $(\mu _{pq}^{i}\oplus \nu _{pq}^{i})^{\alpha ^{i}}= 0$. Since $(\mu _{pq}^{i})^{\alpha ^{i}}= 1$ and $(\nu _{pq}^{i})^{\alpha ^{i}}= 0$, $(\mu _{pq}^{i})^{\alpha ^{i}}\oplus (\nu _{pq}^{i})^{\alpha ^{i}}= 0$. Thus, $(\mu _{pq}^{i}\oplus \nu _{pq}^{i})^{\alpha ^{i}}= (\mu _{pq}^{i})^{\alpha ^{i}}\oplus (\nu _{pq}^{i})^{\alpha ^{i}}$. If $\nu _{pq}^{i}\not = 0$, then $\mu _{pq}^{i}\oplus \nu _{pq}^{i}=\mu _{pq}^{i}\nu _{pq}^{i}\leq \alpha ^{i}$ and so $(\mu _{pq}^{i}\oplus \nu _{pq}^{i})^{\alpha ^{i}}= 0$. Since $(\mu _{pq}^{i})^{\alpha ^{i}}= 0$ and $(\nu _{pq}^{i})^{\alpha ^{i}}= 1$, $(\mu _{pq}^{i})^{\alpha ^{i}}\oplus (\nu _{pq}^{i})^{\alpha ^{i}}= 0$. Thus, $(\mu _{pq}^{i}\oplus \nu _{pq}^{i})^{\alpha ^{i}}= (\mu _{pq}^{i})^{\alpha ^{i}}\oplus (\nu _{pq}^{i})^{\alpha ^{i}}$.

  ᅟ:

  Case 3: For indeterminacy-membership values: The proof can be made by similar way to proof of case 2.

  When it is considered all of the cases, it is concluded that $(\mu \tilde \oplus \nu )^{(\alpha )}\tilde \succeq \mu ^{(\alpha )}\tilde \oplus \nu ^{(\alpha )}$.

• 2. The proof can be made by similar way to proof of statement 1.

1.6 A.6

• 1. The proof will be made for truth, indeterminacy and falsity-memberships values of elements of SVN-matrices.

  ᅟ:

  1. For truth-membership values: There are four cases,

  ᅟ:

  Case 1: $\mu _{pq}^{t}, \nu _{pq}^{t}\geq \alpha ^{t}$.

  ᅟ:

  Subcase 1: Let $\mu _{pq}^{t}> \nu _{pq}^{t}$. Then $\mu _{pq}^{t}\ominus \nu _{pq}^{t}=\mu _{pq}^{t}$. Since $\mu _{pq}^{t}\geq \alpha ^{t}$, $(\mu _{pq}^{t}\ominus \nu _{pq}^{t} )^{\alpha ^{t}}= 1$. Also since $(\mu _{pq}^{t})^{\alpha ^{t}}= 1$ and $(\nu _{pq}^{t})^{\alpha ^{t}}= 1$, $(\mu _{pq}^{t})^{\alpha ^{t}}\ominus (\nu _{pq}^{t})^{\alpha ^{t}}= 0$. Therefore $(\mu _{pq}^{t}\ominus \nu _{pq}^{t})^{\alpha ^{t}}\geq (\mu _{pq}^{t})^{\alpha ^{t}}\ominus (\nu _{pq}^{t})^{\alpha ^{t}}$.

  ᅟ:

  Subcase 2: Let $\mu _{pq}^{t}\leq \nu _{pq}^{t}$. Then $\mu _{pq}^{t}\ominus \nu _{pq}^{t}= 0$. Since 0 ≤ α^t, $(\mu _{pq}^{t}\ominus \nu _{pq}^{t})^{\alpha ^{t}}= 0$ or 1. Also since $(\mu _{pq}^{t})^{\alpha ^{t}}= 1$ and $(\nu _{pq}^{t})^{\alpha ^{t}}= 1$, $(\mu _{pq}^{t})^{\alpha ^{t}}\ominus (\nu _{pq}^{t})^{\alpha ^{t}}= 0$. Therefore $(\mu _{pq}^{t}\ominus \nu _{pq}^{t})^{\alpha ^{t}}\geq (\mu _{pq}^{t})^{\alpha ^{t}}\ominus (\nu _{pq}^{t})^{\alpha ^{t}}$.

  ᅟ:

  Case 2: $\mu _{pq}^{t}, \nu _{pq}^{t}<\alpha ^{t}$.

  ᅟ:

  Subcase 1: Let $\mu _{pq}^{t}> \nu _{pq}^{t}$. Then $\mu _{pq}^{t}\ominus \nu _{pq}^{t}=\mu _{pq}^{t}$. Since $\mu _{pq}^{t}\geq \alpha ^{t}$, $(\mu _{pq}^{t}\ominus \nu _{pq}^{t} )^{\alpha ^{t}}= 1$. Also since $(\mu _{pq}^{t})^{\alpha ^{t}}= 0$ and $(\nu _{pq}^{t})^{\alpha ^{t}}= 0$, $(\mu _{pq}^{t})^{\alpha ^{t}}\ominus (\nu _{pq}^{t})^{\alpha ^{t}}= 0$. Therefore $(\mu _{pq}^{t}\ominus \nu _{pq}^{t})^{\alpha ^{t}}\geq (\mu _{pq}^{t})^{\alpha ^{t}}\ominus (\nu _{pq}^{t})^{\alpha ^{t}}$.

  ᅟ:

  Subcase 2: Let $\mu _{pq}^{t}\leq \nu _{pq}^{t}$. Then $\mu _{pq}^{t}\ominus \nu _{pq}^{t}= 0$. Since 0 ≤ α^t, $(\mu _{pq}^{t}\ominus \nu _{pq}^{t})^{\alpha ^{t}}= 0$ or 1. Also since $(\mu _{pq}^{t})^{\alpha ^{t}}= 0$ and $(\nu _{pq}^{t})^{\alpha ^{t}}= 0$, $(\mu _{pq}^{t})^{\alpha ^{t}}\ominus (\nu _{pq}^{t})^{\alpha ^{t}}= 0$. Therefore, $(\mu _{pq}^{t}\ominus \nu _{pq}^{t})^{\alpha ^{t}}\geq (\mu _{pq}^{t})^{\alpha ^{t}}\ominus (\nu _{pq}^{t})^{\alpha ^{t}}$.

  ᅟ:

  Case 3: Let $\mu _{pq}^{t}< \alpha ^{t}$ and $\nu _{pq}^{t}\geq \alpha ^{t}$. Then, $\mu _{pq}^{t}<\nu _{pq}^{t}$ and $(\mu _{pq}^{t}\ominus \nu _{pq}^{t})^{\alpha ^{t}}= 0$. Also, since $(\mu _{pq}^{t})^{\alpha ^{t}}= 0$ and $(\nu _{pq}^{t})^{\alpha ^{t}}= 1$, $(\mu _{pq}^{t})^{\alpha ^{t}}\ominus (\nu _{pq}^{t})^{\alpha ^{t}}= 0\ominus 1 = 0$. Therefore $(\mu _{pq}^{t}\ominus \nu _{pq}^{t})^{\alpha ^{t}}= (\mu _{pq}^{t})^{\alpha ^{t}}\ominus (\nu _{pq}^{t})^{\alpha ^{t}}$.

  ᅟ:

  Case 4: Let $\mu _{pq}^{t}\geq \alpha ^{t}$ and $\nu _{pq}^{t}<\alpha ^{t}$. Then, $\mu _{pq}^{t}>\nu _{pq}^{t}$ and $(\mu _{pq}^{t}\ominus \nu _{pq}^{t})=\mu _{pq}^{t}$. Since $\mu _{pq}^{t}\geq \alpha ^{t}$, $(\mu _{pq}^{t}\ominus \nu _{pq}^{t})^{\alpha ^{t}}= 1$. Also, since $(\mu _{pq}^{t})^{\alpha ^{t}}= 1$ and $(\nu _{pq}^{t})^{\alpha ^{t}}= 0$, $(\mu _{pq}^{t})^{\alpha ^{t}}\ominus (\nu _{pq}^{t})^{\alpha ^{t}}= 1\ominus 0 = 1$. Therefore $(\mu _{pq}^{t}\ominus \nu _{pq}^{t})^{\alpha ^{t}}= (\mu _{pq}^{t})^{\alpha ^{t}}\ominus (\nu _{pq}^{t})^{\alpha ^{t}}$.

  ᅟ:

  For indeterminacy-membership values: There are four cases,

  ᅟ:

  Case 1: $\mu _{pq}^{i}, \nu _{pq}^{i}> \alpha ^{i}$.

  ᅟ:

  Subcase 1: Let $\mu _{pq}^{i}\geq \nu _{pq}^{i}$. Then $\mu _{pq}^{i}\ominus \nu _{pq}^{i}= 1$. Since 1 ≥ αⁱ, $(\mu _{pq}^{i}\ominus \nu _{pq}^{i} )^{\alpha ^{i}}= 1 \, or \, 0$. Also since $(\mu _{pq}^{i})^{\alpha ^{i}}= 1$ and $(\nu _{pq}^{i})^{\alpha ^{i}}= 1$, $(\mu _{pq}^{i})^{\alpha ^{i}}\ominus (\nu _{pq}^{i})^{\alpha ^{i}}= 1$. Therefore $(\mu _{pq}^{i}\ominus \nu _{pq}^{i})^{\alpha ^{i}}\leq (\mu _{pq}^{i})^{\alpha ^{i}}\ominus (\nu _{pq}^{i})^{\alpha ^{i}}$.

  ᅟ:

  Subcase 2: Let $\mu _{pq}^{i}<\nu _{pq}^{i}$. Then $\mu _{pq}^{i}\ominus \nu _{pq}^{i}=\mu _{pq}^{i}$. Since $\mu _{pq}^{i}> \alpha ^{i}$, $(\mu _{pq}^{i}\ominus \nu _{pq}^{i})^{\alpha ^{i}}= 1$. Also since $(\mu _{pq}^{i})^{\alpha ^{i}}= 1$ and $(\nu _{pq}^{i})^{\alpha ^{i}}= 1$, $(\mu _{pq}^{i})^{\alpha ^{i}}\ominus (\nu _{pq}^{i})^{\alpha ^{i}}= 1$. Therefore $(\mu _{pq}^{i}\ominus \nu _{pq}^{i})^{\alpha ^{i}}= (\mu _{pq}^{i})^{\alpha ^{i}}\ominus (\nu _{pq}^{i})^{\alpha ^{i}}$.

  ᅟ:

  Case 2: $\mu _{pq}^{i}, \nu _{pq}^{i}\leq \alpha ^{i}$.

  ᅟ:

  Subcase 1: Let $\mu _{pq}^{i}\geq \nu _{pq}^{i}$. Then $\mu _{pq}^{i}\ominus \nu _{pq}^{i}= 1$. Since $\mu _{pq}^{i}\leq \alpha ^{i}$, $(\mu _{pq}^{i}\ominus \nu _{pq}^{i} )^{\alpha ^{i}}= 0$. Also since $(\mu _{pq}^{i})^{\alpha ^{i}}= 0$ and $(\nu _{pq}^{i})^{\alpha ^{i}}= 0$, $(\mu _{pq}^{i})^{\alpha ^{i}}\ominus (\nu _{pq}^{i})^{\alpha ^{i}}= 1$. Hence, $(\mu _{pq}^{i}\ominus \nu _{pq}^{i})^{\alpha ^{i}}\leq (\mu _{pq}^{i})^{\alpha ^{i}}\ominus (\nu _{pq}^{i})^{\alpha ^{i}}$.

  ᅟ:

  Subcase 2: Let $\mu _{pq}^{i}<\nu _{pq}^{i}$. Then $\mu _{pq}^{i}\ominus \nu _{pq}^{i}=\mu _{pq}^{i}$. Since $\mu _{pq}^{i}\leq \alpha ^{i}$, $(\mu _{pq}^{i}\ominus \nu _{pq}^{i})^{\alpha ^{i}}= 0$. Also since $(\mu _{pq}^{i})^{\alpha ^{i}}= 0$ and $(\nu _{pq}^{i})^{\alpha ^{i}}= 0$, $(\mu _{pq}^{i})^{\alpha ^{i}}\ominus (\nu _{pq}^{i})^{\alpha ^{i}}= 1$. Therefore $(\mu _{pq}^{i}\ominus \nu _{pq}^{i})^{\alpha ^{i}}\leq (\mu _{pq}^{i})^{\alpha ^{i}}\ominus (\nu _{pq}^{i})^{\alpha ^{i}}$.

  ᅟ:

  Case 3: Let $\mu _{pq}^{i}\leq \alpha ^{i}$ and $\nu _{pq}^{i}>\alpha ^{i}$. Then, $\mu _{pq}^{i}<\nu _{pq}^{i}$ and $\mu _{pq}^{i}\ominus \nu _{pq}^{i}=\mu _{pq}^{i}$. Since $\mu _{pq}^{i}\leq \alpha ^{i}$, $(\mu _{pq}^{i}\ominus \nu _{pq}^{i})^{\alpha ^{i}}= 0$. Also, since $(\mu _{pq}^{i})^{\alpha ^{i}}= 0$ and $(\nu _{pq}^{i})^{\alpha ^{i}}= 1$, $(\mu _{pq}^{i})^{\alpha ^{i}}\ominus (\nu _{pq}^{i})^{\alpha ^{i}}= 0\ominus 1 = 0$. Thus, $(\mu _{pq}^{t}\ominus \nu _{pq}^{t})^{\alpha ^{t}}= (\mu _{pq}^{t})^{\alpha ^{t}}\ominus (\nu _{pq}^{t})^{\alpha ^{t}}$.

  ᅟ:

  Case 4: Let $\mu _{pq}^{i}> \alpha ^{i}$ and $\nu _{pq}^{i}\leq \alpha ^{i}$. Then, $\mu _{pq}^{i}>\nu _{pq}^{i}$ and $(\mu _{pq}^{i}\ominus \nu _{pq}^{i})= 1$. Since 1 ≥ α^t, $(\mu _{pq}^{i}\ominus \nu _{pq}^{i})^{\alpha ^{i}}= 1$ or 0. Also, since $(\mu _{pq}^{i})^{\alpha ^{i}}= 1$ and $(\nu _{pq}^{i})^{\alpha ^{i}}= 0$, $(\mu _{pq}^{i})^{\alpha ^{i}}\ominus (\nu _{pq}^{i})^{\alpha ^{i}}= 1\ominus 0 = 1$. Therefore $(\mu _{pq}^{i}\ominus \nu _{pq}^{i})^{\alpha ^{i}}\leq (\mu _{pq}^{i})^{\alpha ^{i}}\ominus (\nu _{pq}^{i})^{\alpha ^{i}}$.

  ᅟ:

  For falsity-membership values: The proof can be made in similar way to the proof made for indeterminacy-membership values. When it is considered all of cases, it is concluded that $(\mu \tilde \ominus \nu )^{(\alpha )}\tilde \succeq (\mu )^{(\alpha )} \tilde \ominus (\nu )^{(\alpha )}.$

  ᅟ:

  The proof will be made for truth, indeterminacy and falsity-memberships values of elements of SVN-matrices.

  ᅟ:

  1. For truth-membership values: There are four cases,

  ᅟ:

  Case 1: $\mu _{pq}^{t}, \nu _{pq}^{t}\geq \alpha ^{t}$.

  ᅟ:

  Subcase 1: Let $\mu _{pq}^{t}> \nu _{pq}^{t}$. Then $\mu _{pq}^{t}\ominus \nu _{pq}^{t}=\mu _{pq}^{t}$. Since $\mu _{pq}^{t}\geq \alpha ^{t}$, $(\mu _{pq}^{t}\ominus \nu _{pq}^{t} )_{\alpha ^{t}}=\mu _{pq}^{t}$. Also since $(\mu _{pq}^{t})_{\alpha ^{t}}=\mu _{pq}^{t}$ and $(\nu _{pq}^{t})_{\alpha ^{t}}=\mu _{pq}^{t}$, $(\mu _{pq}^{t})_{\alpha ^{t}}\ominus (\nu _{pq}^{t})_{\alpha ^{t}}= 0$. Therefore, $(\mu _{pq}^{t}\ominus \nu _{pq}^{t})_{\alpha ^{t}}\geq (\mu _{pq}^{t})_{\alpha ^{t}}\ominus (\nu _{pq}^{t})_{\alpha ^{t}}$.

  ᅟ:

  Subcase 2: Let $\mu _{pq}^{t}\leq \nu _{pq}^{t}$. Then $\mu _{pq}^{t}\ominus \nu _{pq}^{t}= 0$. Since 0 ≤ α^t, $(\mu _{pq}^{t}\ominus \nu _{pq}^{t})_{\alpha ^{t}}= 0$ or $\mu _{pq}^{t}$. Also since $(\mu _{pq}^{t})_{\alpha ^{t}}=\mu _{pq}^{t}$ and $(\nu _{pq}^{t})_{\alpha ^{t}}=\mu _{pq}^{t}$, $(\mu _{pq}^{t})_{\alpha ^{t}}\ominus (\nu _{pq}^{t})_{\alpha ^{t}}= 0$. Therefore, $(\mu _{pq}^{t}\ominus \nu _{pq}^{t})_{\alpha ^{t}}\geq (\mu _{pq}^{t})_{\alpha ^{t}}\ominus (\nu _{pq}^{t})_{\alpha ^{t}}$.

  ᅟ:

  Case 2: $\mu _{pq}^{t}, \nu _{pq}^{t}<\alpha ^{t}$.

  ᅟ:

  Subcase 1: Let $\mu _{pq}^{t}> \nu _{pq}^{t}$. Then, $\mu _{pq}^{t}\ominus \nu _{pq}^{t}=\mu _{pq}^{t}$. Since $\mu _{pq}^{t}\geq \alpha ^{t}$, $(\mu _{pq}^{t}\ominus \nu _{pq}^{t} )_{\alpha ^{t}}=\mu _{pq}^{t}$. Also since $(\mu _{pq}^{t})_{\alpha ^{t}}= 0$ and $(\nu _{pq}^{t})_{\alpha ^{t}}= 0$, $(\mu _{pq}^{t})_{\alpha ^{t}}\ominus (\nu _{pq}^{t})_{\alpha ^{t}}= 0$. Therefore $(\mu _{pq}^{t}\ominus \nu _{pq}^{t})_{\alpha ^{t}}\geq (\mu _{pq}^{t})_{\alpha ^{t}}\ominus (\nu _{pq}^{t})_{\alpha ^{t}}$.

  ᅟ:

  Subcase 2: Let $\mu _{pq}^{t}\leq \nu _{pq}^{t}$. Then $\mu _{pq}^{t}\ominus \nu _{pq}^{t}= 0$. Since 0 ≤ α^t, $(\mu _{pq}^{t}\ominus \nu _{pq}^{t})_{\alpha ^{t}}= 0$ or $\mu _{pq}^{t}$. Also since $(\mu _{pq}^{t})_{\alpha ^{t}}= 0$ and $(\nu _{pq}^{t})_{\alpha ^{t}}= 0$, $(\mu _{pq}^{t})_{\alpha ^{t}}\ominus (\nu _{pq}^{t})_{\alpha ^{t}}= 0$. Therefore $(\mu _{pq}^{t}\ominus \nu _{pq}^{t})_{\alpha ^{t}}\geq (\mu _{pq}^{t})_{\alpha ^{t}}\ominus (\nu _{pq}^{t})_{\alpha ^{t}}$.

  ᅟ:

  Case 3: Let $\mu _{pq}^{t}< \alpha ^{t}$ and $\nu _{pq}^{t}\geq \alpha ^{t}$. Then, $\mu _{pq}^{t}<\nu _{pq}^{t}$ and $(\mu _{pq}^{t}\ominus \nu _{pq}^{t})= 0$. Since 0 ≤ α^t, $(\mu _{pq}^{t}\ominus \nu _{pq}^{t})_{\alpha ^{t}}= 0$ or $\mu _{pq}^{t}$. Also, since $(\mu _{pq}^{t})_{\alpha ^{t}}= 0$ and $(\nu _{pq}^{t})_{\alpha ^{t}}=\nu _{pq}^{t}$, $(\mu _{pq}^{t})_{\alpha ^{t}}\ominus (\nu _{pq}^{t})_{\alpha ^{t}}= 0\ominus \nu _{pq}^{t}= 0$. Therefore $(\mu _{pq}^{t}\ominus \nu _{pq}^{t})_{\alpha ^{t}}\geq (\mu _{pq}^{t})_{\alpha ^{t}}\ominus (\nu _{pq}^{t})_{\alpha ^{t}}$.

  ᅟ:

  Case 4: Let $\mu _{pq}^{t}\geq \alpha ^{t}$ and $\nu _{pq}^{t}<\alpha ^{t}$. Then, $\mu _{pq}^{t}>\nu _{pq}^{t}$ and $(\mu _{pq}^{t}\ominus \nu _{pq}^{t})=\mu _{pq}^{t}$. Since $\mu _{pq}^{t}\geq \alpha ^{t}$, $(\mu _{pq}^{t}\ominus \nu _{pq}^{t})_{\alpha ^{t}}=\mu _{pq}^{t}$. Also, since $(\mu _{pq}^{t})_{\alpha ^{t}}=\mu _{pq}^{t}$ and $(\nu _{pq}^{t})_{\alpha ^{t}}= 0$, $(\mu _{pq}^{t})_{\alpha ^{t}}\ominus (\nu _{pq}^{t})_{\alpha ^{t}}=\mu _{pq}^{t}\ominus 0=\mu _{pq}^{t}$. Therefore $(\mu _{pq}^{t}\ominus \nu _{pq}^{t})_{\alpha ^{t}}= (\mu _{pq}^{t})_{\alpha ^{t}}\ominus (\nu _{pq}^{t})_{\alpha ^{t}}$.

• 2. For indeterminacy-membership values: There are four cases,

  ᅟ:

  Case 1: $\mu _{pq}^{i}, \nu _{pq}^{i}> \alpha ^{i}$.

  ᅟ:

  Subcase 1: Let $\mu _{pq}^{i}\geq \nu _{pq}^{i}$. Then $\mu _{pq}^{i}\ominus \nu _{pq}^{i}= 1$. Since 1 ≥ αⁱ, $(\mu _{pq}^{i}\ominus \nu _{pq}^{i} )_{\alpha ^{i}}= 1$ or $\mu _{pq}^{i}$. Also since $(\mu _{pq}^{i})_{\alpha ^{i}}= 1$ and $(\nu _{pq}^{i})_{\alpha ^{i}}= 1$, $(\mu _{pq}^{i})_{\alpha ^{i}}\ominus (\nu _{pq}^{i})_{\alpha ^{i}}= 1$. Therefore $(\mu _{pq}^{i}\ominus \nu _{pq}^{i})_{\alpha ^{i}}\leq (\mu _{pq}^{i})_{\alpha ^{i}}\ominus (\nu _{pq}^{i})_{\alpha ^{i}}$.

  ᅟ:

  Subcase 2: Let $\mu _{pq}^{i}<\nu _{pq}^{i}$. Then $\mu _{pq}^{i}\ominus \nu _{pq}^{i}=\mu _{pq}^{i}$. Since $\mu _{pq}^{i}> \alpha ^{i}$, $(\mu _{pq}^{i}\ominus \nu _{pq}^{i})_{\alpha ^{i}}= 1$. Also since $(\mu _{pq}^{i})_{\alpha ^{i}}= 1$ and $(\nu _{pq}^{i})_{\alpha ^{i}}= 1$, $(\mu _{pq}^{i})_{\alpha ^{i}}\ominus (\nu _{pq}^{i})_{\alpha ^{i}}= 1$. Therefore $(\mu _{pq}^{i}\ominus \nu _{pq}^{i})_{\alpha ^{i}}= (\mu _{pq}^{i})_{\alpha ^{i}}\ominus (\nu _{pq}^{i})_{\alpha ^{i}}$.

  ᅟ:

  Case 2: $\mu _{pq}^{i}, \nu _{pq}^{i}\leq \alpha ^{i}$.

  ᅟ:

  Subcase 1: Let $\mu _{pq}^{i}\geq \nu _{pq}^{i}$. Then $\mu _{pq}^{i}\ominus \nu _{pq}^{i}= 1$. Since 1 ≥ αⁱ, $(\mu _{pq}^{i}\ominus \nu _{pq}^{i} )_{\alpha ^{i}}= 1$ or $\mu _{pq}^{i}$. Also since $(\mu _{pq}^{i})_{\alpha ^{i}}=\mu _{pq}^{i}$ and $(\nu _{pq}^{i})_{\alpha ^{i}}=\nu _{pq}^{i}$, $(\mu _{pq}^{i})_{\alpha ^{i}}\ominus (\nu _{pq}^{i})_{\alpha ^{i}}=\mu _{pq}^{i}\ominus \nu _{pq}^{i}= 1$. Therefore $(\mu _{pq}^{i}\ominus \nu _{pq}^{i})_{\alpha ^{i}}\leq (\mu _{pq}^{i})_{\alpha ^{i}}\ominus (\nu _{pq}^{i})_{\alpha ^{i}}$.

  ᅟ:

  Subcase 2: Let $\mu _{pq}^{i}<\nu _{pq}^{i}$. Then $\mu _{pq}^{i}\ominus \nu _{pq}^{i}=\mu _{pq}^{i}$. Since $\mu _{pq}^{i}\leq \alpha ^{i}$, $(\mu _{pq}^{i}\ominus \nu _{pq}^{i})_{\alpha ^{i}}=\mu _{pq}^{i}$. Also since $(\mu _{pq}^{i})_{\alpha ^{i}}=\mu _{pq}^{i}$ and $(\nu _{pq}^{i})_{\alpha ^{i}}=\nu _{pq}^{i}$, $(\mu _{pq}^{i})_{\alpha ^{i}}\ominus (\nu _{pq}^{i})_{\alpha ^{i}}=\mu _{pq}^{i}\ominus \nu _{pq}^{i}=\mu _{pq}^{i}$. Therefore $(\mu _{pq}^{i}\ominus \nu _{pq}^{i})_{\alpha ^{i}}= (\mu _{pq}^{i})_{\alpha ^{i}}\ominus (\nu _{pq}^{i})_{\alpha ^{i}}$.

  ᅟ:

  Case 3: Let $\mu _{pq}^{i}\leq \alpha ^{i}$ and $\nu _{pq}^{i}>\alpha ^{i}$. Then, $\mu _{pq}^{i}<\nu _{pq}^{i}$ and $\mu _{pq}^{i}\ominus \nu _{pq}^{i}=\mu _{pq}^{i}$. Since $\mu _{pq}^{i}\leq \alpha ^{i}$, $(\mu _{pq}^{i}\ominus \nu _{pq}^{i})_{\alpha ^{i}}=\mu _{pq}^{i}$. Also, since $(\mu _{pq}^{i})_{\alpha ^{i}}=\mu _{pq}^{i}$ and $(\nu _{pq}^{i})_{\alpha ^{i}}= 1$, $(\mu _{pq}^{i})_{\alpha ^{i}}\ominus (\nu _{pq}^{i})_{\alpha ^{i}}=\mu _{pq}^{i}\ominus 1 = 1$ or $\mu _{pq}^{i}$. Therefore $(\mu _{pq}^{t}\ominus \nu _{pq}^{t})_{\alpha ^{t}}\leq (\mu _{pq}^{t})_{\alpha ^{t}}\ominus (\nu _{pq}^{t})_{\alpha ^{t}}$.

  ᅟ:

  Case 4: Let $\mu _{pq}^{i}> \alpha ^{i}$ and $\nu _{pq}^{i}\leq \alpha ^{i}$. Then, $\mu _{pq}^{i}>\nu _{pq}^{i}$ and $(\mu _{pq}^{i}\ominus \nu _{pq}^{i})= 1$. Since 1 ≥ α^t, $(\mu _{pq}^{i}\ominus \nu _{pq}^{i})_{\alpha ^{i}}= 1$ or $\mu _{pq}^{i}$. Also, since $(\mu _{pq}^{i})_{\alpha ^{i}}= 1$ and $(\nu _{pq}^{i})_{\alpha ^{i}}=\nu _{pq}^{i}$, $(\mu _{pq}^{i})_{\alpha ^{i}}\ominus (\nu _{pq}^{i})_{\alpha ^{i}}= 1\ominus \nu _{pq}^{i}= 1$. Therefore $(\mu _{pq}^{i}\ominus \nu _{pq}^{i})_{\alpha ^{i}}= (\mu _{pq}^{i})_{\alpha ^{i}}\ominus (\nu _{pq}^{i})_{\alpha ^{i}}$.

  ᅟ:

  3. For falsity-membership values: The proof can be made in similar way to the proof made for indeterminacy-membership values.When it is considered all of cases, it is concluded that $(\mu \tilde \ominus \nu )_{(\alpha )}\tilde \succeq (\mu )_{(\alpha )} \tilde \ominus (\nu )_{(\alpha )}.$

1.7 A.7

• 1. The proof will be made for truth, indeterminacy and falsity membership values of elements of SVN-matrices:

  ᅟ:

  Case 1: For truth-membership values:

  ᅟ:

  Subcase 1: Let $\mu _{pq}^{t},\nu _{pq}^{t}\geq \alpha ^{t}$. Then, $\mu _{pq}^{t}\oplus \nu _{pq}^{t}=\mu _{pq}^{t}\nu _{pq}^{t}\geq \alpha ^{t}$ or $\mu _{pq}^{t}\nu _{pq}^{t}> \alpha ^{t}$.

  If $\mu _{pq}^{t}\nu _{pq}^{t}\geq \alpha ^{t}$, then $(\mu _{pq}^{t}\odot \nu _{pq}^{t})^{\alpha ^{t}}= 1$. Since $(\mu _{pq}^{t})^{\alpha ^{t}}= 1$ and $(\nu _{pq}^{t})^{\alpha ^{t}}= 1$, $(\mu _{pq}^{t})^{\alpha ^{t}}\odot (\nu _{pq}^{t})^{\alpha ^{t}}= 1$. Thus $(\mu _{pq}^{t}\odot \nu _{pq}^{t})^{\alpha ^{t}}=(\mu _{pq}^{t})^{\alpha ^{t}}\odot (\nu _{pq}^{t})^{\alpha ^{t}}$.

  If $\mu _{pq}^{t}\nu _{pq}^{t}< \alpha ^{t}$, then $(\mu _{pq}^{t}\odot \nu _{pq}^{t})^{\alpha ^{t}}= 0$. Since $(\mu _{pq}^{t})^{\alpha ^{t}}= 1$ and $(\nu _{pq}^{t})^{\alpha ^{t}}= 1$, $(\mu _{pq}^{t})^{\alpha ^{t}}\odot (\nu _{pq}^{t})^{\alpha ^{t}}= 1$. Hence $(\mu _{pq}^{t}\odot \nu _{pq}^{t})^{\alpha ^{t}}<(\mu _{pq}^{t})^{\alpha ^{t}}\odot (\nu _{pq}^{t})^{\alpha ^{t}}$.

  ᅟ:

  Subcase 2: Let $\mu _{pq}^{t},\nu _{pq}^{t}< \alpha ^{t}$. Then, $\mu _{pq}^{t}\nu _{pq}^{t}=\mu _{pq}^{t}\odot \nu _{pq}^{t}<\alpha ^{t}$ and so $(\mu _{pq}^{t}\odot \nu _{pq}^{t})^{\alpha ^{t}}= 0$. Since $\mu _{pq}^{t}= 0$, $(\mu _{pq}^{t})^{\alpha ^{t}}\odot (\mu _{pq}^{t})^{\alpha ^{t}}=(\mu _{pq}^{t})^{\alpha ^{t}}(\mu _{pq}^{t})^{\alpha ^{t}}= 0$. Therefore $(\mu _{pq}^{t}\odot \nu _{pq}^{t})^{\alpha ^{t}}=(\mu _{pq}^{t})^{\alpha ^{t}}\odot (\mu _{pq}^{t})^{\alpha ^{t}}$.

  ᅟ:

  Subcase 3: Let $\mu _{pq}^{t}< \alpha ^{t}$ and $\nu _{pq}^{t}\geq \alpha ^{t}$. Then $\mu _{pq}^{t}\odot \nu _{pq}^{t}=\mu _{pq}^{t}\nu _{pq}^{t}\geq \alpha ^{t}$. Therefore $(\mu _{pq}^{t}\odot \nu _{pq}^{t})^{\alpha ^{t}}= 0$. Since $(\mu _{pq}^{t})^{\alpha ^{t}}= 0$, $(\mu _{pq}^{t})^{\alpha ^{t}}\odot (\nu _{pq}^{t})^{\alpha ^{t}}= 0$. Hence, $(\mu _{pq}^{t}\odot \nu _{pq}^{t})^{\alpha ^{t}}= (\mu _{pq}^{t})^{\alpha ^{t}}\odot (\nu _{pq}^{t})^{\alpha ^{t}}$.

  ᅟ:

  Subcase 4: Let $\mu _{pq}^{t}\geq \alpha ^{t}$ and $\nu _{pq}^{t}<\alpha ^{t}$. Then $\mu _{pq}^{t}\odot \nu _{pq}^{t}=\mu _{pq}^{t}\nu _{pq}^{t}< \alpha ^{t}$. Therefore $(\mu _{pq}^{t}\odot \nu _{pq}^{t})^{\alpha ^{t}}= 0$. Since $(\nu _{pq}^{t})^{\alpha ^{t}}= 0$, $(\mu _{pq}^{t})^{\alpha ^{t}}\odot (\nu _{pq}^{t})^{\alpha ^{t}}= 0$. Thus, $(\mu _{pq}^{t}\odot \nu _{pq}^{t})^{\alpha ^{t}}= (\mu _{pq}^{t})^{\alpha ^{t}}\odot (\nu _{pq}^{t})^{\alpha ^{t}}$.

  ᅟ:

  Case 2: For indeterminacy-membership values:

  ᅟ:

  Subcase 1: Let $\mu _{pq}^{i},\nu _{pq}^{i}> \alpha ^{i}$. Then, $\mu _{pq}^{i}\odot \nu _{pq}^{i}=\mu _{pq}^{i}+\nu _{pq}^{i}-\mu _{pq}^{i}+\nu _{pq}^{i}>\alpha ^{i}$ and so $(\mu _{pq}^{i}\odot \nu _{pq}^{i})^{\alpha ^{i}}= 1$. Since $(\mu _{pq}^{i})^{\alpha ^{i}}= 1$ and $(\nu _{pq}^{i})^{\alpha ^{i}}= 1$, $(\mu _{pq}^{i})^{\alpha ^{i}}\odot (\nu _{pq}^{i})^{\alpha ^{i}}= 1 + 1-1 = 1$. Thus $(\mu _{pq}^{i}\odot \nu _{pq}^{i})^{\alpha ^{i}}=(\mu _{pq}^{i})^{\alpha ^{i}}\odot (\nu _{pq}^{i})^{\alpha ^{i}}$.

  ᅟ:

  Subcase 2: Let $\mu _{pq}^{i},\nu _{pq}^{i}\leq \alpha ^{i}$. Then, there are two cases:

  ᅟ:

  1. If $\mu _{pq}^{i}\odot \nu _{pq}^{i}=\mu _{pq}^{i}+\nu _{pq}^{i}-\mu _{pq}^{i}\nu _{pq}^{i}\leq \alpha ^{i}$, then $(\mu _{pq}^{i}\odot \nu _{pq}^{i})^{\alpha ^{i}}= 0$. Since $(\mu _{pq}^{i})^{\alpha ^{i}}= 0$ and $(\nu _{pq}^{i})^{\alpha ^{i}}= 0$, $(\mu _{pq}^{i})^{\alpha ^{i}}\odot (\nu _{pq}^{i})^{\alpha ^{i}}= 0$. Therefore $(\mu _{pq}^{i}\odot \nu _{pq}^{i})^{\alpha ^{i}}=(\mu _{pq}^{i})^{\alpha ^{i}}\odot (\nu _{pq}^{i})^{\alpha ^{i}}$.

  ᅟ:

  2. If $\mu _{pq}^{i}\odot \nu _{pq}^{i}=\mu _{pq}^{i}+\nu _{pq}^{i}-\mu _{pq}^{i}\nu _{pq}^{i}>\alpha ^{i}$, then $(\mu _{pq}^{i}\odot \nu _{pq}^{i})^{\alpha ^{i}}= 1$. Since $(\mu _{pq}^{i})^{\alpha ^{i}}= 0$ and $(\nu _{pq}^{i})^{\alpha ^{i}}= 0$, $(\mu _{pq}^{i})^{\alpha ^{i}}\odot (\nu _{pq}^{i})^{\alpha ^{i}}= 0$. Hence $(\mu _{pq}^{i}\odot \nu _{pq}^{i})^{\alpha ^{i}}>(\mu _{pq}^{i})^{\alpha ^{i}}\odot (\nu _{pq}^{i})^{\alpha ^{i}}$.

  ᅟ:

  Subcase 3: Let $\mu _{pq}^{i}\leq \alpha ^{i}$ and $\nu _{pq}^{i}>\alpha ^{i}$. Then $\mu _{pq}^{i}\odot \nu _{pq}^{i}=\mu _{pq}^{i}+\nu _{pq}^{i}-\mu _{pq}^{i}\nu _{pq}^{i}> \alpha ^{i}$ and so $(\mu _{pq}^{i}\odot \nu _{pq}^{i})^{\alpha ^{i}}= 1$. Since $(\mu _{pq}^{i})^{\alpha ^{i}}= 0$ and $(\nu _{pq}^{i})^{\alpha ^{i}}= 1$, $(\mu _{pq}^{i})^{\alpha ^{i}}\odot (\nu _{pq}^{i})^{\alpha ^{i}}= 0 + 1-0 = 1$. Thus, $(\mu _{pq}^{i}\odot \nu _{pq}^{i})^{\alpha ^{i}}= (\mu _{pq}^{i})^{\alpha ^{i}}\odot (\nu _{pq}^{i})^{\alpha ^{i}}$.

  ᅟ:

  Subcase 4: Let $\mu _{pq}^{i}> \alpha ^{i}$ and $\nu _{pq}^{i}\leq \alpha ^{i}$. Then $\mu _{pq}^{i}\odot \nu _{pq}^{i}=\mu _{pq}^{i}+\nu _{pq}^{i}-\mu _{pq}^{i}\nu _{pq}^{i}> \alpha ^{i}$ and so $(\mu _{pq}^{i}\odot \nu _{pq}^{i})^{\alpha ^{i}}= 1$. Since $(\mu _{pq}^{i})^{\alpha ^{i}}= 1$ and $(\nu _{pq}^{i})^{\alpha ^{i}}= 0$, $(\mu _{pq}^{i})^{\alpha ^{i}}\odot (\nu _{pq}^{i})^{\alpha ^{i}}= 1 + 0-0$. Thus, $(\mu _{pq}^{i}\odot \nu _{pq}^{i})^{\alpha ^{i}}= (\mu _{pq}^{i})^{\alpha ^{i}}\odot (\nu _{pq}^{i})^{\alpha ^{i}}$.

  ᅟ:

  Case 3: For falsity-membership values: The proof can be made by similar way to proof of case 2.When it is considered all of the cases, it is concluded that $(\mu \tilde \odot \nu )^{(\alpha )}\tilde \preceq \mu ^{(\alpha )}\tilde \odot \nu ^{(\alpha )}$.

• 2. The proof can be made by similar way to proof of statement 1.