The concept of q-rung orthopair fuzzy sets generalizes the notions of intuitionistic fuzzy sets and Pythagorean fuzzy sets to describe complicated uncertain information more effectively. Their most dominant attribute is that the sum of the \(q^{th}\) power of the truth-membership and the \(q^{th}\) power of the falsity-membership must be equal to or less than one, so they can broaden the space of uncertain data. This set can adjust the range of indication of decision data by changing the parameter \(q, ~q\geq 1\). In this paper, we define the Hamacher operations of q-rung orthopair fuzzy sets and proved some desirable properties of these operations, such as commutativity, idempotency, and monotonicity. Further, we proved De Morgan’s laws for these operations over complement. Furthermore, we defined the Hamacher scalar multiplication \(({n._{h}}A)\) and Hamacher exponentiation \((A^{\wedge_{h}n})\) operations on q-rung orthopair fuzzy sets and investigated their algebraic properties. Finally, we defined the necessity and possibility operators based on q-rung orthopair fuzzy sets and some properties of Hamacher operations that are considered.
Zadeh fuzzy set (FS) has acquired greater attention by researchers in a wide range of scientific areas, including management sciences, robotics, decision theory and many other disciplines [1]. FSs were further generalized to intuitionistic fuzzy sets (IFSs) by Atanassov [2] in 1983. An IFS is distinguished by a truth-membership (T) and falsity-membership (F) satisfying the condition that the sum of both membership degrees should not exceed one. IF values play an important role in both theoretical and practical progress of IFSs. Applications of IFSs appear in various fields, including medical diagnosis, optimization problems, and decision-making. Recently, Liu et al., [3] introduced and explored various types of centroid transformations of IF values. However, in many practical decision-making problems, the sum of truth-membership and falsity-membership may not be less than one, but the sum of their squares may be less than one. To handle such types of difficulties, Yager [4,5] introduced the novel concept of Pythagorean fuzzy sets (PFSs), which is the generalization of IFSs.
Compensating the constraint that the summation of both membership degrees does not exceed one and that the sum of squares of the membership degrees should not be greater than one makes PFSs more powerful, generalizable, and effective. Zhang and Xu [6] studied various binary operations over PFS and also proposed a decision making algorithm based on PFS. Zeng et al., [7] developed a hybrid method for Pythagorean fuzzy multiple-criteria decision making. Garg [8] defined the new generalized Pythagorean fuzzy information aggregation by using Einstein Operations. A generalized union and a generalized intersection on PFSs can be constructed from a general t-norm and t-conorm. Hamacher operations [9], including Hamacher product and Hamacher sum, are good alternatives to the algebraic product and algebraic sum, respectively. Gao [10] proposed some Pythagorean fuzzy Hamacher prioritized aggregation operators in MADM. In recent years, many researchers studied the Hamacher aggregation operators and their applications [11,12,13,14].
PFSs can deal with various real-life problems more effectively, still there are cases that cannot be handled using PFSs. Take an example: The truth-membership and falsity-membership values suggested by a decision-maker are 0.8 and 0.9, respectively. Then, the problem can never be handled by means of PFSs, as \(\left(0.8\right)^2+\left(0.9\right)^2=1.45> 1\). In order to deal with such types of cases, Yager [15] submitted the idea of q-rung orthopair fuzzy sets (q-ROFSs). A q-ROFS is represented by means of two membership degrees; one is the truth and the other is falsity, with the characteristic that the summation of the \(q^{th}\) power of truth-membership and the \(q^{th}\) power of falsity-membership should not be greater than one. Thus, q-ROFSs extend the concepts of IFSs and PFSs, so that the uncertain information can be dealt with in a widened range. We can see that the IFS and PFS are special cases of q-ROFS. As q-rung increases, the range of processing fuzzy information increases. The acceptable space for different types of fuzzy numbers is shown in the Figure 1. IFS, PFS, and q-ROFS are abbreviations of intuitionistic fuzzy number, Pythagorean fuzzy number, and q-rung orthopair fuzzy number, respectively. Therefore, it can be seen that q-ROFS has stronger capability in dealing with uncertain information than the IFS and PFS.
After that, Liu and Wang [16] developed and applied certain simple weighted operators to aggregate q-ROFSs in decision-making. This set can adjust the range of indication of decision data by changing the parameter \(q, ~q\geq1\). Wei et al., [17] defined q-rung orthopair fuzzy Heronian mean operators in multiple attribute decision-making.
In this paper, we propose the concepts of q-rung orthopair fuzzy Hamacher operations and investigates their desirable properties. This paper is arranged as follows; In Section 2, some basic concepts of q-rung orthopair fuzzy sets and Hamacher operations. In Section 3, we define the Hamacher operations of q-rung orthopair fuzzy sets and studied some desirable properties. In Section 4, on complement of q-rung orthopair fuzzy sets are studied. In Section 5, we constructed Hamacher scalar multiplication \(({n._{h}}A)\) and Hamacher exponentiation \((A^{\wedge_{h}n})\) operations of q-rung orthopair fuzzy set \(A\) and investigated their algebraic properties. In Section 6, some properties of necessity and possibility operators on q-rung orthopair fuzzy sets are proved. In Section 7, comparison analysis of the proposed model with IF and PF models. Section 8 concludes the paper and some future directions are presented in Section 9.
Definition 1. [15] A q-rung orthopair fuzzy set \(A\) on a universe \(X\) is an object of the form \(A=\left\{\left(x, \mu_A(x),\nu_A(x)\right)|x\in X\right\}\), where \(\mu_A(x)\in[0,1]\) is called the degree of membership of \(x\) in \(A\), \(\nu_A(x)\in[0,1]\) is called the degree of non-membership of \(x\) in \(A\), and where \(\mu_A(x)\) and \(\nu_A(x)\) satisfy the following condition; \[0\leq \mu^q_A(x)+\nu^q_A(x)\leq 1 (q\geq 1)\ \ \ \text{for all}\ \ \ x\in X.\]
Definition 2. Let \(A=\left\{\left( x, \mu_A(x),\nu_A(x)\right) | x\in X \right\}\) and \(B=\left\{\left( x, \mu_B(x),\nu_B(x)\right) | x\in X \right\}\) be any two q-rung orthopair fuzzy sets, then we have
Lemma 1. For any two real numbers \(a,b\in [0,1]\), the following inequality holds; \[\left(\dfrac{a^qb^q}{a^q+b^q-a^qb^q}\right)^{1/q}\leq \left(\dfrac{a^q+b^q-2a^qb^q}{1-a^qb^q}\right)^{1/q}.\]
Lemma 2. If \(A\subseteq B\), then for any three real numbers \(a,b,c\in[0,1]\), the following inequalities hold;
Theorem 1. If \( A, B \in q-ROFS(X)\), then \(A\boxtimes_{h} B\subseteq A\boxplus_{h} B\).
Proof. By using Lemma (1), we have \begin{align*} \left(\dfrac{\mu^q_A(x)\mu^q_B(x)}{\mu^q_A(x)+\mu^q_B(x)-\mu^q_A(x)\mu^q_B(x)}\right)^{1/q}&\leq \left(\dfrac{\mu^q_A(x)+\mu^q_B(x)-2\mu^q_A(x)\mu^q_B(x)}{1-\mu^q_A(x)\mu^q_B(x)}\right)^{1/q},\end{align*} and \begin{align*} \left(\dfrac{\nu^q_A(x)+\nu^q_B(x)-2\nu^q_A(x)\nu^q_B(x)}{1-\nu^q_A(x)\nu^q_B(x)}\right)^{1/q} \geq \left(\dfrac{\nu^q_A(x)\nu^q_B(x)}{\nu^q_A(x)+\nu^q_B(x)-\nu^q_A(x)\nu^q_B(x)}\right)^{1/q}\end{align*} for all \(x\in X\).
By Definition (2), it follows that \(A\boxtimes_{h} B\subseteq A\boxplus_{h} B\).
Theorem 2. If \(A\in q-ROFS(X)\), then
Proof.
Theorem 3. If \( A, B, C \in q-ROFS(X)\), then
Theorem 4. If \(A\in q-ROFS(X)\), then
Theorem 5. For \( A, B \in q-ROFS(X)\), if \(A\subseteq B\), then \(A\boxtimes_{h} C\subseteq B\boxtimes_{h} C\).
Proof. Let \(\mu_A(x)\leq \mu_B(x)\) and \(\nu_A(x)\geq \nu_B(x)\) for all \(x \in X\). By using Lemma (2), we have \[\left(\dfrac{\mu^q_A(x)\mu^q_C(x)}{\mu^q_A(x)+\mu^q_C(x)-\mu^q_A(x)\mu^q_C(x)}\right)^{1/q}\leq \left(\dfrac{\mu^q_B(x)\mu^q_C(x)}{\mu^q_B(x)+\mu^q_C(x)-\mu^q_B(x)\mu^q_C(x)}\right)^{1/q}\] and \[\left(\dfrac{\nu^q_B(x)\nu^q_C(x)}{\nu^q_B(x)+\nu^q_C(x)-\nu^q_B(x)\nu^q_C(x)}\right)^{1/q}\geq \left(\dfrac{\nu^q_A(x)\nu^q_C(x)}{\nu^q_A(x)+\nu^q_C(x)-\nu^q_A(x)\nu^q_C(x)}\right)^{1/q}\] for all \(x \in X\). Therefore, \(A\boxtimes_{h} C\subseteq B\boxtimes_{h} C.\)
Theorem 6. For \( A, B \in q-ROFS(X)\), if \(A\subseteq B\), then \(A\boxplus_{h} C\subseteq B\boxplus_{h} C\).
Proof. Let \(\mu_A(x)\leq \mu_B(x)\) and \(\nu_A(x)\geq \nu_B(x)\) for all \(x \in X\). By using Lemma (2), we have \[\left(\dfrac{\mu^q_A(x)+\mu^q_C(x)-2\mu^q_A(x)\mu^q_C(x)}{1-\mu^q_A(x)\mu^q_C(x)}\right)^{1/q}\leq \left(\dfrac{\mu^q_B(x)+\mu^q_C(x)-2\mu^q_B(x)\mu^q_C(x)}{1-\mu^q_B(x)\mu^q_C(x)}\right)^{1/q}\] and \[\left(\dfrac{\nu^q_B(x)+\nu^q_C(x)-2\nu^q_B(x)\nu^q_C(x)}{1-\nu^q_B(x)\nu^q_C(x)}\right)^{1/q}\geq \left(\dfrac{\nu^q_A(x)+\nu^q_C(x)-2\nu^q_A(x)\nu^q_C(x)}{1-\nu^q_A(x)\nu^q_C(x)}\right)^{1/q}.\] Therefore, \(A\boxplus_{h} C\subseteq B\boxplus_{h} C.\)
Theorem 7. If \( A, B \in q-ROFS(X)\), then
Proof.   (i) \begin{align*}(A&\cap B)\boxplus_{h}(A\cup B)\\ &=\left( \min\left\{\mu_A(x),\mu_B(x)\right\},\max\left\{\nu_A(x),\nu_B(x)\right\}\right) \boxplus_{h}\left( \max\left\{\mu_A(x),\mu_B(x)\right\},\min\left\{\nu_A(x),\nu_B(x)\right\}\right)\\ &=\left[ \left(\dfrac{\min\left\{\mu^q_A(x),\mu^q_B(x)\right\}+\max\left\{\mu^q_A(x),\mu^q_B(x)\right\}-2\min\left\{\mu^q_A(x),\mu^q_B(x)\right\}\max\left\{\mu^q_A(x),\mu^q_B(x)\right\}}{1-\min\left\{\mu^q_A(x),\mu^q_B(x)\right\}\max\left\{\mu^q_A(x),\mu^q_B(x)\right\}}\right)^{1/q} , \right.\\ &\;\;\;\;\left.\left(\dfrac{\max\left\{\nu^q_A(x),\nu^q_B(x)\right\}\min\left\{\nu^q_A(x),\nu^q_B(x)\right\}}{\max\left\{\nu^q_A(x),\nu^q_B(x)\right\}+\min\left\{\nu^q_A(x),\nu^q_B(x)\right\}-\max\left\{\nu^q_A(x),\nu^q_B(x)\right\}\min\left\{\nu^q_A(x),\nu^q_B(x)\right\}}\right)^{1/q}\right]\\ &=\left\{\left.\left( x, \left(\dfrac{\mu^q_A(x)+\mu^q_B(x)-2\mu^q_A(x) \mu^q_B(x)}{1-\mu^q_A(x) \mu^q_B(x)}\right)^{1/q},\left(\dfrac{\nu^q_A(x) \nu^q_B(x)}{\nu^q_A(x)+\nu^q_B(x)-\nu^q_A(x) \nu^q_B(x)}\right)^{1/q}\right) \right| x\in X \right\}=A\boxplus_{h} B.\end{align*} (ii)   It can be proved similarly.
Theorem 8. If \( A, B \in q-ROFS(X)\), then
Proof. \begin{align*} \text{(i).}\;\;\;A^C\boxtimes_{h} B^C &=\left\{\left.\left( x, \left(\dfrac{\nu^q_A(x) \nu^q_B(x)}{\nu^q_A(x)+\nu^q_B(x)-\nu^q_A(x) \nu^q_B(x)}\right)^{1/q}, \left(\dfrac{\mu^q_A(x)+\mu^q_B(x)-2\mu^q_A(x) \mu^q_B(x)}{1-\mu^q_A(x) \mu^q_B(x)}\right)^{1/q}\right) \right| x\in X \right\}\\ &=(A\boxplus_{h} B)^C. \end{align*} \begin{align*} \text{(ii).}\;\;\;A^C\boxplus_{h} B^C&=\left\{\left.\left( x, \left(\dfrac{\nu^q_A(x)+\nu^q_B(x)-2\nu^q_A(x) \nu^q_B(x)}{1-\nu^q_A(x) \nu^q_B(x)}\right)^{1/q}, \left(\dfrac{\mu^q_A(x) \mu^q_B(x)}{\mu^q_A(x)+\mu^q_B(x)-\mu^q_A(x) \mu^q_B(x)}\right)^{1/q}\right) \right| x\in X \right\}\\ &=(A\boxtimes_{h} B)^C.\end{align*} \begin{align*} \text{(iii).}\;\;\;(A\boxplus_{h} B)^C&= \left\{\left.\left( x, \left(\dfrac{\nu^q_A(x) \nu^q_B(x)}{\nu^q_A(x)+\nu^q_B(x)-\nu^q_A(x) \nu^q_B(x)}\right)^{1/q}, \left(\dfrac{\mu^q_A(x)+\mu^q_B(x)-2\mu^q_A(x) \mu^q_B(x)}{1-\mu^q_A(x) \mu^q_B(x)}\right)^{1/q}\right) \right| x\in X \right\}\\ A^C\boxplus_{h} B^C&=\left\{\left.\left( x, \left(\dfrac{\nu^q_A(x)+\nu^q_B(x)-2\nu^q_A(x) \nu^q_B(x)}{1-\nu^q_A(x) \nu^q_B(x)}\right)^{1/q}, \left(\dfrac{\mu^q_A(x) \mu^q_B(x)}{\mu^q_A(x)+\mu^q_B(x)-\mu^q_A(x) \mu^q_B(x)}\right)^{1/q}\right) \right| x\in X \right\}.\end{align*} By Lemma (1), we have \[\left(\dfrac{\nu^q_A(x) \nu^q_B(x)}{\nu^q_A(x)+\nu^q_B(x)-\nu^q_A(x) \nu^q_B(x)}\right)^{1/q}\leq \left(\dfrac{\nu^q_A(x)+\nu^q_B(x)-2\nu^q_A(x) \nu^q_B(x)}{1-\nu^q_A(x) \nu^q_B(x)}\right)^{1/q}\] and \[\left(\dfrac{\mu^q_A(x)+\mu^q_B(x)-2\mu^q_A(x) \mu^q_B(x)}{1-\mu^q_A(x) \mu^q_B(x)}\right)^{1/q} \geq \left(\dfrac{\mu^q_A(x) \mu^q_B(x)}{\mu^q_A(x)+\mu^q_B(x)-\mu^q_A(x) \mu^q_B(x)}\right)^{1/q}\] for all \(x\in X\). Hence \((A\boxplus_{h} B)^C\subseteq A^C\boxplus_{h} B^C\). Similarly, we can prove (iv), (v) and (vi).
Based on the Definition (2), Hamacher sum and Hamacher product over two q-ROFSs \(A\) and \(B\) are further indicated as the following operations.
Theorem 9. If \(n\) is any positive number and \(A\) is a q-ROFS of \(X\), then the Hamacher scalar multiplication operation \((._{h})\) is a mapping from \(R^+ \times X\) to \(X\):
Proof. Mathematical induction can be used to prove that the Equation (1) holds for all positive integer \(n\). The Equation (1) is called P(n). Using Definition 2(i) of Hamacher sum (\(A\boxplus_{h} B\)), we have \begin{align*} A._{h} A&= \left\{\left.\left( x, \left(\dfrac{\mu^q_A(x)+\mu^q_A(x)-2\mu^q_A(x)\mu^q_A(x)}{1-\mu^q_A(x)\mu^q_A(x)}\right)^{1/q},\left(\dfrac{\nu^q_A(x)\nu^q_A(x)}{\nu^q_A(x)+\nu^q_A(x)-\nu^q_A(x)\nu^q_A(x)}\right)^{1/q} \hspace{0.2cm}\right) \right| x\in X \right\},\\ &=\left\{\left.\left( x, \left(\dfrac{2\mu^q_A(x)-2a^q_{ij}}{1-a^q_{ij}}\right)^{1/q}, \left(\dfrac{\nu^q_{\mu^q_A(x)}}{2\nu^q_A(x)-\nu^q_{\mu^q_A(x)}}\right)^{1/q} \hspace{0.2cm}\right) \right| x\in X \right\} ,\\ &=\left\{\left.\left( x, \left(\dfrac{2\mu^q_A(x)(1-\mu^q_A(x))}{1-a^6_{ij}}\right)^{1/q}, \left(\dfrac{\nu^q_{\mu^q_A(x)}}{\nu^q_A(x)(2-\nu^q_A(x))}\right)^{1/q}\hspace{0.2cm} \right) \right| x\in X \right\}, \\ &=\left\{\left.\left( x, \left(\dfrac{2\mu^q_A(x)(1-\mu^q_A(x))}{(1-\mu^q_A(x))(1+\mu^q_A(x))}\right)^{1/q}, \left(\dfrac{\nu^q_A(x)}{(2-\nu^q_A(x))}\right)^{1/q} \hspace{0.2cm}\right) \right| x\in X \right\}, \\ &=\left\{\left.\left( x, \left(\dfrac{2\mu^q_A(x)}{1+\mu^q_A(x)}\right)^{1/q}, \left(\dfrac{\nu^q_A(x)}{(2-\nu^q_A(x))}\right)^{1/q}\hspace{0.2cm}\right) \right| x\in X \right\},\\ 2._{h} A&=\left\{\left.\left( x, \left(\dfrac{2\mu^q_A(x)}{1+(2-1)\mu^q_A(x)}\right)^{1/q}, \left(\dfrac{\nu^q_A(x)}{2-(2-1)\nu^q_A(x)}\right)^{1/q}\hspace{0.2cm} \right) \right| x\in X \right\}.\end{align*} Since \(\mu^q_A(x)=(2-1)\mu^q_A(x),\) we have \begin{align*} n._{h} A&=\left\{\left.\left( x, \left(\dfrac{n\mu^q_A(x)}{1+(n-1)\mu^q_A(x)}\right)^{1/q}, \left(\dfrac{\nu^q_A(x)}{n-(n-1)\nu^q_A(x)}\right)^{1/q} \hspace{0.2cm}\right) \right| x\in X \right\}.\end{align*} Hence P(n) holds.
Suppose that Equation (1) holds for \(n=m\), i.e.,
\[ m._{h} A=\underbrace{A\boxplus_{h}…\boxplus_{h} A}_{m}=\left\{\left.\left( x, \left(\dfrac{m\mu^q_A(x)}{1+(m-1)\mu^q_A(x)}\right)^{1/q} , \left(\dfrac{\nu^q_A(x)}{m-(m-1)\nu^q_A(x)}\right)^{1/q}\hspace{0.2cm} \right) \right| x\in X \right\}.\] Then,     \((m+1)._{h}A=((m._{h}A)\boxplus_{h} A)\) \begin{align*}&\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\left[ \left(\dfrac{\dfrac{m\mu^q_A(x)}{1+(m-1)\mu^q_A(x)}+\mu^q_A(x)-2\dfrac{m\mu^q_A(x)}{1+(m-1)\mu^q_A(x)}.\mu^q_A(x)}{1-\dfrac{m\mu^q_A(x)}{1+(m-1)\mu^q_A(x)}.\mu^q_A(x)}\right)^{1/q},\right.\\ &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \;\;\;\;\;\;\;\;\;\;\; \left.\left(\dfrac{\dfrac{\nu^q_A(x)}{m-(m-1)\nu^q_A(x)}.\nu^q_A(x)}{\dfrac{\nu^q_A(x)}{m-(m-1)\nu^q_A(x)}+\nu^q_A(x)-\dfrac{\nu^q_A(x)}{m-(m-1)\nu^q_A(x)}.\nu^q_A(x)}\right)^{1/q}\right]\end{align*} \begin{align*} &=\left\{\left.\left( x, \left(\dfrac{\mu^q_A(x)(m+1)(1-\mu^q_A(x))}{(1+m\mu^q_A(x))(1-\mu^q_A(x))}\right)^{1/q}, \left(\dfrac{(\nu^q_A(x))^2}{\nu^q_A(x)(m+1-m\nu^q_A(x))}\right)^{1/q} \hspace{0.2cm}\right) \right| x\in X \right\} \\ &=\left\{\left.\left( x, \left(\dfrac{(m+1)\mu^q_A(x)}{1+m\mu^q_A(x)}\right)^{1/q}, \left(\dfrac{\nu^q_A(x)}{m+1-m\nu^q_A(x)}\right)^{1/q}\hspace{0.2cm} \right) \right| x\in X \right\} \\ &=\left\{\left.\left( x, \left(\dfrac{(m+1)\mu^q_A(x)}{1+[(m+1)-1]\mu^q_A(x)}\right)^{1/q} , \left(\dfrac{\nu^q_A(x)}{m+1-[(m+1)-1]\nu^q_A(x)}\right)^{1/q} \hspace{0.2cm}\right) \right| x\in X \right\}.\end{align*} So, when \(n=m+1\), \[n._{h} A=\underbrace{A\boxplus_{h}…\boxplus_{h} A}_{n}=\left\{\left.\left( x, \left(\dfrac{n\mu^q_A(x)}{1+(n-1)\mu^q_A(x)}\right)^{1/q} \left(\dfrac{\nu^q_A(x)}{n-(n-1)\nu^q_A(x)}\right)^{1/q}\hspace{0.2cm} \right) \right| x\in X \right\}, \] also holds. Hence, using the induction hypothesis, P(n) holds for any positive integer \(n\).Theorem 10. If \(n\) is any positive integer and \(A\) is a q-ROFS of \(X\), then the Hamacher exponentiation operation \((\wedge_{h})\) is a mapping from \(Z^+ \times X\) to \(X\):
Proof. Mathematical induction can be used to prove that the Equation (2) holds for all positive integer \(n\). The Equation (2) is called P(n). Using Definition 2(ii) of Hamacher product (\(A\boxtimes_{h} B\)), we have \begin{align*} {A^{\wedge_{h}}}^A &=\left\{\left.\left( x, \left(\dfrac{\mu^q_A(x)\mu^q_A(x)}{\mu^q_A(x)+\mu^q_A(x)-\mu^q_A(x)\mu^q_A(x)}\right)^{1/q}, \left(\dfrac{\nu^q_A(x)+\nu^q_A(x)-2\nu^q_A(x)\nu^q_A(x)}{1-\nu^q_A(x)\nu^q_A(x)}\right)^{1/q}\hspace{0.2cm} \right) \right| x\in X \right\}\\ &=\left\{\left.\left( x, \left(\dfrac{\mu^q_{\mu^q_A(x)}}{2\mu^q_A(x)-\mu^q_{\mu^q_A(x)}}\right)^{1/q},\left(\dfrac{2\nu^q_A(x)-2\nu^q_{\mu^q_A(x)}}{1-\nu^q_{\mu^q_A(x)}}\right)^{1/q}\hspace{0.2cm} \right) \right| x\in X \right\}\\ &=\left\{\left.\left( x, \left(\dfrac{\mu^q_{\mu^q_A(x)}}{\mu^q_A(x)(2-\mu^q_A(x))}\right)^{1/q},\left(\dfrac{2\nu^q_A(x)(1-\nu^q_A(x))}{1-\nu^6_{\mu^q_A(x)}}\right)^{1/q}\hspace{0.2cm} \right) \right| x\in X \right\}\\ &=\left\{\left.\left( x, \left(\dfrac{\mu^q_A(x)}{(2-\mu^q_A(x))}\right)^{1/q},\left(\dfrac{2\nu^q_A(x)(1-\nu^q_A(x))}{(1-\nu^q_A(x))(1+\nu^q_A(x))}\right)^{1/q}\hspace{0.2cm} \right) \right| x\in X \right\}\\ &=\left\{\left.\left( x, \left(\dfrac{\mu^q_A(x)}{(2-\mu^q_A(x))}\right)^{1/q},\left(\dfrac{2\nu^q_A(x)}{1+\nu^q_A(x)}\right)^{1/q}\hspace{0.2cm}\right) \right| x\in X \right\},\\ {A^{\wedge_{h}}}^q&=\left\{\left.\left( x, \left(\dfrac{\mu^q_A(x)}{2-(2-1)\mu^q_A(x)}\right)^{1/q},\left(\dfrac{2\nu^q_A(x)}{1+(2-1)\nu^q_A(x)}\right)^{1/q} \hspace{0.2cm}\right) \right| x\in X \right\}. \end{align*} Since \(\mu^q_A(x)=(2-1)\mu^q_A(x)\), we have \begin{align*}{A^{\wedge_{h}}}^n=\left\{\left.\left( x, \left(\dfrac{\mu^q_A(x)}{n-(n-1)\mu^q_A(x)}\right)^{1/q}, \left(\dfrac{n\nu^q_A(x)}{1+(n-1)\nu^q_A(x)}\right)^{1/q} \hspace{0.2cm}\right) \right| x\in X \right\}.\end{align*} Hence P(n) holds.
Suppose that the Equation (2) holds for \(n=m\), i. e.,
\begin{align*}{A^{\wedge_{h}}}^m=\overbrace{A\boxtimes_{h}…\boxtimes_{h} A}^{m}=\left\{\left.\left( x, \left(\dfrac{\mu^q_A(x)}{m-(m-1)\mu^q_A(x)}\right)^{1/q}, \left(\dfrac{m\nu^q_A(x)}{1+(m-1)\nu^q_A(x)}\right)^{1/q}\hspace{0.2cm} \right) \right| x\in X \right\},\end{align*} so, when \(n=m+1\), \begin{align*}{A^{\wedge_{h}}}^{m+1}&=\left\{\left.\left( x, \left(\dfrac{{\mu_A(x)}^2}{m+1-[(m+1)-1]{\mu_A(x)}^2}\right)^{1/q},\left(\dfrac{(m+1)\nu^q_A(x)}{1+[(m+1)-1]\nu^q_A(x)}\right)^{1/q} \hspace{0.2cm} \right) \right| x\in X \right\},\\ {A^{\wedge_{h}}}^n&=\overbrace{A\boxtimes_{h}…\boxtimes_{h} A}^{n}=\left\{\left.\left( x, \left(\dfrac{\mu^q_A(x)}{n-(n-1)\mu^q_A(x)}\right)^{1/q}, \left(\dfrac{n\nu^q_A(x)}{1+(n-1)\nu^q_A(x)}\right)^{1/q}\hspace{0.2cm} \right) \right| x\in X \right\}\end{align*} also holds. Hence, using the induction hypothesis, P(n) holds for any positive integer \(n\).Next, we prove the result of \(({n._{h}}A)\) and \((A^{\wedge_{h}n})\) are also FFMs.
Theorem 11. For any q-ROFS \(A\) and for any positive integer \(n\), \(({n._{h}}A)\) and \((A^{\wedge_{h}n})\) are q-ROFSs.
Proof. Since \( 0\leq \mu^q_A(x) \leq 1,\)   \(0\leq \nu^q_A(x) \leq 1,\)   \( 0\leq \mu^q_A(x)+\nu^q_A(x) \leq 1,\) and \(n>1\), we have \( (n-1)\mu^q_A(x)>-1,\)   \(1+(n-1)\mu^q_A(x)> 0,\) and \(n-(n-1)\nu^q_A(x)=(1-\nu^q_A(x))n+\nu^q_A(x)> \nu^q_A(x)\geq 0.\) Then, it is easy to get \[ \left(\dfrac{n\mu^q_A(x)}{1+(n-1)\mu^q_A(x)}\right)^{1/q}\geq 0,\ \ \ \text{and}\ \ \ \left (\dfrac{\nu^q_A(x)}{n-(n-1)\nu^q_A(x)}\right)^{1/q}\geq 0.\] Considering \(1+(n-1)\mu^q_A(x)=n\mu^q_A(x)+1-\mu^q_A(x)\geq n\mu^q_A(x),\) and \(n-(n-1)\nu^q_A(x)=\nu^q_A(x)+n(1-\nu^q_A(x))\geq \nu^q_A(x),\) we get \[\left(\dfrac{n\mu^q_A(x)}{1+(n-1)\mu^q_A(x)}\right)^{1/q}\leq 1,\ \ \ \text{and}\ \ \ \left(\dfrac{\nu^q_A(x)}{n-(n-1)\nu^q_A(x)}\right)^{1/q}\leq 1.\] For \(\mu^q_A(x)+\nu^q_A(x)\leq 1,\) and \(0\leq \nu^q_A(x)\leq 1-\mu^q_A(x)\), we get \begin{align*} \left(\dfrac{n\mu^q_A(x)}{1+(n-1)\mu^q_A(x)}\right)^{1/q} &+\left(\dfrac{\nu^q_A(x)}{n-(n-1)\nu^q_A(x)}\right)^{1/q}=\left(\dfrac{n\mu^q_A(x)}{1+(n-1)\mu^q_A(x)}\right)^{1/q} +\left(\dfrac{1}{\dfrac{n}{\nu^q_A(x)}-(n-1)}\right)^{1/q}\\ &\leq \left(\dfrac{n\mu^q_A(x)}{1+(n-1)\mu^q_A(x)}\right)^{1/q} +\left(\dfrac{1}{\dfrac{n}{1-\mu^q_A(x)}-(n-1)}\right)^{1/q}=1.\end{align*} Thus, we have \[0\leq \left(\dfrac{n\mu^q_A(x)}{1+(n-1)\mu^q_A(x)}\right)^{1/q}\leq 1,\ \ \ \text{and}\ \ \ 0\leq \left(\dfrac{\nu^q_A(x)}{n-(n-1)\nu^q_A(x)}\right)^{1/q}\leq 1,\] which implies that \[\left(\dfrac{n\mu^q_A(x)}{1+(n-1)\mu^q_A(x)}\right)^{1/q} +\left(\dfrac{\nu^q_A(x)}{n-(n-1)\nu^q_A(x)}\right)^{1/q}\leq 1.\] Similarly, we have \[0\leq \left(\dfrac{\mu^q_A(x)}{n-(n-1)\mu^q_A(x)}\right)^{1/q}\leq 1,\ \ \ \text{and} \ \ \ 0\leq \left(\dfrac{n\nu^q_A(x)}{1+(n-1)\nu^q_A(x)}\right)^{1/q}\leq 1,\] which implies that \[ \left(\dfrac{\mu^q_A(x)}{n-(n-1)\mu^q_A(x)}\right)^{1/q}+\left(\dfrac{n\nu^q_A(x)}{1+(n-1)\nu^q_A(x)}\right)^{1/q}\leq 1.\] Hence, \(({n._{h}})A\) and \((A^{\wedge_{h}n})\) are q-ROFSs.
Theorem 12. If \(A, B \in q-ROFS(X)\), then for any positive integers \(n,n_{1},n_{2}\), we have
Proof. We shall prove (i), (ii), (v) and (iii), (iv), (vi) can be proved analogously.
(i)  By Equations (1) and (2), we have \begin{align*}n_{1}._{h} A&=\left\{\left.\left( x, \left(\dfrac{n_{1}\mu^q_A(x)}{1+(n_{1}-1)\mu^q_A(x)}\right)^{1/q}, \left(\dfrac{\nu^q_A(x)}{n_{1}-(n_{1}-1)\nu^q_A(x)}\right)^{1/q}\hspace{0.2cm}\right) \right| x\in X \right\}=\left( \mu_B(x),\nu_B(x)\right),\end{align*} and \begin{align*} n_{2}._{h} A&=\left\{\left.\left( x, \left(\dfrac{n_{2}\mu^q_A(x)}{1+(n_{2}-1)\mu^q_A(x)}\right)^{1/q}, \left(\dfrac{\nu^q_A(x)}{n_{2}-(n_{2}-1)\nu^q_A(x)}\right)^{1/q}\hspace{0.2cm}\right) \right| x\in X \right\}=\left( \mu_C(x),\nu_C(x)\right).\end{align*} Since \(B\boxplus_{h} C=(n_{1}._{h} A)\boxplus_{h} (n_{2}._{h} A),\) so we get \[B\boxplus_{h} C=\left\{\left. x, \left(\dfrac{\mu^q_B(x)+\mu^q_C(x)-2\mu^q_B(x)\mu^q_C(x)}{1-\mu^q_B(x)\mu^q_C(x)}\right)^{1/q}, \left(\dfrac{\nu^q_B(x)\nu^q_C(x)}{\nu^q_B(x)+\nu^q_C(x)-\nu^q_B(x)\nu^q_C(x)}\right)^{1/q} \right| x\in X \right\}.\] Further, we can get \begin{align*} &\left(\dfrac{\mu^q_B(x)+\mu^q_C(x)-2\mu^q_B(x)\mu^q_C(x)}{1-\mu^q_B(x)\mu^q_C(x)}\right)^{1/q}\\ &\;\;\;\;\;\;\;\;\;\;\;\;=\left(\dfrac{\dfrac{n_{1}\mu^q_A(x)}{1+(n_{1}-1)\mu^q_A(x)}+\dfrac{n_{2}\mu^q_A(x)}{1+(n_{2}-1)\mu^q_A(x)}-2\dfrac{n_{1}\mu^q_A(x)}{1+(n_{1}-1)\mu^q_A(x)}\dfrac{n_{2}\mu^q_A(x)}{1+(n_{2}-1)\mu^q_A(x)}}{1-\dfrac{n_{1}\mu^q_A(x)}{1+(n_{1}-1)\mu^q_A(x)}\dfrac{n_{2}\mu^q_A(x)}{1+(n_{2}-1)\mu^q_A(x)}}\right)^{1/q}\\ &\;\;\;\;\;\;\;\;\;\;\;\;=\left(\dfrac{n_{1}\mu^q_A(x)(1+n_{2}\mu^q_A(x)-\mu^q_A(x))+n_{2}\mu^q_A(x)(1+n_{1}\mu^q_A(x)-\mu^q_A(x))-2n_{1}\mu^q_A(x)n_{2}\mu^q_A(x)}{(1+n_{1}\mu^q_A(x)-\mu^q_A(x))(1+n_{2}\mu^q_A(x)-\mu^q_A(x))-n_{1}\mu^q_A(x)n_{2}\mu^q_A(x)}\right)^{1/q}\\ &\;\;\;\;\;\;\;\;\;\;\;\;=\left(\dfrac{(n_{1}+n_{2})\mu^q_A(x)-(n_{1}+n_{2})\mu^q_A(x)}{1+(n_{1}+n_{2}-2)\mu^q_A(x)-(n_{1}+n_{2}-1)\mu^q_A(x)}\right)^{1/q}\end{align*} \begin{align*} &=\left(\dfrac{(n_{1}+n_{2})\mu^q_A(x)(1-\mu^q_A(x))}{(1+(n_{1}+n_{2}-1)\mu^q_A(x))(1-\mu^q_A(x))}\right)^{1/q}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\\ n_{1}._{h} A&=\left(\dfrac{(n_{1}+n_{2})\mu^q_A(x)}{1+(n_{1}+n_{2}-1)\mu^q_A(x)}\right)^{1/q},\end{align*} and \begin{align*} &\left(\dfrac{\nu^q_B(x)\nu^q_C(x)}{\nu^q_B(x)+\nu^q_C(x)-\nu^q_B(x)\nu^q_C(x)}\right)^{1/q}\\ &\;\;\;\;\;\;\;\;\;\;\;\;=\left(\dfrac{\dfrac{\nu^q_A(x)}{n_{1}-(n_{1}-1)\nu^q_A(x)} \dfrac{\nu^q_A(x)}{n_{2}-(n_{2}-1)\nu^q_A(x)}}{\dfrac{\nu^q_A(x)}{n_{1}-(n_{1}-1)\nu^q_A(x)}+ \dfrac{\nu^q_A(x)}{n_{2}-(n_{2}-1)\nu^q_A(x)}-\dfrac{\nu^q_A(x)}{n_{1}-(n_{1}-1)\nu^q_A(x)} \dfrac{\nu^q_A(x)}{n_{2}-(n_{2}-1)\nu^q_A(x)}}\right)^{1/q}\\ &\;\;\;\;\;\;\;\;\;\;\;\;=\left(\dfrac{\nu^q_A(x)\nu^q_A(x)}{\nu^q_A(x)(n_{2}+\nu^q_A(x)-n_{2}\nu^q_A(x))+\nu^q_A(x)(n_{1}+\nu^q_A(x)-n_{1}\nu^q_A(x))-\nu^q_A(x)\nu^q_A(x)}\right)^{1/q}\\ &\;\;\;\;\;\;\;\;\;\;\;\;=\left(\dfrac{\nu^q_A(x)}{(n_{2}+\nu^q_A(x)-n_{2}\nu^q_A(x))+(n_{1}+\nu^q_A(x)-n_{1}\nu^q_A(x))-\nu^q_A(x)}\right)^{1/q}\\ &\;\;\; n_{2}._{h} A=\left(\dfrac{\nu^q_A(x)}{(n_{1}+n_{2})-(n_{1}+n_{2}-1)\nu^q_A(x)}\right)^{1/q}.\end{align*} Since \[(n_{1}+n_{2})._{h} A =\left\{\left.\left( x, \left(\dfrac{(n_{1}+n_{2})\mu^q_A(x)}{1+(n_{1}+n_{2}-1)\mu^q_A(x)}\right)^{1/q} ,\left(\dfrac{\nu^q_A(x)}{(n_{1}+n_{2})-(n_{1}+n_{2}-1)\nu^q_A(x)}\right)^{1/q}\hspace{0.2cm}\right) \right.| x\in X \right\},\] so we finally get \((n_{1}._{h} A)\boxplus_{h} (n_{2}._{h} A)=(n_{1}+n_{2})._{h} A\).(ii)   By Equations (1) and (2), we have
\[n._{h} A= \left\{\left.\left( x, \left(\dfrac{n\mu^q_A(x)}{1+(n-1)\mu^q_A(x)}\right)^{1/q}, \left(\dfrac{\nu^q_A(x)}{n-(n-1)\nu^q_A(x)}\right)^{1/q}\hspace{0.2cm}\right) \right| x\in X \right\}=\left( \mu_B(x),\nu_B(x)\right),\] and \[n._{h} B= \left\{\left.\left( x, \left(\dfrac{n\mu^q_B(x)}{1+(n-1)\mu^q_B(x)}\right)^{1/q}, \left(\dfrac{\nu^q_B(x)}{n-(n-1)\nu^q_B(x)}\right)^{1/q}\hspace{0.2cm}\right) \right| x\in X \right\} \left( \mu_C(x),\nu_C(x)\right).\] Since \(B\boxplus_{h} C=(n._{h} A)\boxplus_{h}(n._{h} B)\), so we get \[B\boxplus_{h} C=\left\{\left.\left( x, \left(\dfrac{\mu^q_B(x)+\mu^q_C(x)-2\mu^q_B(x)\mu^q_C(x)}{1-\mu^q_B(x)\mu^q_C(x)}\right)^{1/q},\left(\dfrac{\nu^q_B(x)\nu^q_C(x)}{\nu^q_B(x)+\nu^q_C(x)-\nu^q_B(x)\nu^q_C(x)}\right)^{1/q}\hspace{0.2cm}\right) \right| x\in X \right\}.\] Further, we can get \begin{align*} &\left(\dfrac{\mu^q_B(x)+\mu^q_C(x)-2\mu^q_B(x)\mu^q_C(x)}{1-\mu^q_B(x)\mu^q_C(x)}\right)^{1/q}\\ &\;\;\;\;\;\;\;\;\;\;\;\;=\left(\dfrac{\dfrac{n\mu^q_A(x)}{1+(n-1)\mu^q_A(x)}+\dfrac{n\mu^q_B(x)}{1+(n-1)\mu^q_B(x)}-2\dfrac{n\mu^q_A(x)}{1+(n-1)\mu^q_A(x)} \dfrac{n\mu^q_B(x)}{1+(n-1)\mu^q_B(x)}}{1-\dfrac{n\mu^q_A(x)}{1+(n-1)\mu^q_A(x)} \dfrac{n\mu^q_B(x)}{1+(n-1)\mu^q_B(x)}}\right)^{1/q}\end{align*} \begin{align*} &\;\;\;\;\;\;\;\;\;\;\;\;=\left(\dfrac{n\mu^q_A(x)(1+n\mu^q_B(x)-\mu^q_B(x))+n\mu^q_B(x)(1+n\mu^q_A(x)-\mu^q_A(x))-2n\mu^q_A(x)n\mu^q_B(x)}{(1+n\mu^q_B(x)-\mu^q_B(x)) (1+n\mu^q_A(x)-\mu^q_A(x))-n\mu^q_A(x)n\mu^q_B(x)}\right)^{1/q}\\ &\;\;\;(n._{h} A)=\left(\dfrac{n\mu^q_A(x)+n\mu^q_B(x)-2n\mu^q_A(x)\mu^q_B(x)}{1+(n-1)(\mu^q_A(x)+\mu^q_B(x))-(2n-1)\mu^q_A(x)\mu^q_B(x)}\right)^{1/q},\end{align*} and \begin{align*} &\left(\dfrac{\nu^q_B(x)\nu^q_C(x)}{\nu^q_B(x)+\nu^q_C(x)-\nu^q_B(x)\nu^q_C(x)}\right)^{1/q}\\ &\;\;\;\;\;\;\;\;\;\;\;\;=\left(\dfrac{\dfrac{\nu^q_A(x)}{n-(n-1)\nu^q_A(x)} \dfrac{\nu^q_B(x)}{n-(n-1)\nu^q_B(x)}}{\dfrac{\nu^q_A(x)}{n-(n-1)\nu^q_A(x)}+\dfrac{\nu^q_B(x)}{n-(n-1)\nu^q_B(x)}-\dfrac{\nu^q_A(x)}{n-(n-1)\nu^q_A(x)} \dfrac{\nu^q_B(x)}{n-(n-1)\nu^q_B(x)}}\right)^{1/q}\\ &\;\;\;\;\;\;\;\;\;\;\;\;=\left(\dfrac{\nu^q_A(x)\nu^q_B(x)}{\nu^q_A(x)(n+\nu^q_B(x)-n\nu^q_B(x))+\nu^q_B(x)(n+\nu^q_A(x)-n\nu^q_A(x))-\nu^q_A(x)\nu^q_B(x)}\right)^{1/q}\\ &\;\;\;(n._{h} B)=\left(\dfrac{\nu^q_A(x) \nu^q_B(x)}{n(\nu^q_A(x)+\nu^q_B(x))-(2n-1)\nu^q_A(x)\nu^q_B(x)}\right)^{1/q}.\end{align*} Thus, \begin{align*} n._{h}&(A\boxplus_{h} B)\\ &=\left[ \left(\dfrac{n \dfrac{\mu^q_A(x)+\mu^q_B(x)-2\mu^q_A(x)\mu^q_B(x)}{1-\mu^q_A(x)\mu^q_B(x)}}{1+(n-1)\dfrac{\mu^q_A(x)+\mu^q_B(x)-2\mu^q_A(x)\mu^q_B(x)}{1-\mu^q_A(x)\mu^q_B(x)}}\right)^{1/q}, \left(\dfrac{\dfrac{\nu^q_A(x)\nu^q_B(x)}{\nu^q_A(x)+\nu^q_B(x)-\nu^q_A(x)\nu^q_B(x)}}{n-(n-1)\dfrac{\nu^q_A(x)\nu^q_B(x)}{\nu^q_A(x)+\nu^q_B(x)-\nu^q_A(x)\nu^q_B(x)}}\right)^{1/q}\right],\\ &=\left[ \left(\dfrac{n\mu^q_A(x)+n\mu^q_B(x)-2n\mu^q_A(x)\mu^q_B(x)}{1+(n-1)(\mu^q_A(x)+\mu^q_B(x))-(2n-1)\mu^q_A(x)\mu^q_B(x)}\right)^{1/q},\left(\dfrac{\nu^q_A(x) \nu^q_B(x)}{n(\nu^q_A(x)+\nu^q_B(x))-(2n-1)\nu^q_A(x)\nu^q_B(x)}\right)^{1/q} \right]. \end{align*} Comparing above results, we finally get \((n._{h} A)\boxplus_{h} (n._{h} B)=n._{h}(A\boxplus_{h} B)\).(v)  By Equations (1) and (2), we have
\[n_{1}._{h} A=\left\{\left.\left( x, \left(\dfrac{n_{1}\mu^q_A(x)}{1+(n_{1}-1)\mu^q_A(x)}\right)^{1/q},\left(\dfrac{\nu^q_A(x)}{n_{1}-(n_{1}-1)\nu^q_A(x)}\right)^{1/q}\hspace{0.2cm}\right) \right| x\in X \right\}=\left( \mu_B(x),\nu_B(x)\right),\] and \[n_{2}._{h}(n_{1}._{h} A)=\left\{\left.\left( x, \left(\dfrac{n_{2}\mu_B(x)}{1+(n_{2}-1)\mu_B(x)}\right)^{1/q},\left(\dfrac{\nu_B(x)}{n_{2}-(n_{2}-1)\nu_B(x)}\right)^{1/q}\hspace{0.2cm}\right) \right| x\in X \right\}.\] Further we can get \[\left(\dfrac{n_{2}\mu_B(x)}{1+(n_{2}-1)\mu_B(x)}\right)^{1/q}=\left(\dfrac{n_{2} \dfrac{n_{1}\mu^q_A(x)}{1+(n_{1}-1)\mu^q_A(x)}}{1+(n_{2}-1) \dfrac{n_{1}\mu^q_A(x)}{1+(n_{1}-1)\mu^q_A(x)}}\right)^{1/q} =\left(\dfrac{n_{1}n_{2}\mu^q_A(x)}{1+(n_{1}n_{2}-1)\mu^q_A(x)}\right)^{1/q},\] and \[\left(\dfrac{\nu_B(x)}{n_{2}-(n_{2}-1)\nu_B(x)}\right)^{1/q} =\left(\dfrac{\dfrac{\nu^q_A(x)}{n_{1}-(n_{1}-1)\nu^q_A(x)}}{n_{2}-(n_{2}-1) \dfrac{\nu^q_A(x)}{n_{1}-(n_{1}-1)\nu^q_A(x)}}\right)^{1/q}=\left(\dfrac{\nu^q_A(x)}{n_{1}n_{2}-(n_{1}n_{2}-1)\nu^q_A(x)}\right)^{1/q}.\] Since \[n_{1}(n_{2}._{h}) A=\left\{\left.\left( x, \left(\dfrac{n_{1}n_{2}\mu^q_A(x)}{1+(n_{1}n_{2}-1)\mu^q_A(x)}\right)^{1/q} ,\left(\dfrac{\nu^q_A(x)}{n_{1}n_{2}-(n_{1}n_{2}-1)\nu^q_A(x)}\right)^{1/q}\right) \right| x\in X \right\},\] so we finally get \(n_{2}._{h}(n_{1}._{h} A)=(n_{1}n_{2})._{h} A\).Theorem 13. If \( A, B \in q-ROFS(X)\), then for any positive integer \(n\), we have
Proof. We shall prove (ii), (iv) and (i), (iii) can be proved analogously.
Definition 3. For any q-rung orthopair fuzzy set \(A\), the necessity \((\Box)\) and the possibility \((\Diamond)\) operators are defined as follows; \begin{align*}\Box A&=\left\{\left.\left( x, \mu_A(x),\left(1-\mu^q_A(x)\right)^{1/q}\hspace{0.2cm}\right) \right| x\in X \right\},\\ \Diamond A&=\left\{\left.\left( x, \left(1-\nu^q_A(x)\right)^{1/q},\nu_A(x)\hspace{0.2cm}\right) \right| x\in X \right\}.\end{align*}
Theorem 14. If \(A, B\in q-ROFS(X)\), then
Proof. (i) \begin{align*}\Box\left(A\right.&\left.\boxplus_{h} B\right)\\&= \left\{\left.\left( x, \left(\dfrac{\mu^q_A(x)+\mu^q_B(x)-2\mu^q_A(x)\mu^q_B(x)}{1-\mu^q_A(x)\mu^q_B(x)}\right)^{1/q}, \left(1-\dfrac{\mu^q_A(x)+\mu^q_B(x)-2\mu^q_A(x)\mu^q_B(x)}{1-\mu^q_A(x)\mu^q_B(x)}\right)^{1/q}\hspace{0.2cm}\right) \right| x\in X \right\}\\ &=\left\{\left.\left( x, \left(\dfrac{\mu^q_A(x)+\mu^q_B(x)-2\mu^q_A(x)\mu^q_B(x)}{1-\mu^q_A(x)\mu^q_B(x)}\right)^{1/q}, \left(\dfrac{(1-\mu^q_A(x))(1-\mu^q_B(x))}{1-\mu^q_A(x)\mu^q_B(x)}\right)^{1/q}\hspace{0.2cm}\right) \right| x\in X \right\}\\ &=\Box A\boxplus_{h} \Box B.\end{align*} (ii) \begin{align*} \Diamond (A\boxplus_{h} B)&= \left\{\left.\left( x, \left(1-\dfrac{\nu^q_A(x)\nu^q_B(x)}{\nu^q_A(x)+\nu^q_B(x)-\nu^q_A(x)\nu^q_B(x)}\right)^{1/q}, \left(\dfrac{\nu^q_A(x)\nu^q_B(x)}{\nu^q_A(x)+\nu^q_B(x)-\nu^q_A(x)\nu^q_B(x)}\right)^{1/q}\hspace{0.2cm}\right) \right| x\in X \right\}\\ &=\left\{\left.\left( x, \left(\dfrac{\nu^q_A(x)+\nu^q_B(x)-2\nu^q_A(x)\nu^q_B(x)}{\nu^q_A(x)+\nu^q_B(x)-\nu^q_A(x)\nu^q_B(x)}\right)^{1/q},\left(\dfrac{\nu^q_A(x)\nu^q_B(x)}{\nu^q_A(x)+\nu^q_B(x)-\nu^q_A(x)\nu^q_B(x)}\right)^{1/q}\hspace{0.2cm}\right) \right| x\in X \right\}\\ &=\Diamond A\boxplus_{h} \Diamond B.\end{align*} Hence the theorem is proved.
Theorem 15. If \(A, B\in q-ROFS(X)\), then
Proof. (i) \begin{align*}\Box(A&\boxtimes_{h} B)\\ &= \left\{\left.\left( x, \left(\dfrac{\mu^q_A(x)\mu^q_B(x)}{\mu^q_A(x)+\mu^q_B(x)-\mu^q_A(x)\mu^q_B(x)}\right)^{1/q}, \left(1-\dfrac{\mu^q_A(x)\mu^q_B(x)}{\mu^q_A(x)+\mu^q_B(x)-\mu^q_A(x)\mu^q_B(x)}\right)^{1/q}\right) \right| x\in X \right\}\\ &=\left\{\left.\left( x, \left(\dfrac{\mu^q_A(x)\mu^q_B(x)}{\mu^q_A(x)+\mu^q_B(x)-\mu^q_A(x)\mu^q_B(x)}\right)^{1/q}, \left(\dfrac{\mu^q_A(x)+\mu^q_B(x)-2\mu^q_A(x)\mu^q_B(x)}{\mu^q_A(x)+\mu^q_B(x)-\mu^q_A(x)\mu^q_B(x)}\right)^{1/q}\right) \right| x\in X \right\}\\ &=\Box A\boxtimes_{h} \Box B.\end{align*} (ii) \begin{align*}\Diamond(A&\boxtimes_{h} B)\\ &=\left\{\left.\left( x, \left(1-\dfrac{\nu^q_A(x)+\nu^q_B(x)-2\nu^q_A(x)\nu^q_B(x)}{1-\nu^q_A(x)\nu^q_B(x)}\right)^{1/q}, \left(\dfrac{\nu^q_A(x)+\nu^q_B(x)-2\nu^q_A(x)\nu^q_B(x)}{1-\nu^q_A(x)\nu^q_B(x)}\right)^{1/q}\right) \right| x\in X \right\}\\ &=\left\{\left.\left( x, \left(\dfrac{(1-\nu^q_A(x))(1-\nu^q_B(x))}{1-\nu^q_A(x)\nu^q_B(x)}\right)^{1/q}, \left(\dfrac{\nu^q_A(x)+\nu^q_B(x)-2\nu^q_A(x)\nu^q_B(x)}{1-\nu^q_A(x)\nu^q_B(x)}\right)^{1/q}\right) \right| x\in X \right\}\\ &=\Diamond A\boxtimes_{h} \Diamond B.\end{align*} Hence the theorem is proved.
Theorem 16. If \(A, B\in q-ROFS(X)\), then
Proof. (i) \begin{align*}\Box(A&^C\boxplus_{h} B^C)\\ &= \left\{\left.\left( x, \left(\dfrac{\nu^q_A(x)+\nu^q_B(x)-2\nu^q_A(x)\nu^q_B(x)}{1-\nu^q_A(x)\nu^q_B(x)}\right)^{1/q}, \left(1-\dfrac{\nu^q_A(x)+\nu^q_B(x)-2\nu^q_A(x)\nu^q_B(x)}{1-\nu^q_A(x)\nu^q_B(x)}\right)^{1/q}\right) \right| x\in X \right\}\\ &=\left\{\left.\left( x, \left(\dfrac{\nu^q_A(x)+\nu^q_B(x)-2\nu^q_A(x)\nu^q_B(x)}{1-\nu^q_A(x)\nu^q_B(x)}\right)^{1/q}, \left(\dfrac{(1-\nu^q_A(x))(1-\nu^q_B(x))}{1-\nu^q_A(x)\nu^q_B(x)}\right)^{1/q}\right) \right| x\in X \right\},\\ \left(\Box\left(A^C\right.\right.&\boxplus_{h}\left.\left. B^C\right)\right)^{C}\\ &=\left\{\left.\left( x, \left(\dfrac{(1-\nu^q_A(x))(1-\nu^q_B(x))}{1-\nu^q_A(x)\nu^q_B(x)}\right)^{1/q},\left(\dfrac{\nu^q_A(x)+\nu^q_B(x)-2\nu^q_A(x)\nu^q_B(x)}{1-\nu^q_A(x)\nu^q_B(x)}\right)^{1/q} \right) \right| x\in X \right\}\\ &=\Diamond A\boxtimes_{h} \Diamond B.\end{align*} (ii) \begin{align*}\Box(A^C&\boxtimes_{h} B^C)\\ &= \left\{\left.\left( x, \left(\dfrac{\nu^q_A(x)\nu^q_B(x)}{\nu^q_A(x)+\nu^q_B(x)-\nu^q_A(x)\nu^q_B(x)}\right)^{1/q}, \left(1-\dfrac{\nu^q_A(x)\nu^q_B(x)}{\nu^q_A(x)+\nu^q_B(x)-\nu^q_A(x)\nu^q_B(x)}\right)^{1/q}\right) \right| x\in X \right\}\\ &=\left\{\left.\left( x, \left(\dfrac{\nu^q_A(x)\nu^q_B(x)}{\nu^q_A(x)+\nu^q_B(x)-\nu^q_A(x)\nu^q_B(x)}\right)^{1/q}, \left(\dfrac{\nu^q_A(x)+\nu^q_B(x)-2\nu^q_A(x)\nu^q_B(x)}{\nu^q_A(x)+\nu^q_B(x)-\nu^q_A(x)\nu^q_B(x)}\right)^{1/q}\right) \right| x\in X \right\},\end{align*} \begin{align*} \left(\Box\left(A^C\right.\right.&\left.\left.\boxtimes_{h} B^C\right)\right)^{C}\\ &= \left\{\left.\left( x, \left(\dfrac{\nu^q_A(x)+\nu^q_B(x)-2\nu^q_A(x)\nu^q_B(x)}{\nu^q_A(x)+\nu^q_B(x)-\nu^q_A(x)\nu^q_B(x)}\right)^{1/q}, \left(\dfrac{\nu^q_A(x)\nu^q_B(x)}{\nu^q_A(x)+\nu^q_B(x)-\nu^q_A(x)\nu^q_B(x)}\right)^{1/q}\right) \right| x\in X \right\}\\ &=\Diamond A\boxplus_{h} \Diamond B.\end{align*} Hence the theorem is proved.
Theorem 17. If \(A, B\in q-ROFS(X)\), then
Proof. (i) \begin{align*} \Diamond(A^C&\boxplus_{h} B^C)\\ &=\left\{\left.\left( x, \left(1-\dfrac{\mu^q_A(x)\mu^q_B(x)}{\mu^q_A(x)+\mu^q_B(x)-\mu^q_A(x)\mu^q_B(x)}\right)^{1/q},\left(\dfrac{\mu^q_A(x)\mu^q_B(x)}{\mu^q_A(x)+\mu^q_B(x)-\mu^q_A(x)\mu^q_B(x)}\right)^{1/q} \right) \right| x\in X \right\}\\ &=\left\{\left.\left( x, \left(\dfrac{\mu^q_A(x)+\mu^q_B(x)-2\mu^q_A(x)\mu^q_B(x)}{\mu^q_A(x)+\mu^q_B(x)-\mu^q_A(x)\mu^q_B(x)}\right)^{1/q},\left(\dfrac{\mu^q_A(x)\mu^q_B(x)}{\mu^q_A(x)+\mu^q_B(x)-\mu^q_A(x)\mu^q_B(x)}\right)^{1/q}\right) \right| x\in X \right\}\\ &= \left\{\left.\left( x, \left(\dfrac{\mu^q_A(x)\mu^q_B(x)}{\mu^q_A(x)+\mu^q_B(x)-\mu^q_A(x)\mu^q_B(x)}\right)^{1/q},\left(\dfrac{\mu^q_A(x)+\mu^q_B(x)-2\mu^q_A(x)\mu^q_B(x)}{\mu^q_A(x)+\mu^q_B(x)-\mu^q_A(x)\mu^q_B(x)}\right)^{1/q}\right) \right| x\in X \right\}\\ &=\Box A\boxtimes_{h} \Box B. \end{align*} (ii) \begin{align*} \Diamond\left(A^C\right.&\left.\boxtimes_{h} B^C\right)\\ &= \left\{\left.\left( x, \left(1-\dfrac{\mu^q_A(x)+\mu^q_B(x)-2\mu^q_A(x)\mu^q_B(x)}{1-\mu^q_A(x)\mu^q_B(x)}\right)^{1/q}, \left(\dfrac{\mu^q_A(x)+\mu^q_B(x)-2\mu^q_A(x)\mu^q_B(x)}{1-\mu^q_A(x)\mu^q_B(x)}\right)^{1/q}\right) \right| x\in X \right\}\\ &=\left\{\left.\left( x, \left(\dfrac{(1-\mu^q_A(x))(1-\mu^q_B(x))}{1-\mu^q_A(x)\mu^q_B(x)}\right)^{1/q},\left(\dfrac{\mu^q_A(x)+\mu^q_B(x)-2\mu^q_A(x)\mu^q_B(x)}{1-\mu^q_A(x)\mu^q_B(x)}\right)^{1/q} \right) \right| x\in X \right\},\end{align*}\begin{align*} \Diamond(A^C\boxtimes_{h} B^C)^{C} &= \left\{\left.\left( x, \left(\dfrac{\mu^q_A(x)+\mu^q_B(x)-2\mu^q_A(x)\mu^q_B(x)}{1-\mu^q_A(x)\mu^q_B(x)}\right)^{1/q}, \left(\dfrac{(1-\mu^q_A(x))(1-\mu^q_B(x))}{1-\mu^q_A(x)\mu^q_B(x)}\right)^{1/q}\right) \right| x\in X \right\}\\ &=\Box A\boxplus_{h} \Box B.\end{align*} Hence the theorem is proved.
Theorem 18. If \(A\in q-ROFS(X)\), then for any positive integer \(n\), we have
Proof. (i) \begin{align*} \Box(n._{h}A)&=\left\{\left.\left( x, \left(\dfrac{n\mu^q_A(x)}{1+(n-1)\mu^q_A(x)}\right)^{1/q},\left(1-\left(\left(\dfrac{n\mu^q_A(x)}{1+(n-1)\mu^q_A(x)}\right)^{1/q}\hspace{0.2cm}\right)^q\right)^{1/q}\right) \right| x\in X \right\}\end{align*} \begin{align*} &=\left\{\left.\left( x, \left(\dfrac{n\mu^q_A(x)}{1+(n-1)\mu^q_A(x)}\right)^{1/q}, \left(1-\dfrac{n\mu^q_A(x)}{1+(n-1)\mu^q_A(x)}\right)^{1/q}\right) \right| x\in X \right\}\\ &=\left\{\left.\left( x, \left(\dfrac{n\mu^q_A(x)}{1+(n-1)\mu^q_A(x)}\right)^{1/q}, \left(\dfrac{1+(n-1)\mu^q_A(x)-n\mu^q_A(x)}{1+(n-1)\mu^q_A(x)}\right)^{1/q}\right) \right| x\in X \right\}\\ &=\left\{\left.\left( x, \left(\dfrac{n\mu^q_A(x)}{1+(n-1)\mu^q_A(x)}\right)^{1/q}, \left(\dfrac{1+n\mu^q_A(x)-\mu^q_A(x)-n\mu^q_A(x)}{1+(n-1)\mu^q_A(x)}\right)^{1/q}\right) \right| x\in X \right\}\\ &=\left\{\left.\left( x, \left(\dfrac{n\mu^q_A(x)}{1+(n-1)\mu^q_A(x)}\right)^{1/q},\left(\dfrac{1-\mu^q_A(x)}{1+(n-1)\mu^q_A(x)}\right)^{1/q}\right) \right| x\in X \right\},\\ n._{h}(\Box A)&=\left\{\left.\left( x, \left(\dfrac{n\mu^q_A(x)}{1+(n-1)\mu^q_A(x)}\right)^{1/q},\left(\dfrac{\left(\left(1-\mu^q_A(x)\right)^{1/q}\right)^q}{n-(n-1)\left(\left(1-\mu^q_A(x)\right)^{1/q}\right)^q}\right)^{1/q}\right) \right| x\in X \right\}\\ &=\left\{\left.\left( x, \left(\dfrac{n\mu^q_A(x)}{1+(n-1)\mu^q_A(x)}\right)^{1/q},\left(\dfrac{1-\mu^q_A(x)}{n-(n-1)\left(1-\mu^q_A(x)\right)}\right)^{1/q}\right) \right| x\in X \right\}\\ &=\left\{\left.\left( x, \left(\dfrac{n\mu^q_A(x)}{1+(n-1)\mu^q_A(x)}\right)^{1/q},\left(\dfrac{1-\mu^q_A(x)}{1+(n-1)\mu^q_A(x)}\right)^{1/q}\right) \right| x\in X \right\}.\end{align*} Hence, \(\Box(n._{h}A)=n._{h}(\Box A)\).
(ii)
\begin{align*} \Diamond(n._{h}A)&=\left\{\left.\left( x, \left(1-\left(\left(\dfrac{\nu^q_A(x)}{n-(n-1)\nu^q_A(x)}\right)^{1/q}\right)^q\right)^{1/q}, \left(\dfrac{\nu^q_A(x)}{n-(n-1)\nu^q_A(x)}\right)^{1/q}\right) \right| x\in X \right\}\\ &=\left\{\left.\left( x, \left(1-\dfrac{\nu^q_A(x)}{n-(n-1)\nu^q_A(x)}\right)^{1/q}, \left(\dfrac{\nu^q_A(x)}{n-(n-1)\nu^q_A(x)}\right)^{1/q}\right) \right| x\in X \right\}\\ &=\left\{\left.\left( x, \left(\dfrac{n-(n-1)\nu^q_A(x)-\nu^q_A(x)}{n-(n-1)\nu^q_A(x)}\right)^{1/q}, \left(\dfrac{\nu^q_A(x)}{n-(n-1)\nu^q_A(x)}\right)^{1/q}\right) \right| x\in X \right\}\\ &=\left\{\left.\left( x, \left(\dfrac{n-n\nu^q_A(x)+\nu^q_A(x)-\nu^q_A(x)}{n-(n-1)\nu^q_A(x)}\right)^{1/q}, \left(\dfrac{\nu^q_A(x)}{n-(n-1)\nu^q_A(x)}\right)^{1/q}\right) \right| x\in X \right\}\\ &=\left\{\left.\left( x, \left(\dfrac{n(1-\nu^q_A(x))}{n-(n-1)\nu^q_A(x)}\right)^{1/q}, \left(\dfrac{\nu^q_A(x)}{n-(n-1)\nu^q_A(x)}\right)^{1/q}\right) \right| x\in X \right\},\\ n._{h}(\Diamond A)&=\left\{\left.\left( x, \left(\dfrac{n(\left(1-\nu^q_A(x)\right)^{1/q})^q}{1+(n-1)(\left(1-\nu^q_A(x)\right)^{1/q})^q}\right)^{1/q}, \left(\dfrac{\nu^q_A(x)}{n-(n-1)\nu^q_A(x)}\right)^{1/q}\right) \right| x\in X \right\}\\ &=\left\{\left.\left( x, \left(\dfrac{n(1-\nu^q_A(x))}{1+(n-1)(1-\nu^q_A(x))}\right)^{1/q}, \left(\dfrac{\nu^q_A(x)}{n-(n-1)\nu^q_A(x)}\right)^{1/q}\right) \right| x\in X \right\}\\ &=\left\{\left.\left( x, \left(\dfrac{n(1-\nu^q_A(x))}{n-(n-1)\nu^q_A(x)}\right)^{1/q}, \left(\dfrac{\nu^q_A(x)}{n-(n-1)\nu^q_A(x)}\right)^{1/q}\right) \right| x\in X \right\}.\end{align*} Hence, \(\Diamond(n._{h}A)=n._{h}(\Diamond A)\).(iii)
\begin{align*} \Box {A^{\wedge_{h}}}^n&=\left\{\left.\left( x, \left(\dfrac{\mu^q_A(x)}{n-(n-1)\mu^q_A(x)}\right)^{1/q},\left(1-\left(\left(\dfrac{\mu^q_A(x)}{n-(n-1)\mu^q_A(x)}\right)^{1/q}\right)^q\right)^{1/q}\right) \right| x\in X \right\}\\ &=\left\{\left.\left( x, \left(\dfrac{\mu^q_A(x)}{n-(n-1)\mu^q_A(x)}\right)^{1/q},\left(1-\dfrac{\mu^q_A(x)}{n-(n-1)\mu^q_A(x)}\right)^{1/q}\right) \right| x\in X \right\}\\ &=\left\{\left.\left( x, \left(\dfrac{\mu^q_A(x)}{n-(n-1)\mu^q_A(x)}\right)^{1/q},\left(\dfrac{n-(n-1)\mu^q_A(x)-\mu^q_A(x)}{n-(n-1)\mu^q_A(x)}\right)^{1/q}\right) \right| x\in X \right\}\\ &=\left\{\left.\left( x, \left(\dfrac{\mu^q_A(x)}{n-(n-1)\mu^q_A(x)}\right)^{1/q},\left(\dfrac{n-n\mu^q_A(x)+\mu^q_A(x)-\mu^q_A(x)}{n-(n-1)\mu^q_A(x)}\right)^{1/q}\right) \right| x\in X \right\}\\ &=\left\{\left.\left( x, \left(\dfrac{\mu^q_A(x)}{n-(n-1)\mu^q_A(x)}\right)^{1/q},\left(\dfrac{n(1-\mu^q_A(x))}{n-(n-1)\mu^q_A(x)}\right)^{1/q}\right) \right| x\in X \right\},\\ {(\Box A)^{\wedge_{h}}}^n&=\left\{\left.\left( x, \left(\dfrac{\mu^q_A(x)}{n-(n-1)\mu^q_A(x)}\right)^{1/q},\left(\dfrac{n(\left(1-\mu^q_A(x)\right)^{1/q})^q}{n-(n-1)(\left(1-\mu^q_A(x)\right)^{1/q})^q}\right)^{1/q}\right) \right| x\in X \right\}\\ &=\left\{\left.\left( x, \left(\dfrac{\mu^q_A(x)}{n-(n-1)\mu^q_A(x)}\right)^{1/q},\left(\dfrac{n(1-\mu^q_A(x))}{n-(n-1)(1-\mu^q_A(x))}\right)^{1/q}\right) \right| x\in X \right\}\\ &=\left\{\left.\left( x, \left(\dfrac{\mu^q_A(x)}{n-(n-1)\mu^q_A(x)}\right)^{1/q},\left(\dfrac{n(1-\mu^q_A(x))}{n-(n-1)\mu^q_A(x)}\right)^{1/q}\right) \right| x\in X \right\}.\end{align*} Hence, \(\Box {A^{\wedge_{h}}}^n={(\Box A)^{\wedge_{h}}}^n\).(iv)
\begin{align*} \Diamond {A^{\wedge_{h}}}^n&=\left\{\left.\left( x, \left(1-\left(\left(\dfrac{n\nu^q_A(x)}{1+(n-1)\nu^q_A(x)}\right)^{1/q}\right)^q\right)^{1/q}, \left(\dfrac{n\nu^q_A(x)}{1+(n-1)\nu^q_A(x)}\right)^{1/q} \right) \right| x\in X \right\}\\ &=\left\{\left.\left( x, \left(1-\dfrac{n\nu^q_A(x)}{1+(n-1)\nu^q_A(x)}\right)^{1/q}, \left(\dfrac{n\nu^q_A(x)}{1+(n-1)\nu^q_A(x)}\right)^{1/q} \right) \right| x\in X \right\}\\ &=\left\{\left.\left( x, \left(\dfrac{1+(n-1)\nu^q_A(x)-n\nu^q_A(x)}{1+(n-1)\nu^q_A(x)}\right)^{1/q}, \left(\dfrac{n\nu^q_A(x)}{1+(n-1)\nu^q_A(x)}\right)^{1/q} \right) \right| x\in X \right\}\\ &=\left\{\left.\left( x, \left(\dfrac{1-\nu^q_A(x)}{1+(n-1)\nu^q_A(x)}\right)^{1/q}, \left(\dfrac{n\nu^q_A(x)}{1+(n-1)\nu^q_A(x)}\right)^{1/q} \right) \right| x\in X \right\},\\ {(\Diamond A)^{\wedge_{h}}}^n&=\left\{\left.\left( x, \left(\dfrac{\left(\left(1-\nu^q_A(x)\right)^{1/q}\right)^q}{1+(n-1)\left(\left(1-\nu^q_A(x)\right)^{1/q}\right)^q}\right)^{1/q}, \left(\dfrac{n\nu^q_A(x)}{1+(n-1)\nu^q_A(x)}\right)^{1/q} \right) \right| x\in X \right\}\\ &=\left\{\left.\left( x, \left(\dfrac{\left(1-\nu^q_A(x)\right)}{1+(n-1)\left(1-\nu^q_A(x)\right)}\right)^{1/q}, \left(\dfrac{n\nu^q_A(x)}{1+(n-1)\nu^q_A(x)}\right)^{1/q} \right) \right| x\in X \right\}\\ &=\left\{\left( x, \left(\dfrac{1-\nu^q_A(x)}{1+(n-1)\nu^q_A(x)}\right)^{1/q}, \left(\dfrac{n\nu^q_A(x)}{1+(n-1)\nu^q_A(x)}\right)^{1/q} \hspace{0.2cm} \right) | x\in X \right\}.\end{align*} Hence, \(\Diamond {A^{\wedge_{h}}}^n={(\Diamond A)^{\wedge_{h}}}^n\).