## 3 The Notion of the Proportional Dual Hesitant Fuzzy Set
To overcome this issue of ordering and adding, a method for PDHFEs is proposed in which the following two simple steps are performed:  
Step (1): Ordering;  
Arrange the elements in \(pd_{A}=(ph_{A},qg_{A})\) and \(pd_{B}=(ph_{B},qg_{B})\) in decreasing order according to the product values of the assessment and their associated proportions. Step (2): Adding;  
Extended the shorter one by adding the element \((0,0)\) several times until their lengths becomes the same. Definition 12: Let \(A\) and \(B\) be any two PDHFSs on a fixed set \(X\). Then, they are said to be of equal length if their corresponding PDHFEs are of equal length. **Note**: One can generalize the above definition to \(n\) PDHFSs. The following operations are defined for PDHFSs in order to use them in practical decision-making problems:  
Definition 13: Let \(X\) be a universe of discourse, \(pd_{1}\) and \(pd_{2}\) be two PDHFEs of equal length; then, the following operations are defined:  
1. \(\oplus-union\); \(pd_{1}\oplus pd_{2}=\left\langle ph_{1}\oplus ph_{2},qg_{1}\oplus qg_{2}\right\rangle\)
2. \(\oplus-intersection\); \(pd_{1}\otimes pd_{2}=\left\langle ph_{1}\otimes ph_{2},qg_{1}\oplus qg_{2}\right\rangle\)
3. \(\alpha pd_{1}=\left\langle \begin{array}{c}\left\{\left(\gamma_{1}^{\sigma(j) }\gamma_{2}^{\sigma(j)},\tau_{1}^{\sigma(j)}\tau_{2}^{\sigma(j)}\right)\mid j =1,2,0...,l(ph_{1})\right\},\\ \left\{\left(\eta_{1}^{\sigma(t)}+\eta_{2}^{\sigma(t)}-\eta_{1}^{\sigma(t)} \eta_{2}^{\sigma(t)},\tau_{1}^{\sigma(t)}\tau_{2}^{\sigma(t)}\right)\mid t=1, 2,0...,l(qg_{1})\right\},\end{array} \right\rangle;\)
4. \(\left\langle pd_{1}\right\rangle^{\alpha}=\left\langle \begin{array}{c}\left\{ \left(\left(\gamma_{1}^{\sigma(j)}\right)^{\alpha},\tau_{1}^{\sigma(j)}\right) \mid j=1,2,0...,l(ph_{1})\right\},\\ \left\{\left(1-\left(1-\eta_{1}^{\sigma(t)}\right)^{\alpha},\tau_{1}^{\sigma (t)}\right)\mid t=1,2,0...,l(qg_{1})\right\}\end{array} \right\rangle;\)  
where \(\alpha\) is a positive integral, \(\left(\gamma_{1}^{\sigma(j)},\tau_{1}^{\sigma(j)}\right)\) and \(\left(\eta_{1}^{\sigma(t)},\tau_{1}^{\sigma(t)}\right)\) are chosen such that \(\gamma_{1}^{\sigma(j)}\cdot\tau_{1}^{\sigma(j)}\) and \(\eta_{1}^{\sigma(t)}\). \(\tau_{1}^{\
u(t)}\) are the jth and \(t\)th smallest product values in PDHFEs \(pd_{1}\) and \(pd_{2}\), respectively. Remark 3: _Here, we have assumed that all PDHFEs are of equal length. However, in most cases, they are unequal, which produces problems in basic operations and distance measure. Then, in that situation, their lengths are equalized according to the ordering and adding method as discussed above._  
Theorem 1: _Let \(A,B\) and \(C\) be any three PDHFSs on \(X\). Suppose \(\alpha\geq 0\); then:_  
1. \(A\oplus B=B\oplus A\)_,_ \(A\otimes B=B\otimes A\)_;_
2. \(\left(A\oplus B\right)\oplus C=A\oplus\left(B\oplus C\right),\left(A\otimes B \right)\otimes C=A\otimes\left(B\otimes C\right);\)__
3. \(\alpha\left(A\oplus B\right)=\alpha A\oplus\alpha B\)_,_ \(\left(A\otimes B\right)^{\alpha}=A^{\alpha}\otimes B^{\alpha}\)_;_
4. \(\left(A\oplus B\right)^{c}=A^{c}\otimes B^{c},\left(A\otimes B\right)^{c}=A^{ c}\oplus B^{c}\)_._  
**Proof. **\((1):\)  
\[pd_{A}\oplus pd_{B}=\left\langle\left\{\left(\gamma_{A}^{\sigma(j)}+ \gamma_{B}^{\sigma(j)}-\gamma_{A}^{\sigma(j)}\gamma_{B}^{\sigma(j)},\tau_{A}^{ \sigma(j)}\tau_{B}^{\sigma(j)}\right)\mid j=1,2,0...,l(ph_{1})\right\},\right.\] \[\left.\left\{\left(\eta_{A}^{\sigma(t)}\eta_{B}^{\sigma(t)},\tau_ {A}^{\sigma(t)}\tau_{B}^{\sigma(t)}\right)\mid t=1,2,0...,l(qg_{1})\right\} \right\rangle\] \[=\left\langle\left\{\left(\tau_{B}^{\sigma(j)}+\gamma_{A}^{ \sigma(j)}-\gamma_{B}^{\sigma(j)}\tau_{A}^{\sigma(j)},\tau_{B}^{\sigma(j)} \tau_{A}^{\sigma(j)}\right)\mid j=1,2,0...,l(ph_{1})\right\},\right.\] \[\left.\left\{\left(\eta_{B}^{\sigma(t)}\eta_{A}^{\sigma(t)},\tau_ {B}^{\sigma(t)}\tau_{A}^{\sigma(t)}\right)\mid t=1,2,0...,l(qg_{1})\right\}\right\rangle\] \[=pd_{B}\oplus pd_{A}.\]  
Then, \(A\oplus B=B\oplus A\). Similarly, one can show that \(A\otimes B=B\otimes A\).