## 3 Method Description
### Theorems on the Election of a Vector Criterion
** (_On the introduction of a partial criterion) [2,4,5]__If \(\delta_{n}\
eq 0\) exists, then the partial criterion \(F_{N+1}\) can be entered into the base criterion \(F\)_ then and only then, if occurs:  
\[\alpha_{N+1}>Q(F)+\frac{1}{3}\sum_{n=1}^{N}\delta_{n} \tag{15}\]  
**Proof of Theorem 3. ** The extended baseline criterion \(F^{+}\) should increase the value of the functional \(Q\). Therefore,  
\[Q\big{(}F^{+}\big{)}-Q(F)=\frac{1}{N+1}\sum_{n=1}^{N+1}\beta_{n}-\frac{1}{N} \sum_{n=1}^{N}\alpha_{n}\]  
where \(\beta_{n}=\frac{1}{3}[(\alpha_{n,1}-\delta_{n})+\alpha_{n,2}+\alpha_{n,3}]\)\(for\)\(n=\overline{1,\ N}\), \(\beta_{N+1}=\frac{1}{3}(\alpha_{N+1,1}+\alpha_{N+1,2}+\alpha_{N+1,3})=\alpha_{ N+1}\). After substituting the above dependencies and necessary simplifications, we will receive:  
\[Q\big{(}F^{+}\big{)}-Q(F)=\alpha_{N+1}-Q(F)-\frac{1}{3}\sum_{n=1}^{N}\delta_{n}\]  
Hence, \(\alpha_{N+1}>Q(F)+\frac{1}{3}\sum_{n=1}^{N}\delta_{n}\), which ends the proof. The criterion \(F^{+}\) should also fulfill the condition of completeness. This condition is fulfilled if, for each pair of \(x\), \(z\in X\) such that \(F(x)=F(z)\) and the fulfilled condition \(F_{N+1}(x)=F_{N+1}(z)\), the user does not choose \(x\) or \(z\). If \(F_{N+1}(x)>F_{N+1}(z)\) and the user selects \(x\) before \(z\) while maximizing \(F(\cdot)\), then \(F_{N+1}(\cdot)\) shows the user's preferences. The introduction of the \(F_{N+1}\) criterion may result in failure to comply with the consistency condition. If  
\[F_{N+1}\big{(}x^{+}\big{)}\geq F_{N+1}(z^{-}) \tag{16}\]  
then the condition of consistency will be maintained. Taking into account an additional partial criterion will introduce redundancy into the criterion \(F^{+}\), provided that it was not the \(F\) criterion. This redundancy can be avoided only after eliminating one of the criteria \(F_{n}\)_for \(n=\overline{1,\ N}\)_. If no such criterion exists, then no additional criterion can be entered. If \(F\) does not meet the nonredundancy condition, the introduction of redundancy does not matter. If \(\alpha_{N+1}=0\) or \(Q(F)=1\) under the conditions of completeness, consistency, and nonredundancy, then the criterion \(F_{N+1}\) cannot be considered. Let \(\delta=(\delta_{1},\ \delta_{2},\ \ldots,\ \delta_{n},\ \ldots,\ \delta_{N})\) denote the vector of reduction of subjective confidence levels as a result of removing the criterion \(F_{k}\). Therefore, the new, subjective confidence level is \(\alpha_{n,1}+\delta_{n}\). **Theorem 4. **_(On the removal of the partial criterion) [2, 4, 5]_  
_If exists \(\delta_{n}\
eq 0\), then the partial criterion \(F_{k}\) can be removed from criterion \(F\) if and only if_  
\[\alpha_{k}<Q(F)-\frac{1}{3}\sum_{ \begin{array}{c}n=1\\ n\
eq k\end{array} }^{N}\delta_{n} \tag{17}\]  
**Proof of Theorem 4. ** The reduced base criterion \(F^{-}\) should increase the value of the functional \(Q\). Therefore,  
\[Q(F^{-})-Q(F)=\frac{1}{N-1}\sum_{ \begin{array}{c}n=1\\ n\
eq k\end{array} }^{N-1}\beta_{n}-\frac{1}{N}\sum_{n=1}^{N}\alpha_{n}\]where \(\beta_{n}=\frac{1}{3}[(\alpha_{n,1}-\delta_{n})+\alpha_{n,2}+\alpha_{n,3}]\)\(for\)\(n=\overline{1,\)\(N}\), \(\beta_{k}=\frac{1}{3}\Big{(}\alpha_{k,1}+\alpha_{k,2}+\alpha_{k,3}\Big{)}=\alpha_{k}\). After substituting the above dependencies and necessary simplifications, we will receive:  
\[Q\Big{(}F^{+}\Big{)}-Q(F)=-\alpha_{k}+Q(F)-\frac{1}{3}\sum_{n=1}^{N}\delta_{n}>0\]  
Hence, \(\alpha_{k}<Q(F)-\frac{1}{3}\sum_{n=1}^{N}\delta_{n}\), which ends the **proof**. \(\Box\)  
Based on the presented methodology, it is possible to formulate a scheme for formulating a vector criterion when we have a set of scalar criteria that affect the quality of solutions obtained, and thus the effects of decisions.