Handling imprecise evaluations in multiple criteria decision aiding and robust ordinal regression by n-point intervals

Abstract

We consider imprecise evaluation of alternatives in multiple criteria ranking problems. The imprecise evaluations are represented by n-point intervals which are defined by the largest interval of possible evaluations and by its subintervals sequentially nested one in another. This sequence of subintervals is associated with an increasing sequence of plausibility, such that the plausibility of a subinterval is greater than the plausibility of the subinterval containing it. We explain the intuition that stands behind this proposal, and we show the advantage of n-point intervals compared to other methods dealing with imprecise evaluations. Although n-point intervals can be applied in any multiple criteria decision aiding (MCDA) method, in this paper, we focus on their application in robust ordinal regression which, unlike other MCDA methods, takes into account all compatible instances of an adopted preference model, which reproduce an indirect preference information provided by the decision maker. An illustrative example shows how the method can be applied in practice.

Appendix

Proof of Proposition 4.1

• 1. Let $a\in A,$ and $i,k\in \left\{ 1,\ldots ,n\right\}$ such that $i\ge k$; this implies that $g_{j}^{i}(a)\ge g_{j}^{k}(a),\; \forall j=1,\ldots ,m,$ and thus $a\varDelta ^{(i,k)}a$. Therefore $\varDelta ^{(i,k)}$ is reflexive. $\square$

• 2. Let us suppose that $a,b,c\in A$ such that $a\varDelta ^{(i,k)}b$, $b\varDelta ^{(i,k)}c$ and $i\le k$ with $i,k\in \left\{ 1,\ldots ,n\right\}$;

  $$\begin{aligned} \left\{ \begin{array}{lll} a\varDelta ^{(i,k)}b &{} \Leftrightarrow &{} g_{j}^{i}(a)\ge g_{j}^{k}(b), \forall j\in J,\\ \\ b\varDelta ^{(i,k)}c &{} \Leftrightarrow &{} g_{j}^{i}(b)\ge g_{j}^{k}(c), \forall j\in J.\\ \end{array} \right. \end{aligned}$$

  Being $i\le k$ we get that $g_{j}^{k}(b)\ge g_{j}^{i}(b),$ and, consequently:

  $$\begin{aligned}&g_{j}^{i}(a)\ge g_{j}^{k}(b)\ge g_{j}^{i}(b)\ge g_{j}^{k}(c), \forall j=1,\ldots ,m \Rightarrow g_{j}^{i}(a)\ge g_{j}^{k}(c),\\&\quad \forall j\in J \Leftrightarrow a\varDelta ^{(i,k)}c. \end{aligned}$$

  Thus $\varDelta ^{(i,k)}$ is transitive.$\square$

• 3. It follows from points 1 and 2 of this Proposition since a partial preorder is a reflexive and transitive binary relation. $\square$

• 4. Let be $a,b\in A$ and $i,k,r,s\in \left\{ 1,\ldots ,n\right\}$ such that $a\varDelta ^{(i,k)}b$, $r\ge i$ and $k\ge s$. Then

  $$\begin{aligned} \left\{ \begin{array}{lll} a\varDelta ^{(i,k)}b &{} \Leftrightarrow &{} g_{j}^{i}(a)\ge g_{j}^{k}(b),\quad \forall j\in J,\\ \\ r\ge i &{} \Leftrightarrow &{} g_{j}^{r}(a)\ge g_{j}^{i}(a),\quad \forall j\in J, \\ \\ k\ge s &{} \Leftrightarrow &{} g_{j}^{k}(b)\ge g_{j}^{s}(b),\quad \forall j\in J. \end{array} \right. \end{aligned}$$

  From this it follows that:

  $$\begin{aligned}&g_{j}^{r}(a)\ge g_{j}^{i}(a)\ge g_{j}^{k}(b)\ge g_{j}^{s}(b), \;\forall j\in J\Rightarrow g_{j}^{r}(a) \ge g_{j}^{s}(b),\\&\;\quad \forall j\in J\Leftrightarrow a{\varDelta }^{(r,s)}b.\;\; \end{aligned}$$

  $\square$

• 5. Let $a,b,c\in A$, and $i,k,i_1,k_1\in \left\{ 1,\ldots ,n\right\}$ such that $a\varDelta ^{(i,k)}b$, $b\varDelta ^{(i_1,k_1)}c$ and $k\ge i_1$. Then, we have:

  $$\begin{aligned} \left\{ \begin{array}{lll} a\varDelta ^{(i,k)}b &{} \Leftrightarrow &{} g_{j}^{i}(a)\ge g_{j}^{k}(b),\quad \forall j\in J,\\ \\ b\varDelta ^{(i_1,k_1)}c &{} \Leftrightarrow &{} g_{j}^{i_1}(b)\ge g_{j}^{k_1}(c),\quad \forall j\in J,\\ \\ k\ge i_1 &{} \Leftrightarrow &{} g_{j}^{k}(b)\ge g_{j}^{i_1}(b), \quad \forall j\in J. \end{array} \right. \end{aligned}$$

  From this it follows that:

  $$\begin{aligned} g_{j}^{i}(a)\ge g_{j}^{k}(b)\ge g_{j}^{i_1}(b)\ge g_{j}^{k_1}(c),\; \forall j\in J\Rightarrow g_{j}^{i}(a)\ge g_{j}^{k_1}(c),\;\forall j\in J \Leftrightarrow a\varDelta ^{(i,k_1)}c. \end{aligned}$$

  For point 4 of this Proposition, if $r,s\in \left\{ 1,\ldots ,n\right\}$ such that $r\ge i$ and $s\le k_1$, then $a\varDelta ^{(r,s)}c.$ $\square$

• 6. We have said that $\varDelta =\displaystyle \cap _{i=1}^{n}\varDelta ^{(i,i)}$; since $\varDelta ^{(i,i)}$ is a partial preorder for point 3 of this Proposition, and since the intersection of partial preorders is a partial preoder, $\varDelta$ is a partial preorder. $\square$

• 7. Let $a,b,c\in A$, and $i,k\in \left\{ 1,\ldots ,n\right\}$, such that $a\varDelta ^{(i,k)}b$, $b\varDelta c$, $s\ge i$ and $k\ge t$. Then, we have:

  $$\begin{aligned} \left\{ \begin{array}{lll} a\varDelta ^{(i,k)}b &{} \Leftrightarrow &{} g_{j}^{i}(a)\ge g_{j}^{k}(b),\quad \forall j\in J,\\ \\ b\varDelta c &{} \Leftrightarrow &{} g_{j}^{r}(b)\ge g_{j}^{r}(c), \quad \forall j\in J,\quad \forall r=1,\ldots ,n,\\ \\ s\ge i &{} \Leftrightarrow &{} g_{j}^{s}(a)\ge g_{j}^{i}(a), \quad \forall j\in J,\\ \\ k\ge t &{} \Leftrightarrow &{} g_{j}^{k}(c)\ge g_{j}^{t}(c),\quad \forall j\in J.\\ \end{array} \right. \end{aligned}$$

  From this it follows that:

  $$\begin{aligned}&g_{j}^{s}(a)\ge g_{j}^{i}(a)\ge g_{j}^{k}(b)\ge g_{j}^{k}(c)\ge g_{j}^{t}(c),\; \forall j\in J\Rightarrow g_{j}^{s}(a) \ge g_{j}^{t}(c),\\&\quad \;\forall j\in J \Leftrightarrow a\varDelta ^{(s,t)}c. \; \end{aligned}$$

  $\square$

• 8. Let $a,b,c\in A$, and $i,k\in \left\{ 1,\ldots ,n\right\}$, such that $a\varDelta b$, $b\varDelta ^{(i,k)} c$, $s\ge i$ and $k\ge t$. Then, we have:

  $$\begin{aligned} \left\{ \begin{array}{lll} a\varDelta b &{} \Leftrightarrow &{} g_{j}^{r}(a)\ge g_{j}^{r}(b),\quad \forall j\in J,\quad \forall r=1,\ldots ,n,\\ \\ b\varDelta ^{(i,k)}c &{} \Leftrightarrow &{} g_{j}^{i}(b)\ge g_{j}^{k}(c), \quad \forall j\in J,\\ \\ s\ge i &{} \Leftrightarrow &{} g_{j}^{s}(a)\ge g_{j}^{i}(a),\quad \forall j\in J,\\ \\ k\ge t &{} \Leftrightarrow &{} g_{j}^{k}(c)\ge g_{j}^{t}(c),\quad \forall j\in J,\\ \end{array} \right. \end{aligned}$$

  From this it follows that:

  $$\begin{aligned}&g_{j}^{s}(a)\ge g_{j}^{i}(a)\ge g_{j}^{i}(b)\ge g_{j}^{k}(c)\ge g_{j}^{t}(c),\;\quad \forall j\in J\Rightarrow g_{j}^{s}(a)\ge g_{j}^{t}(c),\\&\quad \;\forall j\in J \Leftrightarrow a\varDelta ^{(s,t)}c. \; \end{aligned}$$

  $\square$

Proof of Proposition 4.2

• 1. For all $i\in \left\{ 1,\ldots ,n\right\}$, from point 4 of Proposition 4.1, we have $\varDelta ^{(1,n)}\subseteq \varDelta ^{(i,i)}$ because $i\ge 1$ and $i\le n$; from this follows that $\varDelta ^{(1,n)}\;\subseteq \;\cap _{i=1}^{n}\varDelta ^{(i,i)}=\varDelta$, and this proves the first part of the Proposition.$\square$

  For the same reason, $\forall i\in \left\{ 1,\ldots ,n\right\}$ we have $\varDelta ^{(i,i)}\;\subseteq \;\varDelta ^{(n,1)}$ because $n\ge i$ and $1\le i$. From this it follows that $\varDelta =\cap _{i=1}^{n}\varDelta ^{(i,i)}\;\subseteq \;\varDelta ^{(n,1)}$, and this proves the second part of the Proposition.$\square$

• 2. For all $i,k\in \left\{ 1,\ldots ,n\right\}$, from point 4 of Proposition 4.1, since $1\le i$ and $k\le n$, we have $\varDelta ^{(1,n)}\subseteq \varDelta ^{(i,k)}$ and $\varDelta ^{(i,k)}\subseteq \varDelta ^{(n,1)}.$ In this way we obtain the thesis.$\square$

Proof of Proposition 5.1

• 1. Let $a\in A$ and $i,k\in \left\{ 1,\ldots ,n\right\}$. From the definition, fictitious alternatives $a^{(i)}$ and $a^{(k)}$ are such that $\forall j\in J$, and $\forall r\in \left\{ 1,\ldots ,n\right\} ,$ $g_{j}^{r}\left( a^{(i)}\right) =g_{j}^{i}(a)$, and $g_{j}^{r}\left( a^{(k)}\right) \!=\!g_{j}^{k}(a)$. Since $i\!\ge \!k$, $\forall r\!\in \!\left\{ 1,\ldots ,n\right\}$ and $\forall j\in J$, $g_{j}^{r}(a^{(i)})\ge g_{j}^{r}(a^{(k)})$, and using the monotonicity of marginal value functions $u_{j,r}(\cdot )$, we obtain $\forall r\in \left\{ 1,\ldots ,n\right\}$, $\forall j\in J$ $u_{j}^{r}(g_{j}^{r}(a^{(i)}))\ge u_{j}^{r}(g_{j}^{r}(a^{(k)}))$; adding up with respect to j and r we obtain the thesis.$\square$

• 2. We have seen that:

  $$\begin{aligned} \displaystyle U(a^{(1)})= & {} \sum _{j=1}^{m}\left[ \sum _{i=1}^{n}u_{j}^{i}(g_{j}^{1}(a))\right] ,\;\;\;U(a)=\sum _{j=1}^{m}\left[ \sum _{i=1}^{n}u_{j}^{i}(g_{j}^{i}(a))\right] \\ \displaystyle U(a^{(n)})= & {} \sum _{j=1}^{m}\left[ \sum _{i=1}^{n}u_{j}^{i}(g_{j}^{n}(a))\right] . \end{aligned}$$

  $\forall j\in J,$ and $\forall i\in \left\{ 1,\ldots ,n\right\}$, since $g_{j}^{1}(a)\le g_{j}^{i}(a)\le g_{j}^{n}(a)$ and by monotonicity of marginal value functions $u_{j}^{i}(\cdot )$, we obtain:

  $$\begin{aligned} u_{j}^{i}\left( g_{j}^{1}(a)\right) \le u_{j}^{i}\left( g_{j}^{i}(a)\right) \le u_{j}^{i}\left( g_{j}^{n}(a)\right) \end{aligned}$$

  and therefore adding up with respect to j and i we obtain the thesis, that is

  $$\begin{aligned} U\left( a^{(1)}\right) \le U(a)\le U\left( a^{(n)}\right) . \end{aligned}$$

  $\square$

Proof of Proposition 6.1

• 1. Let $a,b\in A$, such that $a\varDelta b$. This implies that $g_{j}^{i}(a)\ge g_{j}^{i}(b), \forall j\in J, \forall i=1,\ldots ,n.$ We know that, for all $U\in \mathcal{U}$:

  $$\begin{aligned} U(a)=\sum _{j=1}^{m}\left[ \sum _{i=1}^{n}u_{j}^{i}(g_{j}^{i}(a))\right] . \end{aligned}$$

  From monotonicity of marginal value functions $u_{j}^{i}(\cdot )$, we have that $\forall j\in J,$ and $\forall i=1,\ldots ,n,$ $u_{j}^{i}(g_{j}^{i}(a))\ge u_{j}^{i}(g_{j}^{i}(b))$ and adding up with respect to indices j and i, we obtain $U(a)\ge U(b)$ for all compatible value functions, thus we obtain the thesis.$\square$

• 2. Let $a,b\in A$, and $i,k\in \left\{ 1,\ldots ,n\right\} ,$ such that $a\varDelta ^{(i,k)}b$. This implies that $g_{j}^{i}(a)\ge g_{j}^{k}(b), \forall j\in J.$ We know that, for all $U\in \mathcal{U}$:

  $$\begin{aligned} U(a^{(i)})=\sum _{j=1}^{m}\left[ \sum _{r=1}^{n}u_{j}^{r}(g_{j}^{i}(a))\right] ,\;\;\;U(b^{(k)})=\sum _{j=1}^{m}\left[ \sum _{r=1}^{n}u_{j}^{r}(g_{j}^{k}(b))\right] . \end{aligned}$$

  From the monotonicity of marginal value functions $u_{j}^{r}(\cdot )$ we have that $\forall j\in J,$ and $\forall r=1,\ldots ,n,$ $u_{j}^{r}(g_{j}^{i}(a))\ge u_{j}^{r}(g_{j}^{k}(b))$ and adding up with respect to indices j and r we obtain the thesis.$\square$

Proof of Proposition 6.2

• 1. $\forall a,b\in A$, $\forall i,k\in \left\{ 1,\ldots ,n\right\}$, if $a^{(i)}$ is at least as good as $b^{(k)}$ for all compatible value functions $(a\succsim _{(i,k)}^{N}b)$, then there exists at least one compatible value function for which $a^{(i)}$ is at least as good as $b^{(k)}$ $(a\succsim _{(i,k)}^{P}b)$.$\square$

• 2. It follows from point 1 of Proposition 5.1.$\square$

• 3. Let $a,b,c\in A$ and $i,k\in \left\{ 1,\ldots ,n\right\}$, $i\le k$, such that $a\succsim _{(i,k)}^{N}b$ and $b\succsim _{(i,k)}^{N}c$. This means that for all $U\in \mathcal{U}$, $U(a^{(i)})\ge U(b^{(k)})$ and $U(b^{(i)})\ge U(c^{(k)})$. Since $i\le k$, by point 1 of Proposition 5.1 we have that for all $U\in \mathcal{U}$, $U(a^{(i)})\ge U(b^{(k)})\ge U(b^{(i)})\ge U(c^{(k)})$ and, consequently, $U(a^{(i)})\ge U(c^{(k)})$ for all $U\in \mathcal{U}$, that is $a\succsim _{(i,k)}^{N}c$.$\square$

• 4. Let $a,b\in A$, and $i,k\in \left\{ 1,\ldots ,n\right\}$, such that $a\not \succsim _{(i,k)}^{N}b.$ This means that $\exists U\in \mathcal{U}: U(a^{(i)})<U(b^{(k)}).$ Therefore, $b\succsim _{(k,i)}^{P}(a).$ $\square$

• 5. Let $a,b\in A$, and $i,k\in \left\{ 1,\ldots ,n\right\}$ with $i\ge k$ such that $a\not \succsim _{(i,k)}^{P}b$. This means that for all $U\in \mathcal{U}$, $U(b^{(k)})>U(a^{(i)})$. Since $i\ge k$, and from point 1 of Proposition 5.1, we obtain that for all $U\in \mathcal{U},$ $U(b^{(i)})\ge U(b^{(k)})>U(a^{(i)})\ge U(a^{(k)})$; thus for all $U\in \mathcal{U}$, $U(b^{(i)})> U(a^{(k)})$, therefore $b\succsim _{(i,k)}^{N}a$ implying $b\succsim _{(i,k)}^{P}a$ by point 1 of this Proposition. In this way, $\succsim _{(i,k)}^{P}$ is strongly complete.

  Let $a,b,c\in A$, $i,k\in \left\{ 1,\ldots ,n\right\}$, such that $i\ge k$, $a\not \succsim _{(i,k)}^{P}b$ and $b\not \succsim _{(i,k)}^{P}c.$ Then, we have:

  $$\begin{aligned} \left\{ \begin{array}{lll} a\not \succsim _{(i,k)}^{P}b &{} \Leftrightarrow &{} U(a^{(i)})<U(b^{(k)}),\quad \forall U\in \mathcal{U},\\ \\ b\not \succsim _{(i,k)}^{P}c &{} \Leftrightarrow &{} U(b^{(i)})<U(c^{(k)}), \quad \forall U\in \mathcal{U}.\\ \end{array} \right. \end{aligned}$$

  From this and from point 1 of Proposition 5.1 it follows that

  $$\begin{aligned}&U(a^{(i)})<U(b^{(k)})\le U(b^{(i)})<U(c^{(k)}),\;\quad \forall U\in \mathcal{U}\Rightarrow U(a^{(i)}) < U(c^{(k)}), \\&\quad \;\forall U\in \mathcal{U}\Leftrightarrow a\not \succsim _{(i,k)}^{P}c. \end{aligned}$$

  This proves that $\succsim _{(i,k)}^{P}$ is negatively transitive.$\square$

Proof of Proposition 6.3

• 1. Let $a,b\in A$, and $i,k,i_1,k_1\in \left\{ 1,\ldots ,n\right\}$, such that $i_1\ge i$, $k_1\le k$ and $a\succsim _{(i,k)}^{N}b$. Then, from point 1 of Proposition 5.1, we have:

  $$\begin{aligned} \left\{ \begin{array}{lll} a\succsim _{(i,k)}^{N}b &{} \Leftrightarrow &{} U(a^{(i)})\ge U(b^{(k)}),\quad \forall U\in \mathcal{U},\\ \\ i_1\ge i &{} \Rightarrow &{} U(a^{(i_1)})\ge U(a^{(i)}),\quad \forall U\in \mathcal{U},\\ \\ k_1\le k &{} \Rightarrow &{} U(b^{(k_1)})\le U(b^{(k)}),\quad \forall U\in \mathcal{U}.\\ \end{array} \right. \end{aligned}$$

  Thus:

  $$\begin{aligned}&U(a^{(i_1)})\ge U(a^{(i)})\ge U(b^{(k)})\ge U(b^{(k_1)}), \forall U\in \mathcal{U}\Rightarrow U(a^{(i_1)})\ge U(b^{(k_1)}), \\&\forall U\in \mathcal{U}\Leftrightarrow a\succsim _{(i_1,k_1)}^{N}b. \; \end{aligned}$$

  $\square$

• 2. Let $a,b\in A$, and $i,k,i_1,k_1\in \left\{ 1,\ldots ,n\right\}$ such that $i_1\ge i$, $k_1\le k$ and $a\succsim _{(i,k)}^{P}b$. Then, from point 1 of Proposition 5.1, we have:

  $$\begin{aligned} \left\{ \begin{array}{l} a\succsim _{(i,k)}^{P}b \Leftrightarrow \exists U\in \mathcal{U}: U(a^{(i)})\ge U(b^{(k)}),\\ \\ i_1\ge i \Rightarrow U(a^{(i_1)})\ge U(a^{(i)}),\quad \forall U\in \mathcal{U},\\ \\ k_1\le k \Rightarrow U(b^{(k_1)})\le U(b^{(k)}),\quad \forall U\in \mathcal{U}.\\ \end{array} \right. \end{aligned}$$

  Thus:

  $$\begin{aligned}&\exists U\in \mathcal{U}: U(a^{(i_1)})\ge U(a^{(i)})\ge U(b^{(k)})\ge U(b^{(k_1)})\Rightarrow \exists U\in \mathcal{U}: U(a^{(i_1)})\\&\quad \ge U(b^{(k_1)})\Leftrightarrow a\succsim _{(i_1,k_1)}^{P}b. \; \end{aligned}$$

  $\square$

• 3. Let $a,b\in A$ such that $a\succsim _{(1,n)}^{N}b.$ This means that $U(a^{(1)})\ge U(b^{(n)}),$ $\forall U\in \mathcal{U}$. From point 2 of Proposition 5.1, we have that $U(a)\ge U(a^{(1)})\ge U(b^{(n)})\ge U(b),$ $\forall U\in \mathcal{U}$, and thus we obtain $U(a)\ge U(b)$, $\forall U\in \mathcal{U},$ that is $a\succsim ^{N}b.$ In this way we proved that $\succsim _{(1,n)}^{N}\;\subseteq \;\succsim ^{N}$.

  Analogously, $a\succsim ^{N}b$ means that $U(a)\ge U(b),$ $\forall U\in \mathcal{U}$; from point 2 of Proposition 5.1 we obtain $U(a^{(n)})\ge U(a)\ge U(b)\ge U(b^{(1)})$, $\forall U\in \mathcal{U}$, and thus we have $U(a^{(n)})\ge U(b^{(1)})$, $\forall U\in \mathcal{U},$ that is $a\succsim _{(n,1)}^{N}b.$ In this way we proved that $\succsim ^{N}\;\subseteq \;\succsim _{(n,1)}^{N}.$ $\square$

• 4. Let $a,b\in A$ such that $a\succsim _{(1,n)}^{P}b.$ This means that $\exists U\in \mathcal{U}: U(a^{(1)})\ge U(b^{(n)})$. From point 2 of Proposition 5.1 we have:

  $$\begin{aligned} U(a)\ge U(a^{(1)})\ge U(b^{(n)})\ge U(b)\Rightarrow U(a)\ge U(b) \Leftrightarrow a\succsim ^{P}b. \end{aligned}$$

  In this way we proved that $\succsim _{(1,n)}^{P}\;\subseteq \;\succsim ^{P}$.

  Analogously, $a\succsim ^{P}b$ means that $\exists U\in \mathcal{U}: U(a)\ge U(b)$; from point 2 of Proposition 5.1 we obtain:

  $$\begin{aligned} U(a^{(n)})\ge U(a)\ge U(b)\ge U(b^{(1)})\Rightarrow U(a^{(n)})\ge U(b^{(1)})\Leftrightarrow a\succsim _{(n,1)}^{P}b. \end{aligned}$$

  In this way we proved that $\succsim ^{P}\;\subseteq \;\succsim _{(n,1)}^{P}.$ $\square$

Proposition 8.1 provides some further results regarding the imprecise necessary and possible preference relations.

Proposition 8.1

• 1. If $a\succsim _{(i,k)}^{N}b,$ $b\succsim _{(i_1,k_1)}^{N}c,$ and $k\ge i_1$, with $i,k,i_1,k_1\in \left\{ 1,\ldots ,n\right\}$, then $a\succsim _{(r,s)}^{N}c$, for all $r,s\in \left\{ 1,\ldots ,n\right\}$ such that $r\ge i$ and $s\le k_1,$

• 2. If $a\succsim _{(i,k)}^{N}b,$ $b\succsim _{(i_1,k_1)}^{P}c,$ and $k\ge i_1$, with $i,k,i_1,k_1\in \left\{ 1,\ldots ,n\right\}$, then $a\succsim _{(r,s)}^{P}c$, for all $r,s\in \left\{ 1,\ldots ,n\right\}$ such that $r\ge i$ and $s\le k_1,$

• 3. If $a\succsim _{(i,k)}^{P}b,$ $b\succsim _{(i_1,k_1)}^{N}c,$ and $k\ge i_1$, with $i,k,i_1,k_1\in \left\{ 1,\ldots ,n\right\}$, then $a\succsim _{(r,s)}^{P}c$, for all $r,s\in \left\{ 1,\ldots ,n\right\}$ such that $r\ge i$ and $s\le k_1,$

$\square$

Proof

• 1. Let $a,b,c\in A$ and $i,k,i_1,k_1\in \left\{ 1,\ldots ,n\right\}$, such that $a\succsim _{(i,k)}^{N}b$, $b\succsim _{(i_1,k_1)}^{N}c$ and $k\ge i_1$. Then we have:

  $$\begin{aligned} \left\{ \begin{array}{l} a\succsim _{(i,k)}^{N}b \Leftrightarrow U(a^{(i)})\ge U(b^{(k)}),\quad \forall U\in \mathcal{U},\\ \\ b\succsim _{(i_1,k_1)}^{N}c \Leftrightarrow U(b^{(i_1)})\ge U(c^{(k_1)}),\quad \forall U\in \mathcal{U},\\ \\ k\ge i_1 \Rightarrow U(b^{(k)})\ge U(b^{(i_1)}) \end{array} \right. \end{aligned}$$

  From this it follows:

  $$\begin{aligned}&U(a^{(i)})\ge U(b^{(k)})\ge U(b^{(i_1)})\ge U(c^{(k_1)}),\forall U\in \mathcal{U}\Rightarrow U(a^{(i)})\ge U(c^{(k_1)}), \\&\quad \forall U\in \mathcal{U}\Leftrightarrow a\succsim _{(i,k_1)}^{N}c. \end{aligned}$$

  Since $a\succsim _{(i,k_1)}^{N}c$ and $r,s\in \left\{ 1,\ldots ,n\right\}$: $r\ge i$ and $s\le k_1$, from point 1 of Proposition 6.3 we obtain $a\succsim _{(r,s)}^{N}c.$ $\square$

• 2. Let $a,b,c\in A,$ and $i,k,i_1,k_1\in \left\{ 1,\ldots ,n\right\}$, such that $a\succsim _{(i,k)}^{N}b,$ $b\succsim _{(i_1,k_1)}^{P}c,$ and $k\ge i_1.$ Then we have:

  $$\begin{aligned} \left\{ \begin{array}{l} a\succsim _{(i,k)}^{N}b \Leftrightarrow U(a^{(i)})\ge U(b^{(k)}),\quad \forall U\in \mathcal{U},\\ \\ b\succsim _{(i_1,k_1)}^{P}c \Leftrightarrow \exists U\in \mathcal{U}: U(b^{(i_1)})\ge U(c^{(k_1)}),\\ \\ k\ge i_1 \Rightarrow U(b^{(k)})\ge U(b^{(i_1)}),\quad \forall U\in \mathcal{U}. \end{array} \right. \end{aligned}$$

  It follows that:

  $$\begin{aligned}&\exists U\in \mathcal{U}: U(a^{(i)})\ge U(b^{(k)})\ge U(b^{(i_1)})\ge U(c^{(k_1)})\Rightarrow \exists U\in \mathcal{U}: U(a^{(i)})\\&\quad \ge U(c^{(k_1)})\Leftrightarrow a\succsim _{(i,k_1)}^{P}c. \end{aligned}$$

  Since $r,s\in \left\{ 1,\ldots ,n\right\}$: $r\ge i$ and $s\le k_1$, from point 2 of Proposition 6.3 we obtain $a\succsim _{(r,s)}^{P}c.$ $\square$

• 3. Let $a,b,c\in A,$ $i,k,i_1,k_1\in \left\{ 1,\ldots ,n\right\}$ such that $a\succsim _{(i,k)}^{P}b,$ $b\succsim _{(i_1,k_1)}^{N}c,$ and $k\ge i_1.$ We have that:

  $$\begin{aligned} \left\{ \begin{array}{l} a\succsim _{(i,k)}^{P}b \Leftrightarrow \exists U\in \mathcal{U}: U(a^{(i)})\ge U(b^{(k)}),\\ \\ b\succsim _{(i_1,k_1)}^{N}c \Leftrightarrow U(b^{(i_1)})\ge U(c^{(k_1)}), \quad \forall U\in \mathcal{U},\\ \\ k\ge i_1 \Rightarrow U(b^{(k)})\ge U(b^{(i_1)}), \quad \forall U\in \mathcal{U}. \end{array} \right. \end{aligned}$$

  From this it follows:

  $$\begin{aligned}&\exists U\in \mathcal{U}: U(a^{(i)})\ge U(b^{(k)})\ge U(b^{(i_1)})\ge U(c^{(k_1)})\Rightarrow \exists U\in \mathcal{U}: U(a^{(i)}) \\&\quad \ge U(c^{(k_1)})\Leftrightarrow a\succsim _{(i,k_1)}^{P}c. \end{aligned}$$

  Since $r,s\in \left\{ 1,\ldots ,n\right\}$ such that $r\ge i$ and $s\le k_1$, from point 2 of Proposition 6.3 we obtain $a\succsim _{(r,s)}^{P}c.$ $\square$

Proposition 8.2 describes some properties involving the imprecise necessary and possible preference relations together with the classical necessary and possible preference relations.

Proposition 8.2

• 1. Given $a,b,c\in A$, $i\in \left\{ 1,\ldots ,n\right\}$ such that $a\succsim _{(i,n)}^{N} b$ and $b\succsim ^{N}c$, then $a\succsim _{(r,1)}^{N}c$, for all $r\in \left\{ 1,\ldots ,n\right\}$ such that $r\ge i$,

• 2. Given $a,b,c\in A$, $k\in \left\{ 1,\ldots ,n\right\}$ such that $a\succsim ^{N} b$ and $b\succsim _{(1,k)}^{N}c$, then $a\succsim _{(n,r)}^{N}c$, for all $r\in \left\{ 1,\ldots ,n\right\}$ such that $r\le k$,

• 3. Given $a,b,c\in A$, $i\in \left\{ 1,\ldots ,n\right\}$ such that $a\succsim _{(i,n)}^{P} b$ and $b\succsim ^{N}c$, then $a\succsim _{(r,1)}^{P}c$, for all $r\in \left\{ 1,\ldots ,n\right\}$ such that $r\ge i$,

• 4. Given $a,b,c\in A$, $k\in \left\{ 1,\ldots ,n\right\}$ such that $a\succsim ^{N} b$ and $b\succsim _{(1,k)}^{P}c$, then $a\succsim _{(n,r)}^{P}c$, for all $r\in \left\{ 1,\ldots ,n\right\}$ such that $r\le k$,

• 5. Given $a,b,c\in A$, $i\in \left\{ 1,\ldots ,n\right\}$ such that $a\succsim _{(i,n)}^{N} b$ and $b\succsim ^{P}c$, then $a\succsim _{(r,1)}^{P}c$, for all $r\in \left\{ 1,\ldots ,n\right\}$ such that $r\ge i$,

• 6. Given $a,b,c\in A$, $k\in \left\{ 1,\ldots ,n\right\}$ such that $a\succsim ^{P} b$ and $b\succsim _{(1,k)}^{N}c$, then $a\succsim _{(n,r)}^{P}c$, for all $r\in \left\{ 1,\ldots ,n\right\}$ such that $r\le k$.

Proof

• 1. Let $a,b,c\in A$ and $i,r\in \left\{ 1,\ldots ,n\right\}$ such that $a\succsim ^{N}_{(i,n)}b$, $b\succsim ^{N}c$ and $r\ge i$. Then we have:

  $$\begin{aligned} \left\{ \begin{array}{ll} a\succsim ^{N}_{(i,n)}b \Leftrightarrow U(a^{(i)})\ge U(b^{(n)}), &{}\quad \forall U\in \mathcal{U},\\ \\ b\succsim ^{N}c \Leftrightarrow U(b)\ge U(c), &{}\quad \forall U\in \mathcal{U},\\ \\ r\ge i \Rightarrow U(a^{(r)})\ge U(a^{(i)}), &{}\quad \forall U\in \mathcal{U}. \end{array} \right. \end{aligned}$$

  It follows that, for all $U\in \mathcal{U}$, $U(a^{(r)})\ge U(a^{(i)})\ge U(b^{(n)})\ge U(b)\ge U(c)\ge U(c^{(1)})$ where $U(b^{(n)})\ge U(b)$ and $U(c)\ge U(c^{(1)})$ hold by point 2 of Proposition 5.1. Thus, for all $U\in \mathcal{U}$ we obtain $U(a^{(r)})\ge U(c^{(1)})$, and therefore $a\succsim _{(r,1)}^{N}c.$ $\square$ Points 2-6 can be proved analogously.$\square$