A consolidated approach to the axiomatization of outranking relations: a survey and new results

Abstract

Outranking relations such as produced by the Electre I or II or the Tactic methods are based on a concordance and non-discordance principle that leads to declaring that an alternative is “superior” to another, if the coalition of attributes supporting this proposition is “sufficiently important” (concordance condition) and if there is no attribute that “strongly rejects” it (non-discordance condition). Such a way of comparing alternatives is rather natural and does not require a detailed analysis of tradeoffs between the various attributes. However, it is well known that it may produce binary relations that do not possess any remarkable property of transitivity or completeness. The axiomatic foundations of outranking relations have recently received attention. Within a conjoint measurement framework, characterizations of reflexive concordance–discordance relations have been obtained. These relations encompass those generated by the Electre I and II methods, which are non-strict (reflexive) relations. A different characterization has been provided for strict (asymmetric) preference relations such as produced by Tactic. In this paper we briefly review the various kinds of axiomatizations of outranking relations proposed so far in the literature. Then we analyze the relationships between reflexive and asymmetric outranking relations in a conjoint measurement framework, consolidating our previous work. Co-duality plays an essential rôle in our analysis. It allows us to understand the correspondence between the previous characterizations. Making a step further, we provide a common axiomatic characterization for both types of relations. Applying the co-duality operator to concordance–discordance relations also yields a new and interesting type of preference relation that we call concordance relation with bonus. The axiomatic characterization of such relations results directly from co-duality arguments.

Appendices

Appendix 1: Propositions 74 and 76

Proposition 74

If $\mathrel {\mathcal {R}}$ is a complete binary relation on $X$ satisfying $RC 2$, $AC 1$, $AC 2$, $AC 3$, $MM 1$ and $MM 3$, then $\mathrel {\mathcal {R}}$ satisfies $RC 1$.

For proving this proposition, we need the following lemma.

Lemma 75

Let $\mathrel {\mathcal {R}}$ be a binary relation on $X$ satisfying $RC 2_{i}$, $AC 1_{i}$, $AC 2_{i}$, $AC 3_{i}$ and $Maj 1_{i}$, on some attribute $i$. Consider four levels $x,y,z,w \in X_{i}$ such that the pairs $(x,y)$ and $(z,w)$ are not comparable w.r.t. the relation $\mathrel {\succsim _{i}^{*}}$, which we denote by $(x,y) \bowtie (z,w))$. The relative positions of these pairs and the opposite pairs are as follows:

• 1. $[(y,x) \mathrel {\sim _{i}^{*}}(w,z)] \mathrel {\succ _{i}^{*}} [(x,y)\bowtie (z,w)]$,

• 2. furthermore, one of the following configurations holds true:

  • (a) $[(y,z) \mathrel {\sim _{i}^{*}}(y,x) \mathrel {\sim _{i}^{*}}(w,z)] \mathrel {\succ _{i}^{*}} [(x,y)\bowtie (z,w)] \mathrel {\succ _{i}^{*}}(z,y)$

  • (b) $[(w,x) \mathrel {\sim _{i}^{*}}(y,x) \mathrel {\sim _{i}^{*}}(w,z)] \mathrel {\succ _{i}^{*}} [(x,y)\bowtie (z,w)] \mathrel {\succ _{i}^{*}}(x,w)$.

  In the above, the notation $[(x,y)\bowtie (z,w)]$ means that the incomparable pairs $[(x,y)$ and $(z,w)$ have the same relationships with the other pairs listed.

Proof

(of Lemma 75) 1. Let $x,y,z,w \in X_{i}$ be such that the pairs $(x,y)$ and $(z,w)$ are incomparable w.r.t. relation $\mathrel {\succsim _{i}^{*}}$, i.e., we have:

$$\begin{aligned} Not [(x,y) \mathrel {\succsim _{i}^{*}}(z,w) ] \text { and } Not [(z,w) \mathrel {\succsim _{i}^{*}}(x,y) ]. \end{aligned}$$

(30)

In view of Definition 21, this means that there are $a,b,c,d \in X_{-i}$ such that:

$$\begin{aligned} \begin{array}{cc} (x,c) \mathrel {\mathcal {R}}(y,d), &{} Not [(z,c) \mathrel {\mathcal {R}}(w,d) ], \\ (z,a) \mathrel {\mathcal {R}}(w,b), &{} Not [(x,a) \mathrel {\mathcal {R}}(y,b) ], \end{array} \end{aligned}$$

(31)

in other words, $\mathrel {\mathcal {R}}$ does not satisfy $RC 1_{i}$.

Using $RC 2_{i}$ and Lemma 22.2 imply that we have $(y,x) \mathrel {\succsim _{i}^{*}}(w,z)$ and $(w,z) \mathrel {\succsim _{i}^{*}}(y,x)$, yielding:

$$\begin{aligned} (y,x) \mathrel {\sim _{i}^{*}}(w,z) \end{aligned}$$

(32)

The same axiom and lemma entail that $(x,y)$ and $(y,x)$ are comparable w.r.t. $\mathrel {\succsim _{i}^{*}}$, i.e., we must have $(x,y) \mathrel {\succsim _{i}^{*}}(y,x)$ or $(y,x) \mathrel {\succsim _{i}^{*}}(x,y)$. The former is incompatible with $Maj 1_{i}$ as we shall see. Note that $\mathrel {\mathcal {R}}$ satisfies $Maj 1_{i}$ by Lemma 34.1.

2. We show that assuming $(x,y) \mathrel {\succsim _{i}^{*}}(y,x)$ leads to a contradiction. From $(x,y) \mathrel {\succsim _{i}^{*}}(y,x)$, we first derive the following consequences:

• 1. $(z,w) \mathrel {\succ _{i}^{*}}(w,z)$. Assuming $Not [(z,w) \mathrel {\succsim _{i}^{*}}(w,z) ]$ implies, by Lemma 22.2, that $(w,z) \mathrel {\succsim _{i}^{*}}(z,w)$. Hence we would have: $(x,y) \mathrel {\succsim _{i}^{*}}(y,x) \mathrel {\sim _{i}^{*}}(w,z) \mathrel {\succsim _{i}^{*}}(z,w)$. Using the transitivity of $\mathrel {\succsim _{i}^{*}}$ leads to $(x,y) \mathrel {\succsim _{i}^{*}}(z,w)$, a contradiction. The same contradiction can be derived if we suppose $(z,w) \mathrel {\sim _{i}^{*}}(w,z)$.

• 2. $(x,y) \mathrel {\succ _{i}^{*}}(y,x)$. Else, from $(x,y) \mathrel {\sim _{i}^{*}}(y,x)$ we would derive $(z,w) \mathrel {\succ _{i}^{*}}(w,z) \mathrel {\sim _{i}^{*}}(y,x) \mathrel {\sim _{i}^{*}}(x,y)$, from which we deduce $(z,w) \mathrel {\succ _{i}^{*}}(x,y)$, a contradiction.

• 3. $(x,y) \mathrel {\succ _{i}^{*}}(w,z)$. Assuming $Not [(x,y) \mathrel {\succsim _{i}^{*}}(w,z) ]$ implies, by Lemma 22.2, that $(y,x) \mathrel {\succsim _{i}^{*}}(z,w)$. Hence we would have: $(w,z) \mathrel {\sim _{i}^{*}}(y,x) \mathrel {\succsim _{i}^{*}}(z,w)$. Using the transitivity of $\mathrel {\succsim _{i}^{*}}$ leads to $(w,z) \mathrel {\succsim _{i}^{*}}(z,w)$, a contradiction. The same contradiction can be derived if we suppose $(x,y) \mathrel {\sim _{i}^{*}}(w,z)$.

• 4. $(z,w) \mathrel {\succ _{i}^{*}}(y,x)$ is established in a similar way as the previous item.

We thus have the following situation: $(x,y)$ and $(z,w)$ are incomparable differences w.r.t. $\mathrel {\succsim _{i}^{*}}$, both are strictly preferred to $(y,x)$ and $(w,z)$, which are indifferent pairs.

We now use $AC 1_{i}$, $AC 2_{i}$ and $AC 3_{i}$. The main consequence of these axioms is that the relations $\mathrel {\succsim _{i}^{+}}, \mathrel {\succsim _{i}^{-}}$ and $\mathrel {\succsim _{i}^{\pm }}$ are complete (Lemma 47). Moreover, we have, for all $s,t, u, v \in X_{i}$:

$$\begin{aligned} s \mathrel {\succsim _{i}^{+}}t&\Rightarrow (s,u) \mathrel {\succsim _{i}^{*}}(t,u) \end{aligned}$$

(33)

$$\begin{aligned} s \mathrel {\succsim _{i}^{-}}t&\Rightarrow (v,t) \mathrel {\succsim _{i}^{*}}(v,s) \end{aligned}$$

(34)

(direct consequence of $AC 1_{i}, AC 2_{i}$ and the definitions of $\mathrel {\succsim _{i}^{+}}, \mathrel {\succsim _{i}^{-}}$ and $\mathrel {\succsim _{i}^{*}}$).

Consider the pairs $(x,y)$ and $(z,w)$. We claim that there are $u,v \in X_{i}$ such that $(u,v) \mathrel {\succ _{i}^{*}}(x,y)$ and $(u,v) \mathrel {\succ _{i}^{*}}(z,w)$. Furthermore, $(u,v)$ is either $(x,w)$ or $(z,y)$. Observe first that we cannot have:

• 1. $x \mathrel {\succsim _{i}^{\pm }}z$ and $w \mathrel {\succsim _{i}^{\pm }}y$. Else, using (33) and (34), we would have $(x,y) \mathrel {\succsim _{i}^{*}}(z,y) \mathrel {\succsim _{i}^{*}}(z,w)$, a contradiction with the fact that $(x,y)$ and $(z,w)$ are incomparable,

• 2. $z \mathrel {\succsim _{i}^{\pm }}x$ and $y \mathrel {\succsim _{i}^{\pm }}w$. Else, using (33) and (34), we would have $(z,w) \mathrel {\succsim _{i}^{*}}(x,w) \mathrel {\succsim _{i}^{*}}(x,y)$, a contradiction with the fact that $(x,y)$ and $(z,w)$ are incomparable.

Since $\mathrel {\succsim _{i}^{\pm }}$ is complete, we thus have either $[x \mathrel {\succsim _{i}^{\pm }}z$ and $y \mathrel {\succsim _{i}^{\pm }}w]$ or $[z \mathrel {\succsim _{i}^{\pm }}x$ and $w \mathrel {\succsim _{i}^{\pm }}y]$. Consider the former case. Using (33) and (34) yields $(x,w) \mathrel {\succsim _{i}^{*}}(z,w)$ and $(x,w) \mathrel {\succsim _{i}^{*}}(x,y)$. We can have neither $(x,w) \mathrel {\sim _{i}^{*}}(z,w)$ nor $(x,w) \mathrel {\sim _{i}^{*}}(x,y)$, because this would imply that $(x,y)$ and $(z,w)$ are comparable. Our claim is thus proved with $(u,v) = (x,w)$. If the situation was such that $z \mathrel {\succsim _{i}^{\pm }}x$ and $w \mathrel {\succsim _{i}^{\pm }}y$, then we would have that $(z,y) \mathrel {\succsim _{i}^{*}}(z,w)$ and $(z,y) \mathrel {\succsim _{i}^{*}}(x,y)$. The rôle of $(u,v)$ would be played by $(z,y)$. Our claim is proved.

Assume first that $(u,v) = (x,w)$. From $(x,c) \mathrel {\mathcal {R}}(y,d)$ in (31) and $(x,w) \mathrel {\succsim _{i}^{*}}(x,y)$, we derive $(x,c)\mathrel {\mathcal {R}}(w,d)$. Similarly, $(z,a) \mathrel {\mathcal {R}}(w,b)$ and $(x,w) \mathrel {\succsim _{i}^{*}}(z,w)$ entail $(x,a)\mathrel {\mathcal {R}}(w,b)$. This allows us to derive a contradiction with $Maj 1_{i}$. Indeed, we have $(x,c)\mathrel {\mathcal {R}}(w,d)$, $(x,a)\mathrel {\mathcal {R}}(w,b)$ and $(z,a)\mathrel {\mathcal {R}}(w,b)$. Using $Maj 1_{i}$ yields either $(w,a)\mathrel {\mathcal {R}}(z,b)$ or $(z,c)\mathrel {\mathcal {R}}(w,d)$. None of this conclusions holds true. The latter is false by hypothesis (see (31)) and the former cannot be true since $(x,y)\mathrel {\succ _{i}^{*}}(w,z)$ and $Not [(x,a)\mathrel {\mathcal {R}}(y,b) ]$. The case in which we assume $(u,v) = (z,y)$ yields a similar contradiction. As a conclusion, we have established that $(y,x)\mathrel {\succ _{i}^{*}}(x,y)$.

3. We draw the consequences of the fact that $(y,x)\mathrel {\succ _{i}^{*}}(x,y)$, by adapting the ideas that we used in Part 2 of the present proof, under the opposite hypothesis.

The fact that $(y,x) \mathrel {\succsim _{i}^{*}}(x,y)$ entails the following:

• 1. $(w,z) \mathrel {\succ _{i}^{*}}(z,w)$. Assuming $Not [(w,z) \mathrel {\succsim _{i}^{*}}(z,w) ]$ implies, by Lemma 22.2, that $(z,w) \mathrel {\succsim _{i}^{*}}(w,z)$. Hence we would have: $(z,w) \mathrel {\succsim _{i}^{*}}(w,z) \mathrel {\sim _{i}^{*}}(y,x) \mathrel {\succsim _{i}^{*}}(x,y)$. Using the transitivity of $\mathrel {\succsim _{i}^{*}}$ leads to $(z,w) \mathrel {\succsim _{i}^{*}}(x,y)$, a contradiction. The same contradiction arises if we suppose $(z,w) \mathrel {\sim _{i}^{*}}(w,z)$.

• 2. $(y,x) \mathrel {\succ _{i}^{*}}(x,y)$. Else, from $(y,x) \mathrel {\sim _{i}^{*}}(x,y)$ we would derive $(x,y) \mathrel {\sim _{i}^{*}}(y,x) \mathrel {\sim _{i}^{*}}(w,z) \mathrel {\succ _{i}^{*}}(z,w)$, from which we deduce $(x,y) \mathrel {\succ _{i}^{*}}(z,w)$, a contradiction.

• 3. $(y,x) \mathrel {\succ _{i}^{*}}(z,w)$. Assuming $Not [(y,x) \mathrel {\succsim _{i}^{*}}(z,w) ]$ implies, by Lemma 22.2, that $(x,y) \mathrel {\succsim _{i}^{*}}(w,z)$. Hence we would have: $(x,y) \mathrel {\succsim _{i}^{*}}(w,z) \mathrel {\sim _{i}^{*}}(y,x)$. Using the transitivity of $\mathrel {\succsim _{i}^{*}}$ leads to $(x,y) \mathrel {\succsim _{i}^{*}}(y,x)$, a contradiction. The same contradiction can be derived if we suppose $(y,x) \mathrel {\sim _{i}^{*}}(z,w)$.

• 4. $(w,z) \mathrel {\succ _{i}^{*}}(x,y)$ is established in a similar way as the previous item.

We thus have the following situation: $(y,x)$ and $(w,z)$ are incomparable differences w.r.t. $\mathrel {\succsim _{i}^{*}}$. Both are strictly preferred to $(x,y)$ and $(z,w)$, which are indifferent pairs.

Using $AC 1_{i}$, $AC 2_{i}$ and $AC 3_{i}$, we derive exactly the same consequences as in Part 2, i.e., we have either $[x \mathrel {\succsim _{i}^{\pm }}z$ and $y \mathrel {\succsim _{i}^{\pm }}w]$ or $[z \mathrel {\succsim _{i}^{\pm }}x$ and $w \mathrel {\succsim _{i}^{\pm }}y]$.

If $[x \mathrel {\succsim _{i}^{\pm }}z$ and $y \mathrel {\succsim _{i}^{\pm }}w]$, we conclude that $(x,y) \mathrel {\succsim _{i}^{*}}(z,y)$ and $(z,w) \mathrel {\succsim _{i}^{*}}(z,y)$. We can have neither $(x,y) \mathrel {\sim _{i}^{*}}(z,y)$ nor $(z,w) \mathrel {\sim _{i}^{*}}(z,y)$, because this would imply that $(x,y)$ and $(z,w)$ are comparable. Since we have $Not [(z,y) \mathrel {\succsim _{i}^{*}}(x,y) ]$, we deduce that $(y,z) \mathrel {\succsim _{i}^{*}}(y,x)$, using Lemma 22.2. Having $(y,z) \mathrel {\succ _{i}^{*}}(y,x)$ is impossible since this would contradict $Maj 1_{i}$. Indeed, assume that there are $e, f \in X_{-i}$ such that $(y,e) \mathrel {\mathcal {R}}(z,f)$ and $Not [(y,e) \mathrel {\mathcal {R}}(x,f) ]$. Since $(y,x) \mathrel {\succ _{i}^{*}}(z,w)$ and $(z,a) \mathrel {\mathcal {R}}(w,b)$, we get $(y,a) \mathrel {\mathcal {R}}(x,b)$. From $(y,z) \mathrel {\succsim _{i}^{*}}(y,x)$ and $(y,a) \mathrel {\mathcal {R}}(x,b)$, we derive $(y,a) \mathrel {\mathcal {R}}(z,b)$. By (31), we also have $Not [(x,a) \mathrel {\mathcal {R}}(y,b) ]$ . The following configuration is not compatible with $Maj 1_{i}$: $(y,a) \mathrel {\mathcal {R}}(x,b)$, $(y,a) \mathrel {\mathcal {R}}(z,b)$, $(y,e) \mathrel {\mathcal {R}}(z,f)$, $Not [(x,a) \mathrel {\mathcal {R}}(y,b) ]$, $Not [(y,e) \mathrel {\mathcal {R}}(x,f) ]$. We have thus established that $(y,z) \mathrel {\sim _{i}^{*}}(y,x)$. Starting from $Not [(z,y) \mathrel {\succsim _{i}^{*}}(z,w) ]$, one proves similarly that $(y,z) \mathrel {\sim _{i}^{*}}(w,z)$ and we finally have that $(y,z) \mathrel {\sim _{i}^{*}}(w,z) \mathrel {\sim _{i}^{*}}(y,x)$.

In the case in which $[z \mathrel {\succsim _{i}^{\pm }}x$ and $w \mathrel {\succsim _{i}^{\pm }}y]$, one proves in an analogous way that $(x,y) \mathrel {\succ _{i}^{*}}(x,w)$, $(z,w) \mathrel {\succ _{i}^{*}}(x,w)$ and $(w,x) \mathrel {\sim _{i}^{*}}(y,x) \mathrel {\sim _{i}^{*}}(w,z)$.

This concludes the proof of Lemma 75.$\square$

Proof

(of Proposition 74) Since $\mathrel {\mathcal {R}}$ satisfies $MM 1$ and $RC 2$, it satisfies also $Maj 1$ (Lemma 42.2). Let us assume that $\mathrel {\mathcal {R}}$ does not verify $RC 1_{i}$ on some attribute $i$. We shall derive a contradiction from this assumption. If $RC 1_{i}$ is not verified by $\mathrel {\mathcal {R}}$, there exist four levels $x,y,z,w \in X_{i}$ such that $(x,y)$ and $(z,w)$ are incomparable w.r.t. relation $\mathrel {\succsim _{i}^{*}}$, or, in other words, there are $a,b,c,d \in X_{-i}$ such that:

$$\begin{aligned} \begin{array}{cc} (x,c) \mathrel {\mathcal {R}}(y,d) &{} Not [(z,c) \mathrel {\mathcal {R}}(w,d) ] \\ (z,a) \mathrel {\mathcal {R}}(w,b) &{} Not [(x,a) \mathrel {\mathcal {R}}(y,b) ] \end{array} \end{aligned}$$

(35)

Hence $\mathrel {\mathcal {R}}$ is in the conditions of application of Lemma 75. We shall assume that the configuration described in conclusion 2.(a) of the lemma holds true, i.e., we have:

$$\begin{aligned}{}[(y,z) \mathrel {\sim _{i}^{*}}(y,x) \mathrel {\sim _{i}^{*}}(w,z)] \mathrel {\succ _{i}^{*}} [(x,y) \bowtie (z,w)] \mathrel {\succ _{i}^{*}}(z,y). \end{aligned}$$

(36)

Note that case 2.(b) can be dealt with similarly. We leave it to the reader.

Since $\mathrel {\mathcal {R}}$ satisfies $MM 3$ and $RC 2$, it satisfies $M3$ (Lemma 56.1). In the configuration described by (36), $M3_{i}$ implies that the pair $(z,y)$ is a veto. Indeed, assume that there are $e,f \in X_{-i}$ such that $(z,e) \mathrel {\mathcal {R}}(y,f)$. We have $(x,y) \mathrel {\succ _{i}^{*}}(z,y)$, which means there are $g,h \in X_{-i}$ such that $(x,g) \mathrel {\mathcal {R}}(y,h)$ and $Not [(z,g) \mathrel {\mathcal {R}}(y,h) ]$. It holds true that $(y,a) \mathrel {\mathcal {R}}(x,b)$ since, by (35), $(z,a) \mathrel {\mathcal {R}}(w,b)$ and, by (36), $(y,x) \mathrel {\succ _{i}^{*}}(z,w)$. Finally, we have $Not [(x,a) \mathrel {\mathcal {R}}(y,b) ]$ by (35) and $Not [(z,a) \mathrel {\mathcal {R}}(y,b) ]$ since $(x,y) \mathrel {\succ _{i}^{*}}(z,y)$. Gathering the relevant preferences, i.e., $(y,a) \mathrel {\mathcal {R}}(x,b)$, $(x,g) \mathrel {\mathcal {R}}(y,h)$, $(z,e) \mathrel {\mathcal {R}}(y,f)$, $Not [(x,a) \mathrel {\mathcal {R}}(y,b) ]$, $Not [(z,a) \mathrel {\mathcal {R}}(y,b) ]$ and $Not [(z,g) \mathrel {\mathcal {R}}(y,h) ]$, yields a contradiction with $M3_{i}$. We thus have shown that for all $e,f \in X_{-i}$, we have

$$\begin{aligned} Not [(z,e) \mathrel {\mathcal {R}}(y,f) ] \end{aligned}$$

(37)

The fact that $\mathrel {\mathcal {R}}$ is complete enters into play in the following way. Since $\mathrel {\mathcal {R}}$ is complete, (37) entails that for all $e,f \in X_{-i}$, we have $(y,e) \mathrel {\mathcal {R}}(z,f)$. Since (36) tells us that $(y,z) \mathrel {\sim _{i}^{*}}(w,z) \mathrel {\sim _{i}^{*}}(y,x)$, we also have, for all $e,f \in X_{-i}$, $(w,e) \mathrel {\mathcal {R}}(z,f)$ and $(y,e) \mathrel {\mathcal {R}}(x,f)$. In other words, $(y,z), (w,z)$ and $(y,x)$ are bonuses as defined in Sect. 3.6.

The relation $\mathrel {\mathcal {R}}$ induces not only a relation $\mathrel {\succsim _{i}^{*}}$ comparing pairs of levels on $X_{i}$, but also a similar relation $\mathrel {\succsim _{-i}^{*}}$ on the pairs of elements of $X_{-i}$. For $e,f,g,h \in X_{-i}$, we have $(e,f) \mathrel {\succsim _{-i}^{*}} (g,h)$ iff, for all $u,v \in X_{i}$, $[(u,g) \mathrel {\mathcal {R}}(v,h)] \Rightarrow [(u,e) \mathrel {\mathcal {R}}(v,f)]$. The assumption (35) also means that the pairs $(a,b), (c,d) \in X_{-i} \times X_{-i}$ are not comparable w.r.t. $\mathrel {\succsim _{-i}^{*}}$. This relation is transitive by definition and complete iff $RC 1_{i}$ holds.

If $AC 1, AC 2$ and $AC 3$ hold, we claim that there are $g,h \in X_{-i}$ with $(a,b) \mathrel {\succsim _{-i}^{*}} (g,h)$ and $(c,d) \mathrel {\succsim _{-i}^{*}} (g,h)$. $AC 1, AC 2$ and $AC 3$ imply that $\mathrel {\succsim _{j}^{\pm }}$ is a complete weak order for all $j \in N$. We define $g$ (resp.$h$) by specifying its level $g_{j}$ (resp. $h_{j}$) for each $j \ne i$ as follows: for all $j \ne i$,

$$\begin{aligned} g_{j} = \min {\{a_{j},c_{j}\}}= \left\{ \begin{array}{cc} a_{j} &{} \text {if } c_{j} \mathrel {\succsim _{j}^{\pm }} a_{j} \\ c_{j} &{} \text {if } a_{j} \mathrel {\succsim _{j}^{\pm }} c_{j} \end{array} \right. \end{aligned}$$

(38)

$$\begin{aligned} h_{j} = \max {\{b_{j},d_{j}\}}= \left\{ \begin{array}{cc} b_{j} &{} \text {if } b_{j} \mathrel {\succsim _{j}^{\pm }} d_{j} \\ d_{j} &{} \text {if } d_{j} \mathrel {\succsim _{j}^{\pm }} b_{j} \end{array} \right. \end{aligned}$$

(39)

Starting from the trivial $(a,b) \mathrel {\succsim _{-i}^{*}} (a,b)$ and applying repeatedly (33) and (34), using $g$ and $h$, we obtain $(a,b) \mathrel {\succsim _{-i}^{*}} (g,h)$. One proves similarly that $(c,d) \mathrel {\succsim _{-i}^{*}} (g,h)$.

We finish the proof by showing that the above induces a contradiction with $M3_{i}$. We have that $Not [(x,a) \mathrel {\mathcal {R}}(y,b) ]$ entails $Not [(x,g) \mathrel {\mathcal {R}}(y,h) ]$ and $Not [(z,c) \mathrel {\mathcal {R}}(w,d) ]$ entails $Not [(z,g) \mathrel {\mathcal {R}}(w,h) ]$ (since a difference on $X_{-i}$ is substituted by a smaller one w.r.t. $\mathrel {\succsim _{-i}^{*}}$). Since $(y,x)$ is a bonus, we have in particular $(y,g) \mathrel {\mathcal {R}}(x,h)$. By (35), we have $(x,c) \mathrel {\mathcal {R}}(y,d)$ and $(z,a) \mathrel {\mathcal {R}}(w,b)$. Gathering the relevant preferences, i.e., $(y,g) \mathrel {\mathcal {R}}(x,h)$, $(x,c) \mathrel {\mathcal {R}}(y,d)$, $(z,a) \mathrel {\mathcal {R}}(w,b)$, $Not [(x,g) \mathrel {\mathcal {R}}(y,h) ]$, $Not [(z,g) \mathrel {\mathcal {R}}(w,h) ]$ and $Not [(z,c) \mathrel {\mathcal {R}}(w,d) ]$, yields a contradiction with $M3_{i}$.$\square$

The proposition below is another result, besides Proposition 74, showing that $RC 1$ has relationships with the other axioms even though the considered relations here are neither complete nor asymmetric.

Proposition 76

If $\mathrel {\mathcal {R}}$ is a relation on $X$ satisfying $RC 2_{i}$, $AC 1_{i}$, $AC 2_{i}$, $AC 3_{i}$, $Maj 1_{i}$ and $Maj 3_{i}$, for some $i \in N$, then $\mathrel {\mathcal {R}}$ satisfies $RC 1_{i}$.

Proof

Let us assume that $\mathrel {\mathcal {R}}$ does not verify $RC 1_{i}$ on some attribute $i$, i.e., there exist $x,y,z,w \in X_{i}$ and $a,b,c,d \in X_{-i}$ such that:

$$\begin{aligned} \begin{array}{cc} (x,c) \mathrel {\mathcal {R}}(y,d) &{} Not [(z,c) \mathrel {\mathcal {R}}(w,d) ] \\ (z,a) \mathrel {\mathcal {R}}(w,b) &{} Not [(x,a) \mathrel {\mathcal {R}}(y,b) ]. \end{array} \end{aligned}$$

(40)

In other words, the pairs $(x,y)$ and $(z,w)$ are incomparable w.r.t. relation $\mathrel {\succsim _{i}^{*}}$. Therefore, $\mathrel {\mathcal {R}}$ is in the conditions of application of Lemma 75 and we have $[(y,x) \mathrel {\sim _{i}^{*}}(w,z)] \mathrel {\succ _{i}^{*}} [(x,y), (z,w)]$.

The latter is not compatible with $Maj 3_{i}$ as we shall see. Since $(y,x) \mathrel {\succ _{i}^{*}}(z,w)$ and using $(z,a) \mathrel {\mathcal {R}}(w,b)$ in (40), we obtain that $(y,a) \mathrel {\mathcal {R}}(x,b)$. From $(w,z) \mathrel {\sim _{i}^{*}}(y,x)$ and $(y,a) \mathrel {\mathcal {R}}(x,b)$, we derive $(w,a) \mathrel {\mathcal {R}}(z,b)$. We also directly use the four clauses in (40). The following facts contradict $Maj 3_{i}$: $(y,a) \mathrel {\mathcal {R}}(x,b)$, $(w,a) \mathrel {\mathcal {R}}(z,b)$, $(x,c) \mathrel {\mathcal {R}}(y,d)$, $(z,a) \mathrel {\mathcal {R}}(w,b)$, $Not [(x,a) \mathrel {\mathcal {R}}(y,b) ]$ and $Not [(z,c) \mathrel {\mathcal {R}}(w,d) ]$.$\square$

Appendix 2: Examples

The examples below have been checked in order to determine whether they satisfy the following axioms:

$$\begin{aligned}&RC 1,\, RC 2, \, AC 1, \, AC 2,\, AC 3, \, UC ,\, LC ,\,M1,\,M2,\, Maj 1,\, Maj 2,\\&\qquad \qquad MM 1,\, MM 2,\,M3,\, Maj 3,\, MM 3,\, DMM 3. \end{aligned}$$

Those among these axioms that are not satisfied are mentioned below next to the example label. All axioms from the previous list that are not explicitly mentioned are proved to be satisfied. By default, the examples are complete relations. Relations that are asymmetric are explicitly labeled as such, as well as relations that are neither complete nor asymmetric.

Example 77

($Not[RC2_{i}]$) This is example 25 in Bouyssou and Pirlot (2007). Let $N= \lbrace 1,2 \rbrace$ and $X= \lbrace x, y \rbrace \times \lbrace a, b \rbrace$. Let $\mathrel {\mathcal {R}}$ on $X$ be identical to $X^{2}$ except that, $Not [(y,a) \mathrel {\mathcal {R}}(x,a) ]$ and $Not [(y,b) \mathrel {\mathcal {R}}(x,a) ]$. This relation is complete.

It is easy to check that we have:

• $[(x, y), (x, x), (y,y)] \mathrel {\succ _{1}^{*}} (y, x)$ and

• $[(a,b), (b, b)] \mathrel {\succ _{2}^{*}} [(a, a), (b, a)]$.

Using Lemma 22, it is easy to see that $RC1$ and $RC2_{1}$ hold but that $RC2_{2}$ is violated. Using Lemma 8.1 and 8.2 in Bouyssou and Pirlot (2007) it is clear that $UC$ and $LC$ hold so that the same is true for $M1$ and $M2$. As a consequence of Remark 33, we have that $\mathrel {\mathcal {R}}$ satisfies $Maj 1$ and $Maj 2$. Since $M3$ (resp. $Maj 3$) is entailed by $M2$ (resp. $Maj 2$), $\mathrel {\mathcal {R}}$ also satisfies $M3$ (resp. $Maj 3$). Since $\mathrel {\mathcal {R}}$ satisfies $M1$ (resp. $M2$) it satisfies its relaxed version $MM 1$ (resp. $MM 2)$. As $\mathrel {\mathcal {R}}$ satisfies $M2$ it fulfills $M3$ and $MM 3$. As $\mathrel {\mathcal {R}}$ satisfies $M1$ it fulfills $DMM 3$.

Finally, using Lemma 15 in Bouyssou and Pirlot (2007), it is routine to check that we have:

• $x \mathrel {\succ _{1}^{\pm }} y$,

• $a \mathrel {\succ _{2}^{\pm }} b$.

Hence $AC 1$, $AC 2$ and $AC 3$ hold.

Example 78

($Not [ UC _{i}, M1_{i}, Maj 1_{i}, MM 1_{i} ]$) This is Example 33 in Bouyssou and Pirlot (2005b). Also used in the proof of Part 5 of Bouyssou and Pirlot (2007) Lemma 11 and as Example 23 in Bouyssou and Pirlot (2007).

Let $X= \lbrace a, b \rbrace \times \lbrace x, y, z \rbrace$ and $\mathrel {\mathcal {R}}$ on $X$ be identical to the linear order:

$$\begin{aligned} (a,x) \mathrel {\mathcal {R}}(a,y) \mathrel {\mathcal {R}}(a,z) \mathrel {\mathcal {R}}(b,x) \mathrel {\mathcal {R}}(b,y) \mathrel {\mathcal {R}}(b,z), \end{aligned}$$

except that $(a,z)$ and $(b,x)$ are indifferent: $(a,z)\mathrel {\mathcal {R}}(b,x)$ and $(b,x) \mathrel {\mathcal {R}}(a,z)$ both hold true.

This is a complete relation.

We have, abusing notation,

• $(a, b) \mathrel {\succ _{1}^{*}} [(a, a), (b, b)] \mathrel {\succ _{1}^{*}} (b, a)$ and

• $(x, z) \mathrel {\succ _{2}^{*}} [(x, x), (y, y), (z, z), (x, y), (y, z)] \mathrel {\succ _{2}^{*}} [(y, x), (z, x), (z, y)]$,

• $a \mathrel {\succ _{1}^{\pm }} b$ and $x \mathrel {\succ _{2}^{\pm }} y \mathrel {\succ _{2}^{\pm }} z$.

Using Lemma 22, it is easy to check that $\mathrel {\mathcal {R}}$ satisfies $RC 1$, $RC2$, $AC 1$, $AC 2$, $AC 3$.

It is clear that $UC_{1}$, $LC_{1}$ and $LC_{2}$ hold. $UC_{2}$ is violated since we have $(x, y) \mathrel {\succ _{2}^{*}} (y, x)$ and $Not [(x, y) \mathrel {\succsim _{2}^{*}} (x, z) ]$.

Parts 1 and 2 of Lemma 11 in Bouyssou and Pirlot (2007), show that conditions $M1_{1}$ and $M2$ hold. By Part 3 of Lemma 11 in Bouyssou and Pirlot (2007), $M1_{2}$ cannot hold. Using Lemma 34 shows that $Maj 1_{1}$ and $Maj 2$ hold while $Maj 1_{2}$ does not. Using Lemma 42 shows that $MM 1_{1}$ and $MM 2$ hold while $MM 1_{2}$ does not. Since $\mathrel {\mathcal {R}}$ satisfies $M2$ (resp. $Maj 2$, $MM 2$), this implies that $M3$ (resp. $Maj 3$, $MM 3$) also holds. Since $M1_{1}$ holds, $DMM 3_{1}$ holds too. We show that $DMM 3_{2}$ also holds. Assume the contrary. Taking $RC 1_{2}$ into account, this implies that there are $x_{2}, y_{2}, z_{2}, w_{2} \in X_{1}$ such that $(z_{2},w_{2}) \mathrel {\succ _{2}^{*}} (x_{2},y_{2}) \mathrel {\succ _{2}^{*}} (y_{2},x_{2})$. Hence $(z_{2}, w_{2})$ can only be $(x,z)$. The fourth conclusion of $DMM 3_{2}$ is always true since $(u,x) \mathrel {\mathcal {R}}(v,z)$ for all $u,v \in X_{1} = \{a,b\}$.

Example 79

($Not [ RC 1_{i}, AC 2_{i}, LC _{i}, Maj 2_{i}, Maj 3_{1} ]$) This is Example 12 in Bouyssou and Pirlot (2007). Also used in Example 24 in the same paper.

Let $N= \lbrace 1,2,3 \rbrace$ and $X= \lbrace x, y, z, w \rbrace \times \lbrace a, b \rbrace \times \lbrace p, q \rbrace$. Let $\mathrel {\mathcal {R}}$ on $X$ be identical to $X^{2}$ except that, for all $\alpha _{1}, \beta _{1} \in X_{1}$, all $\alpha _{2}, \beta _{2} \in X_{2}$ and all $\alpha _{3}, \beta _{3} \in X_{3}$ the following pairs are missing:

$$\begin{aligned}&Not [(x, a, \alpha _{3}) \mathrel {\mathcal {R}}(y, b, \beta _{3}) ], \quad Not [(z, \alpha _{2}, p) \mathrel {\mathcal {R}}(w, \beta _{2}, q) ],\\&Not [(x, \alpha _{2}, p) \mathrel {\mathcal {R}}(w, \beta _{2}, q) ], \quad Not [(\alpha _{1}, a, p) \mathrel {\mathcal {R}}(\beta _{1}, b, q) ], \end{aligned}$$

There is a total of 25 such pairs that are marked by a cross in Table 1.

Table 1 Relation $\mathrel {\mathcal {R}}$ in Example 79: the missing pairs are marked by a cross

It is not difficult to check that $\mathrel {\mathcal {R}}$ is complete.

For $i \in \{2, 3\}$, it is easy to check that we have:

$$\begin{aligned}&\displaystyle [(b, a),(a,a),(b,b)] \mathrel {\succ _{2}^{*}} (a,b),\\&\displaystyle [(q, p),(p,p),(q,q)] \mathrel {\succ _{3}^{*}} (p,q),\\&\displaystyle b \mathrel {\succ _{2}^{\pm }} a, q \mathrel {\succ _{3}^{\pm }} p, \end{aligned}$$

which shows,

• using Parts 1 and 2 of Lemma 22, that $RC1_{2}$, $RC1_{3}$, $RC2_{2}$ and $RC2_{3}$ hold,

• using Lemma 47, that $AC 1_{2}$, $AC 1_{3}$, $AC 2_{2}$, $AC 2_{3}$, $AC 3_{2}$ and $AC 3_{3}$ hold.

Using Parts 1 and 2 of Lemma 8 in Bouyssou and Pirlot (2007), it is easy to check that $LC_{2}$, $LC_{3}$, $UC_{2}$ and $UC_{3}$ hold. Hence, using Parts 3 and 4 of Lemma 11 in Bouyssou and Pirlot (2007), we know that $M1_{2}$, $M1_{3}$, $M2_{2}$ and $M2_{3}$ hold. Using Lemma 34, we have also $Maj 1_{2}$, $Maj 1_{3}$, $Maj 2_{2}$ and $Maj 2_{3}$.

On attribute $1$, it is easy to check that we have:

$$\begin{aligned}&\displaystyle (c_{1}, d_{1}) \mathrel {\succ _{1}^{*}} (x,y) \text { and}\\&\displaystyle (c_{1}, d_{1}) \mathrel {\succ _{1}^{*}} [(x, w),(z,w)], \end{aligned}$$

for all $(c_{1}, d_{1}) \in \Gamma = \{ (x, x), (x, z),$ $(y, x), (y, y), (y, z), (y, w),$ $(z, x),$ $(z, y),$ $(z, z),$ $(w, x),$ $(w, y),$ $(w, z),$ $(w, w) \}$. The pairs $(x, w)$ and $(z,w)$ are linked by $\mathrel {\sim _{1}^{*}}$. The pairs $(x,y)$ and $(x, w)$ are not comparable in terms of $\mathrel {\succsim _{1}^{*}}$ since $(x,a,p) \mathrel {\mathcal {R}}(y, a, q)$ and $Not [(x,a,p) \mathrel {\mathcal {R}}(w, a, q) ]$, while $(x,a,p) \mathrel {\mathcal {R}}(w, b, p)$ and $Not [(x,a,p) \mathrel {\mathcal {R}}(y, b, p) ]$. Similarly, the pairs $(x,y)$ and $(z, w)$ are not comparable in terms of $\mathrel {\succsim _{1}^{*}}$. This shows, using Part 1 of Lemma 22, that $RC1_{1}$ is violated.

Using Part 2 of Lemma 22, it is easy to see that $RC2_{1}$ holds. Using Part 1 of Lemma 8 in Bouyssou and Pirlot (2007), shows that $UC_{1}$ holds. Hence, using Part 3 of Lemma 11 in Bouyssou and Pirlot (2007), we know that $M1_{1}$ holds.

In view of Part 6 of Lemma 16 in Bouyssou and Pirlot (2005b), $LC _{1}$ does not hold (since this lemma tells us that $RC 2_{1}, UC _{1}$ and $LC _{1}$ entail $RC 1_{1}$. We now check that $M2_{1}$ holds. The two premises of $M2_{1}$ are that $(a_{1},a_{-1}) \mathrel {\mathcal {R}}(b_{1},b_{-1})$ and $(b_{1},c_{-1}) \mathrel {\mathcal {R}}(a_{1},d_{-1})$. The three possible conclusions of $M2_{1}$ are that $(b_{1},a_{-1}) \mathrel {\mathcal {R}}(a_{1},b_{-1})$ or $(c_{1},a_{-1}) \mathrel {\mathcal {R}}(d_{1},b_{-1})$ or $(c_{1},c_{-1}) \mathrel {\mathcal {R}}(d_{1},d_{-1})$.

Suppose first that $(b_{1}, a_{1}) \in \Gamma$. In this case, we have $(b_{1}, a_{1}) \mathrel {\succsim _{1}^{*}} (a_{1}, b_{1})$, so that $(a_{1},a_{-1}) \mathrel {\mathcal {R}}(b_{1},b_{-1})$ implies $(b_{1},a_{-1}) \mathrel {\mathcal {R}}(a_{1},b_{-1})$. Hence, the first conclusion of $M2_{1}$ holds.

Suppose now that $(b_{1}, a_{1}) = (x, y)$.

If $(c_{1}, d_{1})$ is distinct from $(x, w)$ and $(z,w)$, we have $(c_{1}, d_{1}) \mathrel {\succsim _{1}^{*}} (x, y)$, so that $(b_{1},c_{-1}) \mathrel {\mathcal {R}}(a_{1},d_{-1})$ implies $(c_{1},c_{-1}) \mathrel {\mathcal {R}}(d_{1},d_{-1})$ and the third conclusion of $M2_{1}$ holds.

If $(c_{1}, d_{1})=(x, w)$, it is easy to check that there are no $a_{-1}, b_{-1} \in X_{-i}$ such that $(y,a_{-1}) \mathrel {\mathcal {R}}(x, b_{-1})$, $Not [(x,a_{-1}) \mathrel {\mathcal {R}}(y,b_{-1}) ]$ and $Not [(x,a_{-1}) \mathrel {\mathcal {R}}(w,b_{-1}) ]$, so that no violation of $M2_{1}$ is possible in this case. Since $(x, w) \mathrel {\sim _{1}^{*}} (z, w)$, the same is true if $(c_{1}, d_{1})$ $=$ $(z, w)$.

This shows that $M2_{1}$ cannot be violated if $(b_{1}, a_{1}) = (x, y)$. A similar reasoning shows that $M2_{1}$ cannot be violated if $(b_{1}, a_{1}) = (x, w)$ or if $(b_{1}, a_{1}) = (z, w)$. Hence, $M2_{1}$ holds and so does $M3_{1}$.

Using Remark 33, we know that $\mathrel {\mathcal {R}}$ satisfies $Maj 1_{1}$ since $UC _{i}$ entails $Maj 1_{i}$.

Since $\mathrel {\mathcal {R}}$ satisfies $RC 2_{1}$, $M1_{1}$ but not $RC 1_{1}$, it cannot satisfy $Maj 2_{1}$, as a consequence of Lemma 35. Since $\mathrel {\mathcal {R}}$ satisfies $M1$ and $M2$, it also satisfies $MM 1$, $DMM 3$, $MM 2$, $M3$ and $MM 3$.

$Maj 3_{2}$ (resp. $Maj 3_{3}$) holds because $Maj 2_{2}$ (resp. $Maj 2_{3}$) holds but $Maj 3_{1}$ is violated as shown by the following configuration (which also confirms that $Maj 2_{1}$ is violated): $(y,a,p) \mathrel {\mathcal {R}}(x,b,p)$, $(w,a,p) \mathrel {\mathcal {R}}(z,a,p)$, $(w,a,p) \mathrel {\mathcal {R}}(z,a,q)$, $(z,a,p) \mathrel {\mathcal {R}}(w,a,p)$, $Not [(x,a,p) \mathrel {\mathcal {R}}(y,b,p) ]$, $Not [(z,a,p) \mathrel {\mathcal {R}}(w,a,q) ]$.

On attribute 1, it is easy to check that we have:

$$\begin{aligned} \{y, w\} \mathrel {\succ _{1}^{+}} z \mathrel {\succ _{1}^{+}} x. \end{aligned}$$

Hence $AC 1_{1}$ holds. Since $(x, w)$ and $(x, y)$ are not comparable w.r.t. $\mathrel {\succsim _{1}^{*}}$, $y$ and $w$ are not comparable w.r.t. $\mathrel {\succ _{1}^{-}}$, hence $AC 2_{1}$ is violated. It is easy to check, using Lemma 15 in Bouyssou and Pirlot (2007), that $AC3_{1}$ is satisfied.

Remark The co-dual of Example 79 is an asymmetric relation that satisfies all axioms but $RC 1, AC 1, UC$ and $M1$.

Example 80

($Not [ RC 1_{i}, AC 1_{i}, UC _{i}, M1_{i}, DMaj 3_{i} ]$) The co-dual of the relation $\mathrel {\mathcal {R}}$ described in Table 2 is a complete relation satisfying all properties except for $RC 1$, $AC 2$, $M1$ and $DMaj 3$. Note that this relation verifies $DM 3$ and $DMM 3$.

Table 2 Relation $\mathrel {\mathcal {R}}$ in Example 80

Let $X= \{x,y,z\}\times \{a,b\}\times \{p,q\}$ and $\mathrel {\mathcal {R}}$ consist of the set of pairs listed in Table 2. We have to show that $\mathrel {\mathcal {R}}$ satisfies all properties but $RC 1$, $AC 1$, $LC$, $Maj 2$, $Maj 3$. It is easy to check that $\mathrel {\mathcal {R}}$ is asymmetric. As for the comparison of preference differences on each attribute, we have for all $(\alpha , \beta ) \in \Gamma = \{ (x,x), (y,y), (z,z), (x,y), (x,z), (y,z), (z,y)\}$,

• $[(\alpha , \beta )] \mathrel {\succ _{1}^{*}} (y,x)$ and $[(\alpha , \beta )] \mathrel {\succ _{1}^{*}} (z,x)$, while $(y,x)$ and $(z,x)$ are incomparable in terms of $\mathrel {\succsim _{1}^{*}}$,

• $(a,b) \mathrel {\succ _{2}^{*}} [(a,a), (b,b)] \mathrel {\succ _{2}^{*}} (b,a)$,

• $(p,q) \mathrel {\succ _{3}^{*}} [(p,p), (q,q)] \mathrel {\succ _{3}^{*}} (q,p)$.

The upward and downward dominance relations on attributes $2$ and $3$ are as follows:

• $a \mathrel {\succ _{2}^{\pm }} b$,

• $p \mathrel {\succ _{3}^{\pm }} q$.

On attribute $1$, we have:

• $x \mathrel {\succ _{1}^{+}} y$, $x \mathrel {\succ _{i}^{+}}z$,

• $y$ and $z$ are not comparable w.r.t. $\mathrel {\succsim _{i}^{+}}$ since, on the one hand, $zap \mathrel {\mathcal {R}}xaq$ and $Not [yap \mathrel {\mathcal {R}}xaq ]$, and on the other hand, $yap \mathrel {\mathcal {R}}xbp$ and $Not [zap \mathrel {\mathcal {R}}xbp ]$,

• $x \mathrel {\succ _{1}^{-}} y \mathrel {\succ _{i}^{-}}z$.

For $j \in \{2,3\}$, $RC 1_{j}$, $RC 2_{j}$, $AC 1_{j}$, $AC 2_{j}$, $AC 3_{j}$, $UC _{j}$ and $LC _{j}$ are clearly satisfied, implying $M1_{j}$ and $M2_{j}$ (see Remark 26) as well as $Maj 1_{j}$ and $Maj 2_{j}$ (see Remark 33), $MM 1_{j}$ and $MM 2_{j}$. On attribute $1$ it is easy to check that $RC 2_{1}$, $AC 2_{1}$ and $AC 3_{1}$ are verified while $RC 1_{1}$ and $AC 1_{1}$ are violated. Using Lemma 8(1) in Bouyssou and Pirlot (2007), we observe that $UC _{1}$ is satisfied, implying $M1_{1}$, $Maj 1_{1}$ and $MM 1_{1}$. $LC _{1}$ does not hold but $M2_{1}$ does as we shall see.

Using the notation in condition (20), for establishing $M2_{1}$ we only have to consider the cases in which $(y_{1},x) = (y,x)$ or $(y_{1},x) = (z,x)$ (otherwise $(y_{1},x) \mathrel {\succsim _{1}^{*}} (x, y_{1})$ and consequently the first conclusion of (20) is satisfied). Assume that $(y_{1},x) = (y,x)$ and that the first conclusion is not satisfied. This means that either $a_{-1} = ap$ and $b_{-1} = aq$ or $a_{-1} = bp$ and $b_{-1} = bq$. We now distinguish two cases regarding $(z_{i},w_{i})$:

• 1. if $(z_{i}, w_{i}) \ne (z,x)$, the third conclusion is always satisfied because of the second premise and the fact that $(z_{i}, w_{i}) \mathrel {\succsim _{1}^{*}} (y,x)$,

• 2. if $(z_{i}, w_{i}) = (z,x)$ and $a_{-1} = ap$ and $b_{-1} = aq$ or $a_{-1} = bp$ and $b_{-1} = bq$, the second conclusion is satisfied because we have $zap \mathrel {\mathcal {R}}xaq$ and $zbp \mathrel {\mathcal {R}}xbq$.

The case in which $(y_{1}, x) = (z,x)$ is dealt with similarly. Consequently, $\mathrel {\mathcal {R}}$ satisfies $M2_{1}$, $MM 2_{1}$, $M3_{1}$ and $MM 3_{1}$.

For establishing that $Maj 2_{1}$ does not hold, we consider the case in which $(y_{1}, x) = (y,x)$ and use the notation of (22) in Definition 30. In the previous analysis we only need to reconsider the case in which the second conclusion of $M2_{i}$ was used, i.e., when $(z_{i}, w_{i}) = (z,x)$ and $a_{-1} = ap$ and $b_{-1} = aq$ or $a_{-1} = bp$ and $b_{-1} = bq$. We have $xap \mathrel {\mathcal {R}}yaq$, $Not [yap \mathrel {\mathcal {R}}xaq ]$, $xap \mathrel {\mathcal {R}}zaq$, $yap \mathrel {\mathcal {R}}xbp$ and $Not [zap \mathrel {\mathcal {R}}xbp ]$, which means that $Maj 2_{1}$ does not hold. Since $(z,x)$ is no veto (as we have, e.g., $zap \mathrel {\mathcal {R}}xaq$), the latter also shows that $\mathrel {\mathcal {R}}$ does not satisfy $Maj 3_{1}$.

Example 81

($Not [ AC 1_{i}, AC 2_{i} ]$) This is Example 36 in Bouyssou and Pirlot (2005b). Also used in Example 21 in Bouyssou and Pirlot (2007).

Let $X=\lbrace a, b, c, d \rbrace \times \lbrace x, y \rbrace$. We build $\mathrel {\mathcal {R}}$ as the CR in which:

• $a \mathrel {I_{1}} b$, $a \mathrel {P_{1}} c$, $a \mathrel {I_{1}} d$, $b \mathrel {I_{1}} c$, $b \mathrel {P_{1}} d$, $c \mathrel {I_{1}} d$,

• $x \mathrel {P_{2}} y$,

• $\lbrace 1, 2 \rbrace \mathrel {\rhd }\varnothing$, $\lbrace 1,2 \rbrace \mathrel {\triangleq }\lbrace 2 \rbrace$, $\lbrace 1,2 \rbrace \mathrel {\triangleq }\lbrace 1 \rbrace$, $\lbrace 2 \rbrace \mathrel {\triangleq }\lbrace 1 \rbrace$.

Therefore, $\mathrel {\mathcal {R}}$ links any two elements of $X$ except that we have: $(a, x) \mathrel {\mathcal {R}}(c, y)$ but $Not [(c, y) \mathrel {\mathcal {R}}(a, x) ]$ and $(b, x) \mathrel {\mathcal {R}}(d, y)$ but $Not [(d, y) \mathrel {\mathcal {R}}(b, x) ]$. Hence $\mathrel {\mathcal {R}}$ is a complete relation. Since it is a CR, it satisfies $RC 1$, $RC 2$, $UC$, $LC$, $M1$, $M2$, $Maj 1$, $Maj 2$, $MM 1$, $MM 2$, $M3$, $Maj 3$, $MM 3$ and $DMM 3$.

It is easy to see that $AC3$ and $AC1_{2}$ as well as $AC 2_{2}$ hold. $AC1_{1}$ is violated since $(d, y) \mathrel {\mathcal {R}}(a, x)$ and $(c, y) \mathrel {\mathcal {R}}(b, x)$ but neither $(c, y) \mathrel {\mathcal {R}}(a, x)$ nor $(d, y) \mathrel {\mathcal {R}}(b, x)$. $AC 2_{1}$ is also violated (Part 1 of Lemma 27 in Bouyssou and Pirlot 2005b).

Example 82

($Not [AC3_{i} ]$) This is Example 35 in Bouyssou and Pirlot (2005b). Also used as Example 20 in Bouyssou and Pirlot (2007).

Let $X=\lbrace a, b, c, d \rbrace \times \lbrace x, y \rbrace$. We build the CR in which:

• $a \mathrel {P_{1}} b$, $a \mathrel {I_{1}} c$, $a \mathrel {P_{1}} d$, $b \mathrel {I_{1}} c$, $b \mathrel {P_{1}} d$, $c \mathrel {I_{1}} d$,

• $x \mathrel {P_{2}} y$,

• $\lbrace 1, 2 \rbrace \mathrel {\rhd }\varnothing$, $\lbrace 1,2 \rbrace \mathrel {\triangleq }\lbrace 2 \rbrace$, $\lbrace 1,2 \rbrace \mathrel {\triangleq }\lbrace 1 \rbrace$, $\lbrace 2 \rbrace \mathrel {\triangleq }\lbrace 1 \rbrace$.

Therefore, $\mathrel {\mathcal {R}}$ links any two elements of $X$ except that we have: $(a, x) \mathrel {\mathcal {R}}(b, y)$ but $Not [(b, y) \mathrel {\mathcal {R}}(a, x) ]$, $(b, x) \mathrel {\mathcal {R}}(d, y)$ but $Not [(d, y) \mathrel {\mathcal {R}}(b, x) ]$ and $(a, x) \mathrel {\mathcal {R}}(d, y)$ but $Not [(d, y) \mathrel {\mathcal {R}}(a, x) ]$. Hence $\mathrel {\mathcal {R}}$ is a complete relation. Since it is a $CR$, it satisfies $RC 1$, $RC 2$, $UC$, $LC$, $M1$, $M2$, $Maj 1$, $Maj 2$, $MM 1$, $MM 2$, $M3$, $Maj 3$, $MM 3$ and $DMM 3$.

It is easy to see that $AC1$ holds and, hence, $AC2$ (by Part 1 of Lemma 27 in Bouyssou and Pirlot 2005b). One verifies that $AC3_{2}$ holds. $AC3_{1}$ is violated since $(c, y) \mathrel {\mathcal {R}}(a, x)$, $(d, y) \mathrel {\mathcal {R}}(c, x)$ but neither $(b, y) \mathrel {\mathcal {R}}(a, x)$ nor $(d, y) \mathrel {\mathcal {R}}(b, x)$.

Remark. The co-dual of this relation is an asymmetric relation that satisfies all axioms of a CR-AT except $AC 3_{1}$. In particular, it satisfies $M3$ and $Maj 3$ since it satisfies $M2$ and $Maj 2$.

Example 83

($Not [LC_{i}, M2_{i}, Maj 2_{i}, MM 2_{i}, M3_{i}, Maj 3_{i}, MM 3_{i} ]$) This is Example 38 in Bouyssou and Pirlot (2009b). It is used in Remark 16 in Bouyssou and Pirlot (2009a) (but erroneously referred to as Example 39 in Bouyssou and Pirlot 2009b).

Let $X= X_{1} \times X_{2} \times X_{3}$ with $X_{1} = \{x, y, z\}$, $X_{2} = \{a, b\}$ and $X_{3} = \{p, q\}$. Let us consider the relation $\mathrel {\mathcal {R}}$ such that:

$$\begin{aligned} x \mathrel {\mathcal {R}}y \iff \sum _{i=1}^{3} p_{i}(x_{i}, y_{i}) \ge 0, \end{aligned}$$

the functions $p_{i}$ being such that:

$$\begin{aligned} p_{1}(x, y)&= p_{1}(x, z) = p_{1}(y, z) = p_{1}(x, x) = p_{1}(y, y) = p_{1}(z, z) = 4, \\ p_{1}(y, x)&= p_{1}(z, y) = -1, p_{1}(z, x) = -4,\\ p_{2}(a, b)&= 2, p_{2}(a, a) = p_{2}(b, b) = 0, p_{2}(b, a) = -2,\\ p_{3}(a, b)&= 2, p_{3}(p, p) = p_{3}(q, q) = 0, p_{3}(q, p) = -2. \end{aligned}$$

This is a complete relation. Indeed if $\sum _{i=1}^{3} p_{i}(x_{i}, y_{i}) < 0$, then $p_{1}(x,y_{1}) <4$. This implies that $p_{1}(y_{1},x)=4$, hence $\sum _{i=1}^{3} p_{i}(y_{i}, x_{i}) \ge 0$.

It is easily checked that we have (with $(\alpha , \alpha )$ standing for $(x,x)$, $(y,y)$ and $(z,z)$):

$$\begin{aligned}&\displaystyle [(x,y)\mathrel {\sim _{1}^{*}} (x,z)\mathrel {\sim _{1}^{*}} (y,z)\mathrel {\sim _{1}^{*}} (\alpha ,\alpha )] \mathrel {\succ _{1}^{*}} [(y, x) \mathrel {\sim _{1}^{*}} (z, y)] \mathrel {\succ _{1}^{*}} (z, x),\\&\displaystyle x \mathrel {\succ _{1}^{\pm }} y \mathrel {\succ _{1}^{\pm }} z,\\&\displaystyle (a, b) \mathrel {\succ _{2}^{*}} [(a, a) \mathrel {\sim _{2}^{*}} (b, b)] \mathrel {\succ _{2}^{*}} (b, a),\\&\displaystyle a \mathrel {\succ _{2}^{\pm }} b,\\&\displaystyle (p, q) \mathrel {\succ _{3}^{*}} [(p, p) \mathrel {\sim _{2}^{*}} (q, q)] \mathrel {\succ _{2}^{*}} (q, p),\\&\displaystyle p \mathrel {\succ _{3}^{\pm }} q. \end{aligned}$$

This shows that $RC 1$, $RC 2$, $AC 1$, $AC 2$ and $AC 3$ are satisfied. Using Parts 1 of Lemma 8 and Lemma 11 in Bouyssou and Pirlot (2007) shows that $UC$ and $M1$ hold. Similarly, using Parts 2 of Lemma 8 and Lemma 11 in Bouyssou and Pirlot (2007) shows that $\mathrel {\mathcal {R}}$ satisfies $UC_{2}$, $UC_{3}$, $M2_{2}$ and $M2_{3}$, which implies that $M3_{2}$ and $M3_{3}$ hold. Condition $M3_{1}$ is violated since $(x, b, q) \mathrel {\mathcal {R}}(y, a, p)$, $(y, a, q) \mathrel {\mathcal {R}}(x, b, q)$ and $(z, a, p) \mathrel {\mathcal {R}}(x, b, q)$ while $Not [(y, b, q) \mathrel {\mathcal {R}}(x, a, p) ]$, $Not [(z, b, q) \mathrel {\mathcal {R}}(x, a, p) ]$ and $Not [(z, a, q) \mathrel {\mathcal {R}}(x, b, q) ]$.

Hence $M2_{1}$ is violated too. Lemma 11, Part 2, in Bouyssou and Pirlot (2007) implies that $LC_{1}$ is also violated. Using Lemmas 34, 42 and 56, we obtain that $\mathrel {\mathcal {R}}$ satisfies $Maj 1$, $MM 1$, $Maj 2_{2}$, $Maj 2_{3}$, $MM 2_{2}$, $MM 2_{3}$, $Maj 3_{2}$, $Maj 3_{3}$, $MM 3_{2}$, $MM 3_{3}$ but neither $Maj 2_{1}$ nor $Maj 3_{1}$. In view of Lemmas 56, $MM 3_{1}$ is also violated as well as $MM 2_{1}$. Since $\mathrel {\mathcal {R}}$ satisfies $M1$, it fulfills $DMM 3$.

Example 84

(Asymmetric, $Not [ LC _{i}, M2_{i}, M3_{i}, Maj 2_{i}, Maj 3_{i}, MM 2_{i}, MM 3_{i} ]$) This is Example 5 in Bouyssou and Pirlot (2006).

Let $X= \lbrace x, y, z \rbrace \times \lbrace a, b \rbrace \times \lbrace p, q \rbrace$ and $\mathrel {\mathcal {R}}$ on $X$ be as described in Table 3:

Table 3 Relation $\mathrel {\mathcal {R}}$ in Example 84

It is easy to check that $\mathrel {\mathcal {R}}$ is asymmetric. It is not difficult to see that we have, abusing notation,

• $[(x, y), (x, z), (y, z)] \mathrel {\succ _{1}^{*}} [(x, x), (y, y), (z, z), (y, x), (z, y)] \mathrel {\succ _{1}^{*}} (z, x)$,

• $(a, b) \mathrel {\succ _{2}^{*}} [(a, a), (b, b)] \mathrel {\succ _{2}^{*}} (b, a)$, and

• $(p, q) \mathrel {\succ _{3}^{*}} [(p, p), (q, q)] \mathrel {\succ _{3}^{*}} (q, p)$.

This shows that $RC 1$, $RC 2$ and $Maj 1$ hold. It is easy to see that $Maj 2_{2}$ and $Maj 2_{3}$ hold so that $Maj 3_{2}$ and $Maj 3_{3}$ are satisfied. Condition $Maj 3_{1}$ is violated since $(x,a,p)\mathrel {\mathcal {R}}(y,a,p)$, $(x,a,p)\mathrel {\mathcal {R}}(z,a,p)$, $(y,a,p)\mathrel {\mathcal {R}}(x,b,p)$ and $(z,a,p)\mathrel {\mathcal {R}}(x,b,q)$ but neither $(y,a,p)\mathrel {\mathcal {R}}(x,a,p)$ nor $(z,a,p)\mathrel {\mathcal {R}}(x,b,p)$.

Since $RC 1$ and $RC 2$ hold, Lemma 34 implies that $\mathrel {\mathcal {R}}$ satisfies $M1$, $M2_{2}$, $M2_{3}$ but not $M2_{1}$. $M3_{2}$ and $M3_{3}$ hold while $M3_{1}$ is violated (Lemma 56). By Lemma 11 in Bouyssou and Pirlot (2007), $\mathrel {\mathcal {R}}$ satisfies $UC$, $LC _{2}$, $LC _{3}$ but not $LC _{1}$. By Lemmas 42 and 56, we know that $MM 2_{1}$ and $MM 3_{1}$ are violated. $MM 2_{j}$ and $MM 3_{j}$ are satisfied for $j=2,3$. $\mathrel {\mathcal {R}}$ satisfies $MM 1$ so that it also satisfies $DMM 3$.

From relations $\mathrel {\succsim _{i}^{*}}$ described above, we infer the following:

$$\begin{aligned}&\displaystyle x \mathrel {\succ _{1}^{\pm }} y \mathrel {\succ _{1}^{\pm }} z,\\&\displaystyle a \mathrel {\succ _{2}^{\pm }} b,\\&\displaystyle p \mathrel {\succ _{3}^{\pm }} q, \end{aligned}$$

which implies that $\mathrel {\mathcal {R}}$ satisfies $AC 1$, $AC 2$ and $AC 3$.

Example 85

( $Not [ AC 1_{i}, M2_{i}, Maj 2_{i}, MM 2_{i}, LC _{i} ]$) This is Example 36 in Bouyssou and Pirlot (2009b).

Let $X= X_{1} \times X_{2}$ with $X_{1} = \{x, y, z, w \}$ and $X_{2} = \{a, b \}$. We build a CDR on $X$ with:

• $z \mathrel {P_{1}} x$, $z \mathrel {P_{1}} y$, $z \mathrel {P_{1}} w$, $x \mathrel {P_{1}} w$, $x \mathrel {I_{1}} y$, $y \mathrel {I_{1}} w$,

• the relation $\mathrel {V_{1}}$ is empty except that $z \mathrel {V_{1}} y$,

• $b \mathrel {P_{2}} a$,

• the relation $\mathrel {V_{2}}$ is empty,

• $\{1, 2\} \mathrel {\rhd }\varnothing$, $\{1, 2\} \mathrel {\triangleq }\{2\}$, $\{1, 2\} \mathrel {\triangleq }\{1\}$ and $\{1\} \mathrel {\triangleq }\{2\}$.

By construction, $\mathrel {\mathcal {R}}$ is a CDR. Hence, it satisfies $RC 1$, $RC 2$, $M1$, $Maj 1$, $MM 1$, $M3$, $Maj 3$ and $MM 3$ (Theorem 58 and Lemmas 42 and 56). It satisfies $M2_{2}$, $Maj 2_{2}$, $MM 2_{2}$, but not $M2_{1}$ (due to the veto on $X_{1}$), not $Maj 2_{1}$ (by Lemma 34) and not $MM 2_{1}$ (by Lemma 42). Since $\mathrel {\mathcal {R}}$ satisfies $M1$ it also fulfills $DMM 3$. Using Lemma 11 in Bouyssou and Pirlot (2007), we get that $UC$ and $LC _{2}$ are satisfied but $LC _{1}$ is violated.

The relation $\mathrel {\mathcal {R}}$ contains all pairs in $X\times X$ except the following ones:

• $Not [(x, b) \mathrel {\mathcal {R}}(z, a) ]$, $Not [(y, b) \mathrel {\mathcal {R}}(z, a) ]$, $Not [(w, b) \mathrel {\mathcal {R}}(z, a) ]$, $Not [(w, b) \mathrel {\mathcal {R}}(x, a) ]$, due to the fact that $Not [\varnothing \mathrel {\unrhd }\{1, 2\} ]$, and

• $Not [(y, a) \mathrel {\mathcal {R}}(z, a) ]$, $Not [(y, b) \mathrel {\mathcal {R}}(z, b) ]$, $Not [(y, b) \mathrel {\mathcal {R}}(z, a) ]$, $Not [(y, a) \mathrel {\mathcal {R}}(z, b) ]$, due to the fact that $z \mathrel {V_{1}} y$.

One pair is common to these two series of four pairs, so that $\mathrel {\mathcal {R}}$ is equal to $X\times X$ minus the seven distinct pairs in the lists above. It is a complete relation.

On $X_{2}$, it is easy to check that we have $b \mathrel {\succ _{2}^{\pm }} a$, so that $AC 1_{2}$, $AC 2_{2}$ and $AC 3_{2}$ hold.

On $X_{1}$, it is easy to check that $\mathrel {\succsim _{1}^{-}}$ is complete. We indeed have that:

$$\begin{aligned} z \mathrel {\succ _{1}^{-}} x \mathrel {\succ _{1}^{-}} [y \mathrel {\sim _{1}^{-}} w]. \end{aligned}$$

The relation $\mathrel {\succsim _{1}^{+}}$ is not complete. We have $z \mathrel {\succ _{1}^{+}} x$, $x \mathrel {\succ _{1}^{+}} y$ and $x \mathrel {\succ _{1}^{+}} w$ but neither $y\mathrel {\succsim _{1}^{+}} w$ nor $w \mathrel {\succsim _{1}^{+}} y$ since $(y, b) \mathrel {\mathcal {R}}(x, a)$ but $Not [(w, b) \mathrel {\mathcal {R}}(x, a) ]$ and $(w, a) \mathrel {\mathcal {R}}(z, a)$ but $Not [(y, a) \mathrel {\mathcal {R}}(z, a) ]$. This shows that $AC 1_{1}$ is violated. Condition $AC 3_{1}$ holds since $\mathrel {\succsim _{1}^{+}}$ and $\mathrel {\succsim _{1}^{-}}$ are not incompatible.

Example 86

( $Not [ AC 2_{i}, M2_{i}, Maj 2_{i}, MM 2_{i}, LC _{i} ]$)

This is Example 35 in Bouyssou and Pirlot (2009b). It is a slight variation on Example 85 obtained by reversing all relations $\mathrel {S_{i}}$ and $\mathrel {V_{i}}$.

Let $X= X_{1} \times X_{2}$ with $X_{1} = \{x, y, z, w \}$ and $X_{2} = \{a, b \}$. We build a CDR on $X$ with:

• $w \mathrel {P_{1}} z$, $x \mathrel {P_{1}} z$, $y \mathrel {P_{1}} z$, $w \mathrel {P_{1}} x$, $y \mathrel {I_{1}} w$, $y \mathrel {I_{1}} x$ (and all $\mathrel {I_{1}}$ loops),

• $\mathrel {V_{1}}$ is empty except that $y \mathrel {V_{1}} z$,

• $a \mathrel {P_{2}} b$ (and all $\mathrel {I_{2}}$ loops) and the relation $\mathrel {V_{2}}$ is empty,

• $\{1, 2\} \mathrel {\rhd }\varnothing$, $\{1, 2\} \mathrel {\triangleq }\{2\}$, $\{1, 2\} \mathrel {\triangleq }\{1\}$ and $\{1\} \mathrel {\triangleq }\{2\}$.

Observe that $\mathrel {S_{1}}$ is a semiorder (the weak order it induces ranks the elements of $X_{1}$ in the following order: $w, y, x, z$). The relation $\mathrel {V_{1}}$ is a strict semiorder that is included in $\mathrel {P_{1}}$. But $(\mathrel {S_{1}}, \mathrel {U_{1}})$ is not an homogeneous chain of semiorders on $X_{1}$ since the weak order induced by $\mathrel {U_{1}}$ ranks $y$ before $w$, while the weak order induced by $\mathrel {S_{1}}$ does the opposite.

By construction, $\mathrel {\mathcal {R}}$ is a CDR. Hence, it satisfies $RC 1$, $RC 2$, $M1$, $Maj 1$, $MM 1$, $M3$, $Maj 3$ and $MM 3$ (Theorem58 and Lemmas 42 and 56). It satisfies $M2_{2}$, $Maj 2_{2}$, $MM 2_{2}$, but not $M2_{1}$ (due to the veto on $X_{1}$), not $Maj 2_{1}$ (by Lemma 34) and not $MM 2_{1}$ (by Lemma 42). Since $\mathrel {\mathcal {R}}$ satisfies $M1$ it also fulfills $DMM 3$. Using Lemma 11 in Bouyssou and Pirlot (2007), we get that $UC$ and $LC _{2}$ are satisfied but $LC _{1}$ is violated.

The relation $\mathrel {\mathcal {R}}$ contains all pairs in $X\times X$ except the following ones:

• $Not [(z, b) \mathrel {\mathcal {R}}(w, a) ]$, $Not [(z, b) \mathrel {\mathcal {R}}(x, a) ]$, $Not [(z, b) \mathrel {\mathcal {R}}(y, a) ]$, $Not [(x, b) \mathrel {\mathcal {R}}(w, a) ]$, due to the fact that $Not [\varnothing \mathrel {\unrhd }\{1, 2\} ]$, and

• $Not [(z, a) \mathrel {\mathcal {R}}(y, a) ]$, $Not [(z, a) \mathrel {\mathcal {R}}(y, b) ]$, $Not [(z, b) \mathrel {\mathcal {R}}(y, a) ]$, $Not [(z, b) \mathrel {\mathcal {R}}(y, b) ]$, due to the fact that $y \mathrel {V_{1}} z$.

One pair is common to these two series of four pairs, so that $\mathrel {\mathcal {R}}$ is equal to $X\times X$ minus the seven distinct pairs in the lists above. It is clear that $\mathrel {\mathcal {R}}$ is complete.

On $X_{2}$, it is easy to check that we have $a \mathrel {\succ _{2}^{\pm }} b$, so that $AC 1_{2}$, $AC 2_{2}$ and $AC 3_{2}$ hold.

On $X_{1}$, it is easy to check that $\mathrel {\succsim _{1}^{+}}$ is complete. We indeed have that:

$$\begin{aligned}{}[y \mathrel {\sim _{1}^{+}} w] \mathrel {\succ _{1}^{+}} x \mathrel {\succ _{1}^{+}} z. \end{aligned}$$

The relation $\mathrel {\succsim _{1}^{-}}$ is not complete. We have $w \mathrel {\succ _{1}^{-}} x$, $y \mathrel {\succ _{1}^{-}} x$ and $x \mathrel {\succ _{1}^{-}} z$ but neither $y\mathrel {\succsim _{1}^{-}} w$ nor $w \mathrel {\succsim _{1}^{-}} y$ since $(z, a) \mathrel {\mathcal {R}}(w, a)$ but $Not [(z, a) \mathrel {\mathcal {R}}(y, a) ]$ and $(x, b) \mathrel {\mathcal {R}}(y, a)$ but $Not [(x, b) \mathrel {\mathcal {R}}(w, a) ]$. This shows that $AC 2_{1}$ is violated. Condition $AC 3_{1}$ holds since $\mathrel {\succsim _{1}^{+}}$ and $\mathrel {\succsim _{1}^{-}}$ are not incompatible.

Example 87

(Asymmetric, $Not [UC_{i}, M1_{i}, Maj 1_{i}, MM 1_{i} ]$) This is Example 3 in Bouyssou and Pirlot (2006).

Let $X=\lbrace a, b \rbrace \times \lbrace x, y, z \rbrace$ and $\mathrel {\mathcal {R}}$ on $X$ be identical to the strict linear order (abusing notation in an obvious way):

$$\begin{aligned} (a,x) \mathrel {\mathcal {R}}(b,x) \mathrel {\mathcal {R}}(a,y) \mathrel {\mathcal {R}}(b,y) \mathrel {\mathcal {R}}(a,z) \mathrel {\mathcal {R}}(b,z), \end{aligned}$$

except that we have also $(a,y) \mathrel {\mathcal {R}}(b,x)$.

It is easy to see that $\mathrel {\mathcal {R}}$ is asymmetric. We have, abusing notation:

• $(a, b) \mathrel {\succ _{1}^{*}} [(a, a), (b, b)] \mathrel {\succ _{1}^{*}} (b, a)$, and

• $[(x, z), (y, z)] \mathrel {\succ _{2}^{*}} (x, y) \mathrel {\succ _{2}^{*}} [(x, x), (y, y), (z, z)] \mathrel {\succ _{2}^{*}} [(y, x), (z, x), (z, y)]$.

Using Lemma 22, it is easy to check that $\mathrel {\mathcal {R}}$ satisfies $RC 1$ and $RC 2$.

It is clear that $UC_{1}$, $LC_{1}$ and $LC_{2}$ hold. This implies that $M2$, $Maj 2$, $MM 2$, $M3$, $Maj 3$, $MM 3$ hold as well as $M1_{1}$, $Maj 1_{1}$ and $MM 1_{1}$.

$Maj 1_{2}$ is violated since $(a,x)\mathrel {\mathcal {R}}(a,y)$, $(a,x)\mathrel {\mathcal {R}}(a,z)$, $(b,x)\mathrel {\mathcal {R}}(a,z)$ but neither $(a,y)\mathrel {\mathcal {R}}(a,x)$ nor $(b,x)\mathrel {\mathcal {R}}(a,y)$. As a consequence, $UC_{1}$ is also violated. Since $RC 1$ and $RC 2$ hold, Lemmas 34 and 42 imply that $M1_{2}$ and $MM 1_{2}$ are violated.

Since $M1_{1}$ holds, $DMM 3_{1}$ holds too. We show that $DMM 3_{2}$ also holds. Assume the contrary. Taking $RC 1_{2}$ into account, this implies that there are $a, b, z_{2}, w_{2} \in X_{2}$ such that $(z_{2},w_{2}) \mathrel {\succ _{2}^{*}} (a,b) \mathrel {\succ _{2}^{*}} (b,a)$. Hence $(z_{2}, w_{2})$ can only be $(x,z)$ or $(y,z)$. The fourth conclusion of $DMM 3_{2}$ is always true since $(u,x) \mathrel {\mathcal {R}}(v,z)$ and $(u,y) \mathrel {\mathcal {R}}(v,z)$ for all $u,v \in X_{1}=\{a,b\}$.

Using Lemma 47, we have:

• $a \mathrel {\succsim _{1}^{\pm }} b$ and

• $x \mathrel {\succsim _{2}^{\pm }} y \mathrel {\succsim _{3}^{\pm }} z$.

Hence $AC 1$, $AC 2$ and $AC 3$ hold.

Example 88

(Asymmetric, $Not [ AC 1_{i}, M2_{i}, Maj 2_{i}, MM 2_{i}, LC _{i} ]$) This example is an asymmetric variant of Example 85.

Let $X= X_{1} \times X_{2}$ with $X_{1} = \{x, y, z, w \}$ and $X_{2} = \{a, b \}$. We build a CDR on $X$ with:

• $z \mathrel {P^{\circ }_{1}} x$, $z \mathrel {P^{\circ }_{1}} y$, $z \mathrel {P^{\circ }_{1}} w$, $x \mathrel {P^{\circ }_{1}} w$, $y \mathrel {P^{\circ }_{1}} w$,

• the relation $\mathrel {V^{\circ }_{1}}$ is empty except that $z \mathrel {V^{\circ }_{1}} y$,

• $b \mathrel {P^{\circ }_{2}} a$,

• the relation $\mathrel {V^{\circ }_{2}}$ is empty,

• $\{1\} \mathrel {\rhd ^{\circ }}\varnothing$, $\{2\} \mathrel {\rhd ^{\circ }}\{1\}$, $\{1,2\} \mathrel {\rhd ^{\circ }}\varnothing$.

By construction, $\mathrel {\mathcal {R}}$ is an asymmetric CDR. Hence, it satisfies $RC 1$, $RC 2$, $M1$, $Maj 1$, $MM 1$, $M3$, $Maj 3$, $MM 3$ (Theorem 58 and Lemmas 42 and 56). It satisfies $M2_{2}$ but not $M2_{1}$ (due to the veto on $X_{1}$), not $Maj 2_{1}$ (by Lemma 42) and not $MM 2_{1}$ (by Lemma 42). Using Lemma 11 in Bouyssou and Pirlot (2007), we get that $UC$ and $LC_{2}$ are satisfied but $LC _{1}$ is violated.⁷

Since $\mathrel {\mathcal {R}}$ satisfies $M1$ it verifies $DMM 3$.

The relation $\mathrel {\mathcal {R}}$ contains the following pairs in $X\times X$:

• $(x, a) \mathrel {\mathcal {R}}(w, a)$, $(x, b) \mathrel {\mathcal {R}}(x, a)$, $(x, b) \mathrel {\mathcal {R}}(y, a)$, $(x, b) \mathrel {\mathcal {R}}(z, a)$, $(x, b) \mathrel {\mathcal {R}}(w, a)$,

• $(y, a) \mathrel {\mathcal {R}}(w, a)$, $(y, b) \mathrel {\mathcal {R}}(x, a)$, $(y, b) \mathrel {\mathcal {R}}(y, a)$, $(y, b) \mathrel {\mathcal {R}}(w, a)$ (but, due to $z \mathrel {V^{\circ }_{1}} y$, $Not [(y, b) \mathrel {\mathcal {R}}(z, a) ]$),

• $(z, a) \mathrel {\mathcal {R}}(x, a)$, $(z, a) \mathrel {\mathcal {R}}(y, a)$, $(z, a) \mathrel {\mathcal {R}}(w, a)$, $(z, b) \mathrel {\mathcal {R}}(x, a)$, $(z, a) \mathrel {\mathcal {R}}(x, b)$, $(z, b) \mathrel {\mathcal {R}}(y, a)$, $(z, b) \mathrel {\mathcal {R}}(y, b)$, $(z, b) \mathrel {\mathcal {R}}(z, a)$, $(z, b) \mathrel {\mathcal {R}}(w, a)$, $(z, b) \mathrel {\mathcal {R}}(w, b)$,

• $(w, b) \mathrel {\mathcal {R}}(x, a)$, $(w, b) \mathrel {\mathcal {R}}(y, a)$, $(w, b) \mathrel {\mathcal {R}}(z, a)$, $(w, b) \mathrel {\mathcal {R}}(w, a)$.

On $X_{2}$, it is easy to check that we have $b \mathrel {\succ _{2}^{\pm }} a$, so that $AC 1_{2}$, $AC 2_{2}$ and $AC 3_{2}$ hold.

On $X_{1}$, it is easy to check that $\mathrel {\succsim _{1}^{-}}$ is complete. We indeed have that:

$$\begin{aligned} z \mathrel {\succ _{1}^{-}} x \mathrel {\succ _{1}^{-}} y \mathrel {\succ _{1}^{-}} w. \end{aligned}$$

The relation $\mathrel {\succsim _{1}^{+}}$ is not complete. We have $z \mathrel {\succ _{1}^{+}} x$, $x \mathrel {\succ _{1}^{+}} y$ and $x \mathrel {\succ _{1}^{+}} w$ but neither $y\mathrel {\succsim _{1}^{+}} w$ nor $w \mathrel {\succsim _{1}^{+}} y$ since $(y, a) \mathrel {\mathcal {R}}(w, a)$ but $Not [(w, a) \mathrel {\mathcal {R}}(w, a) ]$ and $(w, b) \mathrel {\mathcal {R}}(z, a)$ but $Not [(y, b) \mathrel {\mathcal {R}}(z, a) ]$. This shows that $AC 1_{1}$ is violated. Condition $AC 3_{1}$ holds since $\mathrel {\succsim _{1}^{+}}$ and $\mathrel {\succsim _{1}^{-}}$ are not incompatible.

Each of $\mathrel {P^{\circ }_{1}}$ and $\mathrel {V^{\circ }_{1}}$ is the asymmetric part of some semiorder but these semiorders do not form an homogeneous chain of semiorders (the weak order induced by $\mathrel {P^{\circ }_{1}}$ imposes that $w$ is placed in the last position while that induced by $\mathrel {V^{\circ }_{1}}$ imposes the last position to $y$).

Example 89

(Asymmetric, $Not [ AC 2_{i}, M2_{i}, Maj 2_{i}, LC _{i} ]$) This example is an asymmetric variant of Example 86

Let $X= X_{1} \times X_{2}$ with $X_{1} = \{x, y, z, w \}$ and $X_{2} = \{a, b \}$. We build a CDR on $X$ with:

• $w \mathrel {P^{\circ }_{1}} y$, $w \mathrel {P^{\circ }_{1}} z$, $x \mathrel {P^{\circ }_{1}} z$, $y \mathrel {P^{\circ }_{1}} z$,

• the relation $\mathrel {V^{\circ }_{1}}$ is empty except that $y \mathrel {V^{\circ }_{1}} z$,

• $a \mathrel {P^{\circ }_{2}} b$,

• the relation $\mathrel {V^{\circ }_{2}}$ is empty,

• $\{1\} \mathrel {\rhd ^{\circ }}\varnothing$, $\{2\} \mathrel {\rhd ^{\circ }}\{1\}$.

By construction, $\mathrel {\mathcal {R}}$ is an asymmetric CDR. Hence, it satisfies $RC 1$, $RC 2$, $M1$, $Maj 1$, $MM 1$, $M3$, $Maj 3$, $MM 3$ (Theorem 58 and Lemmas 42 and 56). It satisfies $M2_{2}$ but not $M2_{1}$ (due to the veto on $X_{1}$), not $Maj 2_{1}$ (by Lemma 42) and not $MM 2_{1}$ (by Lemma 42). Using Lemma 11 in Bouyssou and Pirlot (2007), we get that $UC$ and $LC_{2}$ are satisfied but $LC _{1}$ is violated.⁸

Since $\mathrel {\mathcal {R}}$ satisfies $M1$ it verifies $DMM 3$.

The relation $\mathrel {\mathcal {R}}$ contains the following pairs in $X\times X$:

• $(x, a) \mathrel {\mathcal {R}}(x, b)$, $(x, a) \mathrel {\mathcal {R}}(y, b)$, $(x, a) \mathrel {\mathcal {R}}(z, a)$, $(x, a) \mathrel {\mathcal {R}}(z, b)$, $(x, a) \mathrel {\mathcal {R}}(w, b)$, $(x, b) \mathrel {\mathcal {R}}(z, b)$,

• $(y, a) \mathrel {\mathcal {R}}(x, b)$, $(y, a) \mathrel {\mathcal {R}}(y, b)$, $(y, a) \mathrel {\mathcal {R}}(z, a)$, $(y, a) \mathrel {\mathcal {R}}(z, b)$, $(y, a) \mathrel {\mathcal {R}}(w, b)$, $(y, b) \mathrel {\mathcal {R}}(z, b)$,

• $(z, a) \mathrel {\mathcal {R}}(x, b)$, $(z, a) \mathrel {\mathcal {R}}(z, b)$, $(z, a) \mathrel {\mathcal {R}}(w, b)$ (but $Not [(z, a) \mathrel {\mathcal {R}}(y, b) ]$, due to $y \mathrel {V^{\circ }_{1}} z$),

• $(w, a) \mathrel {\mathcal {R}}(x, b)$, $(w, a) \mathrel {\mathcal {R}}(y, a)$, $(w, a) \mathrel {\mathcal {R}}(y, b)$, $(w, a) \mathrel {\mathcal {R}}(z, a)$, $(w, a) \mathrel {\mathcal {R}}(z, b)$, $(w, a) \mathrel {\mathcal {R}}(w, b)$, $(w, b) \mathrel {\mathcal {R}}(y, b)$, $(w, b) \mathrel {\mathcal {R}}(z, b)$.

On $X_{2}$, it is easy to check that we have $a \mathrel {\succ _{2}^{\pm }} b$, so that $AC 1_{2}$, $AC 2_{2}$ and $AC 3_{2}$ hold.

On $X_{1}$, it is easy to check that $\mathrel {\succsim _{1}^{+}}$ is complete. We indeed have that:

$$\begin{aligned} w \mathrel {\succ _{1}^{\pm }} [x \mathrel {\sim _{1}^{+}} y] \mathrel {\succ _{1}^{+}} z. \end{aligned}$$

The relation $\mathrel {\succsim _{1}^{-}}$ is not complete. We have $w \mathrel {\succ _{1}^{-}} x$, $w \mathrel {\succ _{1}^{-}} y$, $x \mathrel {\succ _{1}^{-}} z$ and $y \mathrel {\succ _{1}^{-}} z$ but neither $x \mathrel {\succsim _{1}^{-}} y$ nor $y \mathrel {\succsim _{1}^{-}} x$ since $(z, a) \mathrel {\mathcal {R}}(x, b)$ but $Not [(z, a) \mathrel {\mathcal {R}}(y, b) ]$ and $(w, a) \mathrel {\mathcal {R}}(y, a)$ but $Not [(w, a) \mathrel {\mathcal {R}}(x, a) ]$. This shows that $AC 2_{1}$ is violated. Condition $AC 3_{1}$ holds since $\mathrel {\succsim _{1}^{+}}$ and $\mathrel {\succsim _{1}^{-}}$ are not incompatible.

Each of $\mathrel {P^{\circ }_{1}}$ and $\mathrel {V^{\circ }_{1}}$ is the asymmetric part of some semiorder but these semiorders do not form an homogeneous chain of semiorders (the weak order induced by $\mathrel {P^{\circ }_{1}}$ imposes that $w$ is placed in the first position while that induced by $\mathrel {V^{\circ }_{1}}$ imposes the first position to $y$).

Example 90

(Asymmetric $Not [ RC 1_{i}, M2_{i}, Maj 2_{i}, MM 2_{i}, Maj 3_{i} ]$)

This example was not published before.

Let $X=\{x,y,z,w\}\times \{a,b\}\times \{p,q\}$ and $\mathrel {\mathcal {R}}$ consist of the set of pairs listed in Table 4.

Table 4 Relation $\mathrel {\mathcal {R}}$ in Example 90

It is easy to see that $\mathrel {\mathcal {R}}$ is an asymmetric relation.

As for the comparison of preference differences on each attribute, we have, for all $(\alpha , \beta ) \in \Gamma = \{ (x,x), (y,y),$ $(z,z), (w,w),$ $(x,z), (x,w),$ $(y,x), (y,z),$ (y,w), (z,x),$(w,x), (w,y), (w,z) \}$,

• $(\alpha , \beta ) \mathrel {\succ _{1}^{*}} (x,y) \mathrel {\succ _{1}^{*}} (z,y)$ and $(\alpha , \beta ) \mathrel {\succ _{1}^{*}} (z,w) \mathrel {\succ _{1}^{*}} (z,y)$, while $(x,y)$ and $(z,w)$ are incomparable in terms of $\mathrel {\succsim _{1}^{*}}$,

• $(a,b) \mathrel {\succ _{2}^{*}} [(a,a), (b,b)] \mathrel {\succ _{3}^{*}} (b,a)$,

• $(p,q) \mathrel {\succ _{3}^{*}} [(p,p), (q,q)] \mathrel {\succ _{3}^{*}} (q,p)$.

The upward and downward dominance relations are as follows:

• $y \mathrel {\succ _{1}^{\pm }} w \mathrel {\succ _{1}^{\pm }} x \mathrel {\succ _{1}^{\pm }} z$,

• $a \mathrel {\succ _{2}^{\pm }} b$,

• $p \mathrel {\succ _{3}^{\pm }} q$.

$RC 1_{1}$ does not hold since the pairs $(x,y)$ and $(z,w)$ are not comparable w.r.t. $\mathrel {\succsim _{1}^{*}}$, but $RC 1_{2}$ and $RC 1_{3}$ hold true. For $j \in \{1,2,3\}$, $RC 2_{j}$, $AC 1_{j}$, $AC 2_{j}$, $AC 3_{j}$ are clearly satisfied. For $j=2$ and $j=3$, using Lemma 25, we see that $\mathrel {\mathcal {R}}$ fulfills $M1_{j}$ and $M2_{j}$ hence it satisfies $UC _{j}$ and $LC _{j}$ (by Lemma 11 in Bouyssou and Pirlot 2007), $Maj 1_{j}$ and $Maj 2_{j}$ ( by Lemma 34), $UC _{j}$ and $LC _{j}$ (by  Lemma 11 in Bouyssou and Pirlot 2007), $M3_{j}$, $Maj 3_{j}$, $MM 1_{j}$, $MM 2_{j}$ and $MM 3_{j}$ (since each of the latter is implied by one of the previously established properties of $\mathrel {\mathcal {R}}$ ).

$\mathrel {\mathcal {R}}$ satisfies $M1_{1}$. Assume to the contrary that there are $s,t,u,v \in X_{1}$ and $S,T,U,V \in X_{-1}$ such that: (1) $(s,S)\mathrel {\mathcal {R}}(t,T)$, (2) $(u,U)\mathrel {\mathcal {R}}(v,V)$, (3) $Not [(t,S)\mathrel {\mathcal {R}}(s,T) ]$, (4) $Not [(s,U)\mathrel {\mathcal {R}}(t,V) ]$, (5) $Not [(v,S)\mathrel {\mathcal {R}}(u,T) ]$. Using (1), (3) and Lemma 22.2, we deduce that $(s,t)$ can only be one of the pairs $(y,x), (y,z)$ or $(w,z)$. In all three cases, (2) and (4) cannot both hold true since $(s,t) \mathrel {\succsim _{1}^{*}} (u,v)$, for all $u,v \in X_{1}$, a contradiction.

$\mathrel {\mathcal {R}}$ satisfies $Maj 1_{1}$. Assume to the contrary that there are $s,t,u,v \in X_{1}$ and $S,T,U,V \in X_{-1}$ such that: (1) $(s,S)\mathrel {\mathcal {R}}(t,T)$, (2) $(u,S)\mathrel {\mathcal {R}}(v,T)$, (3) $(u,U)\mathrel {\mathcal {R}}(v,V)$, (4) $Not [(t,S)\mathrel {\mathcal {R}}(s,T) ]$, (5) $Not [(s,U)\mathrel {\mathcal {R}}(t,V) ]$. Using (1), (4) and Lemma 22.2, we deduce that $(s,t)$ can only be one of the pairs $(y,x), (y,z)$ or $(w,z)$. In all three cases, (3) and (5) cannot both hold true since $(s,t) \mathrel {\succsim _{1}^{*}} (u,v)$, a contradiction.

$\mathrel {\mathcal {R}}$ satisfies $M3_{1}$. Assume to the contrary that there are $s,t,u,v \in X_{1}$ and $S,T,U,V,Q,R \in X_{-1}$ such that: (1) $(s,S)\mathrel {\mathcal {R}}(t,T)$, (2) $(t,U)\mathrel {\mathcal {R}}(s,V)$, (3) $(u,Q)\mathrel {\mathcal {R}}(v,R)$, (4) $Not [(t,S)\mathrel {\mathcal {R}}(s,T) ]$, (5) $Not [(u,S)\mathrel {\mathcal {R}}(v,T) ]$, (6) $Not [(u,U)\mathrel {\mathcal {R}}(v,V) ]$. Using (1), (4) and Lemma 22.2, we deduce that $(s,t)$ can only be one of the pairs $(y,x), (y,z)$ or $(w,z)$. If $(s,t)=(y,z)$, (2) never holds true. In case $(s,t)=(y,x)$, (1) and (4) imply $[S=ap \text { and } T= bp]$ or $[S=aq \text { and } T=bq]$. Contradicting (5), we have $(u,S) \mathrel {\mathcal {R}}(v,T)$ for all $u,v \in X_{1}$ except for $(u,v) = (z,y)$, for which (3) does not hold. The case in which $(s,t) = (w,z)$ is dealt with similarly. As a conclusion, $M3_{1}$ holds for $\mathrel {\mathcal {R}}$ .

$\mathrel {\mathcal {R}}$ violates $M2_{1}$. $M2_{1}$ does not hold if we can find $s,t,u,v \in X_{1}$ and $S,T,U,V \in X_{-1}$ such that: (1) $(s,S)\mathrel {\mathcal {R}}(t,T)$, (2) $(t,U)\mathrel {\mathcal {R}}(s,V)$, (3) $Not [(t,S)\mathrel {\mathcal {R}}(s,T) ]$, (4) $Not [(u,S)\mathrel {\mathcal {R}}(v,T) ]$, (5) $Not [(u,U)\mathrel {\mathcal {R}}(v,V) ]$. These 5 conditions can be simultaneously fulfilled by setting: $s=y, t=x, u=z, v=y$ and $S=ap, T=bp, U=ap, V=aq$. Since $MM 2_{i}$ and $RC 2_{i}$ entail $M2_{i}$ (Lemma 42.3), $\mathrel {\mathcal {R}}$ violates $MM 2_{1}$.

$\mathrel {\mathcal {R}}$ violates $Maj 3_{1}$. $Maj 3_{1}$ does not hold if we can find $s,t,u,v \in X_{1}$ and $S,T,U,V,Q,R \in X_{-1}$ such that: (1) $(s,S)\mathrel {\mathcal {R}}(t,T)$, (2) $(v,S)\mathrel {\mathcal {R}}(u,T)$, (3) $(t,U)\mathrel {\mathcal {R}}(s,V)$, (4) $(u,Q)\mathrel {\mathcal {R}}(v,R)$, (5) $Not [(t,S)\mathrel {\mathcal {R}}(s,T) ]$, (6) $Not [(u,U)\mathrel {\mathcal {R}}(v,V) ]$. These 6 conditions can be simultaneously fulfilled by setting: $s=y, t=x, u=z, v=w$ and $S=ap, T=bp, U=ap, V=aq, Q=ap, R=bq$. Since $Maj 2_{i}$ entails $Maj 3_{i}$, $\mathrel {\mathcal {R}}$ also violates $Maj 2_{1}$.

Since $\mathrel {\mathcal {R}}$ satisfies $M1_{1}$ (resp. $M3_{1}$) it satisfies $MM 1_{1}$ (resp. $MM 3_{1}$).

Since $\mathrel {\mathcal {R}}$ satisfies $MM 1_{1}$ it satisfies $DMM 3_{1}$ (by Lemma 66).