Diversity and rights: a social choice-theoretic analysis of the possibility of public reason

Abstract

Public reason liberalism takes as its starting point the deep and irreconcilable diversity we find characterizing liberal societies. This deep and irreconcilable diversity creates problems for social order. One method for adjudicating these conflicts is through the use of rights. This paper is about the ability of such rights to adjudicate disputes when perspectival disagreements—or disagreements over how to categorize objects in the world—obtain. We present both formal possibility and impossibility results for rights structures under varying degrees of perspectival diversity. We show that though perspectival diversity appears to be a troubling problem for the prospect of stable social order, if rights are defined properly then disagreements can likely be resolved in a consistent manner, achieving social cooperation rather than conflict.

Appendix: Proofs

Proposition 1

Let$\left( {F, R,\pi } \right) \in {\Im } \times {\mathcal{R}} \times \varPi$be any social choice problem. Then, $UD\left( {F,P^{*} \left( {F,R,\pi } \right)} \right)$is non-empty if$P^{*} \left( {F,R,\pi } \right)$is acyclic.

Proof of Proposition 1

Let $\left( {F, R,\pi } \right)$ be any social choice problem. Suppose $\left| F \right| = n \in {\mathbb{N}}$ and suppose $P^{*} \left( {F,R,\pi } \right)$ is acyclic. We want to show $UD\left( {F,P^{*} \left( {F,R,\pi } \right)} \right) \ne \emptyset$. Suppose, for a proof by contradiction, that $UD\left( {F,P^{*} \left( {F,R,\pi } \right)} \right) = \emptyset$. Pick any $x_{1} \in F$. Since $UD\left( {F,P^{*} \left( {F,R,\pi } \right)} \right) = \emptyset$, we have $x_{1} \notin UD\left( {F,P^{*} \left( {F,R,\pi } \right)} \right)$. Hence, there exists another social state, say $x_{2} \in F$, such that $x_{2} P^{*} \left( {F,R,\pi } \right)x_{1}$. However, since $UD\left( {F,P^{*} \left( {F,R,\pi } \right)} \right) = \emptyset$, we must have $x_{2} \notin UD\left( {F,P^{*} \left( {F,R,\pi } \right)} \right)$ as well. This implies that there exists another social state, say $x_{3} \in F$, such that $x_{3} P^{*} \left( {F,R,\pi } \right)x_{2} P^{*} \left( {F,R,\pi } \right)x_{1}$. Continue this process $n$ times. Then, we have $x_{n + 1} P^{*} \left( {F,R,\pi } \right)x_{n} P^{*} \left( {F,R,\pi } \right) \ldots x_{2} P^{*} \left( {F,R,\pi } \right)x_{1}$. Since $\left| F \right| = n$, there must exist $j,k \in \left\{ {1, \ldots ,n + 1} \right\}$ such that $x_{k} P^{*} \left( {F,R,\pi } \right)x_{k - 1} P^{*} \left( {F,R,\pi } \right) \ldots x_{j + 1} P^{*} \left( {F,R,\pi } \right)x_{j}$ where $x_{k} = x_{j}$. But, then, we have a cycle, contradicting our assumption that $P^{*} \left( {F,R,\pi } \right)$ is acyclic. Hence, $UD\left( {F,P^{*} \left( {F,R,\pi } \right)} \right) \ne \emptyset$.□

Theorem 1

Under Condition ND (Domain of No Diversity), Condition L (Liberalism) implies Condition NC (No Cycles).

Proof of Theorem 1

Assume ND (Domain of No Diversity) and L (Liberalism). For a proof by contradiction, suppose condition NC (No Cycles) is violated. Then, there exists a social choice problem $\left( {F, R,\pi } \right) \in {\Im } \times {\mathcal{R}} \times {{\Pi }}$ for which the social preference function $P^{*} \left( {F,R,\pi } \right)$ satisfying L produces a cycle of some length $n \in {\mathbb{N}}$. We will argue that this cannot be the case for all $n \in {\mathbb{N}}$.

Suppose $n = 1$. Then, there exists some $x \in F$ such that $xP^{*} \left( {F,R,\pi } \right)x$. Since $xP^{*} \left( {F,R,\pi } \right)x$, there exists an $i \in N$ such that $i$ unconditionally prefers $x^{i}$ to $x^{\text{i}}$ (note that $x^{i}$ and $x^{\text{i}}$ are trivially $i$-variants as $x_{ - i}^{i} = x_{ - i}^{i}$), which implies $x^{i} P_{i} x^{i}$. This contradicts that $P_{i}$ is asymmetric, and, thereby, irreflexive. Hence, the social preference function $P^{*} \left( {F,R,\pi } \right)$ satisfying L cannot produce a cycle of length $n = 1.$

Now, suppose that the social preference function $P^{*} \left( {F,R,\pi } \right)$ satisfying L cannot produce a cycle of length $n = k.$ We wish to show that this implies that the social preference function $P^{*} \left( {F,R,\pi } \right)$ satisfying L cannot produce a cycle of length $n = k + 1.$ So, for a proof by contradiction, suppose that the social preference function $P^{*} \left( {F,R,\pi } \right)$ satisfying L cannot produce a cycle of length $n = {\text{k}},$ but produces a cycle of length $n = k + 1.$ Then, there exists $x_{\left( 1 \right)} , \ldots ,x_{{\left( {k + 1} \right)}} \in F$ such that $x_{\left( 1 \right)} P^{*} \left( {F,R,\pi } \right)x_{\left( 2 \right)} \ldots x_{\left( k \right)} P^{*} \left( {F,R,\pi } \right)x_{{\left( {k + 1} \right)}} P^{*} \left( {F,R,\pi } \right)x_{\left( 1 \right)}$.

Since $x_{\left( 1 \right)} P^{*} \left( {F,R,\pi } \right)x_{\left( 2 \right)}$, there exists an $i \in N$ such that $x_{\left( 1 \right)}^{i}$ and $x_{\left( 2 \right)}^{i}$ are $i$-variants, $i$ unconditionally prefers $x_{\left( 1 \right)}^{i}$ and $x_{\left( 2 \right)}^{i}$, and either

• (i) $\forall j \in N$, $x_{\left( 1 \right)}^{i} = x_{\left( 1 \right)}^{j}$ and $x_{\left( 2 \right)}^{i} = x_{\left( 2 \right)}^{j}$; or

• (ii) $\forall j \in N$ such that $x_{\left( 1 \right)}^{i} \ne x_{\left( 1 \right)}^{j}$ or $x_{\left( 2 \right)}^{i} \ne x_{\left( 2 \right)}^{j}$: $x_{\left( 1 \right)}^{j} R_{j} x_{\left( 2 \right)}^{j}$.

By ND, (ii) cannot happen. So, we have $\forall j \in N$, $x_{\left( 1 \right)}^{i} = x_{\left( 1 \right)}^{j}$ and $x_{\left( 2 \right)}^{i} = x_{\left( 2 \right)}^{j}$.

For any $y \in F$, let $y^{*}$ be the i-variant of $x_{\left( 1 \right)}^{i}$ such that the $i$-component of $y^{*}$ is the $i$-component of $y.$ Since $x_{\left( 1 \right)}^{*i} = x_{\left( 1 \right)}^{i}$ and $x_{\left( 2 \right)}^{*i} = x_{\left( 2 \right)}^{i}$, $x_{\left( 1 \right)}^{i}$ and $x_{\left( 2 \right)}^{i}$ are i-variants, and individual $i$ prefers $x_{\left( 1 \right)i}^{i}$ to $x_{\left( 2 \right)i}^{i}$ unconditionally, we have $x_{\left( 1 \right)}^{i} P_{i} x_{\left( 2 \right)}^{i}$.

Now, since $x_{\left( 2 \right)} P^{*} \left( {F,R,\pi } \right)x_{\left( 3 \right)}$, there exists an individual, say $k \in N$, that $x_{\left( 2 \right)}^{k}$ and $x_{\left( 3 \right)}^{k}$ are $k$-variants, unconditionally prefers $x_{\left( 2 \right)}^{k}$ to $x_{\left( 3 \right)}^{k}$, and either

• (iii) $\forall j \in N$, $x_{\left( 2 \right)}^{k} = x_{\left( 2 \right)}^{j}$ and $x_{\left( 3 \right)}^{k} = x_{\left( 3 \right)}^{j}$; or

• (iv) $\forall j \in N$ such that $x_{\left( 2 \right)}^{k} \ne x_{\left( 2 \right)}^{j}$ or $x_{\left( 3 \right)}^{k} \ne x_{\left( 3 \right)}^{j}$: $x_{\left( 2 \right)}^{j} R_{j} x_{\left( 3 \right)}^{j}$.

Again, by ND, (iv) cannot happen. So, we have $\forall j \in N$, $x_{\left( 2 \right)}^{k} = x_{\left( 2 \right)}^{j}$ and $x_{\left( 3 \right)}^{k} = x_{\left( 3 \right)}^{j}$, and in particular we have $x_{\left( 2 \right)}^{k} = x_{\left( 2 \right)}^{i}$ and $x_{\left( 3 \right)}^{k} = x_{\left( 3 \right)}^{i}$. There are two cases to consider: when $k = i$ and when $k \ne i$. If $k = i$, then, just as before, since $x_{\left( 2 \right)}^{i}$ and $x_{\left( 3 \right)}^{i}$ are $i$-variants and individual $i\left( { = k} \right)$ prefers $x_{\left( 2 \right)i}^{i}$ to $x_{\left( 3 \right)i}^{i}$ unconditionally, we have $x_{\left( 2 \right)}^{i} P_{i} x_{\left( 3 \right)}^{i}$. Since $x_{\left( 2 \right)}^{i} = x_{\left( 2 \right)}^{*i}$ and $x_{\left( 3 \right)}^{i} = x_{\left( 3 \right)}^{*i}$, we have $x_{\left( 2 \right)}^{*i} P_{i} x_{\left( 3 \right)}^{*i}$, which implies $x_{\left( 2 \right)}^{*i} R_{i} x_{\left( 3 \right)}^{*i}$. If $k \ne i$. Then, since $x_{\left( 2 \right)}^{k} = x_{\left( 2 \right)}^{i}$ and $x_{\left( 3 \right)}^{k} = x_{\left( 3 \right)}^{i}$ are $k$-variants, the i-components of $x_{\left( 2 \right)}^{k} = x_{\left( 2 \right)}^{i}$ and $x_{\left( 3 \right)}^{k} = x_{\left( 3 \right)}^{i}$ will be the same. Hence, $x_{\left( 2 \right)}^{*k} = x_{\left( 2 \right)}^{*i} = x_{\left( 3 \right)}^{*i} = x_{\left( 3 \right)}^{*k}$, which implies $x_{\left( 2 \right)}^{*i} R_{i} x_{\left( 3 \right)}^{*i}$. So, in either case, we have $x_{\left( 2 \right)}^{*i} R_{i} x_{\left( 3 \right)}^{*i}$.

By repeating the same argument, we will arrive at $x_{\left( 1 \right)}^{*i} P_{i} x_{\left( 2 \right)}^{*i} R_{i} x_{\left( 3 \right)}^{*i} \ldots R_{i} x_{{\left( {k + 1} \right)}}^{*i} R_{i} x_{\left( 1 \right)}^{*i}$. Since $R_{i}$ is an order, this implies $x_{\left( 1 \right)}^{*i} P_{i} x_{\left( 1 \right)}^{*i}$, which contradicts that $P_{i}$ is asymmetric, and, thereby, irreflexive. Hence, the social preference function $P^{*} \left( {F,R,\pi } \right)$ satisfying L cannot produce a cycle of length $n = k + 1.$

By mathematical induction, we conclude that the social preference function $P^{*} \left( {F,R,\pi } \right)$ satisfying L cannot produce a cycle of any length $n \in {\mathbb{N}}$.□

Theorem 2

Suppose$\left| X \right| \ge 4$and$\ n \ge 4$. Then, there exists no social preference function$\ P^{*} \left( {F, R,\pi } \right)$that satisfies Condition UD (Unrestricted Domain), Condition L (Liberalism), and Condition NC (No Cycles).

Proof of Theorem 2

Let $F = \left\{ {w,x,y,z} \right\}$ and suppose individuals $1$ to $4$ have the following preferences:

$$ z^{1} R_{1} w^{1} P_{1} x^{1} R_{1} y^{1} ;$$

$$ w^{2} R_{2} x^{2} P_{2} y^{2} R_{2} z^{2} ;$$

$$ x^{3} R_{3} y^{3} P_{3} z^{3} R_{3} w^{3} $$

$$ y^{4} R_{4} z^{4} P_{4} w^{4} R_{4} x^{4} $$

where individual 1 prefers $w_{1}^{1}$ to $x_{1}^{1}$ unconditionally; individual 2 prefers $x_{2}^{2}$ to $y_{2}^{2}$ unconditionally; individual 3 prefers $y_{3}^{3}$ to $z_{3}^{3}$ unconditionally; and individual 4 prefers $z_{4}^{4}$ to $w_{4}^{4}$ unconditionally.

For all $j \in N\backslash \left\{ {1,2,3,4} \right\}$, suppose$w^{j} I_{j} x^{j} I_{j} y^{j} I_{j} z^{j}$.Call this profile of individual preferences $R$ and the profile of perspectives $\pi$. By condition UD, the social choice problem $\left( {F,R,\pi } \right)$ is in our domain. Now, assume

$$ w^{1}, x^{1}\,{\rm are}\,1{\rm -variants}; $$

$$ x^{2} , y^{2}\,{\rm are}\,2{\rm -variants};$$

$$ y^{3},\,z^{3}\,{{\text{are}}}\,3{\rm -variants}; $$

$$ z^{4} ,w^{4}\,{{\text{are}}}\,4{\rm -variants} $$

.

Also, suppose

$$ w^{1} = w^{3} \ne w^{2} = w^{4} ;$$

$$ x^{1} = x^{3} \ne x^{2} = x^{4} ;$$

$$ y^{1} = y^{3} \ne {\text{y}}^{2} = y^{4} ;$$

$$ z^{1} = z^{3} \ne z^{2} = z^{4}.$$

So, individuals $1$ and $3$ share the same perspective, while individuals $2$ and $4$ share the same perspectives, and all other individuals share the same perspective that is different from any perspective that individuals $1,2,3$ and $4$ have.

• Since $w^{1}$, $x^{1}$ are 1-variants; individual 1 prefers $w_{1}^{1}$ to $x_{1}^{1}$ unconditionally; $w^{2} R_{2} x^{2}$ (and $x^{1} \ne x^{2} )$; $x^{3} P_{3} w^{3}$ (and $w^{1} = w^{3}$ and $x^{1} = x^{3}$); $w^{4} R_{4} x^{4}$ (and ${\text{w}}^{4} \ne w^{1} )$; and $w^{j} R_{j} x^{j}$ for all $j \in N\backslash \left\{ {1,2,3,4} \right\}$, by L, individual 1 has a liberal right to be socially decisive over $w$ and $x$. Hence, we must have $\varvec{wP}^{\varvec{*}} \left( {\varvec{F},\varvec{R},\varvec{\pi}} \right)\varvec{x}$.

• Since $x^{2}$, $y^{2}$ are 2-variants; individual 2 prefers $x_{2}^{2}$ to $y_{2}^{2}$ unconditionally; $x^{3} R_{3} y^{3}$ (and ${\text{y}}^{2} \ne y^{3} )$; $y^{4} P_{4} x^{4}$ (and $x^{2} = x^{4}$ and $y^{2} = y^{4}$); $x^{1} R_{1} y^{1}$ (and $x^{2} \ne x^{1} )$; and $x^{j} R_{j} y^{j}$ for all $j \in N\backslash \left\{ {1,2,3,4} \right\}$, by L, individual 2 has a liberal right to be socially decisive over $x$ and $y$. Hence, we must have $\varvec{xP}^{\varvec{*}} \left( {\varvec{F},\varvec{R},\varvec{\pi}} \right)\varvec{y}$.

• Since $y^{3}$, $z^{3}$ are 3-variants; individual 3 prefers $y_{3}^{3}$ to $z_{3}^{3}$ unconditionally; $y^{4} R_{4} z^{4}$ (and $z^{3} \ne z^{4} )$; $z^{1} P_{1} y^{1}$ (and $y^{3} = y^{1}$ and $z^{3} = z^{1}$); $y^{2} R_{2} z^{2}$ (and ${\text{y}}^{3} \ne y^{2} )$; and $y^{j} R_{j} z^{j}$ for all $j \in N\backslash \left\{ {1,2,3,4} \right\}$, by L, individual 3 has a liberal right to be socially decisive over $y$ and $z$. Hence, we must have $\varvec{yP}^{\varvec{*}} \left( {\varvec{F},\varvec{R},\varvec{\pi}} \right)\varvec{z}$.

• Since $z^{4}$, $w^{4}$ are 4-variants; individual 4 prefers $z_{4}^{4}$ to $w_{4}^{4}$ unconditionally; $z^{1} R_{1} w^{1}$ (and $w^{4} \ne w^{1} )$; $w^{2} P_{2} z^{2}$ (and $z^{4} = z^{2}$ and $w^{4} = w^{2}$); $z^{3} R_{3} w^{3}$ (and ${\text{z}}^{4} \ne z^{3} )$; and $z^{j} R_{j} w^{j}$ for all $j \in N\backslash \left\{ {1,2,3,4} \right\}$, by L, individual 4 has a liberal right to be socially decisive over $z$ and $w$. Hence, we must have $\varvec{zP}^{\varvec{*}} \left( {\varvec{F},\varvec{R},\varvec{\pi}} \right)\varvec{w}$.

As a result, we have $\varvec{wP}^{\varvec{*}} \left( {\varvec{F},\varvec{R},\varvec{\pi}} \right)\varvec{xP}^{\varvec{*}} \left( {\varvec{F},\varvec{R},\varvec{\pi}} \right)\varvec{yP}^{\varvec{*}} \left( {\varvec{F},\varvec{R},\varvec{\pi}} \right)\varvec{zP}^{\varvec{*}} \left( {\varvec{F},\varvec{R},\varvec{\pi}} \right)\varvec{w}$, a cycle in the social preference relation $\varvec{P}^{\varvec{*}} \left( {\varvec{F},\varvec{R},\varvec{\pi}} \right)$, violating NC.□

Theorem 3

Under Condition RD (Domain of Restricted Perspectival Diversity), Condition L (Liberalism) implies Condition NC (No Cycles).

Proof of Theorem 3

The proof follows the general strategy of the Proof of Theorem 1, but, now, with perspectival diversity, we need to consider additional cases.

Assume RD and L. For a proof by contradiction, suppose NC is violated. Then, there exists a social choice problem $\left( {F, R,\pi } \right) \in {\Im } \times {\mathcal{R}} \times {{\Pi }}$ for which the social preference function $P^{*} \left( {F,R,\pi } \right)$ satisfying L produces a cycle of some length $n \in {\mathbb{N}}$. Again, we will argue that this cannot be the case for all $n \in {\mathbb{N}}$.

Suppose $n = 1$. Then, there exists some $x \in F$ such that $xP^{*} \left( {F,R,\pi } \right)x$. Since $xP^{*} \left( {F,R,\pi } \right)x$, there exists an $i \in N$ such that $i$ unconditionally prefers $x^{i}$ to $x^{\text{i}}$ (note that $x^{i}$ and $x^{\text{i}}$ are trivially $i$-variants as $x_{ - i}^{i} = x_{ - i}^{i}$), which implies $x^{i} P_{i} x^{i}$. This contradicts that $P_{i}$ is asymmetric, and, thereby, irreflexive. Hence, the social preference function $P^{*} \left( {F,R,\pi } \right)$ satisfying L cannot produce a cycle of length $n = 1.$

Now, suppose that the social preference function $P^{*} \left( {F,R,\pi } \right)$ satisfying L cannot produce a cycle of length $n = k.$ We wish to show that this implies that the social preference function $P^{*} \left( {F,R,\pi } \right)$ satisfying L cannot produce a cycle of length $n = k + 1.$ So, for a proof by contradiction, suppose that the social preference function $P^{*} \left( {F,R,\pi } \right)$ satisfying L cannot produce a cycle of length $n = k,$ but produces a cycle of length $n = k + 1.$ Then, there exists $x_{\left( 1 \right)} , \ldots ,x_{{\left( {k + 1} \right)}} \in F$ such that $x_{\left( 1 \right)} P^{*} \left( {F,R,\pi } \right)x_{\left( 2 \right)} \ldots x_{\left( k \right)} P^{*} \left( {F,R,\pi } \right)x_{{\left( {k + 1} \right)}} P^{*} \left( {F,R,\pi } \right)x_{\left( 1 \right)}$.

Since $x_{\left( 1 \right)} P^{*} \left( {F,R,\pi } \right)x_{\left( 2 \right)}$, there exists an $i \in N$ such that $x_{\left( 1 \right)}^{i}$ and $x_{\left( 2 \right)}^{i}$ are $i$-variants, $i$ unconditionally prefers $x_{\left( 1 \right)}^{i}$ and $x_{\left( 2 \right)}^{i}$, and either

• (i) $\forall j \in N$, $x_{\left( 1 \right)}^{i} = x_{\left( 1 \right)}^{j}$ and $x_{\left( 2 \right)}^{i} = x_{\left( 2 \right)}^{j}$; or

• (ii) $\forall j \in N$ such that $x_{\left( 1 \right)}^{i} \ne x_{\left( 1 \right)}^{j}$ or $x_{\left( 2 \right)}^{i} \ne x_{\left( 2 \right)}^{j}$: $x_{\left( 1 \right)}^{j} R_{j} x_{\left( 2 \right)}^{j}$.

Again, for any $y \in F$, let $y^{*}$ be the i-variant of $x_{\left( 1 \right)}^{i}$ such that the $i$-component of $y^{*}$ is the $i$-component of $y.$ Since $x_{\left( 1 \right)}^{*i} = x_{\left( 1 \right)}^{i}$ and $x_{\left( 2 \right)}^{*i} = x_{\left( 2 \right)}^{i}$, $x_{\left( 1 \right)}^{i}$ and $x_{\left( 2 \right)}^{i}$ are i-variants, and individual $i$ prefers $x_{\left( 1 \right)i}^{i}$ to $x_{\left( 2 \right)i}^{i}$ unconditionally, we have $x_{\left( 1 \right)}^{i} P_{i} x_{\left( 2 \right)}^{i}$.

Now, since $x_{\left( 2 \right)} P^{*} \left( {F,R,\pi } \right)x_{\left( 3 \right)}$, there exists an individual, say $k \in N$, that $x_{\left( 2 \right)}^{k}$ and $x_{\left( 3 \right)}^{k}$ are $k$-variants, unconditionally prefers $x_{\left( 2 \right)}^{k}$ to $x_{\left( 3 \right)}^{k}$, and either

• (iii) $\forall j \in N$, $x_{\left( 2 \right)}^{k} = x_{\left( 2 \right)}^{j}$ and $x_{\left( 3 \right)}^{k} = x_{\left( 3 \right)}^{j}$; or

• (iv) $\forall j \in N$ such that $x_{\left( 2 \right)}^{k} \ne x_{\left( 2 \right)}^{j}$ or $x_{\left( 3 \right)}^{k} \ne x_{\left( 3 \right)}^{j}$: $x_{\left( 2 \right)}^{j} R_{j} x_{\left( 3 \right)}^{j}$.

Case 1: $x_{\left( 2 \right)}^{k} = x_{\left( 2 \right)}^{i}$ and $x_{\left( 3 \right)}^{k} = x_{\left( 3 \right)}^{i}$.

Again, there are two cases to consider: when $k = i$ and when $k \ne i$. If $k = i$, then, since $x_{\left( 2 \right)}^{i}$ and $x_{\left( 3 \right)}^{i}$ are $i$-variants and individual $i\left( { = k} \right)$ prefers $x_{\left( 2 \right)i}^{i}$ to $x_{\left( 3 \right)k}^{i}$ unconditionally, we have $x_{\left( 2 \right)}^{i} P_{i} x_{\left( 3 \right)}^{i}$. Since $x_{\left( 2 \right)}^{i} = x_{\left( 2 \right)}^{*i}$ and $x_{\left( 3 \right)}^{i} = x_{\left( 3 \right)}^{*i}$, we have $x_{\left( 2 \right)}^{*i} P_{i} x_{\left( 3 \right)}^{*i}$, which implies $x_{\left( 2 \right)}^{*i} R_{i} x_{\left( 3 \right)}^{*i}$. If $k \ne i$. Then, since $x_{\left( 2 \right)}^{k} = x_{\left( 2 \right)}^{i}$ and $x_{\left( 3 \right)}^{k} = x_{\left( 3 \right)}^{i}$ are $k$-variants, the i-components of $x_{\left( 2 \right)}^{k} = x_{\left( 2 \right)}^{i}$ and $x_{\left( 3 \right)}^{k} = x_{\left( 3 \right)}^{i}$ will be the same. Hence, $x_{\left( 2 \right)}^{*k} = x_{\left( 2 \right)}^{*i} = x_{\left( 3 \right)}^{*i} = x_{\left( 3 \right)}^{*k}$, which implies $x_{\left( 2 \right)}^{*i} R_{i} x_{\left( 3 \right)}^{*i}$. So, in either case, we have $x_{\left( 2 \right)}^{*i} R_{i} x_{\left( 3 \right)}^{*i}$.

Case 2: $x_{\left( 2 \right)}^{k} \ne x_{\left( 2 \right)}^{i}$ or $x_{\left( 3 \right)}^{k} \ne x_{\left( 3 \right)}^{i}$.

Then, $k \ne i$. By RD, $x_{\left( 2 \right)}^{i}$ and $x_{\left( 3 \right)}^{i}$ are $k$-variants (where $k \ne i$). This means that the $i$-component of $x_{\left( 2 \right)}^{i}$ and $x_{\left( 3 \right)}^{i}$ are the same. Hence, we have $x_{\left( 2 \right)}^{*i} = x_{\left( 3 \right)}^{*i}$, implying $x_{\left( 2 \right)}^{*i} R_{i} x_{\left( 3 \right)}^{*i}$

So, in all cases, we have $x_{\left( 2 \right)}^{*i} R_{i} x_{\left( 3 \right)}^{*i}$. So, we have $x_{\left( 2 \right)}^{*i} R_{i} x_{\left( 3 \right)}^{*i}$ and $x_{\left( 1 \right)}^{*i} P_{i} x_{\left( 2 \right)}^{*i}$. (Note that $x_{\left( 1 \right)}^{*i}$, $x_{\left( 2 \right)}^{*i}$, $x_{\left( 3 \right)}^{*i} \in X^{k}$)

By repeating the same argument, we will arrive at $x_{\left( 1 \right)}^{*i} P_{i} x_{\left( 2 \right)}^{*i} R_{i} x_{\left( 3 \right)}^{*i} \ldots R_{i} x_{{\left( {k + 1} \right)}}^{*i} R_{i} x_{\left( 1 \right)}^{*i}$. Since $R_{i}$ is an order, this implies $x_{\left( 1 \right)}^{*i} P_{i} x_{\left( 1 \right)}^{*i}$, which contradicts that $P_{i}$ is asymmetric, and, thereby, irreflexive. Hence, the social preference function $P^{*} \left( {F,R,\pi } \right)$ satisfying L cannot produce a cycle of length $n = k + 1.$

By mathematical induction, we conclude that the social preference function $P^{*} \left( {F,R,\pi } \right)$ satisfying L cannot produce a cycle of any length $\in {\mathbb{N}}$.□