Daunou’s voting rule and the lexicographic assignment of priorities

A study of the history of opinion is a necessary preliminary to the emancipation of the mind. I do not know which makes a man more conservative—to know nothing but the present, or nothing but the past.

John Maynard Keynes, The End of Laissez-Faire, 1926.

Abstract

Pierre Daunou, a contemporary of Borda and Condorcet during the era of the French Revolution and active debates on alternative voting rules, proposed a rule that chooses the strong Condorcet winner if there is one, otherwise eliminates Condorcet losers and uses plurality voting on the remaining candidates. We characterize his rule which combines potentially conflicting desiderata of majoritarianism by ordering them lexicographically. This contribution serves not just to remind ourselves that a 19th-century vintage may still retain excellent aroma and taste, but also to promote a promising general approach to reconcile potentially conflicting desiderata by accommodating them lexicographically. Journal of Economic Literature Classification Nos.: D71, D72.

Appendix

Part A: Insufficiency of the Condorcet loser criterion. We begin with a formal definition of a weak Condorcet loser, a concept that has already been touched upon informally earlier in the text. Let $N \in {\mathcal N}$, $R_N \in {\mathcal R}_N$ and $S \in {\mathcal X}$. A candidate $x \in S$ is a weak Condorcet loser for $R_N$ in S if

$$\begin{aligned} |\{i \in N \mid y R_i x\}| \ge \frac{|N|}{2} \; \text{ for } \text{ all } y \in S \setminus \{x\}. \end{aligned}$$

Note that it is possible for there to be multiple weak Condorcet losers. The set of all weak Condorcet losers for $R_N$ in S is denoted by $WCL(R_N,S)$.

We now define an example to demonstrate that the Condorcet loser criterion is not sufficient to obtain a characterization of Daunou’s voting rule. Consider any population N, any profile $R_N$ and any feasible set S. We construct the set of candidates chosen by f according to the procedure described below.

• (i) In analogy with the first case in the definition of Daunou’s rule, if there is a strong Condorcet winner for $R_N$ in S, choose this candidate and only this candidate.

• (ii) If there is no strong Condorcet winner for $R_N$ in S and there is no strong Condorcet loser for $R_N$ in S, all plurality winners for $R_N$ in the set S are chosen.

• (iii) If there is no strong Condorcet winner for $R_N$ in S and there is a strong Condorcet loser for $R_N$ in S and the set of plurality winners for $R_N$ in $S \setminus CL(R_N,S)$ contains at least one candidate who is not a weak Condorcet loser for $R_N$ in $S \setminus CL(R_N,S)$, the set of plurality winners for $R_N$ in $S \setminus CL(R_N,S)$ who are not weak Condorcet losers is chosen.

• (iv) If there is no strong Condorcet winner for $R_N$ in S and there is a strong Condorcet loser for $R_N$ in S and the set of plurality winners for $R_N$ in $S \setminus CL(R_N,S)$ only contains weak Condorcet losers for $R_N$ in $S \setminus CL(R_N,S)$, the set of plurality winners for $R_N$ in $S \setminus CL(R_N,S)$ is chosen.

The above four cases (i)–(iv) can be defined more formally as follows. Let $N \in {\mathcal N}$, $R_N \in {\mathcal R}_N$ and $S \in {\mathcal X}$, and define $f(R_N,S)$ as follows.

• (i) If $CW(R_N,S) \ne \emptyset$, then $f(R_N,S) = CW(R_N,S)$;

• (ii) If $CW(R_N,S) = CL(R_N,S) = \emptyset$, then $f(R_N,S) = PW(R_N,S)$;

• (iii) If $CW(R_N,S) = \emptyset$ and $CL(R_N,S) \ne \emptyset$ and

  $$\begin{aligned} PW(R_N, S \setminus CL(R_N,S)) \setminus WCL(R_N, S \setminus CL(R_N,S)) \ne \emptyset , \end{aligned}$$

  then $f(R_N,S) = PW(R_N,S\setminus CL(R_N,S)) \setminus WCL(R_N, S \setminus CL(R_N,S))$;

• (iv) If $CW(R_N,S) = \emptyset$ and $CL(R_N,S) \ne \emptyset$ and

  $$\begin{aligned} PW(R_N, S \setminus CL(R_N,S)) \setminus WCL(R_N, S \setminus CL(R_N,S)) = \emptyset , \end{aligned}$$

  then $f(R_N,S) = PW(R_N,S\setminus CL(R_N,S))$.

The voting rule f satisfies the Condorcet loser criterion because a strong Condorcet loser is never chosen. Also, the rule satisfies the axioms of anonymity and neutrality because the labels assigned to the voters and the candidates are irrelevant, and it satisfies the Condorcet winner criterion because a (unique) strong Condorcet winner is always chosen; see case (i). Conditional tops only and conditional reduction monotonicity are satisfied because the axioms are silent in the presence of a strong Condorcet loser, and in case (ii)—the only case that is relevant for the two axioms—there cannot be a violation because the plurality rule applies in all requisite instances. That Condorcet loser independence is violated can be seen by examining the following example. Let $N = \{1,2,3,4\}$, $S = X = \{x,y,z,w,v\}$ and consider the profile $R_N$ given by

$$\begin{aligned}&x R_1 w R_1 z R_1 y R_1 v,&\\&y R_2 z R_2 w R_2 x R_2 v,&\\&z R_3 x R_3 y R_3 w R_3 v,&\\&w R_4 x R_4 z R_4 y R_4 v.&\end{aligned}$$

There is no strong Condorcet winner for $R_N$ in S, candidate v is the unique strong Condorcet loser for $R_N$ in S, and candidates y and w are weak Condorcet losers for $R_N$ in $S \setminus CL(R_N,S) = \{x,y,z,w\}$. The set of plurality winners in $S \setminus CL(R_N,S)$ is $PW(R_N,S \setminus CL(R_N,S)) = PW(\{x,y,z,w\}) = \{x,y,z,w\}$. According to the above definition, it follows that

$$\begin{aligned} f(R_N,S) = f(R_N,\{x,y,z,w,v\}) = \{x,z\} \end{aligned}$$

because case (iii) applies, and

$$\begin{aligned} f(R_N,S \setminus CL(R_N,S)) = f(\{x,y,z,w\}) = \{x,y,z,w\} \end{aligned}$$

because case (ii) applies. Clearly, $f(R_N,S) \ne f(R_N, S \setminus CL(R_N,S))$ so that Condorcet loser independence is violated.

Part B: Independence of the axioms used in Theorem 1. The following examples establish the independence of the axioms used in our characterization. In each of them, the axiom that is violated is indicated.

Anonymity Let $N \in {\mathcal N}$, $R_N \in {\mathcal R}_N$ and $S \in {\mathcal X}$. Define the modified plurality score $mps(R_N,S;x)$ of $x \in S$ for $R_N$ in S by

$$\begin{aligned} mps(R_N,S;x) = \left\{ \begin{array}{ll} |\{ i \in N \setminus \{1\} \mid t(R_i,S) = x\}| + 2 &{} \text{ if } 1 \in N; \\ |\{ i \in N \mid t(R_i,S) = x\}| &{} \text{ if } 1 \not \in N. \end{array} \right. \end{aligned}$$

These scores reflect a special status accorded to voter 1: the top candidate of this voter receives twice the weight of all other voters’ top candidates. A candidate $x \in S$ is a modified plurality winner for $R_N$ in S if

$$\begin{aligned} mps(R_N,S;x) \ge mps(R_N,S;y) \; \text{ for } \text{ all } y \in S. \end{aligned}$$

The set of modified set of plurality winners for $R_N$ is S is denoted by $MPW(R_N,S)$. Now define $f(R_N,S)$ as follows.

• (i) If $CW(R_N,S) \ne \emptyset$, then $f(R_N,S) = CW(R_N,S)$;

• (ii) If $CW(R_N,S) = \emptyset$, then $f(R_N,S) = MPW(R_N,S \setminus CCL(R_N,S))$.

Because voter 1 has preferred status, anonymity is violated. All other axioms are satisfied.

Neutrality Let $N \in {\mathcal N}$, $R_N \in {\mathcal R}_N$ and $S \in {\mathcal X}$. Suppose that $x^*$ is a fixed candidate in X. Now define $f(R_N,S)$ as follows.

• (i) If $CW(R_N,S) \ne \emptyset$, then $f(R_N,S) = CW(R_N,S)$;

• (ii) If $CW(R_N,S) = \emptyset$ and $PW(R_N,S \setminus CCL(R_N,S)) = \{x^*\}$, then

  $$\begin{aligned} f(R_N,S) = \{x^*\}; \end{aligned}$$

• (iii) If $CW(R_N,S) = \emptyset$ and $PW(R_N,S \setminus CCL(R_N,S)) \ne \{x^*\}$, then

  $$\begin{aligned} f(R_N,S) = PW(R_N,S \setminus CCL(R_N,S)) \setminus \{x^*\}. \end{aligned}$$

Neutrality is violated because the candidate $x^*$ is treated worse than the others: this candidate is only chosen if it is the (unique) strong Condorcet winner or, in the absence of a strong Condorcet winner, the unique plurality winner among those who are not cumulative strong Condorcet losers. All other axioms are satisfied.

Condorcet winner criterion Define the voting rule f as follows. For all $N \in {\mathcal N}$, for all $R_N \in {\mathcal R}_N$ and for all $S \in {\mathcal X}$,

$$\begin{aligned} f(R_N,S) = PW(R_N,S \setminus CCL(R_N,S)). \end{aligned}$$

This voting rule violates the Condorcet winner criterion and satisfies all other axioms.

Condorcet loser independence Define the voting rule f as follows. For all $N \in {\mathcal N}$, for all $R_N \in {\mathcal R}_N$ and for all $S \in {\mathcal X}$,

• (i) If $CW(R_N,S) \ne \emptyset$, then $f(R_N,S) = CW(R_N,S)$;

• (ii) If $CW(R_N,S) = \emptyset$, then $f(R_N,S) = PW(R_N,S)$.

This voting rule violates Condorcet loser independence because the Condorcet loser is not removed before applying the plurality rule in part (ii) and, therefore, it is chosen for some profiles. All other axioms are satisfied.

Conditional tops only Define, for all $N \in {\mathcal N}$, for all $R_N \in {\mathcal R}_N$ and for all $S \in {\mathcal X}$,

• (i) If $CW(R_N,S) \ne \emptyset$, then $f(R_N,S) = CW(R_N,S)$;

• (ii) If $CW(R_N,S) = \emptyset$ and

  $$\begin{aligned}&|N| = 2 \text{ and } \text{ there } \text{ exist } \text{ distinct } x,y,z \in S \text{ such } \text{ that }&\nonumber \\&S \setminus CCL(R_N,S) = \{x,y,z\} \text{ and } x R_i z R_i y \text{ and } y R_j z R_j x \text{ for } i,j \in N, \end{aligned}$$

  (7)

  then $f(R_N,S) = \{x,y,z\}$;

• (iii) If $CW(R_N,S) = \emptyset$ and (7) does not apply, then $f(R_N,S) = PW(R_N,S \setminus CCL(R_N,S))$.

This voting rule allows for candidates that are not at the top of any voter’s ranking to be selected in some circumstances and, as such, does not satisfy conditional tops only. The remaining axioms are satisfied, which is straightforward to verify for all of them except perhaps conditional reduction monotonicity. The only way of potentially violating this axiom emerges if the original (non-reduced) profile and the reduced profile are such that one is covered by case (ii) and the other by case (iii); if both are covered by (iii), Daunou’s voting rule applies to both. Case (ii) cannot apply to the original profile: because (ii) requires that $|N| = 2$, any reduced profile has only a single voter—and with a single voter, there trivially is a strong Condorcet winner (this single voter’s top candidate) so that case (i) applies and conditional reduction monotonicity is vacuously satisfied. The only remaining possibility is such that the reduced profile has two voters (case (ii)) and the original non-reduced profile has more than two voters (case (iii)). Because the set $\{x,y,z\}$ is chosen in the reduced profile, the only possibility to satisfy the premise of the axiom is that there are three voters and the additional voter has z as the top candidate. But, in that case, z is the strong Condorcet winner in the original three-voter profile, a contradiction.

For simplicity of exposition, the profiles that are treated differently in this example involve situations with two voters only and, thus, one may be led to suspect that the independence of conditional tops only from the remaining axioms is an artifact of allowing for such small societies. This is not the case. Indeed, the example can be amended so as to use profiles with any even number of voters |N| and assigning the ranking $x R_i z R_i y$ to half of them and the ranking $y R_j z R_j x$ to the other half. Again, any larger profile that reduces to such a profile as in the statement of conditional reduction monotonicity would have to be such that z is a strong Condorcet winner in the original (non-reduced) profile. It follows that conditional tops only is not implied even if smaller societies are excluded from consideration.

Conditional reduction monotonicity Define a voting rule f as follows. For all $N \in {\mathcal N}$, for all $R_N \in {\mathcal R}_N$ and for all $S \in {\mathcal X}$,

• (i) If $CW(R_N,S) \ne \emptyset$, then $f(R_N,S) = CW(R_N,S)$;

• (ii) If $CW(R_N,S) = \emptyset$, then

  $$\begin{aligned} f(R_N,S) = \{ x \in S \setminus CCL(R_N,S) \mid \{i \in N \mid t(R_i,S) = x\} \ne \emptyset \}. \end{aligned}$$

According to case (ii) of this voting rule, all candidates who appear at least once in a top position are chosen. This violates conditional reduction monotonicity because there are profiles such that the elimination of a preference relation with a specific chosen candidate at the top from the profile does not remove this candidate from the chosen set. That the remaining axioms are satisfied is straightforward to verify.

Part C: Illustration of the proof method of Theorem 1. To make the argument employed in the proof of the only-if part of our theorem easier to follow, we provide an informal explanation of the proof structure, along with an illustrative example.

Consider a profile $R_N$ and a feasible set S. First, observe that if there is a strong Condorcet winner, then this candidate must be chosen uniquely; this is an immediate consequence of the Condorcet winner criterion. If there is no strong Condorcet winner, we can, without loss of generality, assume that there are no cumulative strong Condorcet losers; this follows from applying Condorcet loser independence as many times as needed. What remains to be shown is that, in the absence of strong Condorcet winners and of cumulative strong Condorcet losers, the set of candidates chosen by the voting rule must be given by the set of plurality winners.

If the set of plurality winners coincides with the set of candidates S, it follows that each candidate in S has the same plurality-winner score $n^*$ of top positions for the profile under consideration in the set S. Now an auxiliary profile ${\overline{R}}_N$ can be defined that preserves the top candidates of all voters in N and, moreover, is a replica of a completely symmetric profile. By anonymity and neutrality, all candidates in S (and, thus, all plurality winners) must be chosen for the profile ${\overline{R}}_N$. Because the top candidates are the same for $R_N$ and for ${\overline{R}}_N$, conditional tops only implies that these plurality winners must be chosen for $R_N$ as well.

The last (and most subtle) case is obtained if the set of plurality winners is a strict subset of the set of feasible candidates S. Again, let $n^*$ be the plurality-winner score for $R_N$ in S. In contrast to the previous case, now there are candidates in S with a number of top positions assigned to them that is less than $n^*$. To deal with this situation, we first augment the profile $R_N$ by adding as many voters as required to arrive at a profile that is an $n^*$-fold replica of a completely symmetric profile; that this augmentation is well-defined is established in the formal proof. Denote this augmented profile by ${\overline{R}}_{N \cup N'}$, where $N'$ is the set of added voters. By anonymity and neutrality, it follows that all candidates in S must be chosen for the profile ${\overline{R}}_{N \cup N'}$. We then successively reduce the profile ${\overline{R}}_{N \cup N'}$ to a profile ${\overline{R}}_N$ that contains only the original voters in N. Each step in this reduction corresponds to one of the candidates that are in S but not in the set of plurality winners. Applying the axiom of conditional reduction monotonicity in each step, we conclude that, once the profile ${\overline{R}}_N$ is reached, the only candidates that remain chosen are the plurality winners for $R_N$ in S. By construction, the profile ${\overline{R}}_N$ is such that all top candidates are identical to those in $R_N$ and, using conditional tops only, we arrive at the desired conclusion that the chosen set for $R_N$ is the set of plurality winners in S. We also need to show that, at any stage in the reduction process, the set of strong Condorcet winners and the cumulative set of strong Condorcet losers remain empty; this observation is, of course, essential in order to invoke conditional reduction monotonicity. The following example illustrates the proof method just outlined.

Example 1

Let $N = \{1,2,3,4,5,6\}$ and $S = \{x,y,z,w,v\}$, and suppose that the profile $R_N$ is given by

$$\begin{aligned}&x R_1 z R_1 y R_1 w R_1 v,&\\&x R_2 z R_2 y R_2 w R_2 v,&\\&y R_3 v R_3 x R_3 w R_3 z,&\\&y R_4 w R_4 v R_4 x R_4 z,&\\&z R_5 v R_5 x R_5 y R_5 w,&\\&w R_6 z R_6 v R_6 x R_6 y.&\end{aligned}$$

Clearly, $CW(R_N,S) = CCL(R_N,S) = \emptyset$. We have $PW(R_N,S) = \{x,y\}$, $S \setminus PW(R_N,S) = \{z,w,v\}$, and the plurality-winner score is $n^* = 2$. We now add a set of four voters $N' = \{7,8,9,10\}$ with preferences such that z appears $1 = 2 - 1 = n^* - |\{i \in N \mid t(R_i,S) = z\}|$ times at the top of a preference ordering, w is the top candidate in $1 = n^*- 1$ preferences, and v is the best candidate for $2 = n^* - 0$ voters. The reason why the number of added voters is equal to four is that this choice allows us to obtain a replica of a completely symmetric profile. By construction, in the augmented profile, each of the five candidates in S appears $n^* = 2$ times at the top of a preference ordering. Note that the top alternatives of the six initial voters 1 to 6 are unchanged in the augmented profile. We denote this augmented profile by ${\overline{R}}_{N \cup N'} = ({\overline{R}}_N,{\overline{R}}_{N'})$, and it is defined as

$$\begin{aligned}&x {\overline{R}}_1 y {\overline{R}}_1 z {\overline{R}}_1 w {\overline{R}}_1 v,&\\&x {\overline{R}}_2 y {\overline{R}}_2 z {\overline{R}}_2 w {\overline{R}}_2 v,&\\&y {\overline{R}}_3 z {\overline{R}}_3 w {\overline{R}}_3 v {\overline{R}}_3 x,&\\&y {\overline{R}}_4 z {\overline{R}}_4 w {\overline{R}}_4 v {\overline{R}}_4 x,&\\&z {\overline{R}}_5 w {\overline{R}}_5 v {\overline{R}}_5 x {\overline{R}}_5 y,&\\&w {\overline{R}}_6 v {\overline{R}}_6 x {\overline{R}}_6 y {\overline{R}}_6 z,&\\&z {\overline{R}}_7 w {\overline{R}}_7 v {\overline{R}}_7 x {\overline{R}}_7 y,&\\&w {\overline{R}}_8 v {\overline{R}}_8 x {\overline{R}}_8 y {\overline{R}}_8 z,&\\&v {\overline{R}}_9 x {\overline{R}}_9 y {\overline{R}}_9 z {\overline{R}}_9 w,&\\&v {\overline{R}}_{10} x {\overline{R}}_{10} y {\overline{R}}_{10} z {\overline{R}}_{10} w.&\end{aligned}$$

The augmented profile ${\overline{R}}_{N \cup N'}$ is a two-fold replica of a completely symmetric profile. By anonymity and neutrality, it follows that all five candidates must be chosen so that

$$\begin{aligned} f({\overline{R}}_{N \cup N'},S) = S = \{x,y,z,w,v\}. \end{aligned}$$

We now show that, at each stage of the successive reduction process that leads us back to a profile involving the set of voters N, the set of strong Condorcet winners and the cumulative set of strong Condorcet losers remains empty. This allows us to invoke conditional reduction monotonicity in each iteration. To do so, let M be any non-empty subset of $N' = \{7,8,9,10\}$. Observe that, in the profile ${\overline{R}}_{N \cup (N' \setminus M)}$, candidate x appears above candidate y at least $n^* = 2$ times and candidate y appears above candidate x at most $n^* = 2$ times. Analogously, y appears at least twice above z and z appears at most twice above y; z appears at least twice above w and w appears at most twice above z; w appears at least twice above v and v appears at most twice above w; and, finally, v appears at least twice above x and x appears at most twice above v. Thus, there are no strong Condorcet winners and no cumulative strong Condorcet losers in any subprofile of ${\overline{R}}_{N \cup N'}$ that contains N. As a consequence, conditional reduction monotonicity can be applied in the following process of successively eliminating the added voters. We perform this reduction one top candidate at a time in order to iteratively reduce the set of voters back to the original set N.

Let $M_1 = \{9,10\}$. Thus, the profile ${\overline{R}}_{N \cup (N' \setminus M_1)}$ is given by

$$\begin{aligned}&x {\overline{R}}_1 y {\overline{R}}_1 z {\overline{R}}_1 w {\overline{R}}_1 v,&\\&x {\overline{R}}_2 y {\overline{R}}_2 z {\overline{R}}_2 w {\overline{R}}_2 v,&\\&y {\overline{R}}_3 z {\overline{R}}_3 w {\overline{R}}_3 v {\overline{R}}_3 x,&\\&y {\overline{R}}_4 z {\overline{R}}_4 w {\overline{R}}_4 v {\overline{R}}_4 x,&\\&z {\overline{R}}_5 w {\overline{R}}_5 v {\overline{R}}_5 x {\overline{R}}_5 y,&\\&w {\overline{R}}_6 v {\overline{R}}_6 x {\overline{R}}_6 y {\overline{R}}_6 z,&\\&z {\overline{R}}_7 w {\overline{R}}_7 v {\overline{R}}_7 x {\overline{R}}_7 y,&\\&w {\overline{R}}_8 v {\overline{R}}_8 x {\overline{R}}_8 y {\overline{R}}_8 z.&\end{aligned}$$

Because $t({\overline{R}}_i,S) = v$ for all $i \in M_1$, conditional reduction monotonicity implies

$$\begin{aligned} f({\overline{R}}_{N \cup (N' \setminus M_1)},S) = f({\overline{R}}_{N \cup N'},S) \setminus \{v\} = \{x,y,z,w,v\} \setminus \{v\} = \{x,y,z,w\}. \end{aligned}$$

Now let $M_2 = \{8\}$. The profile ${\overline{R}}_{N \cup (N' \setminus (M_1 \cup M_2))}$ is given by

$$\begin{aligned}&x {\overline{R}}_1 y {\overline{R}}_1 z {\overline{R}}_1 w {\overline{R}}_1 v,&\\&x {\overline{R}}_2 y {\overline{R}}_2 z {\overline{R}}_2 w {\overline{R}}_2 v,&\\&y {\overline{R}}_3 z {\overline{R}}_3 w {\overline{R}}_3 v {\overline{R}}_3 x,&\\&y {\overline{R}}_4 z {\overline{R}}_4 w {\overline{R}}_4 v {\overline{R}}_4 x,&\\&z {\overline{R}}_5 w {\overline{R}}_5 v {\overline{R}}_5 x {\overline{R}}_5 y,&\\&w {\overline{R}}_6 v {\overline{R}}_6 x {\overline{R}}_6 y {\overline{R}}_6 z,&\\&z {\overline{R}}_7 w {\overline{R}}_7 v {\overline{R}}_7 x {\overline{R}}_7 y.&\end{aligned}$$

Because $t({\overline{R}}_i,S) = w$ for all $i \in M_2$, conditional reduction monotonicity implies

$$\begin{aligned} f({\overline{R}}_{N \cup (N' \setminus (M_1 \cup M_2))},S) = f({\overline{R}}_{N \cup (N' \setminus M_1)},S) \setminus \{w\} = \{x,y,z,w\} \setminus \{w\} = \{x,y,z\}. \end{aligned}$$

Finally, let $M_3 = \{7\}$. The profile ${\overline{R}}_{N \cup (N' \setminus (M_1 \cup M_2 \cup M_3))} = {\overline{R}}_N$ is given by

$$\begin{aligned}&x {\overline{R}}_1 y {\overline{R}}_1 z {\overline{R}}_1 w {\overline{R}}_1 v,&\\&x {\overline{R}}_2 y {\overline{R}}_2 z {\overline{R}}_2 w {\overline{R}}_2 v,&\\&y {\overline{R}}_3 z {\overline{R}}_3 w {\overline{R}}_3 v {\overline{R}}_3 x,&\\&y {\overline{R}}_4 z {\overline{R}}_4 w {\overline{R}}_4 v {\overline{R}}_4 x,&\\&z {\overline{R}}_5 w {\overline{R}}_5 v {\overline{R}}_5 x {\overline{R}}_5 y,&\\&w {\overline{R}}_6 v {\overline{R}}_6 x {\overline{R}}_6 y {\overline{R}}_6 z.&\end{aligned}$$

Because $t({\overline{R}}_i,S) = z$ for all $i \in M_3$, conditional reduction monotonicity implies

$$\begin{aligned} f({\overline{R}}_N,S)= & {} f({\overline{R}}_{N \cup (N' \setminus (M_1 \cup M_2 \cup M_3))},S) \\= & {} f({\overline{R}}_{N \cup (N' \setminus (M_1 \cup M_2))},S) \setminus \{z\} \\= & {} \{x,y,z\} \setminus \{z\} = \{x,y\} \\= & {} PW(R_N,S). \end{aligned}$$

By conditional tops only (which can be applied because $CW({\overline{R}}_N,S) = CCL({\overline{R}}_N,S) = \emptyset$), it follows that

$$\begin{aligned} f(R_N,S) = f({\overline{R}}_N,S) = \{x,y\} = PW(R_N,S) = f^D(R_N,S), \end{aligned}$$

as desired. $\square$