Premise-based vs conclusion-based collective choice

Abstract

Imagine a group of individuals faces a yes-no type question whose answer is logically determined by multiple premises. There are two salient types of procedures to aggregate individual judgments—the “premise-based way” (PBW) and the “conclusion-based way” (CBW). We derive necessary and sufficient conditions under which two procedures are universally ordered. If (and only if) a decision problem takes a “conjunctive” form, PBW derives a positive collective judgment (i.e., “yes”) whenever CBW does. Furthermore, if we replace “conjunctive” with “disjunctive” in the previous line, PBW derives a negative collective judgment (i.e., “no”) whenever PBW does. These observations highlight the fact that these two procedures are a mathematical dual of each another. Asymptotic properties are also studied. Under classical Condorcetian assumptions, PBW ensures the probability that the voting outcome is correct converges to one as the size of a group tends to infinity, whereas this holds for CBW only if an additional condition is satisfied.

Appendix: omitted proofs

1.1 Preliminary

Let us provide some lattice theoretic tools that will be useful. We fix a natural number N, and consider a Boolean function $\phi : \{0,1\}^N \rightarrow \{0,1\}$. Such a function $\phi$ partitions its domain $\{0,1\}^N$ into $B_1(\phi ) = \{x\in \{0,1\}^N: \phi (x) = 1\}$ and $B_0(\phi ) = \{x\in \{0,1\}^N: \phi (x) = 0\}$. Conversely, any ordered binary partition of $\{0,1\}^N$, say $(B_1,B_0)$, recovers a Boolean function in a way that $\phi (x) = 1$ if and only if $x \in B_1$. Thus, there is a one-to-one correspondence between Boolean functions $\phi$ on $\{0,1\}^N$ and ordered binary partitions of $\{0,1\}^N$.

Several properties of $\phi$ are translated into the properties of the partition $\bigl ( B_1(\phi ), B_0(\phi ) \bigr )$. For example, $\phi$ is non-constant if and only if both $B_1(\phi )$ and $B_0(\phi )$ are non-empty. Also, $\phi$ is monotone if and only if $B_1(\phi )$ is monotone in the sense that $x \ge y$ and $y \in B_1(\phi )$ together imply $x \in B_1(\phi )$. Our Lemma 1 translates the definitions of conjunction, disjunction, and degeneracy into the language of partitions. As a corollary, we see that $\phi$ is degenerate if and only if it is both conjunctive and disjunctive.

A partially ordered set, or poset, is a non-empty set L equipped with a reflexive, transitive, and antisymmetric binary relation $\ge$. For a non-empty subset $S \subseteq L$, denote the supremum and infimum of S by $\bigvee S$ and $\bigwedge S$ if they exist. When S consists of two elements, we write as $\bigvee \{x,y\} = x \vee y$ and $\bigwedge \{x,y\} = x \wedge y$. A poset L is a lattice if $x \wedge y$ and $x \vee y$ exist for all $x,y \in L$. For instance, $\{0,1\}^N$ is a lattice endowed with the usual component-wise relation. A non-empty subset S of a lattice L is called a sublattice (of L) if $x\vee y \in S$ and $x\wedge y \in S$ for any $x,y \in S$, and particularly, it is convex if $z \in S$ whenever there exist $x,y \in S$ such that $x \ge z \ge y$. For example, a closed interval defined by $[\underline{x},\overline{x}] \equiv \{ x\in L: \overline{x}\ge z \ge \underline{x} \}$ with $\overline{x} \ge \underline{x}$ constitutes a convex sublattice of a lattice L.

Lemma 1

The following equivalence results hold for any Boolean function $\phi : \{0,1\}^N \rightarrow \{0,1\}$.

• (i) $\phi$ is conjunctive if and only if $B_1(\phi )$ is a convex sublattice of $\{0,1\}^N$.

• (ii) $\phi$ is disjunctive if and only if $B_0(\phi )$ is a convex sublattice of $\{0,1\}^N$.

• (iii) $\phi$ is degenerate if and only if both $B_1(\phi )$ and $B_0(\phi )$ are convex sublattices of $\{0,1\}^N$.

Proof

Let us prove (i). Suppose that $\phi$ is conjunctive, so that we have $B_1(\phi ) = \{x: (x_i)_{i \in I} = c\}$ for some non-empty set $I \subseteq [N]$ and $c \in \{0,1\}^{I}$. Let $\underline{x} \in \{0,1\}^N$ be the point such that $(\underline{x}_i)_{i \in I} = c$, and $\underline{x}_i =0$ for all $i \notin I$. Similarly, let $\overline{x} \in \{0,1\}^N$ be the point such that $(\overline{x}_i)_{i \in I} = c$, and $\underline{x}_i =1$ for all $i \notin I$. Clearly, $B_1(\phi ) = [\underline{x},\overline{x}]$, thereby $B_1(\phi )$ is a convex sublattice. Conversely, suppose that $B_1(\phi )$ is a convex sublattice. Since $B_1(\phi )$ is finite, using the definition of sublattices iteratively, we have $\underline{x} \equiv \bigwedge B_1(\phi ) \in B_1(\phi )$ and $\overline{x} \equiv \bigvee B_1(\phi ) \in B_1(\phi )$, from which convexity implies $B_1(\phi ) = [\underline{x},\overline{x}]$. Now letting $I = \{i \in [N]: \underline{x}_i = \overline{x}_i\}$ and $c = (\overline{x}_i)_{i \in I}$, we see that $\phi$ is conjunctive. The proof of (ii) is similar.

Let us prove (iii). If $\phi$ is degenerate, there exist $i \in [N]$ and $c \in \{0,1\}$ for which $\phi (x) = 1$ iff $x_i = c$, and hence, $\phi (x) = 0$ iff $x_i = \lnot c$, where $\lnot c$ is an element of $\{0,1\}$ not equal to c. This means that $\phi$ is conjunctive and disjunctive. By (i) and (ii), it follows that both $B_1(\phi )$ and $B_0(\phi )$ are convex sublattices. Conversely, suppose that both $B_1(\phi )$ and $B_0(\phi )$ are convex sublattices, i.e., each of which is characterized by the pair of a non-empty index set and a binary vector, say $(I^1,c^1)$ and $(I^0,c^0)$, respectively. First, we claim that $I^1 = I^0$. Suppose not, there exists $i^* \in I^1 \setminus I^0$. Let $x \in \{0,1\}^N$ be such that $x_{i^*} = \lnot c^1_{i^*}$, while $x_i = \lnot c^0_{i}$ for all $i \in I^0$ (note that such x exists because $i^* \notin I^0$). Then, we see that $x \notin B_1(\phi )$ and $x \notin B_0(\phi )$, but this is a contradiction to that $B_1(\phi )$ and $B_0 (\phi )$ partition $\{0,1\}^N$. Hence, we have $I \equiv I_1 = I_0$. Now, we claim that I is a singleton. Note that the sizes of $B_1(\phi )$ and $B_0(\phi )$ are both equal to $2^{N-|I|}$, while they partition $\{0,1\}^N$ of size $2^N$. Namely,

$$\begin{aligned} 2^{N-|I|} + 2^{N-|I|} = 2^N. \end{aligned}$$

from which $|I| = 1$. Therefore $c^1, c^0 \in \{0,1\}$. In particular, we must have $c^1 \ne c^0$, and thus, the value of $\phi$ is determined by exactly one component, meaning that $\phi$ is degenerate

Proof of Theorem 1

It suffices to prove (i) because (ii) follows from the lattice dual argument.

If part. Suppose that g is conjunctive. Then, together with monotonicity, there exists $J \subseteq [m]$ for which $g(z) = 1$ if and only if $(z_j)_{j \in J} = (1,\ldots ,1)$. Take any ${\varvec{x}}\in X^n$ such that $CBW_{f,g}({\varvec{x}}) = 1$. We claim that $x_j \ge (g(x^1), \ldots , g(x^n))$ for any $j \in J$. Indeed, if i is an individual such that $g(x^i) = 1$, it must be the case $x^i_j = 1$ because g is conjunctive. Then, the monotonicity of f implies

$$\begin{aligned} f(x_j) \ge f(g(x^1),\ldots ,g(x^n)) = CBW_{f,g} ({\varvec{x}}) = 1. \end{aligned}$$

Since the above holds for any $j \in J$, we have $PBW_{f,g}({\varvec{x}}) = g(f(x_1),\ldots ,f(x_m))= 1$. Since $PBW_{f,g}$ and $CBW_{f,g}$ only admit values in $\{0,1\}$, we get $PBW_{f,g} \ge CBW_{f,g}$. The proof of $PBW_{f,g} \ge CBW_{f,g}$ when f is disjunctive is similar.

Only if part. Suppose that g is not conjunctive, and f is not disjunctive. We shall present an example of profiles such that $CBW_{f,g}(\varvec{x}) = 1$, but $PBW_{f,g}(\varvec{x}) = 0$. By Lemma 1, $B_1(g)$ is not a convex sublattice, i.e., there exist distinct $z, z' \in B_1(g)$ such that either (i) $z \wedge z' \notin B_1(g)$, (ii) $z\vee z' \notin B_1(g)$, or (iii) $z'' \notin B_1(g)$ for some $z''$ lies between z and $z'$. Under the monotonicity of g, however, (ii) and (iii) cannot occur. Hence, we must have $z\wedge z' \notin B_1(g)$ for some $z, z' \in B_1(g)$. Similarly, since f is not disjunctive, $B_0(f)$ is not a convex sublattice by Lemma 1. Again by the monotonicity of f, we find distinct $y,y' \in B_0(f)$ such that $y\vee y' \notin B_0(f)$.

Now, define a coalition $U \subseteq [n]$ by $U = \{i\in [n]: (y\vee y')_i = 1\}$, and partition it into $T_1 = \{i\in [n]: y_i = 1\}$ and $T_2 = U \setminus T_1$. Note that $T_1 \ne \emptyset$, since otherwise we would have $y\vee y' = y' \in B_0(f)$, a contradiction. Moreover, $T_1 \subsetneq U$ since $y\vee y' > y$, thereby neither $T_1$ nor $T_2$ is empty. Note that $U = T_1 \cup T_2$ is decisive in the sense that if every $i \in U$ holds the same judgment on an issue, then their common judgment is collectively accepted. On the other hand, neither $T_1$ nor $T_2$ is decisive by themselves. Given these notes, consider a profile ${\varvec{x}} = (x^i)_{i=1}^n \in X^n$ defined as follows

$$\begin{aligned} x^i = {\left\{ \begin{array}{ll} z &{}\text {if } i \in T_1 \\ z' &{}\text {if } i \in T_2 \\ \varvec{0} &{}\text {if } i \in [n] \setminus (T_1 \cup T_2). \end{array}\right. } \end{aligned}$$

By $z,z' \in B_1(g)$, all individuals in $U = T_1 \cup T_2$ accept the conclusion, while those not in U rejects the conclusion. Since U is decisive, the conclusion is collectively accepted under CBW. Now, we claim that the collective judgments on premises are given by $\bigl ( f(x_1),\ldots ,f(x_m) \bigr ) = z \wedge z'$. For any $j \in [m]$, if $(z\wedge z')_j = 1$ then $z_j = z'_j = 1$, and thus every individual in U accepts the jth premise. Hence, it is collectively accepted, i.e., $f(x_j) = (z \wedge z')_j = 1$ as desired. Conversely, $(z\wedge z')_j = 0$ implies either $z_j = 0$ or $z'_j = 0$. In either case, the jth premise is supported by solely $T_1$ or $T_2$, which is not decisive by itself. Hence, $f(x_j) = (z \wedge z')_j = 0$ holds. Therefore, we have shown that $\bigl ( f(x_1),\ldots ,f(x_m) \bigr ) = z \wedge z' \in B_0(g)$, from which the conclusion is rejected under PBW. $\square$

Proof of Theorem 2

Suppose that $C \subseteq \{0,1\}^m$ satisfies Assumption 2 and 3. We want to show that the aggregation of premise judgments is consistent in the sense that $\bigl ( f(x_j) \bigr )_{j=1}^m \in C$ for all $\varvec{x} \in C^n$. To this end, we utilize Theorem 3 in Nehring and Puppe (2008), which establishes the (necessary and) sufficient condition for consistent aggregation of premise judgments. Specifically, in the logic-based model, they show that the aggregation of premise judgments is consistent if aggregation rules satisfy their intersection property. We translate their sufficient condition to be applied for our abstract algebraic model.

To this end, we introduce a dummy variable $*$ to stand for “the jth premise is not in this set.” Then, consider a binary order $\geqslant$ on $\{0,1,*\}$ defined by $0 \geqslant *$ and $1 \geqslant *$, together with all reflexive relations, and extend it to $X^* \equiv \{0,1,*\}^m$ by the componentwise comparison, i.e., $x \geqslant x'$ if and only if $x_j \geqslant x'_j$ for all $j \in [m]$. Note that $X^*$ is a superset of X, and any $x \in X$ is maximal with respect to $\geqslant$. Then, define $D^*$ and $C^*$ by

$$\begin{aligned}&C^* = \bigl \{ x\in X^*: y \geqslant x \text { for some } y \in C \bigr \}, \\&D^* = X^* \setminus C^*. \end{aligned}$$

A profile in $C^*$ is called satisfiable. To interpret it, for example, observe that if $(1,1,*,0) \in C^*$, then either (1, 1, 1, 0) or (1, 1, 0, 0) is in C, or either $\{A_1,A_2,A_3, \lnot A_4\}$ or $\{A_1,A_2,\lnot A_3, A_4\}$ is satisfiable in words of propositional logic. In particular, $\{A_1, A_2, \lnot A_4\}$ is also satisfiable as being a subset of a satisfiable set. On the other hand, we refer to any profile in $D^*$ as unsatisfiable, and particularly, $x \in D^*$ is said to be minimally unsatisfiable if it is a minimal element of $D^*$ with respect to $\geqslant$. The minimally unsatisfiable vectors will play the key role to determine if the consistent aggregation of premises is possible.

Any subset of [n] is called a coalition. Given an aggregation rule $f: \{0,1\}^n \rightarrow \{0,1\}$, the families of winning coalitions for acceptance and rejection are defined as follows:

$$\begin{aligned} \mathcal {W}_{\mathrm{Accept}}&= \bigl \{ \{i: x^i = 1\}: x \in B_1 (f) \bigr \}, \\ \mathcal {W}_{\mathrm{Reject}}&= \bigl \{ \{i: x^i = 0\}: x \in B_0 (f) \bigr \}. \end{aligned}$$

For $x \in X^*$, a selection of winning coalitions associated with x is a family of coalitions $\{W_j\}_{j=1}^m$ satisfying for all $j \in [m]$ that (i) $W_j \in \mathcal {W}_{\mathrm{Accept}}$ if $x_j = 1$, (ii) $W_j \in \mathcal {W}_{\mathrm{Reject}}$ if $x_j = 0$, and (iii) $W_j = [n]$ if $x_j = *$. The following lemma is a direct consequence of Theorem 3 in Nehring and Puppe (2008). $\square$

Lemma 2

The aggregation of premise judgment is consistent if for any minimally unsatisfiable profile $x \in D^*$, and any selection of winning coalitions $\{W_j\}_{j=1}^m$ associated with x, we have

$$\begin{aligned} \bigcap _{j=1}^m W_j \ne \emptyset . \end{aligned}$$

Given this lemma, our remaining task is to check whether the intersection property provided in Lemma 2 is satisfied under Assumption 2 and 3. To this end, we provide Lemma 3 that characterizes minimally unsatisfiable profiles. For $x \in X^*$, the jth component in x is said to be null if $x_j = *$, and non-null otherwise. By the degree of x, we mean the number of non-null components in x.

Lemma 3

Every minimal element of $D^*$ is of degree 2, and it contains both 0 and 1 as its non-null components.

Proof of Lemma 3

Evidently, Assumption 2 excludes any element of degree 1 from $D^*$. Suppose not, there exists a minimal element $x \in D^*$ of degree $k\ge 3$. Assume without loss of generality that x takes the form

$$\begin{aligned} x = \left( \underbrace{x_1,x_2,x_3\ldots ,x_k}_{\text {non-null}},\, \underbrace{*, \ldots , *}_{\text {null}} \right) \in D^*. \end{aligned}$$

Since $k \ge 3$, there are at least two $j, j' \in [k]$ such that $x_j = x_{j'}$. Without loss of generality, assume that $x_j = x_{j'} = 1$, and $j = 1$ and $j' = 2$. Thus, x takes the form

$$\begin{aligned} x = \bigl ( 1,1,x_3\ldots ,x_k, *, \ldots , * \bigr ). \end{aligned}$$

Since x is minimal in $D^*$, any $x' \in X^*$ with $x \geqslant x'$ must belong to $C^*$. In particular, $C^*$ contains the following two elements:

$$\begin{aligned} x'&= \left( *,1,x_3\ldots ,x_k, *, \ldots , * \right) \\ x''&= \left( 1,*,x_3\ldots ,x_k, *, \ldots , * \right) . \end{aligned}$$

By the construction of $C^*$, there exist $y', y'' \in C$ such that $y' \geqslant x'$ and $y'' \geqslant x''$. In particular, we must have $y'_1 = 0$ and $y''_2 = 0$, since otherwise we would have $y' \geqslant x$, which leads to $x \in C^*$, a contradiction. Therefore, $y'$ and $y''$ be such that

$$\begin{aligned} y'&= \left( 0,1,x_3\ldots ,x_k, \overbrace{y'_{k+1}, \ldots , y'_{m}}^{\text {non-null}} \right) \\ y''&= \left( 1,0,x_3\ldots ,x_k, \underbrace{y''_{k+1}, \ldots , y''_{m}}_{\text {non-null}} \right) . \end{aligned}$$

By Assumption 3, we must have $y' \vee y'' \in C$, but since $y' \vee y'' \geqslant x$, it follows that $x \in C^*$, a contradiction. Therefore, we have shown that every minimal element of $D^*$ must be of degree 2. Moreover, Assumption 2 and 3 together imply $\varvec{1}, \varvec{0} \in C^*$. Thus, for any element of degree 2, if both of its non-null components take the same value, it must belong to $C^*$. Therefore, every minimal element of degree 2 has both 1 and 0 as its components. $\square$

Proof of Theorem 2

By Lemma 3, the intersection property in Lemma 2 reduces to the pairwise-intersection property: $W_1 \cap W_0 \ne \emptyset$ for any $W_1 \in \mathcal {W}_{\mathrm{Accept}}$ and $W_0 \in \mathcal {W}_{\mathrm{Reject}}$. Suppose not, this is violated by some $W_1$ and $W_0$. Then, let $x, y \in \{0,1\}^n$ be a profile such that

$$\begin{aligned} x^i = {\left\{ \begin{array}{ll} 1 &{}\text {if } i \in W_1 \\ 0 &{}\text {otherwise} \end{array}\right. } \quad \text { and } \quad y^i = {\left\{ \begin{array}{ll} 0 &{}\text {if } i \in W_0 \\ 1 &{}\text {otherwise}. \end{array}\right. } \end{aligned}$$

By constructions, $f(x^i) = 1$ and $f(x^i) = 0$, while we have $x^i \le y^i$ by $W_1 \cap W_0 = \emptyset$. This is a contradiction to the monotonicity of f. Hence, Lemma 2 concludes that the aggregation of premises is consistent, as desired.

1.2 Proof of Theorem 4 and 5

Since the argument is symmetric, it is enough to prove only Theorem 4 (i) and Theorem 5 (iii). Suppose that $f^c \ge f$ and g is non-constant. Recall the definitions of winning coalitions (used in the proof of Theorem 2), and observe the following facts. The proof is trivial and omitted.

Lemma 4

The winning coalitions have the following properties.

• (i) $W \in \mathcal {W}_{\mathrm{Accept}}$ if and only if $[n] \setminus W \in \mathcal {W}_{\mathrm{Reject}}$.

• (ii) $f^c \ge f$ if and only if $\mathcal {W}_{\mathrm{Accept}} \subseteq \mathcal {W}_{\mathrm{Reject}}$.

• (iii) $f \ge f^c$ if and only if $\mathcal {W}_{\mathrm{Reject}} \subseteq \mathcal {W}_{\mathrm{Accept}}$.

• (iv) f has no veto power if and only if neither $\mathcal {W}_{\mathrm{Accept}}$ nor $\mathcal {W}_{\mathrm{Reject}}$ contains a singleton.

Proof of Theorem 4 (i)

Let us show that $PBW_{f,g} \ge CBW_{f,g}$ when g is conjunctive. Suppose that g is conjunctive, i.e., there exist essential premises $J \subseteq [m]$ and a binary vector $c \in \{0,1\}^J$ such that $g(z) = 1$ if and only if $(z_j)_{j \in J} = (c_j)_{j \in J}$. Take any ${\varvec{x}} \in X_n$ such that $CBW_{f,g}({\varvec{x}}) = 1$. Then, there exists $W_1 \in \mathcal {W}_{\mathrm{Accept}}$ such that $g(x^i) = 1$ for any $i \in W_1$, and $g(x^i) = 0$ for any $i \in [n] \setminus W_1$. Note that Lemma 4 implies that $W_1 \in \mathcal {W}_{\mathrm{Reject}}$. In particular, since g is conjunctive, we have $(x^i_j)_{j \in J} = (c_j)_{j \in J}$ for any $i \in W_1$. That is, the judgment $c_j$ on each essential premise $j \in J$ is supported by the coalition $W_1$, which is winning for both acceptance and rejection. Hence, we have $\bigl ( f(x_j) \bigr )_{j \in J} = (c_j)_{j \in J}$, from which $\bigl (f(x_1), \ldots , f(x_m)\bigr ) \in B_1(g)$. Therefore, $PBW_{f,g}({\varvec{x}}) = 1$.

Next, we claim that if f is monotone and disjunctive and satisfies $f^c \ge f$, then f must be dictatorial. As f is monotone and disjunctive, there exists $I \subseteq [n]$ for which $f(x) = 1$ if and only if $x^i = 1$ for all $i \in [n]$. On the other hand, $f^c \ge f$ implies that $1 \ge f(x) + f(x^c)$, but this can be true only when $|I| = 1$; Otherwise, considering a profile x such that $(x^i, x^{i'}) = (1,0)$ for $i,i' \in I$, we end up with $f(x) + f(x^c) = 1 + 1 = 2$, a contradiction. Hence, f is dictatorial, and thus, the desired inequality is trivially obtained when f is disjunctive.

Necessity of the assumption in Theorem 4. Let us show that the assumption $f^c \ge f$ is indispensable for (i) when g is non-monotonic. That is, assuming that $f^c \not \ge f$, we want to present an aggregation problem such that g is conjunctive, but $PBW_{f,g} \not \ge CBW_{f,g}$. By Lemma 4 and $f^c \not \ge f$, there exists some $W_1 \in \mathcal {W}_{\mathrm{Accept}}$ such that $W_1 \notin \mathcal {W}_{\mathrm{Reject}}$. Assuming that g is conjunctive, the essential premises J are divided into $j \in J_1$ for which $c_j = 1$, and $j \in J_0$ for which $c_j = 0$. In particular, since g is non-monotonic, we must have $J_0 \ne 0$. Then, let $\varvec{x} = (x^i)_{i=1}^n \in X^n$ be a judgment profile with the following construction:

$$\begin{aligned}&x^i = \bigl ( \overbrace{1,\ldots ,1}^{\text {for }j \in J_1},\, \overbrace{0,\ldots ,0}^{\text {for }j \in J_0},\, *,\ldots ,* \bigr ) \quad \text { for } i \in W_1 \\&x^i = \bigl ( \underbrace{0,\ldots ,0}_{\text {for }j \in J_1},\, \underbrace{1,\ldots ,1}_{\text {for }j \in J_0},\, *,\ldots ,* \bigr ) \quad \text { for } i \notin W_1, \end{aligned}$$

where the judgments on unessential premises are arbitrary (written as $*$). Since $g(x^i) = 1$ for every $i \in W_1$, the conclusion is collectively accepted under CBW. On the other hand, $f(x_j) = 0$ for each $j \in J_0$ because $W_1$ is not winning for rejection. Hence, it follows that $\bigl (f(x_1), \ldots , f(x_m)\bigr ) \notin B_1(g)$, from which the conclusion is rejected under PBW. $\square$

Proof of Theorem 5 (iii)

Suppose that $f \ge f^c$, but g is not conjunctive. By Lemma 1, it follows that $B_1(g)$ is not a convex sublattice, which must result in one of the following cases:

• (i) $z \wedge z' \notin B_1(g)$ for some $z, z' \in B_1 (g)$;

• (ii) $z \vee z' \notin B_1(g)$ for some $z, z' \in B_1 (g)$;

• (iii) $z'' \notin B_1(g)$ with some $z, z' \in B_1 (g)$ such that $z< z'' < z'$.

For each of these cases, we shall present an example of individual judgments such that the conclusion is accepted only under CBW. Consider a coalition W that is a minimal element of $\mathcal {W}_{\mathrm{Reject}}$. Fix an arbitrary $i \in W$, and consider a partition

$$\begin{aligned} W = \underbrace{\bigl ( W \setminus \{i\} \bigr )}_{\equiv T_1} \cup \underbrace{\{i\}}_{\equiv T_2}. \end{aligned}$$

Since f has no veto power, $\mathcal {W}_{\mathrm{Reject}}$ does not contain singletons, from which $T_1$ is non-empty. The no veto power condition also implies $T_2 \notin \mathcal {W}_{\mathrm{Accept}}$ and $T_2 \notin \mathcal {W}_{\mathrm{Reject}}$. Moreover, the minimality of $W_0$ implies that $T_1 \notin \mathcal {W}_{\mathrm{Reject}}$. Then, let $T_3 \equiv [n] \setminus (T_1 \cup T_2)$. Note that $W = T_1 \cup T_2 \in \mathcal {W}_{\mathrm{Reject}}$ and $\mathcal {W}_{\mathrm{Reject}} \subseteq \mathcal {W}_{\mathrm{Accept}}$ holds by $f \ge f^c$. Hence, by Lemma 4, we have $T_3 \notin \mathcal {W}_{\mathrm{Reject}}$ and $T_3 \notin \mathcal {W}_{\mathrm{Accept}}$. Also, we can use Lemma 4 to determine whether the union of each pair of coalitions $T_1$, $T_2$, $T_3$ is winning or not. After some investigations, the construction of coalitions is summarized as in Table 1. Now, let us define a profile $\varvec{x} = (x^i)_{i =1}^n \in X^n$ as follows:

$$\begin{aligned} x^i = {\left\{ \begin{array}{ll} {\left\{ \begin{array}{ll} z \wedge z' &{}\text {in case (i)} \\ z \vee z' &{}\text {in case (ii)} \\ z'' &{}\text {in case (iii)} \end{array}\right. } &{} \text {if } i \in T_1 \\ z &{}\text {if } i \in T_2 \\ z' &{}\text {if } i \in T_3. \end{array}\right. } \end{aligned}$$

Observe that the conclusion is supported by the coalition $T_2 \cup T_3$, which is winning for acceptance. Thus, the conclusion is collectively accepted under CBW. On the other hand, the coalition $T_1$ plays the pivotal role in voting on premise-issues, as it forms a winning coalition together with each of $T_2$ or $T_3$ (for both acceptance and rejection). Hence, whenever $z_j$ and $z_{j'}$ differ, the collective judgment on the jth premise must be the one submitted from $T_1$. Therefore, $\bigl ( f(x_j) \bigr )_{j=1}^m$ equals the judgment profile of $T_1$, but this is not in $B_1(g)$. Hence, the conclusion is collectively rejected under PBW. $\square$

Table 1 The construction of coalitions in Theorem 5