An Accuracy Argument in Favor of Ranking Theory

Abstract

Fitelson and McCarthy (2014) have proposed an accuracy measure for confidence orders which favors probability measures and Dempster-Shafer belief functions as accounts of degrees of belief and excludes ranking functions. Their accuracy measure only penalizes mistakes in confidence comparisons. We propose an alternative accuracy measure that also rewards correct confidence comparisons. Thus we conform to both of William James’ maxims: “Believe truth! Shun error!” We combine the two maxims, penalties and rewards, into one criterion that we call prioritized accuracy optimization (PAO). That is, PAO punishes wrong comparisons (preferring the false to the true) and rewards right comparisons (preferring the true to the false). And it requires to prioritize being right und avoiding to be wrong in a specific way. Thus PAO is both, a scoring rule and a decision rule. It turns out that precisely confidence orders representable by two-sided ranking functions satisfy PAO. The point is not to argue that PAO is the better accuracy goal. The point is only that ranking theory can also be supported by accuracy considerations. Thus, those considerations by themselves cannot decide about rational formats for degrees of belief, but are part and parcel of an overall normative assessment of those formats.

Appendix A: Proofs

1.1 A.1 Canonical Order

$k: W \to \mathbb {N} \cup \{\infty \}$ is a ranking mass (or point ranking function) iff it is a total function and k^− 1(0)≠∅. Given a ranking mass, the induced negative ranking function over $\mathcal {A}$ is $\kappa _{k}(A) = \min _{w \in A} k(w)$ if A≠∅ and else $\kappa _{k}(\emptyset )=\infty$. κ_k is indeed a negative ranking function and in fact complete, i.e., for all $S \subseteq \mathcal {A}$, $\kappa _{k}(\bigcup S) =\min _{A \in S} \kappa _{k}(A)$. Conversely, every complete negative ranking function κ over $\mathcal {A}$ has a unique ranking mass k_κ(w) = κ({w}) inducing it.

Definition 8

Let W≠∅ be finite. Given a basic order ≤ over W (i.e., weak order), the canonical two-sided ranking function τ_≤ is recursively defined as follows:

• (1) W₀ = W, $[0]=: \max _{\leq } W_{0}$,

• (2) W_n+ 1 := W_n ∖ [n] (if non-empty, else stop), $[n+1] := \max _{\leq } W_{n+1}$,

• (3) k(w) := x iff w ∈ [x], κ = κ_k,

• (4) $\tau _{\leq }(A)=\kappa (\overline {A}) - \kappa (A)$.

The canonical extension of ≤ to $\mathcal {A}$ is the order ≼ induced by τ_≤: A ≼ B iff τ_≤(A) ≤ τ_≤(B) for all $A,B \in \mathcal {A}$.

Call [m] the m-worlds. It is clear that k is a ranking mass over W (since [0]≠∅, by finiteness and totality), thus κ = κ_k is a negative ranking function and τ_≤ a two-sided ranking function over $\mathcal {A}$, and all are regular. The canonical extension ≼ of ≤ is a weak order, ≤-compatible and clearly representable by a regular two-sided ranking function (namely by τ_≤).

Let $A \in \mathcal {A} \setminus \{\emptyset , W\}$ and n > 0: A is a 0-set, iff A partitions [0] non-voidly (A ∩ [0]≠∅ and $\overline {A} \cap [0] \neq \emptyset$). A is an − n-set iff the smallest world in A is an n-world. A is an n-set iff A is the complement of a − n-set. Note: there are 0-sets iff |[0]| > 1.

Lemma 1

Let ≤ be a basic orderover finite W, ≼itscanonical extension andA_x,A_y x-and y-sets. Then

• (1) for$A \in \mathcal {A} \setminus \{\emptyset , W\}$: A is an x-set iffτ_≤(A) = x,

• (2) for$A_{x}, A_{y} \in \mathcal {A} \setminus \{\emptyset , W\}$:A_x ≼ A_yiffx ≤ y,

• (3) an extension ≼^∗of≤ to$\mathcal {A}$isrepresentable by a regular two-sided ranking function iff≼^∗is≼.

Proof

(1) holds by construction, (2) follows. (3) holds since τ_≤ is, up to order isomorphism, the only regular two-sided ranking function representing ≤. □

Let $n \in \mathbb {N}$: A − n-set A_−n is full iff $[n] \subseteq A_{n}$, it is non-full iff it is not full. An n-set A_n is void iff it is the complement of a full − n-set, it is non-void iff it is not void. $\dot {A}_{-n}, \dot {A}_{n}$ will stand for full − n-sets and void n-sets, and be contrasted with A_−n,A_n, where these are non-full − n-sets and non-void n-sets.

Denote PF(w) := {〈A,B〉 : w ∈ A ∖ B}, resp. PR(w) := {〈A,B〉 : w ∈ B ∖ A}, the set of potentially wrong, resp. right, comparisons inw . In the whole section ≼ is the canonical extension of the basic ≤.

Lemma 2 (Wrongness)

Let ≼be the canonicalextension of the basic ≤.Then

• (1) F(≼,w) = PF(w) ∩{〈A_x,A_y〉 : −n ≤ x < y ≤ n} forw ∈ [n],

• (2) |F(≼,w)| = 0 forw ∈ [0],

• (3) |F(≼,w)| = |F(≼,v)| forallw,v ∈ [n],n ≥ 0.

Proof

Assume ≤,≼ as given. (1) Let 〈A_x,A_y〉∈ F(≼,w) for w ∈ [n]. Then (i) w ∈ A_x ∖ A_y, i.e., 〈A_x,A_y〉∈ PF(w) and (ii) A_x ≺ A_y. By (i) − n ≤ x and y ≤ n. By (ii) x < y (Lemma 1.2). Thus − n ≤ x < y ≤ n. Conversely, let 〈A_x,A_y〉∈ PF(w) ∩{〈A_x,A_y〉 : −n ≤ x < y ≤ n}. Thus w ∈ A_x ∖ A_y and − n ≤ x < y ≤ n. Hence A_x ≺ A_y (Lemma 1.2). (2) − 0 ≤ x < y ≤ 0 is impossible. (3) follows by permutation invariance within classes [n]. □

Lemma 3 (Logical flaws)

Let ≼be the canonicalextension of the basic ≤.Then

• (1) |L(≼,w)| = 0 forw ∈ [0],

• (2) if 〈A_y,A_x〉∈ L_∪(≼,w) forw ∈ [n],then the witnessesA_z,v ∈ [k] aresuch thaty,z < x ≤ k < nand0 < x,

• (3) if 〈A_x,A_y〉∈ L_∩(≼,w) forw ∈ [n],then the witnessesA_z,v ∈ [k] aresuch that − n < −k ≤ x < y,zandx < 0,

• (4) |L(≼,w)| = 0 forw ∈ [1],

• (5) |L(≼,w)| = |L(≼,v)| forallw,v ∈ [n].

Proof

(1) There are no better worlds than 0-worlds (in the sense of ≤ or F), thus there can be no world-witness for flaws. (2) Let 〈A_y,A_x〉∈ L_∪(≼,w) for w ∈ [n]. Thus, according to Definition 7, the witnesses A_z, v ∈ [k] for k < n satisfy (i) A_y ≺ A_x, (ii) A_z ≺ A_x, (iii) w ∈ A_x ∖ A_y, (iv) A_x = A_y ∪ A_z, (v) v∉A_x. k < x contradicts (v). Hence x ≤ k < n. We know x > 0: For suppose (for reductio) x ≤ 0. Since y,z < x, we have − x < −y,−z. But then A_x contains a − x-world and A_y,A_z none, contradicting (iv). (3) follows by a similar reasoning as for (2). (4) is a corollary to (2) and (3). (5) is due to permutation invariance within classes [n]. □

Lemma 4 (Rightness)

Let ≼be the canonicalextension of the basic ≤.Then forw ∈ [n],n ≥ 0:

• (1) R(≼,w) = PR(w) ∖{〈A_y,A_x〉 : −n ≤ x ≤ y ≤ n},

• (2) R(≼,w) = PR(w) ∖{〈A_x,A_y〉 : x = y = 0},ifn = 0,

• (3) |R(≼,w)| = |R(≼,v)| forallw,v ∈ [n],n ≥ 0.

Proof

(1) Let 〈A_y,A_x〉∈ R(≼,w). Then (i) A_y ≺ A_x and (ii) w ∈ A_x ∖ A_y, i.e., 〈A_y,A_x〉∈ PR(w). We show that, given (ii), (i) is equivalent to 〈A_y,A_x〉∉{〈A_y,A_x〉 : −n ≤ x ≤ y ≤ n}. Since w ∈ [n], (ii) implies (a) x ≥−n and (b) y ≤ n. Therefore:

$$\begin{array}{clr} & \langle A_{y}, A_{x}\rangle \notin \{\langle A_{y}, A_{x}\rangle : -n \leq x\leq y \leq n\} &\\ \text{iff} & x< -n \vee y <x \vee n< y & (\text{logic})\\ \text{iff} & y < x & (\text{ii a, b})\\ \text{iff} & A_{y} \prec A_{x} & (\text{canonical } \preceq) \end{array}$$

(2) is a corollary to (1). (3) is due to permutation invariance within classes [n]. □

Definition 2

Let ≼^∗ be a relation over $\mathcal {A}$ and ≼ the canonical extension of the basic ≤.

• ≼^∗agrees weakly with ≼ up to n iff ≼^∗ agrees with the transitive closure of the relation determined by:

  $$ A_{y} \prec A_{z}, $$

  (A.1)
  $$ A_{-t} \prec A_{-n} ~~~ , ~~~ A_{n} \prec A_{t}, $$

  (A.2)
  $$ A_{x} \sim A^{\prime}_{x}, $$

  (A.3)

  where − n ≤ y < z ≤ n, n < t, − n < x < n.

• ≼^∗agrees strongly with ≼ up to n iff ≼^∗ agrees with the transitive closure of the relation determined by Eqs. A.1, A.2, A.3 and

  $$ A_{-n} \sim A^{\prime}_{-n} ~~~,~~~ A_{n} \sim A^{\prime}_{n}. $$

  (A.4)

• ≼^∗agrees with ≼up to n iff agreement is strong in case |[n]| > 1 and agreement is weak in case |[n]| = 1.

Lemma 5

Let ≤ bea basic order over W, ≼ its canonical extension and ≼^∗anotherextension which agrees weakly with ≼ up to n. Then forw ∈ [n + 1] andv ∈ [t] wheret > n + 1,we have |F(≼^∗,v)| > |F(≼,w)|.

Proof

Let ≼^∗,≼,≤,w,v,t as assumed. Define ψ : F(≼,w) → F(≼^∗,v) by

$$\psi(\langle A_{x}, A_{y}\rangle ) = \left\{\begin{array}{ll} \langle A_{x}, A_{y} \rangle & \text{if } v \in A_{x}, v \notin A_{y},\\ \langle A_{x}, (A_{y} \cup \{w\}) \setminus \{v\}\rangle & \text{if } v \in A_{x}, v \in A_{y},\\ \langle (A_{x} \cup \{v\}) \setminus \{w\} , A_{y}\rangle & \text{if } v \notin A_{x}, v \notin A_{y},\\ \langle (A_{x} \cup \{v\}) \setminus \{w\} , (A_{y} \cup \{w\}) \setminus \{v\}\rangle & \text{if } v \notin A_{x}, v \in A_{y}. \end{array}\right.$$

We show: ψ is a total, injective and non-surjective function. Denote $\langle B_{x^{\prime }}, B_{y^{\prime }}\rangle =\psi (\langle A_{x}, A_{y} \rangle )$.

Total function: Let 〈A_x,A_y〉∈ F(≼,w). Thus w ∈ A_x ∖ A_y and − (n + 1) ≤ x < y ≤ n + 1 (Lemma 2.1). By definition of ψ, (i) $v \in B_{x^{\prime }} \setminus B_{y^{\prime }}$. To establish (ii) $B_{x^{\prime }} \prec ^{*} B_{y^{\prime }}$, we show that the type shift$\langle x,y\rangle \mapsto \langle x^{\prime }, y^{\prime }\rangle$ satisfies $x^{\prime } \in \{x, -(n+2)\}, y^{\prime } \in \{y, n+2\}$, where $x^{\prime } =-(n+2)$ iff x = −(n + 1) and A_x contains only w as n + 1-world, and $y^{\prime } = n+2$ iff y = n + 1 and A_y misses only the n + 1-world w among the n + 1-worlds.

Case 1::

the identity involves no shift.

Case 2::

Note − (n + 1) < y ≤ n + 1. Removing the t-world v from A_y (t > n + 1), implies no shift. Adding the n + 1-world w changes the type, iff A_y is an n + 1-set which misses only the n + 1-world w, and in this case y = n + 1 goes to $y^{\prime } =n+2$.

Case 3::

Note − (n + 1) ≤ x < n + 1. Adding the t-world v to A_x (t > n + 1) implies no shift. Removing the n + 1-world w changes the type iff A_x is a − (n + 1)-set which contains only w as n + 1-world, and in this case x = −(n + 1) goes to $x^{\prime }=-(n+2)$.

Case 4::

− n ≤ x < y ≤ n: Combine cases 2 and 3.

This, together with weak agreement up to n, establishes (ii) and thus the totality of ψ.

Injective: For reductio, assume (a) $\langle B_{x^{\prime }}, B_{y^{\prime }}\rangle = \psi (\langle A_{x}, A_{y} \rangle )$ for some 〈A_x,A_y〉∈ F(≼,w) and (b) there is $(A^{*}_{x^{*}}, A^{*}_{y^{*}}) \in F(\preceq , w)$, such that (c) $(A^{*}_{x^{*}}, A^{*}_{y^{*}}) \neq \langle A_{x}, A_{y} \rangle$, but (d) $\langle B_{x^{\prime }}, B_{y^{\prime }}\rangle = \psi ((A^{*}_{x^{*}}, A^{*}_{y^{*}}))$. Note: either $x^{\prime } \geq - (n+1)$, in which case there is no type-shift (cf. cases 1-4), hence $x=x^{\prime }=x^{*}$; or $x^{\prime } = -(n+2)$, in which case (cf. cases 2-4) x = −(n + 1) = x^∗. Similarly for y,y^∗. Thus x = x^∗ and y = y^∗. By (c) either (c1) $A^{*}_{x} \neq A_{x}$ or (c2) $A^{*}_{y} \neq A_{y}$.

(c1). Either (i) $B_{x^{\prime }} = A_{x}$ (iff v ∈ A_x) or (ii) $B_{x^{\prime }}=(A_{x} \cup \{v\}) \setminus \{w\}$ (iff v∉A_x). Similarly, either (i*) $B_{x^{\prime }}= A^{*}_{x}$ (iff $v \in A^{*}_{x}$) or (ii*) $B_{x^{\prime }} = (A^{*}_{x} \cup \{v\}) \setminus \{w\}$ (iff $v \notin A^{*}_{x}$) (equivalently $A^{*}_{x} = (B_{x^{\prime }} \cup \{w\}) \setminus \{v\}$). (i,i*) contradicts (c1). (i, ii*) implies $A_{x} = (A^{*}_{x} \cup \{v\}) \setminus \{w\}$ and thus w∉A_x, contradicting (a). (ii, i*) implies $A^{*}_{x} = (A_{x} \cup \{v\}) \setminus \{w\}$ and thus $w \notin A^{*}_{x}$, contradicting (b). (ii, ii*) implies $A^{*}_{x} =(((A_{x} \cup \{v\}) \setminus \{w\}) \cup \{w\}) \setminus \{v\} = A_{x}$, contradicting (c1). Hence $A^{*}_{x} = A_{x}$.

(c2) follows similarly, with the simultaneous replacements: y/x,w/v,v/w. Hence $A^{*}_{y} =A_{y}$, proving injectivitiy.

Non-Surjective: {v}≺^∗{w} (because ≼^∗ extends ≤) and 〈{v},{w}〉∈ F(≼^∗,v). However, that comparison is not reached by ψ. Else there would be 〈A_x,A_y〉 such that A_x = {w} and A_y = {v}, i.e. x = −(n + 1) and y = −t for t > n + 1. Yet 〈A_x,A_y〉∉F(≼,w), since we do not have {w}≺{v} (because ≼ also extends ≤). □

Lemma 6

Given the same assumptions as in Lemma 5:|L(≼^∗,v)|≥|L(≼,w)|.¹

Proof

Let ≼^∗,≼,≤,w,u,t as described. Define χ : L(≼,w) → L(≼^∗,u) by

$$\chi(\langle A_{y},A_{x} \rangle ) = \left\{\begin{array}{ll} \langle A_{y}, A_{x}\rangle & \text{if } u \in A_{x}, u \notin A_{y},\\ \langle(A_{y} \cup \{w\} )\setminus \{u\}, A_{x}\rangle & \text{if } u \in A_{x}, u \in A_{y},\\ \langle A_{y}, (A_{x} \cup \{u\}) \setminus \{w\}\rangle & \text{if } u \notin A_{x}, u \notin A_{y}. \end{array}\right.$$

If 〈A_y,A_x〉∈ L(≼,w) we have $A_{y} \subseteq A_{x}$. Thus u∉A_x,u ∈ A_y is impossible. We show that χ is a total, injective function. Denote $\langle B_{y^{\prime }}, B_{x^{\prime }} \rangle =\chi (\langle A_{y}, A_{x}\rangle )$. We establish, by a similar reasoning as in Lemma 5: $y^{\prime }=y$ and $x^{\prime }=x$. From Lemma 3.2–3: either (a) z,y < x ≤ k < n + 1 and 0 < x or (b) − (n + 1) < −k ≤ y < x,z and y < 0.

Case 1::

Trivially $y^{\prime } =y, x^{\prime }=x$.

Case 2::

A_ygains a n + 1-world and looses a t-world (t > n + 1). The gain has an effect iff (i) y ≤−(n + 2) or (ii) y = n + 1. (a) excludes (ii), and (i) and A_x = A_z ∪ A_y would imply that A_z ∩ [≤ x] = A_x ∩ [≤ x], i.e., z ≥ x, contrary to the assumption z < x in (a). (b) excludes (i) and (ii). The loss has an effect iff (i) y = −t or (ii) y = t + 1. (a) excludes (ii), and (i) would imply the same contradiction as above, since − t ≤−(n + 2). (b) excludes (i) and (ii). Thus there is no type shift.

Case 3::

A_x gains a t-world and looses a n + 1-world. The gain has an effect iff (i) x ≤−(t + 1) or (ii) x = t. (a) excludes (i) and (ii). (b) excludes (i); and (ii) and y < z imply that there is v ∈ A_z with v ∈ [m] for m < −y and v ∈ A_x (since x = t > n + 1 > m), contradicting A_y = A_x ∩ A_z. The loss has an effect only if (i) x = −(n + 1) or (ii) x = n + 2. (a) excludes (i) and (ii). (b) excludes (i) and (ii) would imply a similar contradiction as above. Thus there is no type shift.

We can therefore write 〈B_y,B_x〉 = ψ(〈A_y,A_x〉). For ∪-flaws, define $B_{z^{\prime }}$ as: Case 1: $B_{z^{\prime }} = A_{z}$. Case 2: $B_{z^{\prime }}= A_{z} \cup \{u\}$. Case 3: $B_{z^{\prime }} = (A_{z} \cup \{u\} )\setminus \{w\}$. One shows $z^{\prime }=z$ (cf. case 2 above, exchanging t and n + 1). For ∩-flaws define $B_{z^{\prime }}$ as : Case 1: $B_{z^{\prime }} = A_{z}$. Case 2: $B_{z^{\prime }}= (A_{z} \cup \{w\}) \setminus \{u\}$. Case 3: $B_{z^{\prime }} = A_{z} \setminus \{u\}$. One shows $z^{\prime }=z$ (cf. case 2 above). Thus in general $z^{\prime } = z$.

Total function: If 〈A_y,A_x〉∈ L_∪(≼,w) has witnesses A_z,v, then 〈B_y,B_x〉∈ L_∪(≼^∗,u) has witnesses B_z,v. I.e., we show: (0) B_y ≺^∗B_x, (1) B_z ≺^∗B_x, (2) u ∈ B_x ∖ B_y, (3) B_x = B_y ∪ B_z and (4) v∉B_x. (0) and (1) hold since z,y < x for 0 < x < n + 1 (weak agreement up to n). (2) holds by definition of χ. (3) Case 1 is clear. Case 2: (A_z ∪{u}) ∪ ((A_y ∪{w}) ∖{u}) = A_x since w ∈ A_x (by assumption) and the removal of u from A_y was compensated with the addition of u to A_z. Case 3: A_y ∪ ((A_z ∪{u}) ∖{w}) = (A_x ∪{u}) ∖{w}, since the addition of u and removal of w from A_x was reproduced for A_z. (4) u ∈ [t] for t > n + 1 is distinct from v ∈ [k], since k < n + 1 < t and thus v∉B_x.

By a dual reasoning: If 〈A_y,A_x〉∈ L_∩(≼,w) has witnesses A_z,v, then 〈B_y,B_x〉∈ L_∩(≼^∗,u) has witnesses B_z,v.

Injective: For reductio, assume (a) 〈B_y,B_x〉 = ψ(〈A_y,A_x〉) for 〈A_y,A_x〉∈ L(≼,w) (i.e. w ∈ A_x ∖ A_y) and (b) there is $\langle A^{*}_{y^{*}}, A^{*}_{x^{*}} \rangle \in L(\preceq , w)$ (i.e. $w \in A^{*}_{x^{*}} \setminus A^{*}_{y^{*}}$), such that (c) $\langle A^{*}_{y^{*}}, A^{*}_{x^{*}} \rangle \neq \langle A_{y}, A_{x}\rangle$ but (d) $\langle B_{y}, B_{x}\rangle = \psi (\langle A^{*}_{y^{*}}, A^{*}_{x^{*}} \rangle )$. We have x^∗ = x and y^∗ = y (c.f. above). By (c) either (c1) $A^{*}_{x} \neq A_{x}$ or (c2) $A^{*}_{y} \neq A_{y}$.

(c1): Either (i) B_x = A_x (iff u ∈ A_x) or (ii) B_x = (A_x ∪{u}) ∖{w} (iff u∉A_x). Similarly, either (i*) $B_{x}= A^{*}_{x}$ (iff $u \in A^{*}_{x}$) or (ii*) $B_{x} = (A^{*}_{x} \cup \{u\}) \setminus \{w\}$ (iff $u \notin A^{*}_{x}$) (equivalently $A^{*}_{x} = (B_{x^{\prime }} \cup \{w\}) \setminus \{u\}$). (i,i*) contradicts (c1). (i,ii*), implies $A_{x} =(A^{*}_{x} \cup \{u\} )\setminus \{w\}$ and thus w∉A_x, contradicting (a). (ii, i*) implies $A^{*}_{x} = (A_{x} \cup \{u\}) \setminus \{w\}$ and thus $w \notin A^{*}_{x}$, contradicting (b). (ii,ii*) implies $A^{*}_{x} = (((A_{x} \cup \{u\}) \setminus \{w\}) \cup \{w\}) \setminus \{u\} = A_{x}$, contradicting (c1). Thus we have a contradiction in each case. Hence $A^{*}_{x} = A_{x}$.

(c2) similarly leads to a contradiction, applying the simultaneous replacements: y/x,w/u,u/w. □

1.2 A.2 Main Theorem

Definition 10

Let ≼^∗ be a relation over $\mathcal {A}$ and ≼ the canonical extension of ≤ (see Definition 8). a(≼^∗) agrees totally with a(≼) up to n iff for all w ∈ [≤ n]:

$$F(\preceq^{*}, w)= F(\preceq, w)~~~, ~~~ L(\preceq^{*}, w)= L(\preceq, w)~~~, ~~~R(\preceq^{*}, w)= R(\preceq, w)$$

a(≼^∗) agrees partially with a(≼) up to n iff a(≼^∗) agrees with a(≼) up to n − 1 and for all w ∈ [n]:

$$F(\preceq^{*}, w)= F(\preceq, w)~~~, ~~~ L(\preceq^{*}, w)= L(\preceq, w)~~~, ~~~|R(\preceq^{*}, w)| \geq |R(\preceq, w)|$$

Clearly, total agreement implies partial agreement.

Definition 11

Let ≼ be a weak order over $\mathcal {A}$ and a(≼) = 〈〈f₁,r₁〉,…,〈f_k,r_k〉〉 an accuracy profile for ≼. For f ∈{f₁,…f_k}, the f -block of a(≼) is the restriction of a(≼) to 〈f_i,r_i〉, where f_i = f.

Note that Definition 6 can now be understood as saying that a(≼^∗) ≥ a(≼) iff (1) for all f the f -block in a(≼^∗) and in a(≼) are equal (if it exists), or (2) there is a first f such that the f -block in a(≼^∗) is unequal to that of a(≼), being either (a) longer or (b) better, i.e., equally long, but for a first i in the block, $r^{*}_{i} > r_{i}$.

Theorem 13

Let ≤ bea basic order over W (finite) and ≼ its canonical extension. Then for ≼^∗atransitive total extension of ≤,a(≼^∗) ≥ a(≼) impliesthat ≼^∗is≼.²

Proof

Assume (I) ≼^∗ is a transitive, total extension of ≤ to $\mathcal {A}$ and (II) a(≼^∗) ≥ a(≼). Thus $a(\preceq ^{*}) = \langle \langle f^{*}_{1},r^{*}_{1}\rangle , \ldots , \langle f^{*}_{k}, r^{*}_{k}\rangle \rangle$ and a(≼) = 〈〈f₁,r₁〉,…,〈f_k,r_k〉〉 satisfy Definition 6. We recursively show profile- and order-agreement up to n, for all n, and thus ≼^∗ is ≼ (cf. Definitions 10 and 9).

Base case: Note that for all v∉[0]: |F(≼^∗,v)| > 0 (≤-compatibility). Thus we must have |F(≼^∗,w)| = 0 and |L(≼^∗,w)| = 0 for all w ∈ [0], else (II) would be violated, i.e., the 0-block would be longer for ≼ (Lemma 2, 3) which contradicts clauses (i)–(iii) of Definition 6. Since the 0-block for ≼^∗ cannot be longer than [0], (II) also implies |R(≼^∗,w)|≥|R(≼,w)| (by (i) and (iii) of Definition 6). Thus a(≼^∗) agreespartially with a(≼) up to 0.

• We show that |F(≼^∗,w)| = 0 for w ∈ [0] implies that for all n,m > 0 all n-sets A_n, all − m-sets A_−m and all 0-sets A₀, $A^{\prime }_{0}$ (which exist only if |[0]| > 1):

  $$A_{-m} \preceq^{*} A_{0} \sim^{*} A^{\prime}_{0} \preceq^{*} A_{n}$$

  • (1) $A_{0} \sim ^{*} A^{\prime }_{0}$: Let A be a 0-set, (i.e., |[0]| > 1). Then there is a witness v_A ∈ [0] ∖ A and w ∈ A ∩ [0], so that minimising F yields {v_A}≼^∗A in w and A ≼^∗{v_A} in v_A, i.e., $\{v_{A}\} \sim ^{*} A$. Therefore $A_{0} \sim ^{*} \{v_{A_{0}}\} \sim ^{*} \{v_{A^{\prime }_{0}}\} \sim ^{*} A^{\prime }_{0}$ (≤-compatibility).

  • (2) A_i ≼^∗A_n, i ≤ 0: $[0] \subseteq A_{n}$, but $[0] \nsubseteq A_{i}$. Hence there is a 0-world v ∈ A_n ∖ A_i and minimising F yields A_i ≼^∗A_n.

  • (3) A_−m ≼^∗A_j, j ≥ 0: A_−m ∩ [0] = ∅ but A_j ∩ [0]≠∅. Minimising F yields A_−m ≼^∗A_j.

• |L(≼^∗,w)| = 0 for w ∈ [0] implies no new constraints.

• |R(≼^∗,w)|≥|R(≼,w)| implies R(≼^∗,w) = R(≼,w). $R(\preceq ^{*}, w) \subseteq PR(w)$ (by definition). Additionally, if (A_x,A_y) ∈ R(≼^∗,w), then A_x ≺^∗A_y and hence not x = y = 0 (due to (1) above and because if |[0]| = 1 there are no 0-sets). Therefore $R(\preceq ^{*}, w) \subseteq R(\preceq , w)$ (Lemma 4). Thus |R(≼^∗,w)| = |R(≼,w)|, and in fact R(≼^∗,w) = R(≼,w). This shows total profile agreement up to 0. Therefore all F-undecided ≼^∗ (not turned into $\sim ^{*}$) are turned into ≺^∗ (Lemma 4), i.e.:

  $$A_{-m} \prec^{*} A_{0} \sim^{*} A^{\prime}_{0} \prec^{*} A_{n}$$

By transitivity, ≼^∗agrees strongly with ≼ up to 0 if |[0]| > 1 and weakly if |[0]| = 1.

Induction step: Suppose a(≼^∗) agrees with a(≼) totally up to n (IH1) and ≼^∗ agrees with ≼ up to n (IH2) (i.e., strongly if |[n]| > 1, and weakly if |[n]| = 1). We lift this to n + 1, by first using in addition to (IH1) only weak order agreement, denoted (IH2*).

By (IH1), X(≼,w) = X(≼^∗,w) for w ∈ [≤ n] and X = F,L,R. We first show that for w ∈ [n + 1] and v ∈ [> n + 1]:

• (1) |F(≼,w)|≤|F(≼^∗,w)|,

• (2) |L(≼,w)|≤|L(≼^∗,w)|,

• (3) |F(≼,w)| < |F(≼^∗,v)|,

• (4) |L(≼,w)|≤|L(≼^∗,v)|.

(1): It suffices to show $F(\preceq , w) \subseteq F(\preceq ^{*}, w)$: Let (A_x,A_y) ∈ F(≼,w) for w ∈ [n + 1]. Then (A_x,A_y) ∈ PF(w) and − (n + 1) ≤ x < y ≤ n + 1 (Lemma 2). Thus A_x ≺^∗A_y (IH2*), i.e., (A_x,A_y) ∈ F(≼^∗,w). (2): It suffices to show $L_{\cup }(\preceq , w) \subseteq L_{\cup }(\preceq ^{*}, w)$, $L_{\cap }(\preceq , w) \subseteq L_{\cap }(\preceq ^{*}, w)$: Let (A_y,A_x) ∈ L_∪(≼,w) for w ∈ [n + 1]. Then (A_y,A_x) ∈ R(≼, [m]), where 0 ≤ m < x ≤ k < n and y < x (Lemma 3). IH2* implies A_y ≺^∗A_x. Thus (A_y,A_x) ∈ L_∪(≼^∗,w) (using the same witnesses). An analogous reasoning applies to for L_∩. (3) follows from Lemma 5, IH2* and ≤-compatibility. (4) follows from Lemma 6 and IH2*.

Thus for w ∈ [n + 1] and v ∈ [> n + 1]:

• |F(≼^∗,w)| + |L(≼^∗,w)|≥|F(≼,w)| + |L(≼,w)|,

• |F(≼^∗,v)| + |L(≼^∗,v)| > |F(≼,w)| + |L(≼,w)|.

Hence: we must have |F(≼^∗,w)| + |L(≼^∗,w)| = |F(≼,w)| + |L(≼,w)|, else the (|F(≼,w)| + |L(≼,w)|)-block would be longer for ≼ than for ≼^∗, contradicting Assumption (II) by Definition 6. (1) and (2) above then imply |F(≼^∗,w)| = |F(≼,w)|, |L(≼^∗,w)| = |L(≼,w)| and in fact F(≼^∗,w) = F(≼,w), L(≼^∗,w) = L(≼,w) and thus |R(≼^∗,w)|≥|R(≼,w)|. This shows partial profile agreement up to n + 1 (using IH1).

By IH2* and transitivity, Eq. A.1 extends from n to n + 1 (set t = n + 1 in Eq. A.1). It thus suffices to establish Eqs. A.2, A.3 for n + 1 and, if |[n + 1]| > 1 (14) for n + 1.

• We show that F(≼^∗,w) = F(≼,w) implies:

  $$ A_{-k} \preceq^{*} A^{\prime}_{-(n+1)} \sim^{*} A_{-(n+1)} \preceq^{*} \dot{A}_{-(n+1)} ~~~,~~~ \dot{A}_{n+1} \preceq^{*} A_{n+1} \sim^{*} A^{\prime}_{n+1} \preceq^{*} A_{k}, $$

  (A.5)

  where k > n + 1, $A^{\prime }_{-(n+1)}, A_{-(n+1)}$ are non-full, $\dot {A}_{-(n+1)}$ is full, $\dot {A}_{n+1}$ is void, and in the case |[n + 1]| > 1 the sets $A^{\prime }_{-(n+1)}, A_{-(n+1)}$ are non-full and $A^{\prime }_{n+1}, A_{n+1}$ are non-void, whereas in the case |[n + 1]| = 1, A_x and $A^{\prime }_{x}$ for x ∈{−(n + 1),n + 1} with the previously required specifications don’t exist and can be discarded. The reasoning for the claimed non-existence is as follows: Assume |[n + 1]| = 1 and consider the case x = −(n + 1). An − (n + 1)-set, contains as smallest world an n + 1 world (by definition). But since there is only one such world (by assumption), such a set must be full. Thus it can only be of the type $\dot {A}_{-(n+1)}$. Hence only $A_{-k} \preceq \dot {A}_{-(n+1)}$ needs to be established (for k > n + 1). The reasoning is similar for x = n + 1.

  Additionally, we also show:

  $$ A_{-n} \preceq^{*} A^{*}_{-n} \sim^{*} A^{\prime}_{-n} \preceq^{*} \dot{A}_{-n} ~~~, ~~~ A_{n} \preceq^{*} A^{*}_{n} \sim^{*} A^{\prime}_{n} \preceq^{*} \dot{A}_{n}, $$

  (A.6)

  where now A_−n,A_n are (n + 1)-empty, $A^{*}_{-n}, A^{*}_{n}, A^{\prime }_{-n}, A^{\prime }_{n}$ are not (n + 1)-empty and not (n + 1)-full and $\dot {A}_{-n}, \dot {A}_{n}$ are n + 1-full. Here a set A is (n + 1)-empty iff A ∩ [n + 1] = ∅ and (n + 1)-full iff $[n+1] \subseteq A$.

  The following establishes Eq. A.5: $A_{n+1} \sim ^{*} A^{\prime }_{n+1}$ is shown as (1) in the base case (replace n + 1/0). A_n+ 1 ≼^∗A_k, A_−k ≼^∗A_−(n+ 1) is shown as (2) and (3) in the base case (replace n + 1/i, k/n,k/m). $\dot {A}_{n+1} \preceq ^{*} A_{n+1}$ is proved as (3) in the base case. $A_{-(n+1)} \preceq ^{*} \dot {A}_{-(n+1)}$ is proved as (2) in the base case. If |[n + 1]| = 1: $A_{-k} \preceq ^{*} \dot {A}_{-(n+1)}, \dot {A}_{n+1} \preceq ^{*} A_{k}$ are also obtained as in the base case (2) and (3).

  The following establishes Eq. A.6: Let |[n]| = 1 (else there is nothing to prove by strong agreement IH2). $A^{*}_{-n} \sim ^{*} A^{\prime }_{-n}, A^{*}_{n} \sim ^{*} A^{\prime }_{n}$ follows as (1) in the base case. $A^{\prime }_{-n} \preceq ^{*} \dot {A}_{-n}, A^{\prime }_{n} \preceq ^{*} \dot {A}_{n}$ follow as (2) in the base case. $A_{-n} \preceq ^{*} A^{*}_{-n}, A_{n} \preceq ^{*} A^{*}_{n}$ follow as (3) in the base case.

• We show that L(≼^∗,w) = L(≼,w) for all w ∈ [n + 1] implies: for all − n sets (full or not) and all n-sets (void or not):

  $$ A_{-n} \sim^{*} A^{\prime}_{-n} ~~~~ , ~~~~ A^{\prime}_{n} \sim^{*} A_{n}. $$

  (A.7)

  And if |[n + 1]| > 1 then for all − (n + 1) sets (full or not) and all n + 1-sets (void or not):

  $$ A_{-(n+1)} \sim^{*} A^{\prime}_{-(n+1)} ~~~~ , ~~~~ A^{\prime}_{n+1} \sim^{*} A_{n+1}. $$

  (A.8)

  Let us establish Eq. A.7: Consider |[n]| = 1 (else Eq. A.7 holds by strong agreement IH2). Assume n > 0 (else there is nothing to prove).

  • (1) $A_{-n} \sim ^{*} A^{\prime }_{-n}$: We first show that $A_{-n} \sim ^{*} \dot {A}_{-n}$ in case A_−n is n + 1-empty and $\dot {A}_{-n}$ is n + 1-full.

    Let $\dot {A}$ be a − n-set which is n + 1-full. Define $A:= \dot {A} \setminus [n+1]$ and $A^{\prime }:= [n+1]$. Thus A is a − n-set which is n + 1-empty and $A^{\prime }$ is a − (n + 1)-set. Then $A \prec ^{*} \dot {A}$ would be a ∪-flaw in any w ∈ [n + 1]. Indeed: (1) $A^{\prime } \prec ^{*} \dot {A}$ by IH2*, (2) $w \in \dot {A} \setminus A$ and (3) $\dot {A} = A \cup A^{\prime }$ and for v ∈ [n − 1] (non-empty, since n > 0), we have (4) $v \notin \dot {A}$. But $(A, \dot {A}) \notin L(\preceq , w)$, hence $(A, \dot {A}) \notin L(\preceq ^{*}, w)$. Thus $\dot {A} \preceq ^{*} A$. Since $A \preceq ^{*} \dot {A}$, we obtain $A \sim ^{*} \dot {A}$.

    Let now $A^{\prime }, \dot {A}$ be two − n sets, such that $\dot {A}$ is n + 1-full and $A^{\prime }$ is not n + 1-empty and not n + 1 full. Then we can find an n + 1-empty witness A (cf. above) such that $A \sim ^{*} \dot {A}$. But $A \preceq ^{*} A^{\prime } \preceq ^{*} \dot {A}$ (by F-minimisation). Thus $A \preceq ^{*} A^{\prime } \preceq ^{*} \dot {A} \sim ^{*} A$. Squeezing yields $A^{\prime } \sim ^{*} \dot {A}$.

    Similarly, let $A, A^{\prime }$ be two − n-sets, such that A is n + 1-empty and $A^{\prime }$ is not n + 1-empty and not n + 1-full. Again, we can find a witness $\dot {A}$ − n-set that is n + 1-full, ($\dot {A} = A \cup [n+1]$), such that ∪-flaw-avoidance yields $A \sim ^{*} \dot {A}$ (choose witness $A^{\prime } = [n+1]$). Hence $A \sim ^{*} \dot {A} \sim ^{*} A^{\prime }$, i.e., $A \sim ^{*} A^{\prime }$ (transitivity). Thus equivalence extends to all − n-sets.

  • (2) $A_{n} \sim ^{*} A^{\prime }_{n}$ follows for all n-sets by a similar reasoning, using ∩-flaw avoidance.

  We have thus extended Eq. A.3 from n to n + 1 (whether |[n]| = 1 or > 1).

  Let us now establish Eq. A.8, provided |[n + 1]| > 1:

  • (1) $A_{n+1} \sim A^{\prime }_{n+1}$: Let $\dot {A}$ be a void n + 1-set and let C₁,C₂ partition [n + 1] (assuming |[n + 1]| > 1). Then $A = \dot {A} \cup C_{1}$ as $A^{\prime }= \dot {A} \cup C_{2}$ are non-void n + 1-sets. Let w ∈ C₁. Then $\dot {A} \prec ^{*} A$ would be a ∩-flaw in w, for witnesses $A^{\prime }$ and v ∈ [n] (using $A \sim ^{*} A^{\prime }$ by Eq. A.5). But $(\dot {A}, A) \notin L(\preceq , w)$. Thus $A \preceq ^{*} \dot {A}$. Since $\dot {A} \preceq ^{*} A$, we obtain $A \sim ^{*} \dot {A}$. By the same reasoning $B \sim ^{*} \dot {B}$ where $\dot {B}$ is another void n + 1-set. But $B \sim ^{*} A$, and hence equivalence extends to all n + 1-sets.

  • (2) $A_{-(n+1)} \sim ^{*} A_{-(n+1)}$ follows by the same reasoning, using ∪-flaws.

  Thus we have also obtained Eq. A.4 for n + 1 (provided |[n + 1]| > 1).

• We show R(≼^∗,w) = R(≼,w). We already have |R(≼^∗,w)|≥|R(≼,w)|. We show $R(\preceq ^{*}, w) \subseteq R(\preceq , w)$. Let (A_y,A_x) ∈ R(≼^∗,w), then (A_y,A_x) ∈ PR(w) and A_y ≺^∗A_x. Hence not x = y for − n ≤ x,y ≤ n (by Eq. A.3). We also obtain not x = y for x ∈{−(n + 1),n + 1}, by (14) established above if |[n + 1]| > 1; and if |[n + 1]| = 1 the sets A_y,A_x are both void or both full and A_y ≺^∗A_x cannot be a right comparison. Thus: not x = y for − (n + 1) ≤ x,y ≤ n + 1. Similarly not − (n + 1) ≤ x < y ≤ (n + 1) (by IH2, i.e., Eqs. A.1 and A.2 for n). But this implies (A_y,A_x) ∈ R(≼,w) (Lemma 4). Hence |R(≼^∗,w)| = |R(≼,w)| and thus R(≼^∗,w) = R(≼,w). This proves total profile agreement up to n + 1.

  R(≼^∗,w) = R(≼,w) then implies that all ≼^∗ obtained by F-minimisation in [n + 1] and left undecided (not turned into $\sim ^{*}$) are now turned into ≺^∗. In particular

  $$A_{-t} \prec^{*} A_{-(n+1)} ~~~~ A_{n+1} \prec^{*} A_{t}~~~~ \text{for } t>n+1$$

  This is Eq. A.2 for n + 1. Therefore ≼^∗agrees with ≼ up to n + 1, where agreement is strong if |[n + 1]| > 1 and weak if |[n + 1]| = 1.

The recursively constructed ≼^∗ over $\mathcal {A} \setminus \{\emptyset , W\}$ satisfies

$$A_{x} \prec^{*} A_{y} \text{ iff } x<y$$

In the last step n (for n maximal), we also obtain by R-maximisation ∅≺^∗A_−n and A_n ≺^∗W, since these cannot be flaws. The transitive closure of ≼^∗ over $\mathcal {A}$ is thus the canonical ≼. □