Abstract
Discussions of the issue of iterated belief revision are commonly accompanied by the presentation of three “concrete” operators: natural, restrained and lexicographic. This raises a natural question: What is so distinctive about these three particular methods? Indeed, the common axiomatic ground for work on iterated revision, the AGM and DarwichePearl postulates, leaves open a whole range of alternative proposals. In this paper, we show that it is satisfaction of an additional principle of “Independence of Irrelevant Alternatives”, inspired by the literature on Social Choice, that unites and sets apart our three “elementary” revision operators. A parallel treatment of iterated belief contraction is also given, yielding a family of elementary contraction operators that includes, besides the wellknown “conservative” and “moderate” operators, a new contraction operator that is related to restrained revision.
Notes
This principle is given as follows:

\((\mathrm {P}^{\ast }_{\scriptscriptstyle \preccurlyeq })\) If ρ_{A}(x,y) = 1 and ρ_{Ψ}(x,y) ≥ 0, then ρ_{Ψ∗A}(x,y) = 1
It was noted, in [3], that, in its unqualified form (there called “\((\mathrm {P}^{\scriptscriptstyle \div }_{\scriptscriptstyle \preccurlyeq })\)”), without the requirement that \(y\notin \min \limits (\preccurlyeq _{\Psi }, W)\), this principle would lead to trouble.

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This research was partially supported by the Australian Government through an ARC Future Fellowship (project number FT160100092) awarded to Jake Chandler.
Appendix A: Proofs
Appendix A: Proofs
Proposition 1
Given \((KM^{\div }_{\preccurlyeq })\), \((\text {IIA}^{\ast }_{\preccurlyeq })\) does not imply any of \((\mathrm {C}{1},{2}^{\ast }_{\preccurlyeq })\)\((\mathrm {C}{4}^{\ast }_{\preccurlyeq })\)
Proof
Let ∗ be defined in such a way that, for all states Ψ, x,y ∈ W and A ∈ L:

\(\min \limits (\preccurlyeq _{\Psi \ast A}, W) = \min \limits (\preccurlyeq _{\Psi }, [\![{A}]\!])\)

If \(x, y\notin \min \limits (\preccurlyeq _{\Psi }, [\![{A}]\!])\), then

If ρ_{Ψ}(x,y) = 1, then ρ_{Ψ∗A}(x,y) = − 1

If ρ_{Ψ}(x,y) = 0, then ρ_{Ψ∗A}(x,y) = 0

Essentially, this operator will set the \(\preccurlyeq _{\Psi }\)minimal Aworlds as \(\preccurlyeq _{\Psi \ast A}\)minimal (thus satisfying AGM) and simply “flip” the remainder of the ordering.
It is easy to see that ∗ satisfies \((\text {IIA}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\): Assume that \(x, y \notin \min \limits (\preccurlyeq _{\Psi }, [\![{A}]\!])\cup \min \limits (\preccurlyeq _{{\varTheta }}, [\![{B}]\!])\), that ρ_{Ψ}(x,y) = ρ_{Θ}(x,y) and that ρ_{A}(x,y) = ρ_{B}(x,y). We need to show that ρ_{Ψ∗A}(x,y) = ρ_{Θ∗B}(x,y). For this, we consider two cases:

(a) Assume ρ_{Ψ}(x,y) = ρ_{Θ}(x,y) = 1. Then ρ_{Ψ∗A}(x,y) = ρ_{Θ∗B}(x,y) = − 1.

(b) Assume ρ_{Ψ}(x,y) = ρ_{Θ}(x,y) = 0. Then ρ_{Ψ∗A}(x,y) = ρ_{Θ∗B}(x,y) = 0.
In either case, ρ_{Ψ∗A}(x,y) = ρ_{Θ∗B}(x,y), as required.
Clearly, however, each of \((\mathrm {C}{1}^{\ast }_{\scriptscriptstyle \preccurlyeq })\) to \((\mathrm {C}{4}^{\ast }_{\scriptscriptstyle \preccurlyeq })\) will be violated. A counterexample is provided in Fig. 5. There, \((\mathrm {C}{1}^{\ast }_{\scriptscriptstyle \preccurlyeq })\) fails because, for instance, y ≺_{Ψ}x but x ≺_{Ψ∗A}y. Regarding \((\mathrm {C}{2}^{\ast }_{\scriptscriptstyle \preccurlyeq })\), we have t ≺_{Ψ}s but s ≺_{Ψ∗A}t. Regarding \((\mathrm {C}{3}^{\ast }_{\scriptscriptstyle \preccurlyeq })\) and \((\mathrm {C}{4}^{\ast }_{\scriptscriptstyle \preccurlyeq })\), we have z ≺_{Ψ}s but s ≺_{Ψ∗A}z. □
Proposition 2
Given \((KM^{\ast }_{\preccurlyeq })\), \((\mathrm {C}{1},{2}^{\ast }_{\preccurlyeq })\)\((\mathrm {C}{4}^{\ast }_{\preccurlyeq })\) do not jointly imply \((\text {IIA}^{\ast }_{\preccurlyeq })\)
Proof
Consider the operator ∗ defined as follows: for all states Ψ and A ∈ L, Ψ ∗ A = Ψ ∗_{L}A if \(\preccurlyeq _{\Psi }\) is a chain (i.e. an antisymmetric TPO, such that, for all x,y ∈ W, if \(x\sim _{\Psi }y\), then x≠y), and Ψ ∗ A = Ψ ∗_{R}A otherwise.
It is easily verified that ∗ satisfies \((\text {KM}^{\ast }_{\preccurlyeq })\) and \((\mathrm {C}{1,2}^{\ast }_{\scriptscriptstyle \preccurlyeq })\)\((\mathrm {C}{4}^{\ast }_{\scriptscriptstyle \preccurlyeq })\). But ∗ does not satisfy \((\text {IIA}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\), as can be seen from the following countermodel: Let W = {x,y,z,w}, [ [A] ] = {z,x}, z ≺_{Ψ}w ≺_{Ψ}x ≺_{Ψ}y and {z,w}≺_{Θ}x ≺_{Θ}y, so that, notably, Ψ ∗ A = Ψ ∗_{L}A but Θ ∗ A = Θ ∗_{R}A. Then \(w, x \notin \min \limits (\preccurlyeq _{\Psi }, [\![{A}]\!])\cup \min \limits (\preccurlyeq _{{\varTheta }}, [\![{A}]\!])\), ρ_{Ψ}(w,x) = ρ_{Θ}(w,x) = 1, but ρ_{Ψ∗A}(w,x) = − 1 and ρ_{Θ∗A}(w,x) = 1. □
Proposition 3
Given \((\text {UD}_{\scriptscriptstyle \preccurlyeq })\), \((\text {IIA}^{\ast }_{\preccurlyeq })\) is equivalent to the conjunction of the following principles of “Independence of Irrelevant Alternatives” with respect to the “Prior” and the “Input”, respectively:

\((\text {IIAP}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\) If \(x,y \notin \min \limits (\preccurlyeq _{\Psi }, [\![{A}]\!])\cup \min \limits (\preccurlyeq _{{\varTheta }}, [\![{A}]\!])\) then, if ρ_{Ψ}(x,y) = ρ_{Θ}(x,y) then ρ_{Ψ∗A}(x,y) = ρ_{Θ∗A}(x,y)

\((\text {IIAI}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\) If \(x,y \notin \min \limits (\preccurlyeq _{\Psi }, [\![{A}]\!])\cup \min \limits (\preccurlyeq _{\Psi }, [\![{B}]\!])\) then, if ρ_{A}(x,y) = ρ_{B}(x,y) then ρ_{Ψ∗A}(x,y) = ρ_{Ψ∗B}(x,y)
Proof
From \((\text {IIA}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\) to \((\text {IIAP}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\) and \((\text {IIAI}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\): simply set Ψ = Θ in the former case and A = B in the latter.
From \((\text {IIAP}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\) and \((\text {IIAI}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\) to \((\text {IIA}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\): Suppose \(x, y \notin \min \limits (\preccurlyeq _{\Psi }, [\![{A}]\!])\cup \min \limits (\preccurlyeq _{{\varTheta }}, [\![{B}]\!])\), ρ_{Ψ}(x,y) = ρ_{Θ}(x,y) and ρ_{A}(x,y) = ρ_{B}(x,y). We need to show ρ_{Ψ∗A}(x,y) = ρ_{Θ∗B}(x,y). Consider a state Φ such that, for all z∉{x,y},z ≺_{Φ}x,z ≺_{Φ}y and ρ_{Φ}(x,y) = ρ_{Ψ}(x,y) = ρ_{Θ}(x,y). Such a state exists by \((\text {UD}_{\scriptscriptstyle \preccurlyeq })\). Since \(x, y \notin \min \limits (\preccurlyeq _{\Psi }, [\![{A}]\!])\), by the construction of \(\preccurlyeq _{\Phi }\), we must have \(x, y \notin \min \limits (\preccurlyeq _{\Phi }, [\![{A}]\!])\). Hence, by \((\text {IIAP}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\), we have (1) ρ_{Ψ∗A}(x,y) = ρ_{Ψ∗A}(x,y). Similarly, since \(x, y \notin \min \limits (\preccurlyeq _{{\varTheta }}, [\![{B}]\!])\), we have \(x, y \notin \min \limits (\preccurlyeq _{\Phi }, [\![{B}]\!])\) and so, by \((\text {IIAP}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\), we have: (2) ρ_{Θ∗B}(x,y) = ρ_{Θ∗B}(x,y). Finally, from the fact that \(x, y \notin \min \limits (\preccurlyeq _{\Phi }, [\![{A}]\!])\cup \min \limits (\preccurlyeq _{\Phi }, [\![{B}]\!])\) and ρ_{A}(x,y) = ρ_{B}(x,y), by \((\text {IIAI}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\), we have: (3) ρ_{Φ∗A}(x,y) = ρ_{Φ∗B}(x,y). The required result then follows from (1), (2) and (3). □
Proposition 4
\((\text {IIAI}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\) does not imply \((\text {IIAP}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\) or vice versa, even in the presence of \((\text {KM}^{\ast }_{\preccurlyeq })\) and \((\mathrm {C}{1,2}^{\ast }_{\scriptscriptstyle \preccurlyeq })\)\((\mathrm {C}{4}^{\ast }_{\scriptscriptstyle \preccurlyeq })\).
Proof
For an operator satisfying \((\text {KM}^{\ast }_{\preccurlyeq })\), \((\mathrm {C}{1,2}^{\ast }_{\scriptscriptstyle \preccurlyeq })\)\((\mathrm {C}{4}^{\ast }_{\scriptscriptstyle \preccurlyeq })\), and \((\text {IIAI}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\) but not \((\text {IIAP}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\), we can consider the operator ∗ introduced in the proof of Proposition 2 above. It was defined as follows: for all states Ψ and A ∈ L, Ψ ∗ A = Ψ ∗_{L}A if \(\preccurlyeq _{\Psi }\) is a chain (i.e. an antisymmetric TPO, such that, for all x,y ∈ W, if \(x\sim _{\Psi }y\), then x≠y), and Ψ ∗ A = Ψ ∗_{R}A otherwise.
We’ve already noted above that it satisfies \((\text {KM}^{\ast }_{\preccurlyeq })\) and \((\mathrm {C}{1,2}^{\ast }_{\scriptscriptstyle \preccurlyeq })\)\((\mathrm {C}{4}^{\ast }_{\scriptscriptstyle \preccurlyeq })\). \((\text {IIAI}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\) is also satisfied, since, for any given state, the same method (i.e. ∗_{L} or ∗_{R}) is used for all sentences and this method satisfies \((\text {IIAI}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\). But ∗ does not satisfy \((\text {IIAP}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\), as can be seen from the countermodel given in the proof of Proposition 2 above. For convenience, we repeat it here. Let W = {x,y,z,w}, [ [A] ] = {z,x}, z ≺_{Ψ}w ≺_{Ψ}x ≺_{Ψ}y and {z,w}≺_{Θ}x ≺_{Θ}y, so that, notably, Ψ ∗ A = Ψ ∗_{L}A but Θ ∗ A = Θ ∗_{R}A. Then \(w, x \notin \min \limits (\preccurlyeq _{\Psi }, [\![{A}]\!])\cup \min \limits (\preccurlyeq _{{\varTheta }}, [\![{A}]\!])\), ρ_{Ψ}(w,x) = ρ_{Θ}(w,x) = 1, ρ_{A}(w,x) − 1, but ρ_{Ψ∗A}(w,x) = − 1 and ρ_{Θ∗A}(w,x) = 1.
For an operator that satisfies \((\text {KM}^{\ast }_{\preccurlyeq })\), \((\mathrm {C}{1,2}^{\ast }_{\scriptscriptstyle \preccurlyeq })\)\((\mathrm {C}{4}^{\ast }_{\scriptscriptstyle \preccurlyeq })\), and \((\text {IIAP}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\) but not \((\text {IIAI}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\), consider the operator ∗ defined as follows: For all states Ψ and A ∈ L, Ψ ∗ A = Ψ ∗_{L}A if \(\lvert [\![{\neg A}]\!]\rvert =1\), and Ψ ∗ A = Ψ ∗_{R}A otherwise.
Again, as for the previous operator, it is easily verified that ∗ satisfies \((\text {KM}^{\ast }_{\preccurlyeq })\), and \((\mathrm {C}{1,2}^{\ast }_{\scriptscriptstyle \preccurlyeq })\)\((\mathrm {C}{4}^{\ast }_{\scriptscriptstyle \preccurlyeq })\). \((\text {IIAP}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\) is also satisfied, since, , for any given sentence, the same method (i.e. ∗_{L} or ∗_{R}) is used for all states and this method satisfies \((\text {IIAI}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\). But ∗ does not satisfy \((\text {IIAI}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\), as can be seen from the following countermodel: Let W = {x,y,z,w}, [ [A] ] = {x,z,w}, [ [B] ] = {x,w} and w ≺_{Ψ}y ≺_{Ψ}x ≺_{Ψ}z. Then \(x, y \notin \min \limits (\preccurlyeq _{\Psi }, [\![{A}]\!])\cup \min \limits (\preccurlyeq _{\Psi }, [\![{B}]\!])\) and ρ_{A}(x,y) = 1 = ρ_{B}(x,y), but ρ_{Ψ∗A}(x,y) = 1, whereas ρ_{Ψ∗B}(x,y) = − 1. □
Proposition 5
Given (KM\(^{\div }_{\preccurlyeq })\), \((\text {IIAP}^{\ast }_{\preccurlyeq })\) entails \((\text {TPOR}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\)
Proof
Assume that \(\preccurlyeq _{\Psi }=\preccurlyeq _{{\varTheta }}\). Consider arbitrary x,y ∈ W. If either x or y is in \(\min \limits (\preccurlyeq _{\Psi }, [\![{A}]\!])\cup \min \limits (\preccurlyeq _{{\varTheta }}, [\![{A}]\!])\), then \(\rho _{\preccurlyeq _{\Psi \ast A}}({x},{y})= \rho _{\preccurlyeq _{{\varTheta }\ast A}}({x},{y})\) by virtue of \((\text {KM}^{\ast }_{\preccurlyeq })\). If \(x, y \notin \min \limits (\preccurlyeq _{\Psi }, [\![{A}]\!])\cup \min \limits (\preccurlyeq _{{\varTheta }}, [\![{A}]\!])\), then \(\rho _{\preccurlyeq _{\Psi \ast A}}({x},{y})= \rho _{\preccurlyeq _{{\varTheta }\ast A}}({x},{y})\) by virtue of \((\text {IIAP}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\). Hence \(\preccurlyeq _{\Psi \ast A}=\preccurlyeq _{{\varTheta }\ast A}\), as required. □
Proposition 6
Given \((KM^{\ast }_{\preccurlyeq })\), \((\mathrm {C}{1,2}^{\ast }_{\scriptscriptstyle \preccurlyeq })\)\((\mathrm {C}{4}^{\ast }_{\scriptscriptstyle \preccurlyeq })\), \((\text {IIAI}^{\ast }_{\preccurlyeq })\) is equivalent to the conjunction of:

(\(\beta 1^{\ast }_{\scriptscriptstyle \preccurlyeq }\)) If \(x\notin \min \limits (\preccurlyeq _{\Psi }, [\![C]\!]), \rho _{A}(x,y)= 1\) and ρ_{Ψ∗A}(x,y) ≤ 0, then ρ_{Ψ∗C}(x,y) ≤ 0

(\(\beta 2^{\ast }_{\scriptscriptstyle \preccurlyeq }\)) If \(x\notin \min \limits (\preccurlyeq _{\Psi }, [\![C]\!]), \rho _{A}(x,y)= 1\) and ρ_{Ψ∗A}(x,y) = − 1, then ρ_{Ψ∗C}(x,y) = − 1
Proof
The proof of this claim closely resembles the proof of Proposition 3 of [8]. For ease of comparison, we use the \(\preccurlyeq \)notation, rather than the ρnotation, so that \((\beta {1}^{\ast }_{\scriptscriptstyle \preccurlyeq })\) and \((\beta {2}^{\ast }_{\scriptscriptstyle \preccurlyeq })\) are presented as follows:

\((\beta {1}^{\ast }_{\scriptscriptstyle \preccurlyeq })\) If \(x \not \in \min \limits (\preccurlyeq _{\Psi }, [\![C]\!])\), x ≺_{A}y, and \(y \preccurlyeq _{\Psi \ast A} x\), then \(y \preccurlyeq _{\Psi \ast C} x\)

\((\beta {2}^{\ast }_{\scriptscriptstyle \preccurlyeq })\) If \(x \not \in \min \limits (\preccurlyeq _{\Psi }, [\![C]\!])\), x ≺_{A}y, and y ≺_{Ψ∗A}x, then y ≺_{Ψ∗C}x
□
We first establish the following lemma:
Lemma 4
Given \((\mathrm {C}{1, 2}^{\ast }_{\scriptscriptstyle \preccurlyeq })\)–\((\mathrm {C}{4}^{\ast }_{\scriptscriptstyle \preccurlyeq })\),

(a) If x ≺_{A}y and ρ_{A}(x,y)≠ρ_{C}(x,y), then, if \(y \preccurlyeq _{\Psi \ast A} x\), then \(y \preccurlyeq _{\Psi \ast C} x\).

(b) If x ≺_{A}y and ρ_{A}(x,y)≠ρ_{C}(x,y), then, if y ≺_{Ψ∗A}x, then y ≺_{Ψ∗C}x.
We simply derive (a), since the proof of (b) is analogous. Assume that x ≺_{A}y and ρ_{A}(x,y)≠ρ_{C}(x,y). In other words: x ∈ [ [A] ],y ∈ [ [¬A] ], and either (i) x ∈ [ [C] ],y ∈ [ [C] ], (ii) x ∈ [ [¬C] ],y ∈ [ [¬C] ] or (iii) x ∈ [ [¬C] ],y ∈ [ [C] ]. Assume that \(y \preccurlyeq _{\Psi \ast A} x\). From this and x ∈ [ [A] ],y ∈ [ [¬A] ], it follows, by \((\mathrm {C}{3}^{\ast }_{\scriptscriptstyle \preccurlyeq })\), that \(y \preccurlyeq _{\Psi } x\). From this, if either (i), (ii) or (iii) hold, then, by \((\mathrm {C}{1, 2}^{\ast }_{\scriptscriptstyle \preccurlyeq })\), \((\mathrm {C}{2}^{\ast }_{\scriptscriptstyle \preccurlyeq })\), and \((\mathrm {C}{4}^{\ast }_{\scriptscriptstyle \preccurlyeq })\), respectively, we have \(y \preccurlyeq _{\Psi \ast C} x\), as required. This completes the proof of Lemma 4.
With this in hand, we can derive each direction of the equivalence:

(a) From \((\text {IIAI}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\) to \((\beta {1}^{\ast }_{\scriptscriptstyle \preccurlyeq })\) and \((\beta {2}^{\ast }_{\scriptscriptstyle \preccurlyeq })\): Regarding \((\beta {1}^{\ast }_{\scriptscriptstyle \preccurlyeq })\), assume \(x \not \in \min \limits (\preccurlyeq _{\Psi }, [\![{C}]\!])\), x ≺_{A}y, and \(y \preccurlyeq _{\Psi \ast A} x\). We need to show that \(y \preccurlyeq _{\Psi \ast C} x\). If ρ_{A}(x,y)≠ρ_{C}(x,y), then the required result follows by principle (a) of Lemma 4. So assume ρ_{A}(x,y) = ρ_{C}(x,y), and hence that x ≺_{C}y. We now establish that \(x, y \notin \min \limits (\preccurlyeq _{\Psi }, [\![{A}]\!])\cup \min \limits (\preccurlyeq _{\Psi }, [\![{C}]\!])\). We already have \(x \not \in \min \limits (\preccurlyeq _{\Psi }, [\![{C}]\!])\). Since, by x ≺_{C}y, it follows that y ∈ [ [¬C] ], we therefore have \(y \not \in \min \limits (\preccurlyeq _{\Psi }, [\![{C}]\!])\). Furthermore, by x ≺_{A}y, it follows that y ∈ [ [¬A] ] and so \(y \not \in \min \limits (\preccurlyeq _{\Psi }, [\![{A}]\!])\). Finally, assume for contradiction that \(x\in \min \limits (\preccurlyeq _{\Psi }, [\![{A}]\!])\). Then \(x\in \min \limits (\preccurlyeq _{\Psi \ast A}, W)\), by \((\text {KM}^{\ast }_{\preccurlyeq })\). Since y ∈ [ [¬A] ], by \((\text {KM}^{\ast }_{\preccurlyeq })\), \(y\notin \min \limits (\preccurlyeq _{\Psi \ast A}, W)\). Hence x ≺_{Ψ∗A}y, contradicting \(y \preccurlyeq _{\Psi \ast A} x\). So we can infer that \(x\notin \min \limits (\preccurlyeq _{\Psi }, [\![{A}]\!])\). With this in hand, we can apply \((\text {IIAI}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\) to derive \(y \preccurlyeq _{\Psi \ast C} x\), as required. The derivation of \((\beta {2}^{\ast }_{\scriptscriptstyle \preccurlyeq })\) is analogous, but using principle (b) of Lemma 4.

(b) From \((\beta {1}^{\ast }_{\scriptscriptstyle \preccurlyeq })\) and \((\beta {2}^{\ast }_{\scriptscriptstyle \preccurlyeq })\) to \((\text {IIAI}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\): Assume that \(x, y \notin \min \limits (\preccurlyeq _{\Psi }, [\![{A}]\!])\cup \min \limits (\preccurlyeq _{\Psi }, [\![{C}]\!])\) and that ρ_{A}(x,y) = ρ_{C}(x,y). We want to show that \(x\preccurlyeq _{\Psi \ast A} y\) iff \(x\preccurlyeq _{\Psi \ast C} y\). By symmetry, it suffices for this to show that \(x\preccurlyeq _{\Psi \ast A} y\) implies \(x\preccurlyeq _{\Psi \ast C} y\). So assume \(x\preccurlyeq _{\Psi \ast A} y\). Since ρ_{A}(x,y) = ρ_{C}(x,y), we have three cases to consider:

(i) \(\rho _{A}({x},{y}){\kern .5pt}={\kern .5pt}\rho _{C}({x},{y}){\kern .5pt}={\kern .5pt}1\): Assume for contradiction that y ≺_{Ψ∗C}x. From this, \(x \not \in \min \limits (\preccurlyeq _{\Psi }, [\![{A}]\!])\) and x ≺_{C}y, it follows by \((\beta {2}^{\ast }_{\scriptscriptstyle \preccurlyeq })\) that y ≺_{Ψ∗A}x, contradicting \(x\preccurlyeq _{\Psi \ast A} y\). Hence \(x\preccurlyeq _{\Psi \ast C} y\), as required.

(ii) ρ_{A}(x,y) = ρ_{C}(x,y) = 0: It follows from this, via \((\mathrm {C}{1, 2}^{\ast }_{\scriptscriptstyle \preccurlyeq })\), that \(x\preccurlyeq _{\Psi \ast A} y\) iff \(x\preccurlyeq _{\Psi } y\) iff \(x\preccurlyeq _{\Psi \ast C} y\). Hence \(x\preccurlyeq _{\Psi \ast C} y\), as required.

(iii) ρ_{A}(x,y) = ρ_{C}(x,y) = − 1: By \((\beta {1}^{\ast }_{\scriptscriptstyle \preccurlyeq })\), it follows, from \(x \not \in \min \limits \) \((\preccurlyeq _{\Psi }, [\![{A}]\!])\), y ≺_{A}x, and \(x\preccurlyeq _{\Psi \ast A} y\), that \(x\preccurlyeq _{\Psi \ast C} y\), as required.

Lemma 2
The conjunction of \((\text {KM}^{\ast }_{\preccurlyeq })\) transitivity of \({\preccurlyeq }_{\Psi \ast A}, (\text {PI}^{\ast }_{\preccurlyeq })\) and \((\text {IIA}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\) implies the following principle of “Zero Symmetry”:

\((\text {ZS}^{\ast }_{\preccurlyeq })\) If \(x,y \notin \min \limits (\preccurlyeq _{\Psi }, [\![{A}]\!])\cup \min \limits (\preccurlyeq _{{\varTheta }}, [\![{B}]\!]), \rho _{\Psi }(x,y)= \rho _{{\varTheta }}(x,y)\) and ρ_{A}(x,y) = −ρ_{B}(x,y) then, ρ_{Ψ∗A}(x,y) = −ρ_{Ψ∗B}(x,y)
Proof
\((\text {KM}^{\ast }_{\preccurlyeq })\) and \((\text {IIA}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\) jointly tell us that that, for a pair of worlds {x,y}, the posterior relative rank ρ_{Ψ∗A}(x,y) is determined by the prior relative rank ρ_{Ψ}(x,y) and input sentence relative rank ρ_{A}(x,y) (although this mapping may be different for different pairs of worlds). With the case in which x or \(y\in \min \limits (\preccurlyeq _{\Psi }, [\![{A}]\!])\) being taken care of by \((\text {KM}^{\ast }_{\preccurlyeq })\), the behaviour of ∗ with respect to {x,y} can therefore be represented in the form of a matrix giving us, for \(x, y\notin \min \limits (\preccurlyeq _{\Psi }, [\![{A}]\!])\), the values of ρ_{Ψ∗A}(x,y) as a function of those of ρ_{Ψ}(x,y) and ρ_{A}(x,y).
Since we assume A to be consistent and hence that \(\min \limits (\preccurlyeq _{\Psi }, [\![{A}]\!])\) is nonempty, we must have strictly more than two, and therefore (since \(\lvert W\rvert =2^{n}\)) at least four, worlds. Let these worlds be x, y, z and w. We will consider the matrices for the pairs 〈x,y〉, 〈y,z〉 and 〈z,x〉 in Table 7.
\((\text {PI}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\) tells us that the value of the central cell in each matrix is 0. To establish \((\text {ZS}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\), we then just need to show that a_{i} = −b_{i}, for 1 ≤ i ≤ 4. We shall simply show that a_{1} = −b_{1}, since the proof strategy is identical for other values of i.
Let Ψ be such that ρ_{Ψ}(x,z) = ρ_{Ψ}(x,y) = ρ_{Ψ}(y,z) = 0. Let A be such that x,z ∈ [ [A] ] but y∉[ [A] ] (with \(x, z\notin \min \limits (\preccurlyeq _{\Psi },[\![{A}]\!])\); this can be ensured by designating the fourth world w to be the sole member of that set) and so ρ_{A}(x,z) = 0. Since ρ_{Ψ}(x,z) = 0, we therefore have ρ_{Ψ∗A}(x,z) = 0. We also have ρ_{A}(x,y) = 1 and ρ_{A}(y,z) = − 1. Therefore ρ_{Ψ∗A}(x,y) = a_{1} and ρ_{Ψ∗A}(y,z) = d_{1}. Assume a_{1}≠ − d_{1} for reductio (so that 〈a_{1},d_{1}〉 is equal to either 〈1, 0〉, 〈0, 1〉, 〈− 1, 0〉, 〈0,− 1〉, 〈1, 1〉, or 〈− 1,− 1〉). It then follows, by transitivity of \(\preccurlyeq _{\Psi \ast A}\), that ρ_{Ψ∗A}(x,z) = 1 or − 1. But this contradicts our earlier finding that ρ_{Ψ∗A}(x,z) = 0. Hence a_{1} = −d_{1}. By similar reasoning, we can establish that e_{1} = −d_{1} (let A be such that y,x ∈ [ [A] ] but z∉[ [A] ], with \(y, x\notin \min \limits (\preccurlyeq _{\Psi },[\![{A}]\!])\) and hence ρ_{A}(y,x) = 0) and e_{1} = −b_{1} (let A be such that z,y ∈ [ [A] ] but x∉[ [A] ], with \(z, y\notin \min \limits (\preccurlyeq _{\Psi },[\![{A}]\!])\), and hence ρ_{A}(z,y) = 0) and hence that a_{1} = −b_{1}, as required. □
Lemma 3
The conjunction of \((\text {KM}^{\ast }_{\preccurlyeq })\), transitivity of \({\preccurlyeq }_{\Psi \ast A}\), \((\text {PI}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\) and \((\text {IIA}^{\ast }_{\preccurlyeq })\) implies the following “Representation Invariance” principle:

\((\text {RI}^{\ast }_{\preccurlyeq })\) ρ_{Ψ∗A}(x,y) = ρ_{Θ∗π(A)}(π(x),π(y)), for any order isomorphism π from \({\preccurlyeq }_{\Psi }\) to \({{\preccurlyeq }_{{\varTheta }}}\)
Proof
We consider 3 cases, depending on x and y’s membership of \(\min \limits (\preccurlyeq _{\Psi }, [\![{A}]\!])\): (a) one of x or y is in the set, (b) both x and y are in the set and (c) neither x nor y are in the set.
For each case, we shall prove the identity of the matrices relating to the pair 〈x,y〉 and revision by A on the one hand, and the pair 〈π(x),π(y)〉 and revision by π(A), on the other. Since, if π is an order isomorphism from \(\preccurlyeq _{\Psi }\) to \(\preccurlyeq _{{\varTheta }}\), then ρ_{Ψ}(x,y) = ρ_{Θ}(π(x),π(y)) and ρ_{A}(x,y) = ρ_{π(A)}(π(x),π(y)), it then follows that ρ_{Ψ∗A}(x,y) = ρ_{Θ∗π(A)}(π(x),π(y)), as required.
Assume (a), so that, for example, \(x \in \min \limits (\preccurlyeq _{\Psi }, [\![{A}]\!])\), \(y \notin \min \limits (\preccurlyeq _{\Psi }, [\![{A}]\!])\) (the other case in analogous). Since we then have ρ_{A}(x,y) ≥ 0, it must be the case that either (i) ρ_{A}(x,y) = 1, or (ii) ρ_{A}(x,y) = 0. Furthermore, if (ii) is the case, it must be the case that ρ_{Ψ}(x,y) = 1. The relevant matrix is then given in Table 8a, with impossible combinations of values indicated by “×”. Since, for all x ∈ W, A ∈ L and order isomorphisms π from \(\preccurlyeq _{\Psi }\) to \(\preccurlyeq _{{\varTheta }}\), we have \(x\in \min \limits (\preccurlyeq _{\Psi }, [\![{A}]\!])\) iff \(\pi (x)\in \min \limits (\preccurlyeq _{\Psi }, [\![{\pi (A)}]\!])\), the same matrix characterises the pair 〈π(x),π(y)〉 under revision by π(A). Assume (b). The reasoning is analogous to the one provided in relation to (a) above, this time with reference to the matrix given in Table 8b. Assume (c). Again, as we have noted above, since we assume A to be consistent and hence that \(\min \limits (\preccurlyeq _{\Psi }, [\![{A}]\!])\) is nonempty, we must have at least 4 worlds. Here we make use of the transitivity of \(\preccurlyeq _{\Psi \ast A}\), much as we did in the proof of Lemma 2, and consider, in addition to 〈x,y〉, the pairs 〈y,z〉 and 〈z,x〉. The matrices for these pairs are given in Table 7 above.
Using the strategy applied in relation to the same case in the proof of Lemma 2, we can show that, for 1 ≤ i ≤ 4, not only a_{i} = −b_{i} and c_{i} = −d_{i}, but also a_{i} = −d_{i} and so a_{i} = c_{i}. So the matrix for all sentences is the same for all pairs of worlds. In particular, the matrix relating to the pair 〈x,y〉 and revision by A on the one hand, is identical to that relating the pair 〈π(x),π(y)〉 and revision by π(A), on the other.
Proposition 7
In the presence of \((\text {UD}_{\scriptscriptstyle \preccurlyeq })\), \((\text {ZS}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\) implies \((\text {IIA}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\).
Proof
Assume \(x, y \notin \min \limits (\preccurlyeq _{\Psi }, [\![{A}]\!])\cup \min \limits (\preccurlyeq _{{\varTheta }}, [\![{B}]\!])\), ρ_{Ψ}(x,y) = ρ_{Θ}(x,y) and ρ_{A}(x,y) = ρ_{B}(x,y). We need to show that ρ_{Ψ∗A}(x,y) = ρ_{Θ∗B}(x,y). By \((\text {UD}_{\scriptscriptstyle \preccurlyeq })\), there will exist a state Ξ and sentence C such that \(x, y \notin \min \limits (\preccurlyeq _{\Psi }, [\![{A}]\!])\cup \min \limits (\preccurlyeq _{\Xi }, [\![{C}]\!])\), ρ_{Ψ}(x,y) = −ρ_{Ξ}(x,y) and ρ_{A}(x,y) = −ρ_{C}(x,y). By \((\text {ZS}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\), we then have ρ_{Ψ∗A}(x,y) = −ρ_{Ξ∗C}(x,y). But since ρ_{Ψ}(x,y) = ρ_{Θ}(x,y) and ρ_{A}(x,y) = ρ_{B}(x,y), we also have ρ_{Θ}(x,y) = −ρ_{Ξ}(x,y) and ρ_{B}(x,y) = −ρ_{C}(x,y). Since \(x, y \notin \min \limits (\preccurlyeq _{{\varTheta }}, [\![{B}]\!])\cup \min \limits (\preccurlyeq _{\Xi }, [\![{C}]\!])\), we can reapply \((\text {ZS}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\), to obtain ρ_{Θ∗B}(x,y) = −ρ_{Ξ∗C}(x,y). Hence ρ_{Ψ∗A}(x,y) = ρ_{Θ∗B}(x,y), as required. □
Proposition 8
\((\text {RI}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\) does not imply either \((\text {IIAI}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\) or \((\text {IIAP}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\) even in the presence \((\text {KM}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\) and \((\mathrm {C}{1}, {2}^{\ast }_{\scriptscriptstyle \preccurlyeq })\)\((\mathrm {C}{4}^{\ast }_{\scriptscriptstyle \preccurlyeq })\)
Proof
We simply need to show that both the operators introduced in the proof of Proposition 4 satisfy \((\text {RI}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\). Indeed, both satisfy \((\text {KM}^{\ast }_{\preccurlyeq })\) and \((\mathrm {C}{1,2}^{\ast }_{\scriptscriptstyle \preccurlyeq })\)\((\mathrm {C}{4}^{\ast }_{\scriptscriptstyle \preccurlyeq })\), but the first does not satisfy \((\text {IIAP}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\) and the second does not satisfy \((\text {IIAI}^{\ast }_{\scriptscriptstyle \preccurlyeq })\).
Regarding the first operator: If π is an order isomorphism between \(\preccurlyeq _{\Psi }\) and \(\preccurlyeq _{{\varTheta }}\), then \(\preccurlyeq _{\Psi }\) is a chain if and only if \(\preccurlyeq _{{\varTheta }}\) is. So we have \(\rho _{\Psi \ast A}({x},{y}) = \rho _{\Psi \ast _{\mathrm {L}} A}(x,y)\) if and only if we have \(\rho _{{\varTheta }\ast \pi (A)}({\pi (x)},{\pi (y)}) = \rho _{{\varTheta }\ast _{\mathrm {L}} \pi (A)}({\pi (x)},{\pi (y)})\) and similarly regarding ∗_{R}. But we have also noted that ∗_{L} and ∗_{R} both satisfy \((\text {KM}^{\ast }_{\preccurlyeq })\) and \((\mathrm {C}{1,2}^{\ast }_{\scriptscriptstyle \preccurlyeq })\)\((\mathrm {C}{4}^{\ast }_{\scriptscriptstyle \preccurlyeq })\). Furthermore, since they also satisfy \((\text {PI}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\) (because they satisfy \((\mathrm {C}{1,2}^{\ast }_{\scriptscriptstyle \preccurlyeq })\); see Lemma 1) and \((\text {IIA}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\), we know from Lemma 3 that they also satisfy \((\text {RI}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\).
Regarding the second operator: If π is an order isomorphism between \(\preccurlyeq _{\Psi }\) and \(\preccurlyeq _{{\varTheta }}\), then \(\lvert [\![{\neg A}]\!]\rvert =1\) iff \(\lvert [\![{\pi (\neg A)}]\!]\rvert =1\). So we have \(\rho _{\Psi \ast A}({x},{y}) = \rho _{\Psi \ast _{\mathrm {L}} A}({x},{y})\) iff \(\rho _{\Psi \ast \pi (A)}({\pi (x)},{\pi (y)}) = \rho _{\Psi \ast _{\mathrm {L}} \pi (A)}({\pi (x)},{\pi (y)})\) and similarly regarding ∗_{R}. The reasoning to establish \((\text {RI}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\) then proceeds as above. □
Proposition 9
\((\text {RI}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\) entails \((\text {TPOR}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\).
Proof
Let \(\preccurlyeq _{\Psi }=\preccurlyeq _{{\varTheta }}\) and the order isomorphism π be such that π(x) = x. It follows by \((\text {RI}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\) that ρ_{Ψ∗A}(x,y) = ρ_{Θ∗π(A)}(π(x),π(y)) = ρ_{Θ∗A}(x,y). □
Proposition 10
Given AGM, \((\text {PI}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\) is equivalent to:

(PI^{∗} ) If [Ψ∗B] = Cn(B), then [(Ψ∗A) ∗ A ∧ B] = [Ψ∗A ∧ B] and [(Ψ∗A) ∗¬A ∧ B] = [Ψ ∗¬A ∧ B]
Proof
From \((\text {PI}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\) to (PI^{∗}): Assume that [Ψ∗B] = Cn(B) but, for reductio, [(Ψ ∗ A) ∗ A ∧ B]≠[Ψ∗A ∧ B] (the case of [(Ψ ∗ A) ∗¬A ∧ B] = [Ψ∗¬A ∧ B] is analogous). We consider two cases:

(i) \([{\Psi \ast A\wedge B}] \nsubseteq [{({\Psi }\ast A)\ast A\wedge B}]\), so \( [\![[{({\Psi }\ast A)\ast A\wedge B}]]\!] \nsubseteq [\![[{({\Psi }\ast A)\ast A\wedge B}]]\!]\): Then, since we assume A ∧ B to be consistent, \(\exists x\in \min \limits (\preccurlyeq _{\Psi \ast A}, [\![{A\wedge B}]\!])\setminus \min \limits (\preccurlyeq _{\Psi }, [\![{A\wedge B}]\!])\). Let \(y\in \min \limits (\preccurlyeq _{\Psi }, [\![{A\wedge B}]\!])\). On the one hand, since x,y ∈ [ [B] ] and [Ψ∗B] = Cn(B), we have \(x, y\in \min \limits (\preccurlyeq _{\Psi }, [\![{B}]\!])\), by \((\text {KM}^{\ast }_{\preccurlyeq })\), and so \(x\sim _{\Psi } y\). On the other hand, since \(y\in \min \limits (\preccurlyeq _{\Psi }, [\![{A\wedge B}]\!])\), and x ∈ [ [A ∧ B] ] but \(x\notin \min \limits (\preccurlyeq _{\Psi }, [\![{A\wedge B}]\!])\), we have y ≺_{Ψ}x. Contradiction.

(ii) \([{({\Psi }\ast A)\ast A\wedge B}]\nsubseteq [{\Psi \ast A\wedge B}]\) and so \([\![[{\Psi \ast A\wedge B}]]\!]\nsubseteq [\![[({\Psi }\ast A)\ast A\wedge B]]\!]\): Then, since we assume A ∧ B to be consistent, \(\exists x\in \min \limits (\preccurlyeq _{\Psi }, [\![{A\wedge B}]\!]) \setminus \min \limits (\preccurlyeq _{\Psi \ast A}, [\![{A\wedge B}]\!])\). Let \(y\in \min \limits (\preccurlyeq _{\Psi \ast A}, [\![{A\wedge B}]\!])\). As above, since x,y ∈ [ [B] ] and [Ψ∗B] = Cn(B), we have \(x\sim _{\Psi } y\). Since x,y ∈ [ [A] ], it follows from this, by \((\text {PI}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\), that \(x\sim _{\Psi \ast A} y\). But since \(y\in \min \limits (\preccurlyeq _{\Psi \ast A}, [\![{A\wedge B}]\!])\), and x ∈ [ [A ∧ B] ] but \(x\notin \min \limits (\preccurlyeq _{\Psi \ast A}, [\![{A\wedge B}]\!])\), we also have y ≺_{Ψ∗A}x. Contradiction.
From (PI^{∗}) to \((\text {PI}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\): Assume \(x\sim _{\Psi } y\). Then \(\min \limits (\preccurlyeq _{\Psi }, [\![{x\vee y}]\!])=[\![{x\vee y}]\!]\) and so [Ψ∗x ∨ y] = Cn(x ∨ y). By (PI^{∗}), letting B = x ∨ y, we then have:

(a) [(Ψ ∗ A) ∗ A ∧ (x ∨ y)] = [Ψ∗A ∧ (x ∨ y)]

(b) [(Ψ ∗ A) ∗¬A ∧ (x ∨ y)] = [Ψ∗¬A ∧ (x ∨ y)]
We consider two cases:

(i) Assume x,y ∈ [ [A] ]. Then A ∧ (x ∨ y) ≡ x ∨ y. So by (a), [(Ψ ∗ A) ∗ x ∨ y] = [Ψ∗x ∨ y] = Cn(x ∨ y). Since \(x\sim _{\Psi \ast A} y\) iff [(Ψ ∗ A) ∗ x ∨ y] = Cn(x ∨ y), it then follows that \(x\sim _{\Psi \ast A} y\), as required.

(ii) Assume x,y ∈ [ [¬A] ]. Then we obtain \(x\sim _{\Psi \ast A} y\) by the same reasoning as above, this time using (b).
□
Proposition 11
Given AGM and \((\mathrm {C}{1,2}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\), \((\text {IIA}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\),\((\text {IIAP}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\) and \((\text {IIAI}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\) are respectively equivalent to:

\((\text {IIA}_{\preccurlyeq })\) If ¬C ∈ [Ψ ∗ A] ∩ [Θ ∗ B],A ≡_{C}B and Ψ and Θ agree modulo C, then so do Ψ ∗ A and Θ ∗ B

\((\text {IIAP}_{\preccurlyeq })\) If ¬C ∈ [Ψ ∗ A] ∩ [Θ ∗ A], then, if Ψ and Θ agree modulo C , so do Ψ ∗ A and Θ ∗ A

\((\text {IIAI}_{\preccurlyeq })\) If ¬C ∈ [Ψ ∗ A] ∩ [Ψ ∗ B] and A ≡_{C}B, then Ψ ∗ A and Ψ ∗ B agree modulo C
Proof
We simply prove the equivalence of (IIA^{∗}) and \((\text {IIA}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\), since the remaining equivalences are established in an analogous manner. We first note that \((\mathrm {C}{1,2}^{\ast }_{\scriptscriptstyle \preccurlyeq })\) gives us the result that, if ρ_{Ψ}(x,y) = ρ_{Θ}(x,y) and x,y ∈ [ [A ∧¬B] ] or x,y ∈ [ [¬A ∧ B] ], then ρ_{Ψ∗A}(x,y) = ρ_{Θ∗B}(x,y). “Removing” these two cases from the condition ρ_{A}(x,y) = ρ_{B}(x,y) leaves us with four cases: (i) x ∈ [ [A ∧ B] ] and y ∈ [ [¬A ∧¬B] ], (ii) y ∈ [ [A ∧ B] ] and x ∈ [ [¬A ∧¬B] ], (iii) x,y ∈ [ [A ∧ B] ], or (iv) x,y ∈ [ [¬A ∧¬B] ]. In view of this, \((\text {IIA}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\) is equivalent to the following weaker principle in the presence of \((\mathrm {C}{1,2}^{\ast }_{\scriptscriptstyle \preccurlyeq })\):

If \(x, y \notin \min \limits (\preccurlyeq _{\Psi }, [\![{A}]\!])\cup \min \limits (\preccurlyeq _{{\varTheta }}, [\![{B}]\!])\), then, if ρ_{Ψ}(x,y) = ρ_{Θ}(x,y) and (i) x ∈ [ [A ∧ B] ] and y ∈ [ [¬A ∧¬B] ], (ii) y ∈ [ [A ∧ B] ] and x ∈ [ [¬A ∧¬B] ], (iii) x,y ∈ [ [A ∧ B] ], or (iv) x,y ∈ [ [¬A ∧¬B] ], then ρ_{Ψ∗A}(x,y) = ρ_{Θ∗B}(x,y)
We first show that this weaker principle can alternatively be presented as follows:

If \([\![{C}]\!]\cap (\min \limits (\preccurlyeq _{\Psi }, [\![{A}]\!])\cup \min \limits (\preccurlyeq _{{\varTheta }}, [\![{B}]\!]) )= \varnothing \) and, for all x,y ∈ [ [C] ], ρ_{Ψ}(x,y) = ρ_{Θ}(x,y) and either (i) x ∈ [ [A ∧ B] ] and y ∈ [ [¬A ∧¬B] ], (ii) y ∈ [ [A ∧ B] ] and x ∈ [ [¬A ∧¬B] ], (iii) x,y ∈ [ [A ∧ B] ], or (iv) x,y ∈ [ [¬A ∧¬B] ], then, for all x,y ∈ [ [C] ], ρ_{Ψ∗A}(x,y) = ρ_{Θ∗A}(x,y)
Going from the second principle to the first, the entailment is obvious, since the latter is just the special case in which [ [C] ] = {x,y}. Going the other way, assume that \([\![{C}]\!]\cap (\min \limits (\preccurlyeq _{\Psi }, [\![{A}]\!])\cup \min \limits (\preccurlyeq _{{\varTheta }}, [\![{B}]\!])) = \varnothing \) and, for all x,y ∈ [ [C] ], ρ_{Ψ}(x,y) = ρ_{Θ}(x,y) and (i) x ∈ [ [A ∧ B] ] and y ∈ [ [¬A ∧¬B] ], (ii) y ∈ [ [A ∧ B] ] and x ∈ [ [¬A ∧¬B] ], (iii) x,y ∈ [ [A ∧ B] ], or (iv) x,y ∈ [ [¬A ∧¬B] ]. Assume further that x,y ∈ [ [C] ], for arbitrary x,y ∈ W. From these assumptions, we have \(x, y \notin \min \limits (\preccurlyeq _{\Psi }, [\![{A}]\!])\cup \min \limits (\preccurlyeq _{{\varTheta }}, [\![{B}]\!])\), ρ_{Ψ}(x,y) = ρ_{Θ}(x,y) and (i) x ∈ [ [A ∧ B] ] and y ∈ [ [¬A ∧¬B] ], (ii) y ∈ [ [A ∧ B] ] and x ∈ [ [¬A ∧¬B] ], (iii) x,y ∈ [ [A ∧ B] ], or (iv) x,y ∈ [ [¬A ∧¬B] ]. Given this, by \((\text {IIA}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\), it follows that ρ_{Ψ∗A}(x,y) = ρ_{Θ∗B}(x,y), as required.
Next we establish syntactic equivalents for the various semantic properties figuring in this principle.
Clearly \([\![{C}]\!]\cap (\min \limits (\preccurlyeq _{\Psi }, [\![{A}]\!])\cup \min \limits (\preccurlyeq _{{\varTheta }}, [\![{B}]\!]) )= \varnothing \) iff ¬C ∈ [Ψ∗A] ∩ [Θ ∗ B]. Furthermore, it is easy to see that conditions (i) to (iv) hold for all x,y ∈ [ [C] ] iff A ≡_{C}B.
Finally, we can show that ρ_{Ψ}(x,y) = ρ_{Θ}(x,y) for all x,y ∈ [ [C] ] iff Ψ and Θ agree modulo C (and obviously similarly regarding Ψ ∗ A and Θ ∗ B).
Indeed, assume that (1) ∀B ∈ L, [Ψ∗B ∧ C] = [Θ ∗ B ∧ C], but, for reductio, that (2) ∃x,y ∈ [ [C] ], such that ρ_{Ψ}(x,y)≠ρ_{Θ}(x,y). Where B ∈ L is such that [ [B ∧ C] ] = {x,y}, it follows from (2) that \(\min \limits (\preccurlyeq _{\Psi }, [\![{B\wedge C}]\!])\neq \min \limits (\preccurlyeq _{{\varTheta } ,} [\![{B\wedge C}]\!])\). But from (1), we have \(\min \limits (\preccurlyeq _{\Psi }, [\![{B\wedge C}]\!])=\min \limits (\preccurlyeq _{{\varTheta }}, [\![{B\wedge C}]\!])\). Contradiction.
Going the other way, assume that (1) ∀x,y ∈ [ [C] ], ρ_{Ψ}(x,y) = ρ_{Θ}(x,y), but, for reductio, that (2) ∃B ∈ L such that [Ψ∗B ∧ C]≠[Θ ∗ B ∧ C]. From (2), \(\min \limits (\preccurlyeq _{\Psi }, [\![{B\wedge C}]\!])\neq \min \limits (\preccurlyeq _{{\varTheta }}, [\![{B\wedge C}]\!])\). From this, either there exists x ∈ W such that \(x\in \min \limits (\preccurlyeq _{\Psi }, [\![{B\wedge C}]\!])\) but \(x\notin \min \limits (\preccurlyeq _{{\varTheta }}, [\![{B\wedge C}]\!])\), or exists x ∈ W such that \(x\in \min \limits (\preccurlyeq _{{\varTheta }}, [\![{B\wedge C}]\!])\) but \(x\notin \min \limits (\preccurlyeq _{\Psi }, [\![{B\wedge C}]\!])\). Assume the former (the other case is analogous). Let \(y\in \min \limits (\preccurlyeq _{{\varTheta }}, [\![{B\wedge C}]\!])\). From this, we have y ≺_{Θ}x but \(x\preccurlyeq _{\Psi } y\) and hence ρ_{Ψ}(x,y)≠ρ_{Θ}(x,y). However, since x,y ∈ [ [C] ], this contradicts (1). □
Proposition 12
Given AGM and \((\mathrm {C}{1, 2}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\), \((\text {ZS}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\) is equivalent to:

(ZS^{∗} ) If ¬C ∈ [Ψ ∗ A] ∩ [Θ ∗ B],A ≡_{C}¬B and Ψ and Θ are in opposition modulo C, then so are Ψ∗A and Θ ∗ B
Proof
The proof is somewhat similar to the one given in relation to Proposition 11. We first note that \((\mathrm {C}{1,2}^{\ast }_{\scriptscriptstyle \preccurlyeq })\) gives us the result that, if ρ_{Ψ}(x,y) = −ρ_{Θ}(x,y) and x,y ∈ [ [A ∧ B] ] or x,y ∈ [ [¬A ∧¬B] ], then ρ_{Ψ∗A}(x,y) = −ρ_{Θ∗B}(x,y). Note that in these two cases, we have ρ_{A}(x,y) = −ρ_{B}(x,y) = 0. “Removing” the cases from the condition ρ_{A}(x,y) = −ρ_{B}(x,y) leaves us with two cases: (i) x ∈ [ [A ∧¬B] ] and y ∈ [ [¬A ∧ B] ] and (ii) y ∈ [ [A ∧¬B] ] and x ∈ [ [¬A ∧ B] ].
In view of this, \((\text {ZS}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\) is equivalent to the following weaker principle in the presence of \((\mathrm {C}{1,2}^{\ast }_{\scriptscriptstyle \preccurlyeq })\):

If \(x, y \notin \min \limits (\preccurlyeq _{\Psi }, [\![{A}]\!])\cup \min \limits (\preccurlyeq _{{\varTheta }}, [\![{B}]\!])\), ρ_{Ψ}(x,y) = −ρ_{Θ}(x,y) and (i) x ∈ [ [A ∧¬B] ] and y ∈ [ [¬A ∧ B] ] or (ii) y ∈ [ [A ∧¬B] ] and x ∈ [ [¬A ∧ B] ], then ρ_{Ψ∗A}(x,y) = −ρ_{Θ∗B}(x,y)
We first show that this weaker principle can alternatively be presented as follows:

If \([\![{C}]\!]\cap (\min \limits (\preccurlyeq _{\Psi }, [\![{A}]\!])\cup \min \limits (\preccurlyeq _{{\varTheta }}, [\![{B}]\!])) = \varnothing \) and, for all x,y ∈ [ [C] ], ρ_{Ψ}(x,y) = −ρ_{Θ}(x,y) and (i) x ∈ [ [A ∧¬B] ] and y ∈ [ [¬A ∧ B] ] or (ii) y ∈ [ [A ∧¬B] ] and x ∈ [ [¬A ∧ B] ], then, for all x,y ∈ [ [C] ], ρ_{Ψ∗A}(x,y) = −ρ_{Θ∗B}(x,y)
Going from the second principle to the first, the entailment is obvious, since the latter is just the special case in which [ [C] ] = {x,y}. Going the other way, assume that \([\![{C}]\!]\cap (\min \limits (\preccurlyeq _{\Psi }, [\![{A}]\!])\cup \min \limits (\preccurlyeq _{{\varTheta }}, [\![{B}]\!]) = \varnothing \) and, for all x,y ∈ [ [C] ], (i) x ∈ [ [A ∧¬B] ] and y ∈ [ [¬A ∧ B] ] or (ii) y ∈ [ [A ∧¬B] ] and x ∈ [ [¬A ∧ B] ], and ρ_{Ψ}(x,y) = −ρ_{Θ}(x,y). From these assumptions, we have \(x, y \notin \min \limits (\preccurlyeq _{\Psi }, [\![{A}]\!])\cup \min \limits (\preccurlyeq _{{\varTheta }}, [\![{B}]\!])\), ρ_{Ψ}(x,y)= −ρ_{Θ}(x,y) and ρ_{A}(x,y) = −ρ_{B}(x,y). Given this, by \((\text {ZS}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\), it follows that ρ_{Ψ∗A}(x,y) = −ρ_{Θ∗B}(x,y), as required.
Next we establish syntactic equivalents for the various semantic properties figuring in this principle.
As we have noted above, in the proof of Proposition 11, \([\![{C}]\!]\cap (\min \limits (\preccurlyeq _{\Psi }, [\![{A}]\!])\cup \min \limits (\preccurlyeq _{{\varTheta }}, [\![{B}]\!]) )= \varnothing \) iff ¬C ∈ [Ψ∗A] ∩ [Θ ∗ B]. Furthermore, (i) and (ii) hold for all x,y ∈ [ [C] ] iff A ≡_{C}¬B.
Finally, we can show that ρ_{Ψ}(x,y) = −ρ_{Θ}(x,y) for all x,y ∈ [ [C] ] iff Ψ and Θ are in opposition modulo C (and obviously similarly regarding Ψ ∗ A and Θ ∗ B).
From left to right: Assume that (1) ∀x,y ∈ [ [C] ], ρ_{Ψ}(x,y) = −ρ_{Θ}(x,y) but, for reductio, that ∃B,D ∈ L such that B is quasicomplete, \(B\wedge C\nvdash D\), D ∈ [Ψ∗B ∧ C] but ¬D∉[Θ ∗ B ∧ C]. Since \(B\wedge C\nvdash D\), there exists x ∈ [ [B ∧ C ∧¬D] ]. Let \(y\in \min \limits (\preccurlyeq _{\Psi }, [\![{B\wedge C}]\!])\) (we can assume that such a y exists, since \(B\wedge C\nvdash D\) and hence \([\![{B\wedge C}]\!]\neq \varnothing \)). Since D ∈ [Ψ∗B ∧ C] and x ∈ [ [B ∧ C ∧¬D] ], we have y ≺_{Ψ∗B∧C}x. Therefore, since x,y ∈ [ [B ∧ C] ], by \((\mathrm {C}{1, 2}^{\ast }_{\scriptscriptstyle \preccurlyeq })\), it follows that y ≺_{Ψ}x, so that ρ_{Ψ}(x,y) = − 1. Given our assumption that ρ_{Ψ}(x,y) = −ρ_{Θ}(x,y), we then recover ρ_{Θ}(x,y) = 1. From this, given that x,y ∈ [ [B ∧ C] ], we have ρ_{Θ∗B∧C}(x,y) = 1, by \((\mathrm {C}{1, 2}^{\ast }_{\scriptscriptstyle \preccurlyeq })\). Since B is quasicomplete and x,y ∈ [ [B ∧ C] ], we also have [ [B ∧ C] ] = {x,y}. It then follows that ¬D ∈ [Θ ∗ B ∧ C]. Contradiction.
From right to left: Assume that (1) for all B,D ∈ L such that B is quasicomplete and \(B\wedge C\nvdash D\), if D ∈ [Ψ∗B ∧ C], then ¬D ∈ [Θ ∗ B ∧ C], but, for reductio, that (2) ∃x,y ∈ [ [C] ], such that ρ_{Ψ}(x,y)≠ − ρ_{Θ}(x,y). Where B ∈ L is such that [ [B] ] = {x,y}, and D ∈ L is such that x ∈ [ [D] ] but y∉[ [D] ], it follows from (1) that x ≺_{Ψ∗B∧C}y and y ≺_{Θ∗B∧C}x. By \((\mathrm {C}{1, 2}^{\ast }_{\scriptscriptstyle \preccurlyeq })\), since x,y ∈ [ [B ∧ C] ], we then have x ≺_{Ψ}y and y ≺_{Θ}x and hence ρ_{Ψ}(x,y) = −ρ_{Θ}(x,y). Contradiction. □
Proposition 13
Given AGM \((\text {RI}^{\ast }_{\preccurlyeq })\) is equivalent to

\((\text {RI}^{\scriptscriptstyle \ast }_{\scriptscriptstyle })\) \(B {\kern .5pt}\in {\kern .5pt} [({\Psi \ast A}) \ast C]\ \mathit {iff}\ \iota (B){\kern .5pt} \in {\kern .5pt} [({{\varTheta }\ast \iota (A)})\ast \iota (C)]\), for any cbelief isomorphism ι from Ψ to Θ
Proof For convenience, we first recall the definition of \((\text {RI}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\):

\((\text {RI}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\) ρ_{Ψ∗A}(x,y) = ρ_{Θ∗π(A)}(π(x),π(y)), for any order isomorphism π from \( \preccurlyeq _{\Psi }\) to \(\preccurlyeq _{{\varTheta }}\)
Let I be the set of all cbelief isomorphisms between Ψ and Θ and π the set of all order isomorphisms between the corresponding TPOs. We will show that there exists a bijection f between π and I such that if ι = f(π), then the biconditional of (RI^{∗}) holds for ι iff the equality of \((\text {RI}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\) holds for π. It then follows from this that the biconditional holds for all ι in I iff the equality holds for all π in π.
We divide the proof into two lemmas. The first establishes that a particular relation f is a bijection. The second shows that f has the required property stated above. The first lemma, then, is:
Lemma 5
The relation \(f\subseteq {\Pi }\times I\), such that, ∀A ∈ L, ∀π ∈π, ∀ι ∈ I, f(π,ι) iff [ [ι(A)] ] = {x ∈ W∣∃y ∈ [ [A] ], such that x = π(y)}, is a bijection from π to I.
In other words f maps π onto the unique ι (modulo logical equivalence) such that the set of models of the image, under ι, of a sentence A is the set of images, under π, of the models of A. Since f is a bijection, we write f(π) for the unique ι (modulo logical equivalence) such that f(π,ι).
To establish Lemma 5, we first note that [18, Proposition 6.2] already prove the following first of two sublemmas, where \({\Sigma }\supseteq I\) is the set of all belief amount preserving symbol translations (i.e. permutations of L satisfying properties (i)–(iii) of Definition 9) σ on L and \({\varGamma }\supseteq {\Pi }\) is the set of all permutations γ of W:
Sublemma 1
The relation \(g\subseteq {\Pi }\times {\varSigma }\), such that, ∀A ∈ L, ∀γ ∈Γ, ∀σ ∈Σ, f(γ,σ) iff [ [σ(A)] ] = {x ∈ W∣∃y ∈ [ [A] ], such that x = γ(y)}, is a bijection from Γ to Σ.
In view of this, it therefore simply remains to be shown that ι is in I iff its preimage under g is in π, since it then follows from this, and the fact that g is a bijection between Γ and Σ, that f is a bijection between π and I:
Sublemma 2
Where ι = g(π), the following are equivalent

(1) For all x,y ∈ W, ρ_{Ψ}(x,y) = ρ_{Θ}(π(x),π(y))

(2) For all A,B ∈ L, B ∈ [Ψ∗A] iff ι(B) ∈ [Θ ∗ ι(A)]
From (1) to (2): Assume ι = g(π) and (1). We need to establish (2), which, given \((\text {KM}^{\ast }_{\preccurlyeq })\), we can reformulate in terms of minimal sets as: \(\min \limits (\preccurlyeq _{\Psi },[\![{A}]\!])\subseteq [\![{B}]\!]\) iff \(\min \limits (\preccurlyeq _{{\varTheta }},[\![{\iota (A)}]\!])\subseteq [\![{\iota (B)}]\!]\). We simply derive the left to right direction of (2), since the other direction is established in an analogous manner. Assume that \(\min \limits (\preccurlyeq _{\Psi },[\![{A}]\!])\subseteq [\![{B}]\!]\) but, for reductio, that \(\min \limits (\preccurlyeq _{{\varTheta }},[\![{\iota (A)}]\!])\nsubseteq [\![{\iota (B)}]\!]\), so that \(\exists x\in \min \limits (\preccurlyeq _{{\varTheta }},[\![{\iota (A)}]\!])\cap [\![{\neg \iota (B)}]\!]\). By the definitions of ι and g, it follows that π^{− 1}(x) ∈ [ [A] ] ∩ [ [¬B] ]. Let \(y\in \min \limits (\preccurlyeq _{\Psi },[\![{A}]\!])\). Since \(\min \limits (\preccurlyeq _{\Psi },[\![{A}]\!])\subseteq [\![{B}]\!]\), we then have y ≺_{Ψ}π^{− 1}(x). By (1), we then have π(y) ≺_{Θ}x. Since y ∈ [ [A] ], it follows by the definition of g that π(y) ∈ [ [ι(A)] ] and so, given π(y) ≺_{Θ}x, that \(x\notin \min \limits (\preccurlyeq _{{\varTheta }},[\![{\iota (A)}]\!])\) after all: contradiction.
From (2) to (1): Assume that ι = g(π) and that (2) holds. Given \((\text {KM}^{\ast }_{\preccurlyeq })\), we can again reformulate (2) in terms of minimal sets as: \(\min \limits (\preccurlyeq _{\Psi },[\![{A}]\!])\subseteq [\![{B}]\!]\) iff \(\min \limits (\preccurlyeq _{{\varTheta }},[\![{\iota (A)}]\!])\subseteq [\![{\iota (B)}]\!]\). Where [ [A] ] = {x,y}, this gives us, in view of the definition of g:

\(\min \limits (\preccurlyeq _{\Psi }, \{x, y\})\subseteq [\![{B}]\!]\) iff \(\min \limits (\preccurlyeq _{{\varTheta }}, \{\pi (x), \pi (y)\})\subseteq [\![{\iota (B)}]\!]\)
By the definition of g, for all x ∈ W, x ∈ [ [B] ] iff π(x) ∈ [ [ι(B)] ]. Hence ρ_{Ψ}(x,y) = ρ_{Θ}(π(x),π(y)), i.e. (1). This completes the proof of Sublemma 2 and hence of Lemma 5.
We finally derive our second lemma towards the proof of our main result, which tells us that f has the required property, namely that, if ι = f(π), then the biconditional in (RI^{∗}) holds for ι, iff the equality in \((\text {RI}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\) holds for π:
Lemma 6
Where ι = f(π), the following are equivalent

(1) For all x,y ∈ W and A ∈ L, ρ_{Ψ∗A}(x,y) = ρ_{Θ∗π(A)}(π(x),π(y))

(2) For all A,B,C ∈ L, B ∈ [(Ψ ∗ A) ∗ C] iff ι(B) ∈ [(Θ ∗ ι(A)) ∗ ι(C)]
The proof of this is identical to that of Sublemma 2, save for the fact that we need to note that, as we defined the extension of π to L, Θ ∗ π(A) = Θ ∗ ι(A).
The conjunction of Lemmas 5 and 6 then establishes the required result. □
Proposition 15
Lexicographic revision is the only elementary revision operator that satisfies \((\text {IIAP}+^{\ast }_{\preccurlyeq })\)
Proof
We show that, given \((\text {KM}^{\ast }_{\preccurlyeq })\) and \((\text {UD}_{\scriptscriptstyle \preccurlyeq })\), \((\mathrm {IIAP+}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\) entails Recalcitrance (i.e. if ρ_{A}(x,y) = 1, then ρ_{Ψ∗A}(x,y) = 1), which characterises lexicographic revision in the presence of \((\mathrm {C}{1, 2}^{\ast }_{\scriptscriptstyle \preccurlyeq })\). This was already established as Fact 2.2 (a) in [15]. Let ρ_{A}(x,y) = 1, so that x ∈ [ [A] ] and y ∈ [ [¬A] ]. Then, by \((\text {UD}_{\scriptscriptstyle \preccurlyeq })\), for any state Ψ, there will exist a state Θ such that ρ_{Θ}(x,y) = ρ_{Ψ}(x,y) and \(x\in \min \limits (\preccurlyeq _{{\varTheta }}, [\![{A}]\!])\) (and, since y ∈ [ [¬A] ], \(y\notin \min \limits (\preccurlyeq _{{\varTheta }}, [\![{A}]\!])\)). But by \((\text {KM}^{\ast }_{\preccurlyeq })\), if \(x\in \min \limits (\preccurlyeq _{{\varTheta }}, [\![{A}]\!])\) but \(y\notin \min \limits (\preccurlyeq _{{\varTheta }}, [\![{A}]\!])\), then x ≺_{Θ∗A}y. So, by \((\mathrm {IIAP+}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\), x ≺_{Ψ∗A}y, as required. □
Proposition 16
Lexicographic revision is the only elementary revision operator that satisfies \((\text {IIAI}+^{\ast }_{\preccurlyeq })\)
Proof
\((\mathrm {IIAI+}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\), in conjunction with \((\mathrm {C}{1, 2}^{\ast }_{\scriptscriptstyle \preccurlyeq })\)–\((\mathrm {C}{4}^{\ast }_{\scriptscriptstyle \preccurlyeq })\), can be shown to entail a principle that we have called “\((\beta {1 +}^{\ast }_{\scriptscriptstyle \preccurlyeq })\)” in previous work [6]. Indeed, in the proof of Proposition 6 above, we established the equivalence between \((\text {IIAI}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\) and the conjunction of \((\beta {1}^{\ast }_{\scriptscriptstyle \preccurlyeq })\) and \((\beta {2}^{\ast }_{\scriptscriptstyle \preccurlyeq })\) using only \((\mathrm {C}{1, 2}^{\ast }_{\scriptscriptstyle \preccurlyeq })\)–\((\mathrm {C}{4}^{\ast }_{\scriptscriptstyle \preccurlyeq })\). This proof can be adapted to establish a strengthening of Proposition 3 in [8], in which, unlike in the original, the principle of “Independence” \((\mathrm {P}^{\ast }_{\scriptscriptstyle \preccurlyeq })\) is not appealed to. This yields the following: In the presence of \((\text {KM}^{\ast }_{\preccurlyeq })\) and \((\mathrm {C}{1, 2}^{\ast }_{\scriptscriptstyle \preccurlyeq })\)–\((\mathrm {C}{4}^{\ast }_{\scriptscriptstyle \preccurlyeq })\), Booth & Meyer’s strengthening of \((\text {IIAI}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\) is equivalent to the conjunction of what [6] call “\((\beta {1 +}^{\ast }_{\scriptscriptstyle \preccurlyeq })\)” and “\((\beta {2 +}^{\ast }_{\scriptscriptstyle \preccurlyeq })\)”.
Furthermore, we also showed in [6] (see Corollary 1 there) that \((\beta {1 +}^{\ast }_{\scriptscriptstyle \preccurlyeq })\) characterises lexicographic revision, given \((\text {KM}^{\ast }_{\preccurlyeq })\) and \((\mathrm {C}{1, 2}^{\ast }_{\scriptscriptstyle \preccurlyeq })\)\((\mathrm {C}{2}^{\ast }_{\scriptscriptstyle \preccurlyeq })\). □
Proposition 17
Lexicographic revision is the only elementary revision operator that satisfies \((\text {ZS}{+}^{\ast }_{\preccurlyeq })\)
Proof
We show that, given \((\text {KM}^{\ast }_{\preccurlyeq })\) and \((\mathrm {C}{3}^{\ast }_{\scriptscriptstyle \preccurlyeq })\), \((\mathrm {ZS+}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\) entails Recalcitrance (i.e. if ρ_{A}(x,y) = 1, then ρ_{Ψ∗A}(x,y) = 1), which characterises lexicographic revision in the presence of \((\mathrm {C}{1, 2}^{\ast }_{\scriptscriptstyle \preccurlyeq })\). Assume for reductio that Recalcitrance fails, so that there exists Ψ, x,y ∈ W, A ∈ L such that

(1) ρ_{A}(x,y) = 1

(2) ρ_{Ψ∗A}(x,y)≠ 1
From (1) and (2), by \((\mathrm {C}{3}^{\ast }_{\scriptscriptstyle \preccurlyeq })\), we must have:

(3) ρ_{Ψ}(x,y)≠ 1
From (1), we have:

(4) ρ_{¬A}(x,y) = − 1
From (1), it follows that y ∈ [ [¬A] ] and x≠y, and so by \((\text {UD}_{\scriptscriptstyle \preccurlyeq })\) there exists a state Θ such that:

(5) \(y\in \min \limits (\preccurlyeq _{{\varTheta }}, [\![{\neg A}]\!])\)

(6) ρ_{Θ}(x,y) = −ρ_{Ψ}(x,y)
From (1), (4), (3) and (6), by \((\mathrm {ZS+}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\):

(7) ρ_{Θ∗¬A}(x,y)≠ − 1
However, by AGM, given (4) and (5), we have ρ_{Θ∗¬A}(x,y) = − 1, directly contradicting (7). □
Proposition 18
Given AGM ÷_{R} is characterised by the following property:
Proof
We first recall the definition of ÷_{R}:

If x ∈ or \(y \in \min \limits (\preccurlyeq _{\Psi }, W)\cup \min \limits (\preccurlyeq _{\Psi }, [\![{\neg A}]\!])\), then:
$$ \rho_{\Psi\div_{\mathrm{R}} A}({x},{y})=\begin{cases} 1 \text{, if } x \in \text{ and } y\notin \min(\preccurlyeq_{\Psi}, W)\cup \min(\preccurlyeq_{\Psi}, [\![{\neg A}]\!]) \\ 0\text{, if } x\in \text{ and } y \in \min(\preccurlyeq_{\Psi}, W)\cup \min(\preccurlyeq_{\Psi}, [\![{\neg A}]\!]) \\ 1\text{, if } x\notin \text{ and } y \in \min(\preccurlyeq_{\Psi}, W)\cup \min(\preccurlyeq_{\Psi}, [\![{\neg A}]\!]) \end{cases} $$ 
If x∉ and \(y \notin \min \limits (\preccurlyeq _{\Psi }, W)\cup \min \limits (\preccurlyeq _{\Psi }, [\![{\neg A}]\!])\), then:
$$ \rho_{\Psi\div_{\mathrm{R}} A}({x},{y})=\begin{cases} \rho_{\Psi}({x},{y})\text{, if }\rho_{\Psi}({x},{y})\neq 0\\ \rho_{\neg A}({x},{y})\text{, if }\rho_{\Psi}({x},{y})= 0\end{cases} $$
Next, with the help of (KM\(^{\div }_{\preccurlyeq }\)), we translate the two conditionals on the right hand side of the equality of the characteristic syntactic property into the following statements about minimal sets:

(1) If \(\min \limits (\preccurlyeq _{\Psi }, W)\cup \min \limits (\preccurlyeq _{\Psi }, [\![{\neg B}]\!])\subseteq [\![{A\vee B}]\!]\), then \(\min \limits (\preccurlyeq _{({\Psi }\div A)\div B}, W)=\min \limits (\preccurlyeq _{\Psi \div A}, W)\cup \min \limits (\preccurlyeq _{\Psi \div A}, [\![{A\wedge \neg B}]\!])\)

(2) If \(\min \limits (\preccurlyeq _{\Psi }, W)\cup \min \limits (\preccurlyeq _{\Psi }, [\![{\neg B}]\!])\nsubseteq [\![{A\vee B}]\!]\), then \(\min \limits (\preccurlyeq _{({\Psi }\div A)\div B}, W)=\min \limits (\preccurlyeq _{\Psi \div A}, W)\cup \min \limits (\preccurlyeq _{\Psi \div A}, [\![{\neg A\wedge \neg B}]\!])\)
From the semantic characterisation to the syntactic one: We first note that, given (KM\(^{\div }_{\preccurlyeq }\)), the consequent of (1) is equivalent to \(\min \limits (\preccurlyeq _{\Psi \div A}, [\![{\neg B}]\!])  \min \limits (\preccurlyeq _{\Psi \div A}, W) = \min \limits (\preccurlyeq _{\Psi }, [\![{A\wedge \neg B}]\!]) \min \limits (\preccurlyeq _{\Psi \div A}, W)\), while the consequent of (2) is equivalent to \(\min \limits (\preccurlyeq _{\Psi \div A}, [\![{\neg B}]\!])  \min \limits (\preccurlyeq _{\Psi \div A}, W) = \min \limits (\preccurlyeq _{\Psi }, [\![{\neg A\wedge \neg B}]\!]) \min \limits (\preccurlyeq _{\Psi \div A}, W)\). We then split the proof into two obvious parts:

(A) If ÷ = ÷_{R}, then (1): Assume \(\min \limits (\preccurlyeq _{\Psi }, W)\cup \min \limits (\preccurlyeq _{\Psi }, [\![{\neg B}]\!])\subseteq [\![{A\vee B}]\!]\), so that there does not exist \(x\in \min \limits (\preccurlyeq _{\Psi },[\![{\neg B}]\!])\cap [\![{\neg A}]\!]\). Assume for reductio that \(\min \limits (\preccurlyeq _{\Psi \div A}, [\![{\neg B}]\!]) \min \limits (\preccurlyeq _{\Psi \div A}, W)\nsubseteq [\![{A}]\!]\), so that there exists \(x\in \min \limits (\preccurlyeq _{\Psi \div A}, [\![{\neg B}]\!])\cap [\![{\neg A}]\!]\) and \(x\notin \min \limits (\preccurlyeq _{\Psi \div A}, W)\). As we have noted, from our initial assumption, it must be the case that \(x\notin \min \limits (\preccurlyeq _{\Psi }, [\![{\neg B}]\!])\cap [\![{\neg A}]\!]\) and hence, since x ∈ [ [A] ], \(x\notin \min \limits (\preccurlyeq _{\Psi }, [\![{\neg B}]\!])\). So, since x ∈ [ [¬B] ], there exists y ∈ [ [¬B] ] such that ρ_{Ψ}(y,x) = 1. Given the semantic definition of ÷_{R}, the fact that \(x\notin \min \limits (\preccurlyeq _{\Psi \div A}, W)\) then suffices to ensure that ρ_{Ψ÷A}(y,x) = 1 (indeed, if \(y\in \min \limits (\preccurlyeq _{\Psi \div A}, W)\), then this follows from the first conditional of the definition, and if \(y\notin \min \limits (\preccurlyeq _{\Psi \div A}, W)\), then, since ρ_{Ψ}(y,x)≠ 0, the second conditional tells us that ρ_{Ψ÷A}(y,x) = ρ_{Ψ}(y,x) = 1), and hence, since y ∈ [ [¬B] ], \(x\notin \min \limits (\preccurlyeq _{\Psi \div A}, [\![{\neg B}]\!])\cap [\![{\neg A}]\!]\). Contradiction. We can therefore conclude that \(\min \limits (\preccurlyeq _{\Psi \div A}, [\![{\neg B}]\!])\min \limits (\preccurlyeq _{\Psi \div A}, W)\subseteq [\![{A}]\!]\). From this, it follows that \(\min \limits (\preccurlyeq _{\Psi \div A}, [\![{\neg B}]\!]) \min \limits (\preccurlyeq _{\Psi \div A}, W)= \min \limits (\preccurlyeq _{\Psi \div A}, [\![{A\wedge \neg B}]\!])\min \limits (\preccurlyeq _{\Psi \div A}, W)\). Since ÷_{R} satisfies \((\mathrm {C}{1,2}^{\scriptscriptstyle \div }_{\scriptscriptstyle \preccurlyeq })\), it follows that \(\min \limits (\preccurlyeq _{\Psi \div A}, [\![{A\wedge \neg B}]\!])= \min \limits (\preccurlyeq _{\Psi }, [\![{A\wedge \neg B}]\!])\) and hence we finally have \(\min \limits (\preccurlyeq _{\Psi \div A}, [\![{\neg B}]\!]) \min \limits (\preccurlyeq _{\Psi \div A}, W)= \min \limits (\preccurlyeq _{\Psi }, [\![{A\wedge \neg B}]\!])\min \limits (\preccurlyeq _{\Psi \div A}, W)\), as required.

(B) If ÷ = ÷_{R}, then (2): We first note that, like ÷_{P}, ÷_{R} satisfies the following property, which could be considered the analogue for contraction of the property \((\mathrm {P}^{\ast }_{\scriptscriptstyle \preccurlyeq })\), satisfied by ∗_{L} and ∗_{R}:

\((\text {wP}^{\scriptscriptstyle \div }_{\scriptscriptstyle \preccurlyeq })\) If \(y\notin \min \limits (\preccurlyeq _{\Psi }, W)\), ρ_{¬A}(x,y) = 1 and ρ_{Ψ}(x,y) ≥ 0, then ρ_{Ψ÷A}(x,y) = 1^{Footnote 1}
Now assume \(\min \limits (\preccurlyeq _{\Psi }, W)\cup \min \limits (\preccurlyeq _{\Psi }, [\![{\neg B}]\!])\nsubseteq [\![{A\vee B}]\!]\), so that there exists \(x\in \min \limits (\preccurlyeq _{\Psi }, W)\cup \min \limits (\preccurlyeq _{\Psi }, [\![{\neg B}]\!])\), such that x ∈ [ [¬A ∧¬B] ]. If \(x\in \min \limits (\preccurlyeq _{\Psi }, W)\), since x ∈ [ [¬B] ], we have \(x\in \min \limits (\preccurlyeq _{\Psi }, [\![{\neg B}]\!])\). So, either way, \(x\in \min \limits (\preccurlyeq _{\Psi }, [\![{\neg B}]\!])\). Assume for reductio that \(\min \limits (\preccurlyeq _{\Psi \div A}, [\![{\neg B}]\!]) \min \limits (\preccurlyeq _{\Psi \div A}, W)\nsubseteq [\![{\neg A}]\!]\), so that there exists \(y\in \min \limits (\preccurlyeq _{\Psi \div A}, [\![{\neg B}]\!])\min \limits (\preccurlyeq _{\Psi \div A}, W)\), such that y ∈ [ [A] ] and therefore, since x ∈ [ [¬A] ], ρ_{¬A}(x,y) = 1. Since \(x\in \min \limits (\preccurlyeq _{\Psi }, [\![{\neg B}]\!])\) and y ∈ [ [¬B] ], we have ρ_{Ψ}(x,y) ≥ 0. Since, by assumption, \(y\notin \min \limits (\preccurlyeq _{\Psi \div A}, W)\), it follows, by (KM\(^{\div }_{\preccurlyeq }\)), that \(y\notin \min \limits (\preccurlyeq _{\Psi }, W)\). Given, \(y\notin \min \limits (\preccurlyeq _{\Psi }, W)\), ρ_{¬A}(x,y) = 1, and ρ_{Ψ}(x,y) ≥ 0, we can now apply \((\text {wP}^{\scriptscriptstyle \div }_{\scriptscriptstyle \preccurlyeq })\), to obtain ρ_{Ψ÷A}(x,y) = 1. Since x,y ∈ [ [¬B] ], we then have \(y\notin \min \limits (\preccurlyeq _{\Psi \div A}, [\![{\neg B}]\!])\). Contradiction. We can therefore conclude that \(\min \limits (\preccurlyeq _{\Psi \div A}, [\![{\neg B}]\!])\min \limits (\preccurlyeq _{\Psi \div A}, W)\subseteq [\![{\neg A}]\!]\). From this, it follows that \(\min \limits (\preccurlyeq _{\Psi \div A}, [\![{\neg B}]\!]) \min \limits (\preccurlyeq _{\Psi \div A}, W)= \min \limits (\preccurlyeq _{\Psi \div A}, [\![{\neg A\wedge \neg B}]\!])\min \limits (\preccurlyeq _{\Psi \div A}, W)\). Since ÷_{R} satisfies \((\mathrm {C}{1,2}^{\scriptscriptstyle \div }_{\scriptscriptstyle \preccurlyeq })\), it follows that \(\min \limits (\preccurlyeq _{\Psi \div A}, [\![{\neg A\wedge \neg B}]\!])= \min \limits (\preccurlyeq _{\Psi }, [\![{\neg A\wedge \neg B}]\!])\) and hence we finally have \(\min \limits (\preccurlyeq _{\Psi \div A}, [\![{\neg B}]\!]) \min \limits (\preccurlyeq _{\Psi \div A}, W)= \min \limits (\preccurlyeq _{\Psi }, [\![{\neg A\wedge \neg B}]\!])\min \limits (\preccurlyeq _{\Psi \div A}, W)\), as required.

From the syntactic characterisation to the semantic one: Assume x ∈ or \(y \in \min \limits (\preccurlyeq _{\Psi }, W)\cup \min \limits (\preccurlyeq _{\Psi }, [\![{\neg A}]\!])\). Then the required result follows simply by (KM\(^{\div }_{\preccurlyeq }\)). So assume that x∉ and \(y \notin \min \limits (\preccurlyeq _{\Psi }, W)\cup \min \limits (\preccurlyeq _{\Psi }, [\![{\neg A}]\!])\). We divide the remainder of the proof into two obvious parts:

(A) Proof that, if ρ_{Ψ}(x,y)≠ 0, then ρ_{Ψ÷A}(x,y) = ρ_{Ψ}(x,y): It suffices to show that, if ρ_{Ψ}(x,y) = 1, then ρ_{Ψ÷A}(x,y) = 1 (establishing that, if ρ_{Ψ}(x,y) = − 1, then ρ_{Ψ÷A}(x,y) = − 1, is analogous). So assume that ρ_{Ψ}(x,y) = 1. We first note that the required conclusion that ρ_{Ψ÷A}(x,y) = 1 holds iff \(y\notin \min \limits (\preccurlyeq _{({\Psi }\div A)\div \neg x\wedge \neg y}, W)\): Indeed, by (KM\(^{\div }_{\preccurlyeq }\)), we have \(\min \limits (\preccurlyeq _{({\Psi }\div A)\div \neg x\wedge \neg y}, W)= \min \limits (\preccurlyeq _{\Psi \div A}, W)\cup \min \limits (\preccurlyeq _{\Psi \div A}, \{x, y\})\). Since \(\{x, y\}\subseteq W\), it then follows from this that \(y\in \min \limits (\preccurlyeq _{({\Psi }\div A)\div \neg x\wedge \neg y}, W)\) iff \(y\in \min \limits (\preccurlyeq _{\Psi \div A}, \{x, y\})\). Since, trivially, we have ρ_{Ψ÷A}(x,y) = 1 iff \(y\notin \min \limits (\preccurlyeq _{\Psi \div A}, \{x, y\})\), the required result then follows. We then split the proof into two cases:

(a) Assume \( \min \limits (\preccurlyeq _{\Psi }, W)\cup \min \limits (\preccurlyeq _{\Psi }, \{x, y\})\subseteq [\![{A\vee (\neg x\wedge \neg y)}]\!]\). Then, by (KM\(^{\div }_{\preccurlyeq }\)) and the syntactic characteristic principle (setting B = ¬x ∧¬y) we have:
$$ \begin{array}{@{}rcl@{}} &&\min(\preccurlyeq_{({\Psi}\div A)\div\neg x\wedge \neg y}, W)\\ &&\quad\quad\quad\quad\quad\quad\quad= \min(\preccurlyeq_{\Psi\div A}, W)\cup \min(\preccurlyeq_{\Psi\div \neg A\vee (\neg x\wedge \neg y)}, W) \end{array} $$By assumption, \(y \!\notin \! \min \limits (\preccurlyeq _{\Psi \div A}, W)\). So it remains to be shown that \(y\!\notin \! \min \limits (\preccurlyeq _{\Psi \div \neg A\vee (\neg x\wedge \neg y)}, W)\). By (KM\(^{\div }_{\preccurlyeq }\)), \( \min \limits (\preccurlyeq _{\Psi \div \neg A\vee (\neg x\wedge \neg y)},\) \(W) = \min \limits (\preccurlyeq _{\Psi }, W)\cup \min \limits (\preccurlyeq _{\Psi }, [\![{A\wedge (x\vee y)}]\!])\). Since we have already also assumed that \(y\notin \min \limits (\preccurlyeq _{\Psi }, W)\), we are just left with verifying that \(y\notin \min \limits (\preccurlyeq _{\Psi }, [\![{A\wedge (x\vee y)}]\!])\). So assume for reductio that \(y\in \min \limits (\preccurlyeq _{\Psi }, [\![{A\wedge (x\vee y)})]\!]\), so that y ∈ [ [A] ] and, if x ∈ [ [A] ], then ρ_{Ψ}(x,y) ≤ 0. From \(\min \limits (\preccurlyeq _{\Psi }, W)\cup \min \limits (\preccurlyeq _{\Psi }, \{x, y\})\subseteq [\![{A\vee (\neg x\wedge \neg y)}]\!]\) and ρ_{Ψ}(x,y) = 1, it follows that x ∈ [ [A] ]. Hence ρ_{Ψ}(x,y) ≤ 0. But this contradicts our assumption that ρ_{Ψ}(x,y) = 1. So \(y\notin \min \limits (\preccurlyeq _{\Psi }, [\![{A\wedge (x\vee y)})]\!]\) and we therefore have \(y\notin \min \limits (\preccurlyeq _{({\Psi }\div A)\div \neg x\wedge \neg y}, W)\), as required.

(b) Assume \(\min \limits (\preccurlyeq _{\Psi }, W)\cup \min \limits (\preccurlyeq _{\Psi }, \{x, y\})\nsubseteq [\![{A\vee (\neg x\wedge \neg y)}]\!]\). Then the reasoning is similar to the above, except that we proceed by showing that \(y\notin \min \limits (\preccurlyeq _{\Psi }, [\![{\neg A\wedge (x\vee y)}]\!])\).


(B) Proof that, if ρ_{Ψ}(x,y) = 0, then ρ_{Ψ÷A}(x,y) = ρ_{¬A}(x,y): Assume that ρ_{Ψ}(x,y) = 0. We again split the proof into two cases:

(a) Assume \(\min \limits (\preccurlyeq _{\Psi }, W)\cup \min \limits (\preccurlyeq _{\Psi }, \{x, y\})\subseteq [\![{A\vee (\neg x\wedge \neg y)}]\!]\). Then x,y ∈ [ [A] ] and so ρ_{¬A}(x,y) = 0. So we need to establish that ρ_{Ψ÷A}(x,y) = 0, or equivalently \(x, y\in \min \limits (\preccurlyeq _{({\Psi }\div A)\div \neg x\wedge \neg y}, W)\). It also follows, by (KM\(^{\div }_{\preccurlyeq }\)) and the syntactic characteristic principle (setting B = ¬x ∧¬y), that:
$$ \begin{array}{@{}rcl@{}} &&\min(\preccurlyeq_{({\Psi}\div A)\div \neg x\wedge \neg y}, W) \\ &&\qquad\quad\quad\quad\qquad= \min(\preccurlyeq_{\Psi\div A}, W)\cup \min(\preccurlyeq_{\Psi\div (\neg A\vee (\neg x\wedge \neg y)}, W) \end{array} $$As we have already noted, by (KM\(^{\div }_{\preccurlyeq }\)), \( \min \limits (\preccurlyeq _{\Psi \div (\neg A\vee (\neg x\wedge \neg y)}, W) = \min \limits (\preccurlyeq _{\Psi }, W)\cup \min \limits (\preccurlyeq _{\Psi }, [\![{A\wedge (x\vee y)})]\!]\). Furthermore, it follows from x,y ∈ [ [A] ] and ρ_{Ψ}(x,y) = 0 that \(x, y\in \min \limits (\preccurlyeq _{\Psi }, [\![{A\wedge (x\vee y)})]\!]\). Hence \(x, y\in \min \limits (\preccurlyeq _{({\Psi }\div A)\div \neg x\wedge \neg y}, W)\), as required.

(b) Assume \(\min \limits (\preccurlyeq _{\Psi }, W)\cup \min \limits (\preccurlyeq _{\Psi }, \{x, y\})\nsubseteq [\![{A\vee (\neg x\wedge \neg y)}]\!]\). Then, by (KM\(^{\div }_{\preccurlyeq }\)) and the syntactic characteristic principle (setting B = ¬x ∧¬y):
$$ \begin{array}{@{}rcl@{}} &&\min(\preccurlyeq_{({\Psi}\div A)\div\neg x\wedge \neg y}, W)\\ &&\quad\quad\quad\quad\quad\quad\quad= \min(\preccurlyeq_{\Psi\div A}, W)\cup \min(\preccurlyeq_{\Psi\div A\vee (\neg x\wedge \neg y)}, W) \end{array} $$From the fact that \(\min \limits (\preccurlyeq _{\Psi }, \{x, y\})\nsubseteq [\![{A\vee (\neg x\wedge \neg y)}]\!]\) and hence, since ρ_{Ψ}(x,y) = 0, that \(\{x, y\}\nsubseteq [\![{A}]\!]\), we are left with two possibilities to consider:

(i) Assume x,y ∈ [ [¬A] ]. Then ρ_{¬A}(x,y) = 0. So we need to establish that ρ_{Ψ÷A}(x,y) = 0, or equivalently \(x, y\in \min \limits (\preccurlyeq _{({\Psi }\div A)\div \neg x\wedge \neg y}, W)\). By (KM\(^{\div }_{\preccurlyeq }\)), it follows that we have \( \min \limits (\preccurlyeq _{\Psi \div A\vee (\neg x\wedge \neg y)}, W) = \min \limits (\preccurlyeq _{\Psi }, W)\cup \min \limits (\preccurlyeq _{\Psi }, [\![{\neg A\wedge (x\vee y)}]\!])\). Since x,y ∈ [ [¬A] ] and ρ_{Ψ}(x,y) = 0, we have \(x, y\in \min \limits (\preccurlyeq _{\Psi }, [\![{\neg A\wedge (x\vee y)}]\!])\) and so, finally, we recover \(x, y\in \min \limits (\preccurlyeq _{({\Psi }\div A)\div \neg x\wedge \neg y}, W)\), as required.

(ii) Assume x ∈ [ [A] ], y ∈ [ [¬A] ] (the other case, in which y ∈ [ [A] ], x ∈ [ [¬A] ] is analogous). Then ρ_{¬A}(x,y) = − 1. So we need to establish that ρ_{Ψ÷A}(x,y) = − 1, or equivalently \(x\notin \min \limits (\preccurlyeq _{({\Psi }\div A)\div \neg x\wedge \neg y}, W)\). We already know that \(x\notin \min \limits (\preccurlyeq _{\Psi \div A}, W)\). So it remains to be established that \(x\notin \min \limits (\preccurlyeq _{\Psi \div A\vee (\neg x\wedge \neg y)}, W)\). By (KM\(^{\div }_{\preccurlyeq }\)), \(\min \limits (\preccurlyeq _{\Psi \div (A\vee (\neg x\wedge \neg y)}, W) = \min \limits (\preccurlyeq _{\Psi }, W)\cup \min \limits (\preccurlyeq _{\Psi }, [\![{\neg A\wedge (x\vee y)}]\!])\). We have already assumed that \(x\notin \min \limits (\preccurlyeq _{\Psi }, W)\) and, since x ∈ [ [A] ], \(x\notin \min \limits (\preccurlyeq _{\Psi }, [\![{\neg A\wedge (x\vee y)}]\!])\). We therefore recover \(x\notin \min \limits (\preccurlyeq _{({\Psi }\div A)\div \neg x\wedge \neg y}, W)\), as required.


□
Proposition 19
Where i ∈ { P, R, N }, it is not the case that for any state Ψ and sentences A, B ∈ L there exists C ∈ L such that [(Ψ÷_{i}A) ÷_{i}B] = [Ψ ÷_{i}C]
Proof
We shall say that a rank associated with \(\preccurlyeq _{\Psi }\) is an equivalence class generated by the indifference relation \(\sim _{\Psi }\). All three operators are such that, if A ∈ [Ψ], then the number of ranks associated with \(\preccurlyeq _{\Psi \div A}\) is equal to n − 1, where n is the number of ranks associated with \(\preccurlyeq _{\Psi }\) (and equal to n if A∉[Ψ]). We can then find a countermodel by considering Ψ and A,B ∈ L such that A ∈ [Ψ] and B ∈ [Ψ÷A]: we will then have n − 2 ranks associated with \(\preccurlyeq _{({\Psi }\div A)\div B}\) and therefore no C ∈ L such that \(\preccurlyeq _{\Psi \div C}=\preccurlyeq _{({\Psi }\div A)\div B}\). Figure 6 depicts a countermodel relating to all three operators. □
Proposition 21
Let 〈i,j〉∈{〈N,N〉,〈R,R〉}. Then, if ∗ is defined from ÷_{j} using \((\mathrm {iLIRC_{\preccurlyeq }})\), then ∗ = ∗_{i}.
Proof
Let 〈i,j〉∈{〈N,N〉,〈R,R〉}. We need to show that the following equality holds: \(\rho _{\Psi \ast _{i} A}({x},{y}) = \rho _{({\Psi } \div _{j} \neg A) \ast _{\mathrm {N}} A}({x},{y})\).
We first note that our operators satisfy certain preservation conditions for strict preference. In particular ∗_{N} and ∗_{R} satisfy:

(SPPres\(^{\ast }_{\preccurlyeq }\)) If \(y\notin \min \limits (\preccurlyeq _{\Psi }, [\![{A}]\!])\) and x ≺_{Ψ}y, then x ≺_{Ψ∗A}y
whereas ÷_{N} and ÷_{R} satisfy:

(SPPres\(^{\div }_{\preccurlyeq }\)) If \(y\notin \min \limits (\preccurlyeq _{\Psi }, [\![{A}]\!])\) and x ≺_{Ψ}y, then x ≺_{Ψ÷¬A}y
We divide the proof into two main cases:

(1) Assume x ∈ or y ∈ \(\min \limits (\preccurlyeq _{\Psi }, [\![{A}]\!])\). By \((\text {KM}^{\ast }_{\preccurlyeq })\), it follows that \(\min \limits (\preccurlyeq _{\Psi \ast _{i} A}, W)=\min \limits (\preccurlyeq _{\Psi }, [\![{A}]\!])\). By \((\mathrm {C}{1,2}^{\scriptscriptstyle \div }_{\scriptscriptstyle \preccurlyeq })\), we have \(\min \limits (\preccurlyeq _{\Psi }, [\![{A}]\!])= \min \limits (\preccurlyeq _{\Psi \div _{j} \neg A}, [\![{A}]\!])\). Finally, again by \((\text {KM}^{\ast }_{\preccurlyeq })\), we recover the result that \( \min \limits (\preccurlyeq _{\Psi \div _{j} \neg A}, [\![{A}]\!])=\min \limits (\preccurlyeq _{({\Psi }\div _{j} \neg A)\ast _{\mathrm {N}} A}, W)\). These 3 equalities then yield \(\min \limits (\preccurlyeq _{\Psi \ast _{i} A}, W)=\min \limits (\preccurlyeq _{({\Psi }\div _{j} \neg A)\ast _{\mathrm {N}} A}, W)\). By \((\text {KM}^{\ast }_{\preccurlyeq })\), again, we also have \(\min \limits (\preccurlyeq _{\Psi }, [\![{A}]\!])=\min \limits (\preccurlyeq _{\Psi \ast _{i} A}, W)\). Hence:
$$ \min(\preccurlyeq_{\Psi}, [\![{A}]\!])=\min(\preccurlyeq_{\Psi\ast_{i} A}, W)= \min(\preccurlyeq_{({\Psi}\div_{j} \neg A)\ast_{\mathrm{N}} A}, W) $$With this in hand: \(\rho _{\Psi \ast _{i} A}({x},{y}) = \rho _{({\Psi }\div _{j} \neg A)\ast _{\mathrm {N}} A}({x},{y})=1\), if x ∈ and y∉ \(\min \limits (\preccurlyeq _{\Psi }, [\![{A}]\!])\), \(\rho _{\Psi \ast _{i} A}({x},{y}) = \rho _{({\Psi }\div _{j} \neg A)\ast _{\mathrm {N}} A}({x},{y})=0\), if x,y ∈ \(\min \limits (\preccurlyeq _{\Psi }, [\![{A}]\!])\), and \(\rho _{\Psi \ast _{i} A}({x},{y}) = \rho _{({\Psi }\div _{j} \neg A)\ast _{\mathrm {N}} A}({x},{y})=1\), if x∉ and y ∈ \(\min \limits (\preccurlyeq _{\Psi }, [\![{A}]\!])\).

(2) Assume x,y∉ \(\min \limits (\preccurlyeq _{\Psi }, [\![{A}]\!])\).

(a) Assume x ∈ or y ∈ \(\min \limits (\preccurlyeq _{\Psi }, W)\).

(i) Assume \(x,y\in \min \limits (\preccurlyeq _{\Psi }, W)\). Then ρ_{Ψ}(x,y) = 0. Since x,y∉ \(\min \limits (\preccurlyeq _{\Psi }, [\![{A}]\!])\), we must also have x,y ∈ [ [¬A] ] and hence ρ_{A}(x,y) = 0. By \((\mathrm {C}{1,2}^{\ast }_{\scriptscriptstyle \preccurlyeq })\), it then follows that we have \(\rho _{\Psi \ast _{i} A}({x},{y})=\rho _{({\Psi } \div _{j} \neg A) \ast _{\mathrm {N}} A}({x},{y})=0\).

(ii) Assume that x ∈ and y∉ \(\min \limits (\preccurlyeq _{\Psi }, W)\) (the remaining case is analogous). Then ρ_{Ψ}(x,y) = 1. Furthermore, since \(y\notin \min \limits (\preccurlyeq _{\Psi }, [\![{A}]\!])\), by (SPPres\(^{\ast }_{\preccurlyeq }\)) and (SPPres\(^{\div }_{\preccurlyeq }\)), \(\rho _{\Psi \ast _{i} A}({x},{y}) = \rho _{\Psi \div _{j} \neg A}({x},{y}) =1\). By reapplying (SPPres\(^{\ast }_{\preccurlyeq }\)) again, we then obtain \(\rho _{({\Psi } \div _{j} \neg A) \ast _{\mathrm {N}} A}({x},{y})=1\) and we are done.


(b) Assume x∉ and y∉ \(\min \limits (\preccurlyeq _{\Psi }, W)\). We know from the definitions of these operators that, for x,y∉ \(\min \limits (\preccurlyeq _{\Psi }, W)\cup \min \limits (\preccurlyeq _{\Psi }, [\![{A}]\!])\), we have \(\rho _{\Psi \div _{j} \neg A}({x},{y}) = \rho _{\Psi \ast _{i} A}({x},{y})\). Therefore \(\rho _{({\Psi }\div _{j} \neg A)\ast _{\mathrm {N}} A}({x},{y})=\rho _{({\Psi }\ast _{i} A)\ast _{\mathrm {N}} A}({x},{y})\). But, since, by \((\text {KM}^{\ast }_{\preccurlyeq })\), \(\rho _{({\Psi }\ast _{i} A)\ast _{\mathrm {N}} A}({x},{y})\subseteq [\![{A}]\!]\), it follows, by the definition of ∗_{N}, that \(\rho _{({\Psi }\ast _{i} A)\ast _{\mathrm {N}} A}({x},{y}) = \rho _{\Psi \ast _{i} A}({x},{y})\). We can therefore conclude that \(\rho _{\Psi \ast _{i} A}({x},{y})=\rho _{({\Psi }\div _{j} \neg A)\ast _{\mathrm {N}} A}({x},{y})\), as required.

□
Proposition 22
If ∗ is defined from ÷_{P} via \((\mathrm {iLIRC_{\preccurlyeq }})\) then ∗≠∗_{L}.
Proof
We provide a countermodel in Fig. 7. □
Proposition 23
If 〈i,j〉∈{〈L,P〉,〈N,N〉,〈R,R〉} then ∗_{i} and ÷_{j} jointly satisfy \((\mathrm {iLI\ast _{\preccurlyeq }})\).
Proof
Let 〈i,j〉∈{〈L,P〉,〈N,N〉,〈R,R〉}. We need to show that \(\rho _{\Psi \ast _{i} A}({x},{y}) = \rho _{({\Psi }\div _{j} \neg A)\ast _{i} A}({x},{y})\).
Regarding the case in which 〈i,j〉 = 〈L,P〉: Assume ρ_{A}(x,y) = 1. Then, since it is a property of lexicographic revision that, if ρ_{A}(x,y) = 1, then \(\rho _{\Psi \ast _{\mathrm {L}} A}({x},{y})=1\), it follows that \(\rho _{\Psi \ast _{\mathrm {L}} A}({x},{y}) = \rho _{({\Psi }\div _{\mathrm {P}} \neg A)\ast _{\mathrm {L}} A}({x},{y})=1\). Assume ρ_{A}(x,y) = 0. Then, by \((\mathrm {C}{1,2}^{\ast }_{\scriptscriptstyle \preccurlyeq })\), \(\rho _{\Psi \ast _{\mathrm {L}} A}({x},{y}) = \rho _{({\Psi }\div _{\mathrm {P}} \neg A)\ast _{\mathrm {L}} A}({x},{y})=\rho _{\Psi }({x},{y})\).
Regarding the case in which 〈i,j〉∈{〈N,N〉,〈R,R〉}, we divide the proof into two main cases:

(1) Assume x ∈ or y ∈ \(\min \limits (\preccurlyeq _{\Psi }, [\![{A}]\!])\). The proof proceeds as in the corresponding case in the proof of Proposition 21 above.

(2) Assume x,y∉ \(\min \limits (\preccurlyeq _{\Psi }, [\![{A}]\!])\).

(a) Assume x ∈ or y ∈ \(\min \limits (\preccurlyeq _{\Psi }, W)\). Again, the proof proceeds as in the corresponding case in the proof of Proposition 21.

(b) Assume x∉ and y∉ \(\min \limits (\preccurlyeq _{\Psi }, W)\). We know from the definitions of these operators that, for x,y∉ \(\min \limits (\preccurlyeq _{\Psi }, W)\cup \min \limits (\preccurlyeq _{\Psi }, [\![{A}]\!])\), we have \(\rho _{\Psi \div _{j} \neg A}({x},{y}) = \rho _{\Psi \ast _{i} A}({x},{y})\). Since ∗_{R} and ∗_{N} both satisfy \((\text {KM}^{\ast }_{\preccurlyeq })\) and \((\text {IIA}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\) and the latter tell us that ρ_{Ψ}(x,y) and ρ_{A}(x,y) jointly determine ρ_{Ψ∗A}(x,y), it follows that \(\rho _{({\Psi }\div _{j} \neg A)\ast _{i} A}({x},{y})=\rho _{({\Psi }\ast _{i} A)\ast _{i} A}({x},{y})\). For the final step, we note that ∗_{R} and ∗_{N} both satisfy:

\((\text {Idem}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\) ρ_{(Ψ∗A)∗A}(x,y) = ρ_{Ψ∗A}(x,y)
Indeed, in the presence of \((\text {KM}^{\ast }_{\preccurlyeq })\), \((\text {IIAP}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\) tells us that the posterior relative rank ρ_{Ψ∗A}(x,y) of a pair 〈x,y〉 of worlds is determined by the prior relative rank ρ_{Ψ}(x,y), with the nature of this mapping depending, for nonminimal Aworlds, on the relevant proposition and pair of worlds. \((\text {Idem}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\) is then the requirement that, for any given value of ρ_{A}(x,y), if the row for value v points to value w, then the row for w also points to w. That this condition holds for ∗_{R} and ∗_{N} can be seen from Table 2. (Note, in passing, that \((\text {Idem}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\) can more generally be shown to be derivable from \((\text {KM}^{\ast }_{\preccurlyeq })\), \((\text {IIAP}^{\scriptscriptstyle \ast }_{\scriptscriptstyle \preccurlyeq })\) and \((\mathrm {C}{1,2}^{\ast }_{\scriptscriptstyle \preccurlyeq })\)–\((\mathrm {C}{4}^{\ast }_{\scriptscriptstyle \preccurlyeq })\) and is hence also satisfied by ∗_{L}.)
Hence \(\rho _{({\Psi }\ast _{i} A)\ast _{i} A}({x},{y}) = \rho _{\Psi \ast _{i} A}({x},{y})\) and we can conclude that \(\rho _{\Psi \ast _{i} A}({x},{y})=\rho _{({\Psi }\div _{j} \neg A)\ast _{\mathrm {N}} A}({x},{y})\), as required.

□
Proposition 24
Let 〈i,j〉∈{〈L,P〉,〈N,N〉,〈R,R〉}. Then if ÷ defined from ∗_{i} by \(\preccurlyeq _{\Psi \div A} = {\preccurlyeq }_{\Psi }{\oplus _{\mathrm {TQ2}}}{\preccurlyeq }_{{\Psi }{{\ast }_{i}}\neg A}\), then ÷ = ÷_{j}.
Proof
The result is obvious from the definitions of the various elementary contraction operators in Definition 10 and elementary revision operators in Definition 2. ⊕_{TQ2} combination of \(\preccurlyeq _{\Psi }\) and \(\preccurlyeq _{\Psi \ast _{i} \neg A}\) yields the result that the minimal equivalence class under \(\preccurlyeq _{\Psi \div A}\), is given by \(\min \limits (\preccurlyeq _{\Psi }, W)\cup \min \limits (\preccurlyeq _{\Psi \ast _{i} \neg A},W)\), which is equal to \( \min \limits (\preccurlyeq _{\Psi \div _{j} \neg A},W)\). Regarding the subsequent equivalence classes, ⊕_{TQ2} combination gives us the result that, for x∉ and \(y \notin \min \limits (\preccurlyeq _{\Psi }, W)\cup \min \limits (\preccurlyeq _{\Psi \ast _{i} \neg A},W)\), \(\rho _{\Psi \div A}({x},{y})=\rho _{\Psi \ast _{i} \neg A}({x},{y})\). But as we noted in Section 5.1, for the same x and y, we have \(\rho _{\Psi \div _{j} A}({x},{y})=\rho _{\Psi \ast _{i} \neg A}({x},{y})\) and so \(\rho _{\Psi \div A}({x},{y})=\rho _{\Psi \div _{j} A}({x},{y})\). □
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Chandler, J., Booth, R. Elementary Belief Revision Operators. J Philos Logic 52, 267–311 (2023). https://doi.org/10.1007/s10992022096726
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DOI: https://doi.org/10.1007/s10992022096726
Keywords
 Belief revision
 Belief contraction
 Iterated belief change
 Irrelevant alternatives
 Social choice