Blockage Revision

Abstract

Blockage revision is a version of descriptor revision, i.e. belief change in which a belief set K is changed with inputs whose success conditions are metalinguistic expressions containing the belief predicate $\mathfrak {B}$. This is a highly general framework that allows a single revision operator $\circ$ to take inputs corresponding to sentential revision ($K\circ \mathfrak {B}p$), contraction ($K\circ \lnot \mathfrak {B}p$) as well as more complex and composite operations. In blockage revision, such an operation is based on a relation $\rightharpoondown$ of blockage among the set of potential outcomes. $X\rightharpoondown Y$ signifies that if X satisfies the success condition of a belief change, then Y cannot be its outcome. The properties of blockage revision are investigated, and conditions on the blocking relation are specified that characterize various properties of the resulting operation of change.

Appendix: Proofs

Definition 2

(Hansson 2014a) For any belief set X, we denote by $\Pi _X$ the descriptor $\{\mathfrak {B}p \mid p \in X\} \cup \{\lnot \mathfrak {B}p \mid p \notin X\}$.

Definition 3

For any set $\mathbb {X}$ of belief sets and any descriptor $\Psi$:

$$\begin{aligned} \llbracket \Psi \rrbracket _\mathbb {X}=\{X\in \mathbb {X}\mid X\Vdash \Psi \}. \end{aligned}$$

The index can be omitted if no ambiguity follows from doing so, i.e. we can then write $\llbracket \Psi \rrbracket$ instead of $\llbracket \Psi \rrbracket _\mathbb {X}$.

Lemma 1

If a binary relation satisfies non-occlusion then it satisfies acyclicity. If it satisfies acyclicity then it satisfies asymmetry. If it satisfies asymmetry then it satisfies irreflexivity.

Proof of Lemma 1:

Left to the reader. $\square$

Lemma 2

$K \Vdash \Psi \veebar \Xi$ holds if and only if either $K \Vdash \Psi$ or $K \Vdash \Xi$ holds.

Proof of Lemma 2:

See Hansson (2014b, p. 85). $\square$

Proof of Observation 1:

For one direction, let $K\in \mathbb {X}$ and let $\rightharpoondown$ satisfy irreflexivity within $\mathbb {X}\setminus \{K\}$. It follows directly from $K\in \mathbb {X}$ and Definition 1 that the outcome set is a subset of $\mathbb {X}$. We also have to show that each element of $\mathbb {X}$ is an element of the outcome set. For each $X\in \mathbb {X}\setminus \{K\}$, $\{X\}$ is the set of $\Pi _X$-satisfying elements of $\mathbb {X}$, and due to the irreflexivity of $\rightharpoondown , X$ is the unique unblocked element of $\{X\}$, thus $K\circ \Pi _X=X$. Furthermore, $K=K\circ \lnot \mathfrak {B}{\scriptstyle \top }$ (where ${\scriptstyle \top }$ is a tautology) due to clause (ii) of Definition 1. Thus all elements of $\mathbb {X}$ are elements of the outcome set.

For the other direction, we assume that either (i) $K\notin \mathbb {X}$ or (ii) $X\rightharpoondown X$ for some $X\in \mathbb {X}\setminus \{K\}$. In the first case it follows from $K\circ \lnot \mathfrak {B}{\scriptstyle \top }=K$ that the outcome set has an element that is not in $\mathbb {X}$. In the second case $K\circ \Psi \ne X$ for all $\Psi$, thus X is not in the outcome set although it is in $\mathbb {X}$. $\square$

Proof of Observation 2:

Part 1: Let $\mathbb {X}=\{K,X,Y\}$ and:

$$\begin{aligned} \rightharpoondown =\{\langle K,X\rangle ,\langle K,Y\rangle ,\langle X,Y\rangle ,\langle Y,X\rangle \}. \end{aligned}$$

Then $K\circ \Pi _X\Vdash \Pi _X\veebar \Pi _Y$ but $K\circ (\Pi _X\veebar \Pi _Y)\nVdash \Pi _X\veebar \Pi _Y$.

Part 2: Let $\mathbb {X}=\{K,X,Y\}$ and $\rightharpoondown =\{\langle K,X\rangle ,\langle X,Y\rangle ,\langle Y,K\rangle \}$. Then $K\circ (\Pi _K\veebar \Pi _X\veebar \Pi _Y)=K$ and $K\circ ((\Pi _K\veebar \Pi _X\veebar \Pi _Y)\cup (\Pi _K\veebar \Pi _Y))=K\circ (\Pi _K\veebar \Pi _Y)=Y$, thus $K\circ (\Pi _K\veebar \Pi _X\veebar \Pi _Y)\Vdash \Pi _K\veebar \Pi _Y$ but $K\circ (\Pi _K\veebar \Pi _X\veebar \Pi _Y)\ne K\circ ((\Pi _K\veebar \Pi _X\veebar \Pi _Y)\cup (\Pi _K\veebar \Pi _Y))$.

Part 3: Let $\mathbb {X}=\{K,X,Y\}$ and $\rightharpoondown =\{\langle K,X\rangle ,\langle X,Y\rangle ,\langle Y,K\rangle \}$. Then $K\circ (\Pi _K\veebar \Pi _X\veebar \Pi _Y)\Vdash \Pi _K\veebar \Pi _Y$ and $K\circ (\Pi _K\veebar \Pi _Y)\Vdash \Pi _K\veebar \Pi _X\veebar \Pi _Y$ but $K\circ (\Pi _K\veebar \Pi _X\veebar \Pi _Y)\ne K\circ (\Pi _K\veebar \Pi _Y)$.

Part 4: Let $\mathbb {X}=\{K,X\}$ and let $\rightharpoondown =\{\langle X,K\rangle \}$. Then $K\Vdash \Pi _K\veebar \Pi _X$ but $K\circ (\Pi _K\veebar \Pi _X)=X$. $\square$

Proof of Observation 3:

Part 1, peripheral cumulativity: Let $K\circ \Psi \ne K\ne K\circ (\Psi \cup \Xi )$ and $K\circ \Psi \Vdash \Xi$. It follows from Definition 1 and $K\ne K\circ \Psi$ that $K\circ \Psi \Vdash \Psi$, thus $K\circ \Psi \Vdash \Psi \cup \Xi$, i.e. $K\circ \Psi \in \llbracket \Psi \cup \Xi \rrbracket$. It also follows from $K\ne K\circ \Psi$ that $K\circ \Psi$ is unblocked within $\llbracket \Psi \rrbracket$, and since $\llbracket \Psi \cup \Xi \rrbracket \subseteq \llbracket \Psi \rrbracket$ it is then also unblocked within $\llbracket \Psi \cup \Xi \rrbracket$. Thus $K\circ \Psi$ is $\Psi \cup \Xi$-satisfying and unblocked within $\llbracket \Psi \cup \Xi \rrbracket$. Since $K\ne K\circ (\Psi \cup \Xi )$ there is exactly one belief set with that property, namely $K\circ (\Psi \cup \Xi )$. It follows that $K\circ \Psi =K\circ (\Psi \cup \Xi )$.

Part 1, peripheral disjunctive identity: Since $K\ne K\circ \Psi$, all elements of $\llbracket \Psi \rrbracket \setminus \{K\circ \Psi \}$ are blocked within $\llbracket \Psi \rrbracket$, and similarly all elements of $\llbracket \Xi \rrbracket \setminus \{K\circ \Psi \}$ are blocked within $\llbracket \Xi \rrbracket$. Since $\llbracket \Psi \veebar \Xi \rrbracket =\llbracket \Psi \rrbracket \cup \llbracket \Xi \rrbracket$, all elements of $\llbracket \Psi \veebar \Xi \rrbracket \setminus \{K\circ \Psi \}$ are blocked within $\llbracket \Psi \veebar \Xi \rrbracket$. Since $K\circ \Psi$ is unblocked both within $\llbracket \Psi \rrbracket$ and within $\llbracket \Xi \rrbracket$, it is unblocked within $\llbracket \Psi \veebar \Xi \rrbracket$. Thus $K\circ \Psi$ is the only unblocked element within $\llbracket \Psi \veebar \Xi \rrbracket$, thus $K\circ \Psi =K\circ (\Psi \veebar \Xi )$.

Part 2, peripheral cumulativity: Let $\mathbb {X}=\{K,X,Y,Z\}$ and let $\circ$ be based on a monoselective choice function $\widehat{C}$ such that $\widehat{C}(\{X,Y,Z\})=\{Y\}$ and $\widehat{C}(\{Y,Z\})=\{Z\}$. Then $K\circ (\Pi _X\veebar \Pi _Y\veebar \Pi _Z)=Y$ and $Y\Vdash \Pi _Y\veebar \Pi _Z$ but $K\circ ((\Pi _X\veebar \Pi _Y\veebar \Pi _Z)\cup (\Pi _Y\veebar \Pi _Z))=K\circ (\Pi _Y\veebar \Pi _Z)=Z$.

Part 2, peripheral disjunctive identity: Let $\mathbb {X}=\{K,X,Y,Z\}$ and let $\circ$ be based on a monoselective choice function $\widehat{C}$ such that $\widehat{C}(\{X,Y\})=\widehat{C}(\{Y,Z\})=Y$ and $\widehat{C}(\{X,Y,Z\})=X$. Then $K\circ (\Pi _X\veebar \Pi _Y) = Y, K\circ (\Pi _Y\veebar \Pi _Z) = Y$, and $K\circ ((\Pi _X\veebar \Pi _Y)\veebar (\Pi _Y\veebar \Pi _Z)) = X$. $\square$

Proof of Theorem 1:

From regularity to peripheral non-occlusion: Let $K\notin \llbracket \Psi \rrbracket$ and $X\in \llbracket \Psi \rrbracket$. Due to Observation 1 and our assumption that $\rightharpoondown$ is irreflexive there is some $\Xi$ with $X=K\circ \Xi$. Then $K\circ \Xi \Vdash \Psi$ and regularity yields $K\circ \Psi \Vdash \Psi$. From this and $K\circ \Psi \ne K$ it follows according to Definition 1 that $K\circ \Psi$ is an unblocked element within $\llbracket \Psi \rrbracket$.

From regularity to peripheral weak connectivity: Let $X, Y \in \mathbb {X}, X\ne Y$ and $X\ne K\ne Y$. Due to Observation 1 and our assumption that $\rightharpoondown$ is irreflexive there is some $\Xi$ with $X=K\circ \Xi$. Thus $K\circ \Xi \Vdash \Pi _X\veebar \Pi _Y$. Regularity yields $K\circ (\Pi _X\veebar \Pi _Y)\Vdash \Pi _X\veebar \Pi _Y$, thus $K\circ (\Pi _X\veebar \Pi _Y)\in \{X,Y\}$, from which it follows that either $X\rightharpoondown Y$ or $Y\rightharpoondown X$.

From peripheral non-occlusion and peripheral weak connectivity to regularity: Let $K\circ \Xi \Vdash \Psi$.

First case, $K\nVdash \Psi$: Due to peripheral non-occlusion it follows from $K\circ \Xi \Vdash \Psi$ and $K\nVdash \Psi$ that $\llbracket \Psi \rrbracket$ has at least one non-blocked element. It follows from peripheral weak connectivity that it has at most one such element. Due to clause (i) of Definition 1, that element is equal to $K\circ \Psi$, thus $K\circ \Psi \Vdash \Psi$.

Second case, $K\Vdash \Psi$: According to Definition 1, $K\circ \Psi$ is either an element of $\llbracket \Psi \rrbracket$ or equal to K. In both cases, $K\circ \Psi \Vdash \Psi$. $\square$

Proof of Observation 4:

For one direction, let $\circ$ be a monoselective descriptor revision. Then $\circ$ satisfies regularity, and we can conclude from Theorem 1 that $\rightharpoondown$ satisfies peripheral non-occlusion and peripheral weak connectivity.

For the other direction, let $\rightharpoondown$ satisfy peripheral non-occlusion and peripheral weak connectivity, and let $\circ$ be the revision operator generated from $\rightharpoondown$. It follows from Theorem 1 that $\circ$ satisfies regularity. Let $\mathbb {X}=\{X\mid (\exists \Psi )(X=K\circ \Psi )\}$ and let $\widehat{C}$ be a function such that (a) when $\varnothing \ne \mathbb {Y}\subseteq \mathbb {X}$, then $\widehat{C}(\mathbb {Y})=K\circ \Psi$ for all $\Psi$ such that $\{X\mid X\Vdash \Psi \}=\mathbb {Y}$ and (b) otherwise $\widehat{C}(\mathbb {Y})$ is undefined. We need to show (1) that $\widehat{C}$ is a monoselective choice function on $\mathbb {X}$, and (2) that $\circ$ is the monoselective descriptor revision on $\mathbb {X}$ that is based on $\widehat{C}$.

For (1) it is sufficient to show that $\widehat{C}$ is indeed a function. This can be done by noting that since $\circ$ is a blockage relation it satisfies the uniformity postulate:

If $K\circ \Xi \Vdash \Psi$ iff $K\circ \Xi \Vdash \Psi '$ for all $\Xi$, then $K\circ \Psi =K\circ \Psi '$

For (2) we need to consider the two cases referred to in Sect. 1. In case (i) there is some $X \in \mathbb {X}$ with $X \Vdash \Psi$. It follows directly from regularity that $K\circ \Psi$ is in this case indeed identical to $\widehat{C}(\{X\in \mathbb {X}\mid X\Vdash \Psi \})$. In case (ii) there is no $X \in \mathbb {X}$ with $X \Vdash \Psi$. It follows from Definition 1 that $K\circ \Psi =K$, as required. $\square$

Proof of Theorem 2:

From cumulativity to peripheral non-occlusion: Suppose to the contrary that $K\notin \llbracket \Psi \rrbracket \ne \varnothing$ and $\llbracket \Psi \rrbracket$ has no unblocked element. Let $X\in \llbracket \Psi \rrbracket$. Then $K\circ \Psi =K, K\circ \Psi \Vdash \Pi _K\veebar \Pi _X$, and $K\circ (\Psi \cup (\Pi _K\veebar \Pi _X))=K\circ \Pi _X=X$, contrary to cumulativity.

From cumulativity to peripheral weak connectivity: Suppose to the contrary that there are $X,Y\in \mathbb {X}$ such that $X\ne Y\ne K\ne X$ and . Then $K\circ (\Pi _X\veebar \Pi _Y)=K$, thus $K\circ (\Pi _X\veebar \Pi _Y)\Vdash \Pi _K\veebar \Pi _Y$, but $K\circ ((\Pi _X\veebar \Pi _Y)\cup (\Pi _K\veebar \Pi _Y))=K\circ \Pi _Y=Y$, contrary to cumulativity.

From cumulativity to top adjacency: Let $\circ$ be a blockage revision based on $\rightharpoondown$, and furthermore suppose that top adjacency does not hold. We are going to show that cumulativity does not hold. Let $X,Y\in \mathbb {X}\setminus \{K\}$ and $X\rightharpoondown Y\rightharpoondown K$. Since top adjacency does not hold, if $K\rightharpoondown X$ then , and furthermore, if then . We therefore have the following two cases:

Case 1, $K\rightharpoondown X$ and : Then there is no unblocked element within $\{K,X,Y\}$ but there is a unique unblocked element within $\{K,Y\}$, namely Y. It follows that $K\circ (\Pi _K\veebar \Pi _X\veebar \Pi _Y)=K$, thus $K\circ (\Pi _K\veebar \Pi _X\veebar \Pi _Y)\Vdash \Pi _K\veebar \Pi _Y$, but $K\circ ((\Pi _K\veebar \Pi _X\veebar \Pi _Y)\cup (\Pi _K\veebar \Pi _Y))=Y$, contrary to cumulativity.

Case 2, and : We have $X\rightharpoondown Y$ and by applying peripheral non-occlusion (that we have just proved) to $\{X,Y\}$ we obtain . Thus $\{K,X,Y\}$ has a unique unblocked element namely X, whereas $\{K,X\}$ has the unblocked elements K and X. It follows that $K\circ (\Pi _K\veebar \Pi _X\veebar \Pi _Y)=X$, thus $K\circ (\Pi _K\veebar \Pi _X\veebar \Pi _Y)\Vdash \Pi _K\veebar \Pi _X$, but $K\circ ((\Pi _K\veebar \Pi _X\veebar \Pi _Y)\cup (\Pi _K\veebar \Pi _X))=K$, contrary to cumulativity.

Thus in both cases, if top adjacency does not hold, then neither does cumulativity.

From peripheral non-occlusion, peripheral weak connectivity, and top adjacency to cumulativity: Let $K\circ \Psi \Vdash \Xi$.

Case 1, $\llbracket \Psi \rrbracket =\varnothing$: Then $\llbracket \Psi \cup \Xi \rrbracket =\varnothing$, and we have $K\circ \Psi =K$ and $K\circ (\Psi \cup \Xi )=K$.

Case 2, $K\notin \llbracket \Psi \rrbracket \ne \varnothing$: It follows from peripheral non-occlusion and peripheral weak connectivity that $\llbracket \Psi \rrbracket$ has exactly one unblocked element, and due to Definition 1 that element is equal to $K\circ \Psi$. It follows that $K\circ \Psi \Vdash \Psi$ and we already have $K\circ \Psi \Vdash \Xi$, so $K\circ \Psi \Vdash \Psi \cup \Xi$. Since $K\circ \Psi$ is unblocked within $\llbracket \Psi \rrbracket$, it is also unblocked within its subset $\llbracket \Psi \cup \Xi \rrbracket$. It follows from $K\notin \llbracket \Psi \rrbracket$ and $\llbracket \Psi \cup \Xi \rrbracket \subseteq \llbracket \Psi \rrbracket$ that $K\notin \llbracket \Psi \cup \Xi \rrbracket$. It follows from peripheral non-occlusion and peripheral weak connectivity that $\llbracket \Psi \cup \Xi \rrbracket$ has exactly one unblocked element, and then $K\circ \Psi$ is that element. Due to Definition 1, $K\circ \Psi =K\circ (\Psi \cup \Xi )$.

Case 3, $K\in \llbracket \Psi \rrbracket$ and $K\ne K\circ \Psi$: It follows from Definition 1 that $K\circ \Psi$ is the unique unblocked element of $\llbracket \Psi \rrbracket$. Since $K\circ \Psi \Vdash \Xi$ we have $K\circ \Psi \in \llbracket \Psi \cup \Xi \rrbracket$. Since $K\circ \Psi$ is unblocked within $\llbracket \Psi \rrbracket$, it is also unblocked within its subset $\llbracket \Psi \cup \Xi \rrbracket$.

Case 3A, $K\notin \llbracket \Psi \cup \Xi \rrbracket$: It follows from peripheral non-occlusion and peripheral weak connectivity that $\llbracket \Psi \cup \Xi \rrbracket$ has exactly one unblocked element, and then $K\circ \Psi$ is that element. Due to Definition 1, $K\circ \Psi =K\circ (\Psi \cup \Xi )$.

Case 3B, $K\in \llbracket \Psi \cup \Xi \rrbracket$: Since $K\circ \Psi$ is the unique unblocked element of $\llbracket \Psi \rrbracket$, there is some $X\in \llbracket \Psi \rrbracket$ with $X\rightharpoondown K$. Since $K\circ \Psi$ is unblocked within $\llbracket \Psi \rrbracket$ we also have and . Due to peripheral weak connectivity, $K\circ \Psi \rightharpoondown X$. We conclude from top adjacency that $K\circ \Psi \rightharpoondown K$.

Next, let $Z\in \llbracket \Psi \cup \Xi \rrbracket \setminus \{K,K\circ \Psi \}$. Since $K\circ \Psi$ is unblocked within $\llbracket \Psi \cup \Xi \rrbracket$ it follows from peripheral weak connectivity that $K\circ \Psi \rightharpoondown Z$. Thus $K\circ \Psi$ is the only unblocked element within $\llbracket \Psi \cup \Xi \rrbracket$, thus $K\circ \Psi =K\circ (\Psi \cup \Xi )$.

Case 4, $K\in \llbracket \Psi \rrbracket$ and $K=K\circ \Psi$: It follows from $K\circ \Psi \Vdash \Xi$ that $K\in \llbracket \Psi \cup \Xi \rrbracket$.

Case 4A, K is unblocked within $\llbracket \Psi \cup \Xi \rrbracket$: If K is the only unblocked element within $\llbracket \Psi \cup \Xi \rrbracket$, then $K=K\circ (\Psi \cup \Xi )$ due to clause (i) of Definition 1. If it is one of at least two unblocked elements within $\llbracket \Psi \cup \Xi \rrbracket$, then $K=K\circ (\Psi \cup \Xi )$ due to clause (ii) of the same definition.

Case 4B, K is blocked within $\llbracket \Psi \cup \Xi \rrbracket$: Due to peripheral non-occlusion and peripheral weak connectivity there is some $Y\in \llbracket \Psi \cup \Xi \rrbracket \setminus \{K\}$ such that for all $X\in \llbracket \Psi \cup \Xi \rrbracket \setminus \{K,Y\}$. It follows that $K\circ (\Psi \cup \Xi )$ is either Y or K. We are going to show that it is not Y. Suppose that it is. Then clearly .

Case 4Ba, : Since K is blocked within $\llbracket \Psi \cup \Xi \rrbracket$ there is then some $X\in \llbracket \Psi \cup \Xi \rrbracket \setminus \{K,Y\}$ such that $X\rightharpoondown K$. We then have and $X\rightharpoondown K$, contrary to top adjacency.

Case 4Bb, $Y\rightharpoondown K$: Due to peripheral non-occlusion and peripheral weak connectivity there is some $Z\in \llbracket \Psi \rrbracket \setminus \{K\}$ such that for all $V\in \llbracket \Psi \rrbracket \setminus \{K,Z\}$. Since K is blocked (within $\llbracket \Psi \cup \Xi \rrbracket$ and therefore also) within $\llbracket \Psi \rrbracket$ and $K\circ \Psi \ne Z, Z$ is blocked within $\llbracket \Psi \rrbracket$, thus $K\rightharpoondown Z$. Since by assumption . We then have and $K\rightharpoondown Z$, contrary to top adjacency.

Thus, in neither subcase can $K\circ (\Psi \cup \Xi )$ be equal to Y. We can conclude that $K\circ (\Psi \cup \Xi )=K$, thus $K\circ \Psi =K\circ (\Psi \cup \Xi )$ in case 4B as well. $\square$

Proof of Theorem 3:

From cumulativity to regularity: Directly from Theorems 1 and 2.

From cumulativity to reciprocity : Let $K \circ \Psi \Vdash \Xi$ and $K \circ \Xi \Vdash \Psi$. Then cumulativity yields $K \circ \Psi = K \circ (\Psi \cup \Xi ) = K \circ \Xi$.

From regularity and reciprocity to cumulativity: Let $K \circ \Psi \Vdash \Xi$. There are two cases.

Case (i), $K \circ \Psi \nVdash \Psi$: Regularity yields $K \circ (\Psi \cup \Xi ) \nVdash \Psi$, thus $K \circ (\Psi \cup \Xi ) \nVdash \Psi \cup \Xi$. It follows from Definition 1 that $K \circ \Psi = K = K \circ (\Psi \cup \Xi )$.

Case (ii), $K \circ \Psi \Vdash \Psi$: Then $K \circ \Psi \Vdash \Psi \cup \Xi$. Regularity yields $K \circ (\Psi \cup \Xi ) \Vdash \Psi \cup \Xi$. We thus have $K \circ \Psi \Vdash \Psi \cup \Xi$ and $K \circ (\Psi \cup \Xi ) \Vdash \Psi$, and reciprocity yields $K \circ \Psi = K\circ ( \Psi \cup \Xi )$. $\square$

Proof of Observation 5:

Part 1: Let $\mathbb {X}=\{K,X,Y\}$ and $\rightharpoondown =\{\langle X,Y\rangle , \langle Y,K\rangle \}$. It follows from Theorem 1 that regularity is satisfied. We have $K\circ (\Pi _K\veebar \Pi _X\veebar \Pi _Y) = X$ and $K\circ (\Pi _K\veebar \Pi _X)=K$, thus $K\circ (\Pi _K\veebar \Pi _X\veebar \Pi _Y)\Vdash \Pi _K\veebar \Pi _X$ and $K\circ (\Pi _K\veebar \Pi _X)\Vdash \Pi _K\veebar \Pi _X\veebar \Pi _Y$ but $K\circ (\Pi _K\veebar \Pi _X\veebar \Pi _Y) \ne K\circ (\Pi _K\veebar \Pi _X)$, which shows that reciprocity does not hold.

Part 2: Let $\mathbb {X}=\{K,X,Y\}$ and $\rightharpoondown =\{\langle K,X\rangle , \langle K,Y\rangle \}$. In order to show that reciprocity holds it is sufficient to that there are no $\Psi$ and $\Xi$ such that either (1) $K\circ \Psi =K, K\Vdash \Xi , K\circ \Xi =X$, and $X\Vdash \Psi$, (2) $K\circ \Psi =K, K\Vdash \Xi , K\circ \Xi =Y$, and $Y\Vdash \Psi$, or (3) $K\circ \Psi =X, X\Vdash \Xi , K\circ \Xi =Y$, and $Y\Vdash \Psi$.

Suppose that (1) holds. Due to our construction of $\circ$ it follows from $K\circ \Xi =X$ that $X\Vdash \Xi , K\nVdash \Xi$, and $Y\nVdash \Xi$. But we also have $K\Vdash \Xi$, thus (1) does not hold. A symmetrical proof shows that (2) does not hold. Suppose that (3) holds. It then follows from $K\circ \Psi =X$ that $X\Vdash \Psi , K\nVdash \Psi$, and $Y\nVdash \Psi$, but we also have $Y\Vdash \Psi$, so that this case is impossible as well. Thus there are no $\Psi$ and $\Xi$ that satisfy either (1), (2), or (3), thus reciprocity holds. It follows from $K\circ \Pi _X=X, X\Vdash \Pi _X\veebar \Pi _Y$, and $K\circ (\Pi _X\veebar \Pi _Y)=K$ that regularity does not hold. $\square$

Proof of Theorem 4:

From confirmation to near-superiority: Let $X\rightharpoondown K$. Confirmation yields $K\circ (\Pi _K\veebar \Pi _X)=K$, which would not be the case if . Thus $K\rightharpoondown X$.

If also follows from confirmation that $K\circ (\Pi _K\veebar \Pi _X\veebar \Pi _Y)=K$. Since $X\rightharpoondown K, K\circ (\Pi _K\veebar \Pi _X\veebar \Pi _Y)$ cannot follow from clause (i) of Definition 1, so it must be based on clause (ii). We already have $K\rightharpoondown X$, thus Y must be blocked by either K or X.

From near-superiority to confirmation: Let $K\Vdash \Psi$. If $\llbracket \Psi \rrbracket$ does not have exactly one unblocked element, then clause (ii) of Definition 1 yields $K\circ \Psi =K$. It remains to treat the case when $\llbracket \Psi \rrbracket$ has exactly one unblocked element. Suppose that element is not K. Then there is some $X\in \llbracket \Psi \rrbracket$ with $X\rightharpoondown K$. It follows from near-superiority that all elements of $\llbracket \Psi \rrbracket$ are blocked (either by K or by X). This contradicts the assumption that the only unblocked element of $\llbracket \Psi \rrbracket$ is not K. We conclude that it is K and that therefore $K\circ \Psi =K$ in this case as well. $\square$

Proof of Observation 6:

Parts 1 and 2 are left to the reader.

Part 3: Let $\mathbb {X}=\{K,X,Y\}$ and $\rightharpoondown =\{\langle X,Y\rangle , \langle X,K\rangle , \langle Y,K\rangle \}$.

Part 4: Let $\mathbb {X}=\{K,X,Y\}$ and $\rightharpoondown =\{\langle X,Y\rangle , \langle Y,X\rangle , \langle Y,K\rangle , \langle K,Y\rangle \}$.

Proof of Observation 7:

Let $\mathbb {X}=\{K,X,Y\}$ and $\rightharpoondown =\{\langle X,Y\rangle , \langle X,K\rangle , \langle Y,K\rangle \}$. It follows from Theorem 2 that $\circ$ satisfies cumulativity and from $K\circ (\Pi _K\veebar \Pi _X)=X$ that it does not satisfy confirmation. $\square$

Proof of Observation 8:

Part 1: Let $\mathbb {X}=\{K,X,Y\}$ and $\rightharpoondown =\{\langle K,X\rangle , \langle K,Y\rangle \}$. It follows from Theorem 4 that $\circ$ satisfies confirmation and from $K\circ \Pi _X=X, X\Vdash \Pi _X\veebar \Pi _Y$, and $K\circ (\Pi _X\veebar \Pi _Y)=K$ that it does not satisfy regularity.

Part 2: Let $\mathbb {X}=\{K,X,Y,Z,V\}$ and:

$$\begin{aligned} \rightharpoondown =\{\langle K,X\rangle , \langle K,Y\rangle , \langle K,Z\rangle , \langle K,V\rangle ,\langle X,Z\rangle ,\langle Z,Y\rangle ,\langle Y,V\rangle ,\langle V,X\rangle \}. \end{aligned}$$

It follows from Theorem 4 that $\circ$ satisfies confirmation. However, $K\circ (\Pi _X\veebar \Pi _Y\veebar \Pi _Z)=X$ and $K\circ (\Pi _X\veebar \Pi _Y\veebar \Pi _V)=Y$. Since $K\circ (\Pi _X\veebar \Pi _Y\veebar \Pi _Z)\Vdash \Pi _X\veebar \Pi _Y\veebar \Pi _V$ and $K\circ (\Pi _X\veebar \Pi _Y\veebar \Pi _V)\Vdash \Pi _X\veebar \Pi _Y\veebar \Pi _Z$, it follows that reciprocity is not satisfied. $\square$

Proof of Theorem 5:

From relational revision to blockage revision: Let $\rightharpoondown$ be the strict part of $\leqq$. Note that near-superiority and top adjacency are satisfied vacuously. (Since $\leqq$ is antisymmetric, $\rightharpoondown$ is irreflexive. Since it also has the property that $K\leqq X$ for all $X\in \mathbb {X}$, there is no X with $X\rightharpoondown K$.)

From blockage revision to relational revision: It was shown in Hansson (2014a) that $\circ$ is a relational descriptor revision if it satisfies extensionality (If $\Psi \dashv \Vdash \Psi '$ then $K \circ \Psi = K \circ \Psi '$), closure ($K\circ \Psi = \mathrm {Cn}(K\circ \Psi )$), relative success ($K \circ \Psi \Vdash \Psi$ or $K \circ \Psi = K$), regularity, cumulativity, and confirmation. It follows straight-forwardly from Definition 1 that the first three of these conditions are satisfied. That the last three are satisfied follows from Theorems 1, 2, and 4. $\square$