Selective Base Revisions

Abstract

Belief Revision addresses the problem of rationally incorporating pieces of new information into an agent’s belief state. In the AGM paradigm, the most used framework in Belief Revision, primacy is given to the new information, which is fully incorporated into the agent’s belief state. However, in real situations, one may want to reject the new information or only accept a part of it. A constructive model called Selective Revision was proposed to meet this need but, as in the AGM framework, focused on belief sets (sets closed under logical consequence). In this paper we adapt the selective revision operators, that were proposed for belief sets, to the belief base context, obtaining a model in which an agent’s epistemic state is represented by a belief base and that allows the acceptance of only part of the new information. We present several representation theorems for selective base revision operators based on different base revision operators.

Appendix Proofs

Lemma 1

Let A be a belief base and $\circledast$ be a selective revision operator on A based on a revision operator ∗¹ and a transformation function f that satisfies lower boundary. If $\alpha \in A \circledast \alpha$, then f(α) = α.

Proof

Assume that $\alpha \in A \circledast \alpha$. Hence α ∈ A ∗ f(α). By ∗ inclusion it follows that α ∈ A ∪{f(α)}. Thus either α ∈ A or α = f(α). It holds that f satisfies lower boundary, thus in both cases it holds that f(α) = α. □

Proof of Observation 7

Statements 1, 2 follow trivially. 3. Assume that $\alpha \in A \circledast \alpha$. Let $\beta \in A\setminus (A \circledast \alpha )$. By weak relevance there exists some $A^{\prime }$ such that $A\circledast \alpha \subseteq A^{\prime }\subseteq A\cup \{\alpha \}, A^{\prime }\not \vdash \perp$ but $A^{\prime }\cup \{\beta \}\vdash \perp$. It holds that $\alpha \in A^{\prime }$. Let $B=A^{\prime }\setminus \{\alpha \}$. Hence $B\subseteq A$. On the other hand, since $A^{\prime }\not \vdash \perp$ it follows that B ∪{α}⊯ ⊥. Thus B⊯¬α. From $A^{\prime }\cup \{\beta \}\vdash \perp$ it follows that (B ∪{α}) ∪{β}⊩⊥. Hence, by deduction B ∪{β}⊩¬α. 4. Assume that $\alpha \in A \circledast \alpha$. Let $\beta \in A\setminus (A \circledast \alpha )$. Assume by reductio ad absurdum that $A\circledast \alpha \vdash \neg \alpha \vee \beta$. Hence $A\circledast \alpha \vdash \beta$. By weak relevance, there exists some $A^{\prime }$ such that $A \circledast \alpha \subseteq A^{\prime }\subseteq A \cup \{\alpha \}, A^{\prime }\not \vdash \perp$ but $A^{\prime }\cup \{\beta \}\vdash \perp$. Contradiction, since every set that contains $A\circledast \alpha$ implies β. 5. Assume that $\alpha \in A \circledast \alpha$. Let $\beta \in A\cap Cn(A\circledast \alpha \cap A)$ and assume by reductio ad absurdum that $\beta \not \in A\circledast \alpha$. Then, by weak disjunctive elimination, $A\circledast \alpha \not \vdash \neg \alpha \vee \beta$. On the other hand, from $\beta \in Cn(A\circledast \alpha \cap A)$ it follows by monotony of Cn that $A\circledast \alpha \vdash \beta$. Thus $A\circledast \alpha \vdash \neg \alpha \vee \beta$. Contradiction. Therefore, $\beta \in A\circledast \alpha$. Hence $A\cap Cn(A\circledast \alpha \cap A) \subseteq A\circledast \alpha$. 6. Consider that $\alpha \in A \circledast \alpha$ and A⊯¬α. Assume by reductio ad absurdum that $A\cup \{\alpha \}\not \subseteq A\circledast \alpha$. From $\alpha \in A\circledast \alpha$ it follows that there exists some β such that β ∈ A and $\beta \not \in A\circledast \alpha$. Thus, by weak core-retainment, there exists $A^{\prime }\subseteq A$ such that $A^{\prime }\not \vdash \neg \alpha$ and $A^{\prime }\cup \{\beta \}\vdash \neg \alpha$. Contradiction, since $A^{\prime }\cup \{\beta \}\subseteq A$ and A⊯¬α. Therefore, $A\cup \{\alpha \} \subseteq A\circledast \alpha$. □

Proof of Observation 8

1. Stability: Let α ∈ A. f satisfies lower boundary. Thus f(α) = α. Hence $A \circledast \alpha =A * f(\alpha )=A * \alpha$. Thus, by ∗ success it follows that $\alpha \in A \circledast \alpha$. Weak inclusion: Let $\alpha \in A \circledast \alpha$. Thus f(α) = α (by Lemma 1). Hence $A \circledast \alpha =A * f(\alpha )=A * \alpha$. From which it follows by ∗ inclusion, that $A \circledast \alpha \subseteq A \cup \{\alpha \}$. 2. Let A ⊩ α. f satisfies strong lower boundary. Thus f(α) = α. Hence $A \circledast \alpha =A * f(\alpha )=A * \alpha$. Thus by ∗ success it follows that $\alpha \in A \circledast \alpha$. 3. Let α⊯ ⊥. Hence ⊯¬α. f satisfies consistency preservation. Thus ⊯¬f(α). Hence f(α)⊯ ⊥. Therefore, by ∗ consistency, it follows that A∗f(α)⊯ ⊥, from which it follows by definition of $\circledast$ that $A \circledast \alpha \not \vdash \perp$. 4. It holds that ⊯¬f(α). Hence f(α)⊯ ⊥, from which it follows, by ∗ consistency, that A ∗ f(α)⊯ ⊥. Thus $A\circledast \alpha \not \vdash \perp$. 5. Assume that A⊯ ⊥. It holds that f satisfies inherited consistency, thus ⊯¬f(α). By definition of $\circledast$ it follows that $A \circledast \alpha =A {*} f(\alpha )$. From which it follows by ∗ consistency that $A \circledast \alpha \not \vdash \perp$. 6. Assume that A⊯¬α and $\alpha \in A \circledast \alpha$. From the latter it follows, by Lemma 1, that f(α) = α. Thus A⊯¬f(α). By ∗ vacuity it follows that $A{*} f(\alpha )\subseteq A\cup \{f(\alpha )\}$. Thus $A\circledast \alpha \subseteq A\cup \{\alpha \}$. 7. Assume that A⊯¬α. f satisfies weak maximality, thus f(α) = α. Hence, by ∗ vacuity, it follows that $A\cup \{\alpha \}\subseteq A * \alpha = A {*} f(\alpha )=A \circledast \alpha$. 8. It holds that $A \circledast \alpha =A * f(\alpha )$. By ∗ success it follows that $f(\alpha ) \in A * f(\alpha )=A \circledast \alpha$. On the other hand, f satisfies idempotence, thus f(f(α)) = f(α). Therefore, A∗f(α) = A∗f(f(α)). By definition of $\circledast$ it follows that $A {*} f(f(\alpha ))=A \circledast f(\alpha )$. Hence $f(\alpha ) \in A \circledast \alpha =A \circledast f(\alpha )$. 9. As showed in the previous item, it holds that $f(\alpha ) \in A \circledast \alpha =A \circledast f(\alpha )$. On the other hand, f satisfies implication, thus $\vdash \alpha \rightarrow f(\alpha )$. 10. Let $\alpha \in A \circledast \alpha$. By Lemma 1 it follows that f(α) = α. Hence $A \circledast \alpha =A * \alpha$. Let β ∈ A and $\beta \not \in A \circledast \alpha$. Thus, by ∗ relevance, it follows that there is some $A^{\prime }$ such that $A \circledast \alpha \subseteq A^{\prime }\subseteq A \cup \{\alpha \}, A^{\prime }\not \vdash \perp$ but $A^{\prime }\cup \{\beta \}\vdash \perp$. 11. The proof that $\circledast$ satisfies weak core-retainment is similar to the proof presented for weak relevance. 12. Assume that ∗ satisfies weak relative closure. It holds by definition of $\circledast$ that $A\cap Cn(A\cap A\circledast \alpha )=A\cap Cn(A\cap A* f(\alpha ))$. By ∗ weak relative closure it holds that $A\cap Cn(A\cap A{*} f(\alpha ))\subseteq A{*} f(\alpha )$. From which it follows, by definition of $\circledast$, that $A\cap Cn(A\cap A\circledast \alpha )\subseteq A\circledast \alpha$. 13. Let $\alpha \in A\circledast \alpha$ and $\beta \in A \setminus (A \circledast \alpha )$. From the former it follows by Lemma 1 that f(α) = α and from the latter that β ∈ A ∖ (A ∗ f(α)). By ∗ disjunctive elimination it follows that A ∗ f(α)⊯¬f(α) ∨ β. Hence $A \circledast \alpha \not \vdash \neg f(\alpha ) \vee \beta$. Therefore, $A \circledast \alpha \not \vdash \neg \alpha \vee \beta$. 14. Consider that it holds for all subsets $A^{\prime }$ of A that $A^{\prime }\cup \{\alpha \} \vdash \perp$ if and only if $A^{\prime }\cup \{\beta \}\vdash \perp$. Assume that $\alpha \in A \circledast \alpha$. By Lemma 1 it follows that f(α) = α. On the other hand, f satisfies uniform identity, thus f(β) = β. By definition of $\circledast$ it holds that $A \circledast \beta =A * f(\beta )$. By ∗ success it follows that $f(\beta )\in A \circledast \beta$. Thus $\beta \in A \circledast \beta$. By symmetry of the case it holds that if $\beta \in A \circledast \beta$, then $\alpha \in A \circledast \alpha$. Hence it holds that $\alpha \in A \circledast \alpha$ if and only if $\beta \in A \circledast \beta$. 15. Let $\alpha \in A \circledast \alpha$. Consider that it holds for all subsets $A^{\prime }$ of A that $A^{\prime }\cup \{\alpha \} \vdash \perp$ if and only if $A^{\prime }\cup \{\beta \}\vdash \perp$. It holds that $\circledast$ satisfies uniform success (as showed in the previous item), thus $\beta \in A \circledast \beta$. On the other hand, from $\alpha \in A \circledast \alpha$ it follows that f(α) = α (by Lemma 1). By symmetry of the case it follows that f(β) = β. On the other hand, by ∗ uniformity, it follows that A ∩ A∗α = A ∩ A∗β. Thus $A\cap A \circledast \alpha =A\cap A {*} f(\alpha )=A\cap A {*} \alpha =A\cap A {*} \beta =A\cap A {*} f(\beta )=A\cap A \circledast \beta$. 16. Let $\vdash \alpha \leftrightarrow \beta$. Assume that $\alpha \in A \circledast \alpha$. From $\alpha \in A \circledast \alpha$ it follows that f(α) = α (by Lemma 1). It holds that f satisfies equivalence propagation. Thus f(β) = β. On the other hand, by definition of $\circledast$, it follows that $A \circledast \beta =A {*} f(\beta )$. By ∗ success it holds that f(β) ∈ A∗f(β), from which it follows that $\beta \in A \circledast \beta$. Hence: - Extensional success: By symmetry of the case, if $\beta \in A \circledast \beta$ then $\alpha \in A \circledast \alpha$. Thus $\alpha \in A \circledast \alpha$ holds if and only if $\beta \in A \circledast \beta$ holds. - Weak relative extensionality: From $\beta \in A\circledast \beta$ it follows, by Lemma 1, that f(β) = β. On the other hand by ∗ relative extensionality it follows that A ∩ A ∗ α = A ∩ A ∗ β. Thus $A\cap A\circledast \alpha =A\cap A* f(\alpha )=A\cap A* \alpha =A\cap A* \beta =A\cap A* f(\beta )=A\cap A\circledast \beta$. □

Proof of Theorem 9

(a) implies (b): We will first define f and ∗. Let $f:{\mathscr{L}}\longrightarrow {\mathscr{L}}$ be such that:

$$ f(\alpha) = \left\{ \begin{array}{ll} \alpha & \text{if } \alpha \in A \circledast \alpha \\ r(\alpha) & \text{otherwise}\\ \textrm{where } r \textrm{ is a function from } \mathcal{L} \textrm{ to } \mathcal{L} \textrm{ such that }\\ r(\alpha) \in A \circledast \alpha, A \circledast \alpha=A \circledast r(\alpha) \\\textrm{ and } \vdash \alpha \rightarrow r(\alpha). \end{array} \right. $$

Let

$$ A * \alpha = \left\{ \begin{array}{ll} A \circledast \alpha & \text{if } \alpha \in A \circledast \alpha\\ A*^{\prime}\alpha& \text{otherwise}\\ \textrm{where } *^{\prime} \textrm{is any revision operator.} \end{array} \right. $$

We need to show that: (i) f is a (well defined) function that satisfies the properties listed in (b); (ii) ∗ satisfies inclusion, success and consistency; (iii) $A\circledast \alpha =A * f(\alpha )$, for all α. (i) To prove that f is a (well defined) function we must show that for all $\alpha \in {\mathscr{L}}$ there exists $\alpha ^{\prime } \in {\mathscr{L}}$ such that $f(\alpha )=\alpha ^{\prime }$ and that, if α₁ = α₂, then f(α₁) = f(α₂). Let $\alpha \in {{\mathscr{L}}}$. If $\alpha \in A\circledast \alpha$, then f(α) = α. If $\alpha \not \in A\circledast \alpha$, then $f(\alpha )=r(\alpha )=\alpha ^{\prime }$, for some $\alpha ^{\prime }$ such that $\alpha ^{\prime }\in A\circledast \alpha =A\circledast \alpha ^{\prime }$ and $\vdash \alpha \rightarrow \alpha ^{\prime }$. Such $\alpha ^{\prime }$ exists since $\circledast$ satisfies proxy success. Assume now that α₁ = α₂. If $\alpha _1 \in A\circledast \alpha _1$, then $\alpha _2 \in A\circledast \alpha _2$. Thus f(α₁) = α₁ = α₂ = f(α₂). If $\alpha _1 \not \in A\circledast \alpha _1$, then $\alpha _2 \not \in A\circledast \alpha _2$. Thus f(α₁) = r(α₁) and f(α₂) = r(α₂). r is a function. Thus, from α₁ = α₂ it follows that f(α₁) = f(α₂). That f satisfies lower boundary follows by definition of f and $\circledast$ stability. That f satisfies implication follows by definition of f. To show that f satisfies idempotence: case 1) $\alpha \in A\circledast \alpha$. Thus f(α) = α. Hence f(f(α)) = f(α). case 2) $\alpha \not \in A\circledast \alpha$. Thus f(α) = r(α) and $r(\alpha ) \in A\circledast r(\alpha )$. Hence, by definition of f, it follows that f(r(α)) = r(α). From the latter and f(α) = r(α) it follows that f(f(α)) = f(α). (ii) That ∗ satisfies success follows trivially by definition of ∗. That ∗ satisfies consistency follows from the fact that both $\circledast$ and ${*}^{\prime }$ satisfy consistency. Next we show that ∗ satisfies inclusion. If $\alpha \in A\circledast \alpha$, then by $\circledast$ weak inclusion, it follows that $A\circledast \alpha \subseteq A\cup \{\alpha \}$. From which it follows, by definition of ∗, that $A{*} \alpha \subseteq A\cup \{\alpha \}$. If $\alpha \not \in A\circledast \alpha$, then $A{*} \alpha =A{*}^{\prime } \alpha$. It holds that ${*}^{\prime }$ satisfies inclusion thus $A{*} \alpha \subseteq A\cup \{\alpha \}$. (iii) We will now prove that $A\circledast \alpha =A* f(\alpha )$. case 1) $\alpha \not \in A\circledast \alpha$. By definition of f it holds that $A\circledast \alpha =A\circledast f(\alpha )$ and $f(\alpha ) \in A\circledast \alpha$. Hence $f(\alpha ) \in A\circledast f(\alpha )$. Thus, by definition of ∗, it follows that $A{*} f(\alpha )=A\circledast f(\alpha )$, from which it follows that $A\circledast \alpha =A{*} f(\alpha )$. case 2) $\alpha \in A\circledast \alpha$. Hence f(α) = α and $A{*} \alpha =A\circledast \alpha$. Thus $A{*} f(\alpha )=A\circledast \alpha$. (b) implies (a): It holds that f satisfies implication, thus it also satisfies consistency preservation. That $\circledast$ satisfies the postulates listed in condition (a) of the theorem follows from Observation 8. □

Proof of Theorem 10

(a) implies (b): We will use essentially, for this part of the proof, the same constructions as in the proof of Theorem 9. The only difference is that, in this proof, the operator $*^{\prime }$ used in the definition of ∗ is a partial meet revision instead of a revision operator as in the proof of former theorem. We need to show that: (i) f is a (well-defined) function that satisfies the properties listed in (b); (ii) ∗ satisfies inclusion, success, consistency, uniformity and relevance (by Observation 3); (iii) $A\circledast \alpha =A * f(\alpha )$, for all α. It holds that $*^{\prime }$ is a partial meet revision. Thus, according to Observation 3 it satisfies inclusion, success, consistency, uniformity and relevance. (i) The proof that f is a (well-defined) function that satisfies lower boundary, idempotence and implication is similar to the one presented for Theorem 9. We will now show that f satisfies uniform identity. Assume that for all subsets $A^{\prime }\subseteq A$ it holds that $A^{\prime }\cup \{\alpha \} \vdash \perp$ if and only if $A^{\prime }\cup \{\beta \}\vdash \perp$. Let f(α) = α. By definition of f it holds that $f(\alpha )\in A \circledast \alpha$. Thus $\alpha \in A \circledast \alpha$. By $\circledast$ uniform success it follows that $\beta \in A \circledast \beta$. Thus f(β) = β. By symmetry of the case it follows that if f(β) = β, then f(α) = α. Hence it holds that f(α) = α if and only if f(β) = β. (ii) That ∗ satisfies success, consistency and inclusion follows as in the proof of Theorem 9. That ∗ satisfies relevance follows from the definition of ∗, $\circledast$ weak relevance and $*^{\prime }$ relevance. To prove that ∗ satisfies uniformity assume that for all subsets $A^{\prime }\subseteq A$, it holds that $A^{\prime }\cup \{\alpha \} \vdash \perp$ if and only if $A^{\prime }\cup \{\beta \}\vdash \perp$. By $\circledast$uniform success it follows that $\alpha \in A \circledast \alpha$ if and only if $\beta \in A \circledast \beta$. We will consider two cases: case 1) $\alpha \in A \circledast \alpha$. Then $\beta \in A \circledast \beta$. Thus $A * \alpha =A \circledast \alpha$ and $A * \beta =A \circledast \beta$. From which it follows by $\circledast$ weak uniformity that A ∩ A∗α = A ∩ A∗β. case 2) $\alpha \not \in A \circledast \alpha$. Then $\beta \not \in A \circledast \beta$. Thus $A {*} \alpha =A {*}^{\prime } \alpha$ and $A {*} \beta =A {*}^{\prime } \beta$. Hence by ${*}^{\prime }$ uniformity it follows that A ∩ A∗α = A ∩ A∗β. (iii) This follows as in the proof of Theorem 9. (b) implies (a): It holds that f satisfies implication, thus it also satisfies consistency preservation. That $\circledast$ satisfies the postulates listed in condition (a) of the theorem follows from Observation 8. □

Proof of Theorem 11

(a) implies (b): We will use essentially, for this part of the proof, the same constructions as in the proof of Theorem 9. The only difference is that, in this proof, the operator ${*}^{\prime }$ used in the definition of ∗ is a kernel revision instead of a revision operator as in the proof of that theorem (hence, according to Observation 4, ${*}^{\prime }$ satisfies inclusion, success, consistency, uniformity and core-retainment). The proof, for this part, is very similar to the one presented for Theorem 10. The only difference is that we need to show that ∗ satisfies core-retainment instead of relevance. That ∗ satisfies core-retainment follows from the definition of ∗, $\circledast$ weak core-retainment and ${*}^{\prime }$ core-retainment. (b) implies (a): It holds that f satisfies implication, thus it also satisfies consistency preservation. That $\circledast$ satisfies the postulates listed in condition (a) of the theorem follows from Observation 8. □

Proof of Theorem 12

(a) implies (b): We will use essentially, for this part of the proof, the same constructions as in the proof of Theorem 9 but in this case the operator $*^{\prime }$, used in the definition of ∗, is a smooth kernel revision. The proof, for this part, is very similar to the one presented for Theorem 11. The only difference is that additionally we need to show that ∗ satisfies weak relative closure. This follows from the definition of ∗ since $\circledast$ and ${*}^{\prime }$ satisfy weak relative closure.

(b) implies (a): It holds that f satisfies implication, thus it also satisfies consistency preservation. That $\circledast$ satisfies the postulates listed in condition (a) of the theorem follows from Observation 8. □

Proof of Theorem 13

(a) implies (b): Once more, for this part of the proof, we will use a similar construction to the one used in the proof of Theorem 9 but in this case the operator ${*}^{\prime }$, in the definition of ∗, is a basic AGM-generated base revision operator instead of a revision operator as in the proof of that theorem. We need to show that:

(i) f is a (well-defined) function that satisfies the properties listed in (b); (ii) ∗ satisfies success, consistency, inclusion, vacuity, relative extensionality and disjunctive elimination (by Observation 6); (iii) $A\circledast \alpha =A * f(\alpha )$, for all α.

It holds that $*^{\prime }$ is a basic AGM-generated base revision. Thus, according to Observation 6 it satisfies success, consistency, inclusion, vacuity, relative extensionality and disjunctive elimination.

(i) The proof that f is a (well-defined) function that satisfies lower boundary, idempotence and implication is similar to the one presented for Theorem 9. It remains to show that f satisfies equivalence propagation.

Assume that $\vdash \alpha \leftrightarrow \beta$. Let f(α) = α. By definition of f it holds that $f(\alpha )\in A \circledast \alpha$. Thus $\alpha \in A \circledast \alpha$. By $\circledast$ extensional success it follows that $\beta \in A \circledast \beta$. From which it follows that f(β) = β. By symmetry of the case it follows that if f(β) = β, then f(α) = α. Hence it holds that f(α) = α if and only if f(β) = β. (ii) The proof that ∗ satisfies success, consistency and inclusion follows as in the proof of Theorem 9. Vacuity: Follows from the definition of ∗, $\circledast$ weak vacuity and $*^{\prime }$ vacuity. Disjunctive elimination: Follows from the definition of ∗, $\circledast$ weak disjunctive elimination and $*^{\prime }$ disjunctive elimination. Relative extensionality: Assume that $\vdash \alpha \leftrightarrow \beta$. By $\circledast$ extensional success, $\alpha \in A \circledast \alpha$ holds if and only if $\beta \in A \circledast \beta$ holds. We will consider two cases: case 1) $\alpha \in A \circledast \alpha$. Then $\beta \in A \circledast \beta$. Thus $A * \alpha =A \circledast \alpha$ and $A {*} \beta =A \circledast \beta$. Hence, by $\circledast$ weak relative extensionality, it follows that A ∩ A∗α = A ∩ A∗β. case 2) $\alpha \not \in A \circledast \alpha$. Then $\beta \not \in A \circledast \beta$. Thus $A {*} \alpha =A {*}^{\prime } \alpha$ and $A {*} \beta =A {*}^{\prime } \beta$. Hence by ${*}^{\prime }$ relative extensionality it follows that A ∩ A∗α = A ∩ A∗β. (iii) This follows as in the proof of Theorem 9. (b) implies (a): It holds that f satisfies implication, thus it also satisfies consistency preservation. That $\circledast$ satisfies the postulates listed in condition (a) of the theorem follows from Observation 8. □

Proof of Theorem 15

From (b) to (a) follows by Observation 8, since strong lower bounding implies lower bounding and implication implies consistency preservation. From (a) to (b) we use the same construction as in the proofs of Theorems 9 to 13. Since strong stability implies stability we only need to prove that the transformation function f satisfies strong lower bounding. This follows by definition of f and by $\circledast$ strong stability. □

Proof of Theorem 16

From (b) to (a) follows by Observation 8. From (a) to (b) we use essentially the same construction as in the proofs of Theorems 9 to 15 removing the condition $\vdash \alpha \rightarrow r(\alpha )$ in the definition of f. The existence of such function follows from weak proxy success. The proof is similar to the ones presented for those theorems, the main difference is the proof that f satisfies consistency preservation. Assume that ⊯¬α. If $\alpha \in A\circledast \alpha$, then f(α) = α, from which it follows that ⊯¬f(α). Consider now that $\alpha \not \in A\circledast \alpha$. By definition of f it follows that $f(\alpha )\in A \circledast \alpha$, from which it follows, by $\circledast$ consistency, that $A \circledast \alpha \not \vdash \perp$. Therefore f(α)⊯ ⊥. Hence ⊯¬f(α). □

Proof of Theorem 17

From (b) to (a) follows by Observation 8 (having in mind that if f satisfies implication, then it also satisfies consistency preservation). From (a) to (b) we use the same construction as in the proofs of Theorems 9 to 16. We only need to prove that f satisfies inherited consistency. This follows by definition of f and by $\circledast$ strong consistency. □

Proof of Theorem 18

From (b) to (a) follows by Observation 8 (having in mind that if f satisfies implication, then it also satisfies consistency preservation). From (a) to (b) we will use the same construction as in the proofs of Theorems 9 to 16. It holds that $\circledast$ satisfies consistency (since it satisfies strong consistency). Thus, we only need to prove that f satisfies consistency. This follows by definition of f and by $\circledast$ strong consistency. □