We introduce a few notations that are used in some proofs for convenience. Given a formulaµand a profile K=hϕ1, . . . , ϕni:
– Kµ denotes the (possibly empty) vector formed of the bases from the profile Kwhich are consistent withµ, listed in the same order;
– K∧µ denotes the profileK∧µ =hϕ1∧µ, . . . , ϕn∧µi;
– K∧µµ is a shortcut for (K∧µ)µ;
– K◦Dµ denotes the profileK◦Dµ = hϕ1◦Dµ, . . . , ϕn◦Dµi, where ◦D is the drastic revision operator.
For instance, if µ = a∧b and K = ha,¬a∨ ¬b, bi, then Kµ = ha, bi, K∧µ = ha∧b,⊥, a∧bi,K∧µµ =ha∧b, a∧biandK◦Dµ =ha∧b, a∧b, a∧bi.
Proposition 1.Let∆be an extended merging operator satisfying (Inc), (IC0), (IC1), (IC5) and (IC6). Then for every profile K, for every formula µ and for every m≥0,
∆µ(K t h⊥im)≡∆µ(K).
Proof: Let∆ be a merging operator satisfying (Inc), (IC0), (IC1), (IC5) and (IC6). Let K be a profile, µ be a formula and assume that µ is consistent (if µis inconsistent, then the proof trivially follows from (IC0)). By (Inc) we have
∆µ(h⊥im)≡µ. Yet, by (IC0) we have∆µ(K)|=µ. Hence,∆µ(K)∧∆µ(h⊥im)≡
∆µ(K), which is consistent by (IC1). Then by (IC5) and (IC6), we get∆µ(K)∧
∆µ(h⊥im)≡∆µ(K t h⊥im). Hence,∆µ(K t h⊥im)≡∆µ(K). ut Proposition 2. An extended merging operator ∆ is an EIC merging opera-tor iff there exists an extended syncretic assignment associating every profile K with a total preorder ≤K such that for every formula µ, mod(∆µ(K)) = min(mod(µ),≤K).
Proof: (Only If ) Let∆ be an EIC merging operator. Let us consider the as-signment mapping every profileK to a binary relation ≤K over worlds, defined for all worlds I, J as I ≤K J iff I |= ∆ϕ{I,J}(K). The proof of Theorem 11 from [15] can be directly adapted to profiles here to show that ≤K is a total preorder, that mod(∆µ(K)) = min(mod(µ),≤K), and that it satisfies condi-tions 1 - 6 of a syncretic assignment. And by (Inc), for all worlds I, J, we have
∆ϕ{I,J}(h⊥in) ≡ϕ{I,J}, so I 'h⊥in J. Hence, condition 0 of an extended syn-cretic assignment (cf. Definition 13) is satisfied.
(If ) Consider an extended syncretic assignment mapping every profile K to a preorder ≤K over worlds, and define the extended merging operator ∆ by
mod(∆µ(K)) =
min(mod(µ),≤K), for every profile K and every formula µ. One can directly adapt the proof of Theorem 11 from [15] to show that∆ satisfies (IC0)-(IC8).
Yet by (IC0) we have ∆µ(h⊥in) |=µ, and by condition 0 of an extended syn-cretic assignment and by definition of ∆, we have that for every worldI |=µ, I|=∆µ(h⊥in). Hence,∆satisfies (Inc). Therefore,∆is an EIC merging
opera-tor. ut
Proposition 3. Let ∆ be a merging operator and ∆0 its canonical extension.
Then ∆0 satisfies (Inc), and for x ∈ {0, . . . ,8}, if ∆ satisfies (ICx) then ∆0 satisfies (ICx).
Proof:The proof is obvious: the fact that the canonical extension∆0of a merg-ing operator satisfies (Inc) each one of the IC postulates can be verified by
definition of ∆0. ut
Proposition 4. Let ∆E[∆] be the expansion-based rationalizing operator in-duced by ∆. For x∈ {0, . . . ,6}, if ∆ satisfies (ICx) then∆E[∆] satisfies (ICx).
This is not true in the general case forx∈ {7,8}.
Proof:Let∆E be an expansion-based rationalizing merging operator and∆be its characterizing merging operator. In the following,Kdenotes any p-consistent profilehϕ1, . . . , ϕni,µis any formula, ϕ1, ϕ2 are two consistent bases such that ϕ1|=µandϕ2 |=µ, and K∧µ denotes the profilehϕ1∧µ, . . . , ϕn∧µi. Assume
∆ satisfies:
– (IC0): then ∆Eµ(K) ≡∆µ(K∧µ), and ∆µ(K∧µ) |= µ by (IC0), so ∆E also satisfies (IC0).
– (IC1): assumeµis consistent. Then∆Eµ(K)≡∆µ(K∧µ), which is consistent by (IC1), so∆E also satisfies (IC1).
– (IC2): assumeK ∧µis consistent. Then∆Eµ(K)≡∆µ(K∧µ). YetK∧µ∧µ≡ K ∧µ, thusK∧µ∧µis consistent, so by (IC2)∆µ(K∧µ)≡ K∧µ∧µ≡ K ∧µ.
We get that∆Eµ(K)≡ K ∧µ, so ∆E also satisfies (IC2).
– (IC3): sinceK1 ≡ K2, we have K∧µ1 ≡ K2∧µ, from which it follows that∆E also satisfies (IC3).
– (IC4): sinceϕ1 |=µand ϕ2 |=µ, we have ∆Eµ(hϕ1, ϕ2i)≡∆µ(hϕ1, ϕ2i), so
∆E also satisfies (IC4).
– (ICx),x∈ {5,6}: we have ∆Eµ(K1)∧∆Eµ(K2)≡∆µ(K∧µ1 )∧∆µ(K∧µ2 ), and
∆Eµ(K1t K2) ≡ ∆µ(K∧µ1 t K∧µ2 ). Hence, since ∆ satisfies (ICx), ∆E also satisfies (ICx).
The following example shows that (IC7) (respectively, (IC8)) is not necessarily satisfied by∆Eeven if∆satisfies it. Let∆dH,Σbe the merging operator based on the Hamming distance and the sum function, and let∆Ebe the expansion-based rationalizing merging operator it characterizes. We know that ∆dH,Σ satisfies (IC7) and (IC8) (see [15], Theorem 20). Letµ1=>,µ2=a∨b andϕ1=¬a:
– Letϕ2 =a. On the one hand, ∆Eµ
1(hϕ1, ϕ2i)≡∆E>(h¬a, ai)≡∆d>H,Σ(h¬a, ai)≡ >, so∆Eµ1(hϕ1, ϕ2i)∧µ2≡a∨b. On the other hand,∆Eµ1∧µ2(hϕ1, ϕ2i)
≡∆Ea∨b(h¬a, ai)≡∆da∨bH,Σ(h¬a∧(a∨b), a∧(a∨b)i)≡∆da∨bH,Σ(h¬a∧b, ai)≡b.
So we get that∆Eµ1(hϕ1, ϕ2i)∧µ26|=∆Eµ1∧µ2(hϕ1, ϕ2i), which means that∆E does not satisfy (IC7).
– Letϕ2 =a∧ ¬b. One the one hand,∆Eµ1(hϕ1, ϕ2i)≡ ∆E>(h¬a, a∧ ¬bi)≡
∆d>H,Σ(h¬a, a∧¬bi)≡ ¬b, so∆Eµ1(hϕ1, ϕ2i)∧µ2≡a∧¬b. On the other hand,
∆Eµ
1∧µ2(hϕ1, ϕ2i) ≡ ∆Ea∨b(h¬a∧(a∨b), a∧ ¬b∧(a∨b)i) ≡ ∆da∨bH,Σ(h¬a∧ b, a∧ ¬bi) ≡ a∨b. So we get that ∆Eµ
1(hϕ1, ϕ2i)∧µ2 is consistent, and
∆Eµ1∧µ2(hϕ1, ϕ2i)6|=∆Eµ1(hϕ1, ϕ2i)∧µ2, which means that∆E does not sat-isfy (IC8).
u t Proposition 5. For any aggregation function f which satisfies (symmetry), (composition) and (decomposition), ∆dD,f is an expansion-based rationalizing merging operator, induced by itself.
Proof:We use the notations introduced at the beginning of this appendix.
Letf be an aggregation function which satisfies (symmetry), (composition) and (decomposition). We need to prove that∆dD,f is an expansion-based ratio-nalizing merging operator induced by itself, i.e., that for every formula µ and every p-consistent profile K, ∆dµD,f(K) ≡ ∆dµD,f(K∧µ). Let µ be a formula, K be a profile. According to Definition 4 we need to prove that for all worlds I, J |=µ, fϕ∈K(dD(I, ϕ)) ≤fϕ∈K(dD(J, ϕ)) if and only iffϕ∈K∧µ(dD(I, ϕ))≤ fϕ∈K∧µ(dD(J, ϕ)).
LetI, J be two worlds such thatI, J|=µ. For every baseϕsuch thatϕ∧µ is inconsistent, we have thatI, J 6|=ϕ, thusdD(I, ϕ) =dD(J, ϕ) = 1. Hence, by (symmetry), (composition) and (decomposition) off, we get that
fϕ∈K(dD(I, ϕ))≤fϕ∈K(dD(J, ϕ)) if and only if
fϕ∈Kµ(dD(I, ϕ))≤fϕ∈Kµ(dD(J, ϕ)). (2) Note that every base ϕ from Kµ is consistent with µ, i.e., ϕ∧µ is consistent.
Moreover, for any worldL|=µ and every baseϕsuch thatϕ∧µ is consistent we have that:
– ifL|=ϕ, thenL|=ϕ∧µ, so dD(L, ϕ) =dD(L, ϕ∧µ) = 0;
– ifL6|=ϕ, thenL6|=ϕ∧µ, so dD(L, ϕ) =dD(L, ϕ∧µ) = 1.
Yet I, J |= µ, so for every base ϕ such that ϕ∧µ is consistent we have that dD(I, ϕ) = dD(I, ϕ ∧ µ) and dD(J, ϕ) = dD(J, ϕ ∧ µ). This means that fϕ∈Kµ(dD(I, ϕ)) =fϕ∈K∧µ
µ (dD(I, ϕ)) andfϕ∈Kµ(dD(J, ϕ)) =fϕ∈K∧µ
µ (dD(J, ϕ)).
Hence, we get that
fϕ∈Kµ(dD(I, ϕ))≤fϕ∈Kµ(dD(J, ϕ)) if and only if fϕ∈K∧µ
µ (dD(I, ϕ))≤fϕ∈K∧µ
µ (dD(J, ϕ)). (3)
Now, for every inconsistent base ϕ we have dD(I, ϕ) = dD(J, ϕ) = e. So by (symmetry), (composition) and (decomposition) off, we get that
fϕ∈K∧µ
µ (dD(I, ϕ))≤fϕ∈K∧µ
µ (dD(J, ϕ)) if and only if
fϕ∈K∧µ(dD(I, ϕ))≤fϕ∈K∧µ(dD(J, ϕ)). (4) Equations 2, 3 and 4 together show thatfϕ∈K(dD(I, ϕ))≤fϕ∈K(dD(J, ϕ)) if and only iffϕ∈K∧µ(dD(I, ϕ))≤fϕ∈K∧µ(dD(J, ϕ)), from which we can conclude that
∆dµD,f(K)≡∆dµD,f(K∧µ). Therefore, ∆dD,f is an expansion-based rationalizing
merging operator induced by itself. ut
Proposition 6. For any aggregation function f which satisfies (symmetry), (composition) and (decomposition),∆dD,f is a revision-based rationalizing merg-ing operator, induced by itself and h◦1, . . . ,◦niwith◦1=· · ·=◦n=◦D, where
◦Dis the drastic (KM) revision operator.
Proof:We use the notations introduced at the beginning of this appendix.
Letf be an aggregation function which satisfies (symmetry), (composition) and (decomposition). Let us prove that∆dD,f satisfies (Ind-◦). We need to show
that one can associate with each agentia KM revision operator◦isuch that for every formula µ and every profile hϕ1, . . . , ϕni, ∆µ(hϕ1, . . . , ϕni) ≡∆µ(hϕ1◦1 µ, . . . , ϕn◦nµi). Let us set ◦1 =· · ·=◦n =◦D, the drastic revision operator, let µ be a formula and K be a profile. According to Definition 4 we need to prove that for all worlds I, J |= µ, fϕ∈K(dD(I, ϕ)) ≤ fϕ∈K(dD(J, ϕ)) if and only if fϕ∈K◦D µ(dD(I, ϕ)) ≤ fϕ∈K◦D µ(dD(J, ϕ)). Let I, J be two worlds such that I, J |= µ. Note that Equations 2 and 3 from the proof of Proposition 5 can be identically proved, so we only need to show that fϕ∈K∧µ
µ (dD(I, ϕ)) ≤ fϕ∈K∧µ
µ (dD(J, ϕ)) if and only if fϕ∈K◦D µ(dD(I, ϕ)) ≤ fϕ∈K◦D µ(dD(J, ϕ)). Yet one can easily verify that the profileK◦D,µconsists of bases from the profileK∧µµ and an arbitrary number of bases equal to µ. And sinceI, J|=µ, we have that Equations 2 and 3 from the proof of Proposition 5, together with Equation 5 above show that fϕ∈K(dD(I, ϕ)) ≤ fϕ∈K(dD(J, ϕ)) if and only if fϕ∈K◦D µ(dD(I, ϕ)) ≤ fϕ∈K◦D µ(dD(J, ϕ)), from which we can conclude that
∆dµD,f(K)≡∆dµD,f(K◦Dµ). Therefore,∆dD,f satisfies (Ind-◦). ut Proposition 7.Let∆R be a revision-based rationalizing merging operator in-duced by ∆ and ◦1, . . . ,◦n. For x ∈ {0, . . . ,6}, if ∆ satisfies (ICx) then ∆R satisfies (ICx). This is not true in the general case for x∈ {7,8}.
Proof:We use the notations introduced at the beginning of this appendix.
Let∆Rbe a revision-based rationalizing merging operator, induced by and∆ andh◦1, . . . ,◦ni. In the following,Kdenotes any p-consistent profilehϕ1, . . . , ϕni, which is consistent by (IC1), so∆R also satisfies (IC1).
– (IC2): assumeVK ∧µis consistent. This means that for each agenti,ϕi∧µ is consistent. Since each revision operator◦isatisfies (R2), we haveϕi◦iµ≡ ϕi∧µfor each agenti. So∆Rµ(K)≡∆µ(hK1◦1µ, . . . , Kn◦nµi)≡∆µ(K∧µ).
– (IC4): for every agenti, every revision operator◦i satisfies (R2), soϕi|=µ impliesϕi◦iµ≡ϕi∧µ≡ϕi. Hence, sinceϕ1|=µandϕ2|=µ, we have that
∆Rµ(hϕ1, ϕ2i)≡∆µ(hϕ1◦1µ, ϕ2◦2µi)≡∆µ(hϕ1, ϕ2i), so ∆R also satisfies (IC4).
– (ICx),x ∈ {5,6}: we have∆Rµ(K1)∧∆Rµ(K2) ≡∆µ(K◦µ1 )∧∆µ(K◦µ2 ), and
∆Rµ(K1t K2) ≡ ∆µ(K◦µ1 t K◦µ2 ). Hence, since ∆ satisfies (ICx), ∆R also satisfies (ICx).
The example introduced in the proof of Proposition 4 - which shows that (IC7) (respectively, (IC8)) is not necessarily satisfied by ∆E even if ∆ satisfies it -can also be used here to show that (IC7) (respectively, (IC8)) is not necessarily satisfied by∆Reven if∆satisfies it, with the following settings: consider again
∆dH,Σ be the merging operator based on the Hamming distance and the sum function, and additionally define ◦1 =· · · =◦n =◦D, where ◦D is the drastic (KM) revision operator; then let∆Rbe the revision-based rationalizing merging operator induced by∆dH,Σandh◦1, . . . ,◦ni. Letµ1=>,µ2=a∨bandϕ1=¬a.
Then, one can easily verify thatϕ2=a(resp.ϕ2=a∧ ¬b) provides a
counter-example for (IC7) (resp. (IC8)). ut
Proposition 8. Let ∆U be an update-based rationalizing merging operator induced by ∆ and 1, . . . ,n. For x∈ {0,1,3,4,5,6}, if∆ satisfies (ICx) then
∆U satisfies (ICx). This is not true in the general case forx∈ {2,7,8}.
Proof:Let∆U be an update-based rationalizing merging operator, induced by and∆andh1, . . . ,ni. We use the notations introduced at the beginning of this appendix, and in addition,Kµdenotes the profilehK11µ, . . . , Knnµi. In the following,K denotes any p-consistent profilehϕ1, . . . , ϕni,µis any formula and ϕ1, ϕ2 are two consistent bases such thatϕ1|=µandϕ2|=µ.
Assume∆satisfies:
– (IC0):∆Uµ(K)≡∆µ(hK11µ, . . . , Knnµi), and∆µ(hK11µ, . . . , Knnµi)|= µby (IC0), so∆U also satisfies (IC0).
– (IC1): assumeµis consistent. We have∆Uµ(K)≡∆µ(hK11µ, . . . , Knnµi), which is consistent by (IC1), so∆U also satisfies (IC1).
– (IC3): we have K1 ≡ K2, yet each update operator i satisfies (U4), so Kµ1 ≡ Kµ2 , from which it follows that∆U also satisfies (IC3).
– (IC4): for every agenti, every update operator i satisfies (U2), soϕi |=µ impliesϕiiµ≡ϕi∧µ≡ϕi. Hence, sinceϕ1|=µandϕ2|=µ, we have that
∆Uµ(hϕ1, ϕ2i)≡∆µ(hϕ11µ, ϕ22µi)≡∆µ(hϕ1, ϕ2i), so∆U also satisfies (IC4).
– (ICx),x ∈ {5,6}: we have ∆Uµ(K1)∧∆Uµ(K2) ≡∆µ(Kµ1 )∧∆µ(Kµ2 ), and
∆Uµ(K1t K2) ≡ ∆µ(Kµ1 t Kµ2 ). Hence, since ∆ satisfies (ICx), ∆U also satisfies (ICx).
The following example shows that (IC2) is not necessarily satisfied by∆U even if ∆ satisfies it. Let ∆ be any merging operator satisfying (IC2), and let 1 =
· · ·=n=D, where D is the drastic (KM) update operator defined as ϕDµ=
ϕ ifϕ|=µ, µ otherwise.
Let∆U be the update-based rationalizing merging operator induced by ∆ and h1, . . . ,ni. We have∆Ua(hbi)≡∆a(hbDai)≡∆a(hai), which is equivalent to a since ∆ satisfies (IC2). So ∆Ua(hbi) 6≡a∧b, which means that ∆U does not satisfy (IC2).
The example introduced in the proof of Proposition 4 - which shows that (IC7) (respectively, (IC8)) is not necessarily satisfied by∆E even if∆satisfies it - can also be used here to show that (IC7) (respectively, (IC8)) is not necessarily satisfied by∆U even if∆satisfies it, with the following settings: consider again
∆dH,Σ be the merging operator based on the Hamming distance and the sum function, and additionally define 1 =· · · =n =D, where D is the drastic (KM) update operator defined as above; then let ∆U be the update-based ra-tionalizing merging operator induced by ∆dH,Σ and h◦1, . . . ,◦ni. Let µ1 =>, µ2 =a∨band ϕ1=¬a. Then, one can easily verify that setting ϕ2 =a(resp.
ϕ2=a∧ ¬b) provides a counter-example for (IC7) (resp. (IC8)). ut Proposition 9. If ∆ is a merging operator satisfying (IC3), then ∆ satisfies (Ind) if and only if∆satisfies (IIA).
Proof:Let∆be a merging operator satisfying (IC3). (If ) Assume that∆ sat-isfies (IIA). Letϕ1, . . . , ϕn benbelief bases andµbe a formula. We havehϕ1∧ µ, . . . , ϕn∧µi ≡ hϕ1∧µ∧µ, . . . , ϕn∧µ∧µi. From (IIA) we get∆µ(hϕ1, . . . , ϕni)≡
∆µ(hϕ1∧µ, . . . , ϕn∧µi). Hence,∆ satisfies (Ind).
(Only If ) Assume that∆satisfies (Ind). Letϕ1, . . . , ϕn, ϕ01, . . . , ϕ0n be 2nbelief bases,µbe a formula, and assume thathϕ1∧µ, . . . , ϕn∧µi ≡ hϕ01∧µ, . . . , ϕ0n∧µi.
From (IC3) we get that ∆µ(hϕ1 ∧µ, . . . , ϕn ∧µi) ≡ ∆µ(hϕ01 ∧µ, . . . , ϕ0n ∧ µi). Yet from (Ind) we have ∆µ(hϕ1 ∧ µ, . . . , ϕn ∧ µi) ≡ ∆µ(hϕ1, . . . , ϕni) and ∆µ(hϕ01∧µ, . . . , ϕ0n∧µi)≡ ∆µ(hϕ01, . . . , ϕ0ni). Hence, ∆µ(hϕ1, . . . , ϕni)≡
∆µ(hϕ01, . . . , ϕ0ni). Therefore,∆ satisfies (IIA). ut Proposition 10. Let ∆ be a merging operator satisfying (IC3). If ∆ is an expansion-based rationalizing merging operator, then it satisfies (Ind).
Proof:Let∆be a merging operator satisfying (IC3). Let us assume that∆is an expansion-based rationalizing merging operator and let us show that it satisfies (Ind). What we need to prove is that for every formula µ and every profile hϕ1, . . . , ϕni, ∆µ(hϕ1∧µ, . . . , ϕn ∧µi) ≡ ∆µ(hϕ1, . . . , ϕni). Yet ∆ is induced by a merging operator∆0, so for every formulaµand every profile hϕ1, . . . , ϕni,
∆µ(hϕ1, . . . , ϕni)≡∆0µ(hϕ1∧µ, . . . , ϕn∧µi). Letµbe a formula andhϕ1, . . . , ϕni be a profile. We have:
∆µ(hϕ1∧µ, . . . , ϕn∧µi)≡∆0µ(hϕ1∧µ∧µ, . . . , ϕn∧µ∧µi)
≡∆0µ(hϕ1∧µ, . . . , ϕn∧µi)
≡∆µ(hϕ1, . . . , ϕni).
This concludes the proof. ut
Proposition 11.Let∆be a merging operator satisfying (IC3). If∆is a revision-based rationalizing merging operator, then it satisfies (Ind-◦).
Proof:Let us first emphasize an intermediate result. Let ϕ be a base,µ be a formula and◦be a revision operator satisfying (R1) and (R2). By (R1)ϕ◦µ|=µ.
Since (ϕ◦µ)∧µis consistent, by (R2) (ϕ◦µ)◦µ≡(ϕ◦µ)∧µ≡ϕ◦µ. That is,
(ϕ◦µ)◦µ≡ϕ◦µ. (6)
Now, let ∆ be a merging operator satisfying (IC3). Let us assume that ∆ is a revision-based rationalizing merging operator and let us show that it satisfies (Ind-◦). What we need to prove is that one can associate with each agentia KM revision operator◦isuch that for every formulaµand every profilehϕ1, . . . , ϕni,
∆µ(hϕ1◦1µ, . . . , ϕn◦nµi)≡∆µ(hϕ1, . . . , ϕni). Yet∆ is induced by a merging operator ∆0 and h◦1, . . . ,◦ni, where each ◦i is a KM revision operator. So for every formula µ and every profile hϕ1, . . . , ϕni, ∆µ(hϕ1, . . . , ϕni) ≡∆0µ(hϕ1◦1 µ, . . . , ϕn◦nµi). Letµbe a formula andhϕ1, . . . , ϕnibe a profile. We have:
∆µ(hϕ1◦1µ, . . . , ϕn◦nµi)≡∆0µ(h(ϕ1◦1µ)◦1µ, . . . ,(ϕn◦nµ)◦nµi)
≡∆0µ(hϕ1◦1µ, . . . , ϕn◦nµi) (from Equation 6)
≡∆µ(hϕ1, . . . , ϕni).
This concludes the proof. ut
Proposition 12. Let ∆ be a merging operator satisfying (IC2) and (Ind-◦).
Then every revision operator◦iconsidered in (Ind-◦) is the revision operator◦∆ corresponding to∆ in the sense of Definition 8.
Proof:Let ∆ be a merging operator satisfying (IC2) and (Ind-◦). This means that one can associate with each agentia KM revision operator◦isuch that for every formulaµand every profile hϕ1, . . . , ϕni,
∆µ(hϕ1, . . . , ϕni)≡∆µ(hϕ1◦1µ, . . . , ϕn◦nµi). (7) Let us first prove that for every KM revision operator◦i involved in Equation 7 above, we have◦i=◦∆. Let◦i be any revision operator involved in Equation 7, ϕbe a belief base andµbe a formula. By (R1) we haveϕ◦iµ|=µ, so by (Ind-◦) and (IC2), we get ∆µ(hϕi)≡ ∆µ(hϕ◦iµi) ≡ ϕ◦iµ. Hence for every revision
operator◦i, we have◦i=◦∆. ut
Proposition 13. Let∆be an EIC merging operator satisfying (Ind-◦). Then every revision operator◦i is◦D, the drastic revision operator.
Proof: Let ∆ be an EIC merging operator satisfying (Ind-◦). From Proposi-tion 13 we know that every revision operator ◦i is ◦∆, the revision operator corresponding to∆. Let us prove that ◦∆ =◦D, i.e.,◦∆ is the drastic revision operator. On the one hand, we know that the faithful assignment associated with ◦∆ coincides with the syncretic assignment associated with ∆, restricted to singleton profiles. So since ∆ is an EIC merging operator, by Theorem 1 it
corresponds to an assignment satisfying conditions 1 - 6 (see Definition 3). On the other hand, ◦D corresponds to a faithful assignment defined such that for all worldsI, J, L, ifL /∈ {I, J}thenI'ϕ{L}J. Towards a contradiction, assume that◦∆6=◦D. This means that there exist three distinct worldsI, J, Lsuch that (i) I <ϕ{I} J <ϕ{I} L, i.e., there exists a total preorder<ϕ{I} with at least 3 distinct levels. Letϕ≡ϕ{J,L}. By conditions 1 and 2, we have (ii)J 'ϕL <ϕI.
Now:
(a) By (i), we have J <ϕ{I} Land by (ii), we haveJ 'ϕL. Thus by condition 6 of a syncretic assignment, we getJ <hϕ{I},ϕiL.
(b) By (R2),ϕ◦∆ ϕ{I,J} ≡ ϕ{J}, andϕ{I}◦∆ϕ{I,J} ≡ϕ{I}. Then by
(Ind-◦), ∆ϕ{I,J}(hϕ, ϕ{I}i) ≡∆ϕ{I,J}(hϕ{J}, ϕ{I}i). Hence, by (IC0), (IC1) and (IC4), we get∆ϕ{I,J}(hϕ, ϕ{I}i)≡ϕ{I,J}. This means that I'hϕ{I},ϕiJ. (c) Using a similar reasoning as for (b) withLinstead ofJ, we getI'hϕ{I},ϕiL.
We have that (a), (b) and (c) together lead to a contradiction, which means that
◦∆=◦D. ut
Proposition 14. Let∆ be an EIC merging operator.∆ satisfies (Ind) if and only if∆satisfies (Ind-◦).
Proof:We use the notations introduced at the beginning of this appendix.
Let∆ be an EIC merging operator. Let us first prove that for every profile K, every formula µ and all m ≥ 0, ∆µ(K) ≡ ∆µ(K t hµim). So let K be a profile, µ be a formula and let m ≥ 0. If µ is inconsistent, then the proof trivially follows from Proposition 1. So assume thatµis consistent. By (IC2) we have that∆µ(hµim)≡µ. Yet, by (IC0) we have∆µ(K)|=µ. Hence, ∆µ(K)∧
∆µ(hµim)≡∆µ(K), which is consistent by (IC1). Then by (IC5) and (IC6), we get∆µ(K)∧∆µ(hµim)≡∆µ(K t hµim). Hence,
∆µ(K)≡∆µ(K t hµim). (8)
Now,
∆µ(K∧µ)≡∆µ(K∧µµ t h⊥im) by (IC3)
≡∆µ(K∧µµ ) from Proposition 1
≡∆µ(K∧µµ t hµim) from Equation 8
≡∆µ(K◦Dµ) by (IC3) and by definition of◦D, wheremis the number of bases fromK that are inconsistent withµ.
Hence, we get that ∆ satisfies (Ind) if and only if for every profile K and every formula µ,∆µ(K)≡∆µ(K∧µ) if and only if for every profile Kand every formulaµ,∆µ(K)≡∆µ(K◦Dµ) if and only if∆ satisfies (Ind-◦).
This concludes the proof. ut
Proposition 15. Let ∆ be a merging operator satisfying (IC3). If ∆ is an update-based rationalizing merging operator, then it satisfies (Ind-).
Proof:Let us first emphasize an intermediate result. Let ϕ be a base,µ be a formula andbe an update operator satisfying (U1) and (U2). By (U1)ϕµ|=µ.
So by (U2),
(ϕµ)µ≡ϕµ. (9)
Now, let ∆ be a merging operator satisfying (IC3). Let us assume that ∆ is an update-based rationalizing merging operator and let us show that it satisfies (Ind-). What we need to prove is that one can associate with each agentia KM revision operatorisuch that for every formulaµand every profilehϕ1, . . . , ϕni,
∆µ(hϕ11µ, . . . , ϕnnµi)≡∆µ(hϕ1, . . . , ϕni). Yet∆ is induced by a merging operator ∆0 and h1, . . . ,ni, where each i is a KM revision operator. So for every formula µ and every profile hϕ1, . . . , ϕni, ∆µ(hϕ1, . . . , ϕni) ≡∆0µ(hϕ11
µ, . . . , ϕnnµi). Letµbe a formula andhϕ1, . . . , ϕnibe a profile. We have:
∆µ(hϕ11µ, . . . , ϕnnµi)≡∆0µ(h(ϕ11µ)1µ, . . . ,(ϕnnµ)nµi)
≡∆0µ(hϕ11µ, . . . , ϕnnµi) (from Equation 9)
≡∆µ(hϕ1, . . . , ϕni).
This concludes the proof. ut
Proposition 16.There is no merging operator satisfying (IC2) and (Ind-).
Proof:Towards a contradiction, assume that there exists a merging operator∆ satisfying (IC2) and (Ind-). By (Ind-) (cf. Definition 22) we can associate with every agentia KM update operatori such that for every profilehϕ1, . . . , ϕni,
∆µ(hϕ1, . . . , ϕni)≡∆µ(hϕ11µ, . . . , ϕnnµi). (10) Let us first show that every KM update operatori involved in Equation 10 above is the drastic update operator, i.e, i = D, where the drastic update operatorD is defined for every belief baseϕand every formula µas:
ϕDµ=
ϕ ifϕ|=µ, µ otherwise.
According to Theorem 4, any KM update operator corresponds to a faith-ful assignment associating every world I with a preorder ≤ϕ{I} such that for every base ϕ and every formula µ, mod(ϕµ) = S
I|=ϕmin(mod(µ),≤ϕ{I}).
Then, one can first remark that the drastic update operatorD corresponds to a faithful assignment defined such that for all distinct worldsI, J, L, I6≤ϕ{L}J (andJ 6≤ϕ{L}Iby symmetry). Indeed, if one would have thatJ <ϕ{L}Ifor some distinct worlds I, J, L, then we would get by Theorem 4 thatϕ{L}Dϕ{I,J}≡ ϕ{J}, which contradicts the definition ofD.
So now we want to show that for every KM update operator i involved in Equation 10 above, we havei =D. Towards a contradiction, assume that for some KM update operator i involved in Equation 10,i is not the dras-tic update operator, i.e., i 6= D. That is, we assume that there exist three distinct worlds I, J, L such that J <ϕi{L} I. On the one hand, by (U1), we
have ϕ{L}j ϕ{I,J} |= ϕ{I,J}, and by (U3), ϕ{L}jϕ{I,J} is consistent. Yet J <ϕi
{L} I by hypothesis, thus (i) ϕ{L}iϕ{I,J} ≡ ϕ{J}. On the other hand, by (U2), we have (ii) ϕ{I} iϕ{I,J} ≡ ϕ{I}. Yet by (U8) we have ϕ{I,L}i
ϕ{I,J} ≡(ϕ{I}iϕ{I,J})∨(ϕ{L}iϕ{I,J}). Hence, by (i) and (ii) we get that (iii)ϕ{I,L}iϕ{I,J}≡ϕ{I}∨ϕ{J}≡ϕ{I,J}. Now, from (Ind-) and (iii) we get that (iv)∆ϕ{I,J}(hϕ{I,L}i)≡∆ϕ{I,J}(hϕ{I,L}iϕ{I,J}i)≡∆ϕ{I,J}(hϕ{I,J}i). Yet by (IC2), we get that∆ϕ{I,J}(hϕ{I,L}i)≡ϕ{I} and ∆ϕ{I,J}(hϕ{I,J}i)≡ϕ{I,J}, which contradicts (iv). This proves that every KM update operator i involved in Equation 10 above is the drastic update operatorD.
Now, letI, J, Lbe three distinct worlds. On the one hand,∆ϕ{I,J}(hϕ{J,L}i) = ϕ{J} by (IC2). On the other hand,∆ϕ{I,J}(hϕ{J,L}Dϕ{I,J}i), which is equiv-alent to ∆ϕ{I,J}(hϕ{I,J}i) by definition of D, which is equivalent toϕ{I,J} by (IC2). Hence, we have that∆ϕ{I,J}(hϕ{J,L}i)6≡∆ϕ{I,J}(hϕ{J,L}Dϕ{I,J}i). This contradicts the initial assumption that ∆ satisfies (Ind-), from which we can conclude that there is no merging operator∆satisfying both (IC2) and (Ind-).
u t Proposition 17. Every mapping associating with every profile K a preorder
≤Kover worlds and satisfying conditions (1), (2) and (F) also satisfies condition (4).
Proof:Consider a mapping associating with every profileKa preorder≤Kover worlds and satisfying conditions (1), (2) and (F). We need to prove that condition (4) is satisfied, that is, letϕ1, ϕ2 be two consistent belief bases, letIbe a world such thatI|=ϕ1 and let us prove that there exists a worldJ such thatJ |=ϕ2 andJ ≤hϕ1,ϕ2iI. This is trivially true if I|=ϕ2, so assumeI6|=ϕ2. LetJ be a world such thatJ |=ϕ2. IfJ |=ϕ1, then conditions 1 and 2 together imply that J <hϕ1,ϕ2iI, so condition 4 is satisfied. Now, assume J 6|=ϕ1. Then conditions 1 and 2 together imply that I <ϕ1 J andJ <ϕ2 I. So by condition F, we get I'hϕ1,ϕ2iJ, thus condition 4 is satisfied. ut Proposition 18. Every mapping associating with every profileKa preorder≤K over worlds and satisfying conditions (1), (2), (6) and (F) maps every singleton belief profilehϕito a unique total preorder≤ϕover worlds defined for all worlds I, J and every belief baseϕasI <ϕJ if and only ifI|=ϕandJ 6|=ϕ.
Proof:(If )Direct from condition 2.
(Only If ) Let ϕ be a belief base. The result is trivial if ϕ in inconsistent, so assume that ϕ is consistent. Let us first show that if I 6|=ϕ and J 6|=ϕ, then I 'hϕi J. Let I 6|=ϕ, J 6|=ϕ and assume by contradiction that I <hϕi J. Let L|=ϕ. By condition 2 of a syncretic assignement we haveL <hϕiJ,J <ϕ{J}I and J <ϕ{J} L. Thus by condition F we get I '{ϕ,ϕ{J}} J andJ '{ϕ,ϕ{J}} L.
Thus I '{ϕ,ϕ{J}} L. Since by condition 2, L <hϕi I, then by condition 6 we getI <ϕ{J} L. Yet by conditions 1 and 2 we haveI 'ϕ{I,L}L <ϕ{I,L} J. Since J <ϕ{J} I, I <ϕ{I,L} J, J <ϕ{J} L and L <ϕ{I,L} J, by condition F we get I '{ϕ{J},ϕ{I,L}} J and J '{ϕ{J},ϕ{I,L}} L. Thus I '{ϕ{J},ϕ{I,L}} L. But since
I <ϕ{J}LandI'ϕ{I,L}L, by condition 6 we get I <{ϕ{J},ϕ{I,L}}L, leading to a contradiction. Thus we do not haveI <hϕiJ. By a similar reasoning, we can prove that we do not have J <hϕiIeither. Hence, we get
IfI6|=ϕandJ 6|=ϕ, thenI'hϕiJ. (11) In the case whereI |=ϕandJ |=ϕthen by condition 1 we directly obtain I 'hϕi J. Now, let I, J be two interpretations and assume that I <hϕi J. We need to show that I |= ϕ and J 6|= ϕ. Assume towards a contradiction that I 6|= ϕ. If J |= ϕ, then by condition 2 we get J <hϕi I; if J 6|= ϕ, then by Equation 11 we getI 'hϕi J. Both cases contradict the assumption I <hϕi J. Hence, we have I |=ϕ. Now, assume towards a contradiction thatJ |= ϕ. By condition 1 we get that I 'hϕi J, which contradicts the assumption I <hϕi J.
Hence,J 6|=ϕ. This concludes the proof. ut
Proposition 19. Let≤K be the preorder over worlds associated with a profile K by a filtering syncretic assignment. We have
I <KJ iff |I(K)|>|J(K)|.
We first prove the following lemma:
Lemma 1. Every assignment satisfying conditions 5 and F satisfies the follow-ing condition:
If I <KJ, then I≤KthϕiJ.
Proof:We prove it by recursion on the sizenofK:
• Base case (n= 1): let I <hϕi J. IfI ≤hϕ0iJ, then by condition 5 we get I≤hϕ,ϕ0iJ. IfJ <hϕ0iI, then by condition F we getI'hϕ,ϕ0iJ.
• Let n ≥ 1 and assume that for every K such that |K| = n, we have I <KJ ⇒I≤KthϕiJ or equivalently I <KthϕiJ ⇒I≤KJ. LetKbe a profile such that |K| = n+ 1 and assume I <K J. If I ≤hϕi J, then by condition 5 we get I≤KthϕiJ. Then assumeJ <hϕiI. Since I <K J, by condition 5 there exists a belief baseϕ0 ∈ Ksuch thatI <hϕ0iJ, and by the recursion hypothesis, we also get I ≤K\hϕ0i J. Since J <hϕi I and I <hϕ0i J, by condition F we get I'hϕ,ϕ0iJ. Hence, by condition 5, we getI≤KthϕiJ.
u t We now prove Proposition 19.
Proof:We prove the equivalence stated in Proposition 19 by recursion on the sizenofK:
•Base case (n= 1) : by Proposition 18,I <hϕiJ iffI|=ϕandJ 6|=ϕ, this leads directly to|I(K)|>|J(K)|.
•Base case (n= 2) : let us show thatI <hϕ,ϕ0iJiff (I≤hϕiJ andI <hϕ0iJ) or (I <hϕiJ andI≤hϕ0iJ).
(Only If ) Direct from condition 6.
(If ) Let I <hϕ,ϕ0i J. By contradiction, assume J <hϕi I. Assuming J ≤hϕ0i I (respectively I <hϕ0iJ) contradictsI <hϕ,ϕ0iJ by condition 5 (respectively by condition F). Hence,I≤hϕiJ. In a symmetric way, we can show thatI≤hϕ0iJ. Now, assumingI 'hϕi J and I 'hϕ0i J contradictsI <hϕ,ϕ0i J by condition 5.
Hence, (I≤hϕiJ andI <hϕ0iJ) or (I <hϕiJ andI≤hϕ0iJ).
Since we deal with singleton profiles (|hϕi|=|hϕ0i|= 1), we can apply Proposi-tion 18 and directly get|I(hϕ, ϕ0i)|>|J(hϕ, ϕ0i)|.
• Letn ≥2 and assume that for every profileK, |K|= k, k ≤n, we have I <KJ iff|I(K)|>|J(K)|. LetKbe a profile such that|K|=n+ 1. We need to show thatI <KJ iff|I(K)|>|J(K)|. Assume first thatI <KJ and let us show that |I(K)|>|J(K)|. Letϕ∈ K:
- Assume I <hϕi J. Since I <K J, then by Lemma 1 we have I ≤K\hϕi J. By recursion hypothesis, we have |I(hϕi)| > |J(hϕi)| and |I(K \ hϕi)| ≥
|J(K \ hϕi)|. Hence,|I(K)|>|J(K)|.
- AssumeI 'hϕi J. SinceI <K J, then by condition 5, we have I <K\hϕiJ. By recursion hypothesis, we have |I(hϕi)| = |J(hϕi)| and |I(K \ hϕi)| >
|J(K \ hϕi)|. Hence,|I(K)|>|J(K)|.
- Assume J <hϕi I. Since I <K J, then by condition 5 there exists a belief baseϕ0∈ Ksuch thatI <hϕ0iJ. By condition F, we getI'hϕ,ϕ0iJ. And by condition 5, we have necessarilyI <K\hϕ,ϕ0iJ. By recursion hypothesis, we have|I(hϕ, ϕ0i)|=|J(hϕ, ϕ0i)|and|I(K \ hϕ, ϕ0i)|>|J(K \ hϕ, ϕ0i)|. Hence,
|I(K)|>|J(K)|.
Assume now that|I(K)|>|J(K)| and let us show thatI <KJ. By hypothesis, there exists a belief base ϕ∈ K such thatI |=ϕandJ 6|=ϕ, so by condition 2 I <hϕi J. We get |I(K \ hϕi)| ≥ |J(K \ hϕi)|, and by recursion hypothesis,
Assume now that|I(K)|>|J(K)| and let us show thatI <KJ. By hypothesis, there exists a belief base ϕ∈ K such thatI |=ϕandJ 6|=ϕ, so by condition 2 I <hϕi J. We get |I(K \ hϕi)| ≥ |J(K \ hϕi)|, and by recursion hypothesis,