**2.5 Completeness of** *PBL<sup>ω</sup>*

We shall show that the inference system of *PBL<sup>ω</sup>* provides a complete axiomatization for probabilistic belief with respect to a probabilistic model. To achieve this aim, it suffices to prove that every *PBLω*-consistent set is satisfiable with respect to a probabilistic model. We prove this by using a general technique that works for a wide variety of probabilistic modal logic. We construct a special structure *PM* called a canonical structure for *PBLω*. *PM* has a state *sV* corresponding to every maximal *PBLω*-consistent set *V* and the following property holds: (*PM*,*sV*) |= *ϕ* iff *ϕ* ∈ *V*.

We need some definitions before giving the proof of the completeness. Given an inference system of *PBLω*, we say a set of formulas Γ is a consistent set with respect to *LPBL<sup>ω</sup>* exactly if false is not provable from Γ. A set of formulas Γ is a maximal consistent set with respect to *<sup>L</sup>PBL<sup>ω</sup>* if (1) it is *PBLω*-consistent, and (2) for all *<sup>ϕ</sup>* in *<sup>L</sup>PBL<sup>ω</sup>* but not in <sup>Γ</sup>, the set <sup>Γ</sup> ∪ {*ϕ*} is not *PBLω*-consistent.

**Definition 2.7** The probabilistic model *PM* with respect to *PBL<sup>ω</sup>* is (*S*, *P*1, ..., *Pn*, *π*).

(1) *S* = {Γ|Γ is a maximal consistent set with respect to *PBLω*};

(2) *Pi* maps every element of *S* to a probability space: *Pi*(Γ)=(*S*, *Xi*,Γ, *μi*,Γ), where *Xi*,<sup>Γ</sup> = {*X*(*ϕ*)|*ϕ* is a formula of *PBLω*}, here *X*(*ϕ*) = {Γ� |*ϕ* ∈ Γ� }; *μi*,<sup>Γ</sup> is a probability assignment: *Xi*,<sup>Γ</sup> → [0, 1], and *μi*,Γ(*X*(*ϕ*)) = *sup*({*a*|*Bi*(*a*, *ϕ*) ∈ Γ});

(3) *π* is a truth assignment as follows: for any atomic formula *p*, *π*(*p*, Γ) = *true* ⇔ *p* ∈ Γ.

**Lemma 2.4** *S* is a nonempty set.

*Proo f* . Since the rules and axioms of *PBL<sup>ω</sup>* are consistent, *S* is nonempty.

**Lemma 2.5** *Xi*,<sup>Γ</sup> satisfies the conditions of Definition 2.2.

*sup*({*am*|*Bi*(*am*, *ϕ*)}), hence Γ<sup>0</sup> � *Bi*(*c*, *ϕ*). This proof can be transferred to a new proof *D* of falsity from ∪*n*<*k*Γ*<sup>n</sup>* which does not apply *Rule* 5, this reduces to the case that the proof does not apply *Rule* 5. If this proof does apply *Rule* 6, since our construction of Γ<sup>0</sup> ensures: for any *ϕ* ∈ Σ there is some *ci*,*<sup>ϕ</sup>* ∈ [0, 1], ∪*ϕ*∈Σ({*Bi*(*a*, *ϕ*)|0 ≤ *a* ≤ *ci*,*ϕ*} ∪ {¬*Bi*(*b*, *ϕ*)|1 ≥ *b* > *ci*,*ϕ*}) ⊆ Γ0, hence if (∪*n*<*k*Γ*n*) ∪ (∪*ϕ*∈Σ({*Bi*(*a*, *ϕ*)|0 ≤ *a* ≤ *ai*,*ϕ*} ∪ {¬*Bi*(*b*, *ϕ*)|1 ≥ *b* > *ai*,*ϕ*})) � *ψ* for any *ai*,*<sup>ϕ</sup>* ∈ [0, 1] then (∪*n*<*k*Γ*n*)=(∪*n*<*k*Γ*n*) ∪ Γ<sup>0</sup> � (∪*n*<*k*Γ*n*) ∪ (∪*ϕ*∈Σ({*Bi*(*a*, *ϕ*)|0 ≤ *a* ≤ *ci*,*ϕ*} ∪ {¬*Bi*(*b*, *ϕ*)|1 ≥ *b* > *ci*,*ϕ*})) � *ψ*. This proof can be transferred to a new proof *E* of falsity from ∪*n*<*k*Γ*<sup>n</sup>* which does not apply *Rule* 6, this reduces to the case that the proof does not apply *Rule* 6. Therefore a proof of falsity from ∪*n*<*k*Γ*<sup>n</sup>* can be transferred to a proof without

Probabilistic Belief Logics for Uncertain Agents 27

The proof of that Γ is consistent is similar to the above proof of that ∪*n*<*k*Γ*<sup>n</sup>* is consistent. We claim that <sup>Γ</sup> is maximal, for suppose *<sup>ψ</sup>* <sup>∈</sup> *<sup>L</sup>PBL<sup>ω</sup>* and *<sup>ψ</sup>* <sup>∈</sup>/ <sup>Γ</sup>, since *<sup>ψ</sup>* must appear in our sequence, say as *ψk*, here we assume *k* is a successor ordinal, the case of limit ordinal *k* can be proved similarly. If Γ*<sup>k</sup>* ∪ {*ψk*} were *PBLω*-consistent, then our construction would guarantee that *ψ<sup>k</sup>* ∈ Γ*k*<sup>+</sup>1. Hence *ψ<sup>k</sup>* ∈ Γ. Because *ψ<sup>k</sup>* = *ψ* ∈/ Γ, it follows that Γ*<sup>k</sup>* ∪ {*ψ*} is not

By the above discussion, we have a maximal *PBLω*-consistent set Γ such that Δ ⊆ Γ.

**Lemma 2.8** Let *Proi*,Γ(*ϕ*) = {*a*|*Bi*(*a*, *ϕ*) is in Γ}, then *sup*(*Proi*,Γ(*ϕ*)) ∈ *Proi*,Γ(*ϕ*).

**Lemma 2.9** If *A* ∈ *Xi*,Γ, then 0 ≤ *μi*,Γ(*A*) ≤ 1. Furthermore, *μi*,Γ(∅) = 0, *μi*,Γ(*S*) = 1. *Proo f* . By the construction of model, it is obvious that if *A* ∈ *Xi*,<sup>Γ</sup> then 0 ≤ *μi*,Γ(*A*) ≤ 1.

**Lemma 2.7** For any Γ, *Pi*(Γ) is well defined, i.e., for any *S* ∈ *Xi*,Γ, the value of *μi*,Γ(*S*) is unique. *Proo f* . It suffices to prove the following claim: if *S*1, *S*<sup>2</sup> ∈ *Xi*,<sup>Γ</sup> and *S*<sup>1</sup> = *S*2, then *μi*,Γ(*S*1) = *μi*,Γ(*S*2). Since *S*1, *S*<sup>2</sup> ∈ *Xi*,Γ, by the construction of *Xi*,Γ, there are *ϕ* and *ψ* such that *S*<sup>1</sup> = *X*(*ϕ*) and *S*<sup>2</sup> = *X*(*ψ*). Assume *S*<sup>1</sup> = *S*2, then *X*(*ϕ*) = *X*(*ψ*). It is clear that � *ϕ* ↔ *ψ*, suppose not, w.l.o.g, assume � *ϕ* → *ψ* does not hold, then by Lemma 2.6, {*ϕ*, ¬*ψ*} is consistent and

Furthermore, by rule: � *ϕ* → *ψ* ⇒� *Bi*(*a*, *ϕ*) → *Bi*(*a*, *ψ*), we have � *Bi*(*a*, *ϕ*) ↔ *Bi*(*a*, *ψ*), so

*Proo f* . Suppose *Proi*,Γ(*ϕ*) = {*a*|*Bi*(*a*, *ϕ*) is in Γ}, therefore Γ � *Bi*(*an*, *ϕ*) for all *an* ∈ *Proi*,Γ(*ϕ*), by *Rule* 5 of *PBLω*, Γ � *Bi*(*a*, *ϕ*), where *a* = *supn*∈*M*({*an*}) = *sup*(*Proi*,Γ(*ϕ*)), so we get

By rule: � *ϕ* ⇒� *Bi*(1, *ϕ*), therefore we have *μi*,Γ(*S*) = *μi*,Γ(*X*(*true*)) = 1 as desired. By axiom: *Bi*(0, *ϕ*), we get *Bi*(0, *f alse*), so *μi*,Γ(∅) = *μi*,Γ(*X*(*f alse*)) ≥ 0. By rule: � ¬(*ϕ* ∧ *ψ*) ⇒� *Bi*(*a*, *ϕ*) ∧ *Bi*(*b*, *ψ*) → *Bi*(*a* + *b*, *ϕ* ∨ *ψ*), where *a* + *b* ≤ 1, we have *μi*,Γ(*S*) = *μi*,Γ(*X*(*true* ∨ *f alse*)) ≥ *μi*,Γ(*X*(*true*)) + *μi*,Γ(*X*(*f alse*)) = *μi*,Γ(*S*) + *μi*,Γ(∅), since *μi*,Γ(*S*) = 1, 1 ≥ 1 +

**Lemma 2.10** If *A*<sup>1</sup> and *A*<sup>2</sup> are disjoint members of *Xi*,Γ, then *μi*,Γ(*A*<sup>1</sup> ∪ *A*2) = *μi*,Γ(*A*1) +

*Proo f* . Suppose *Ai* = *X*(*ϕi*), and � ¬(*ϕ*<sup>1</sup> ∧ *ϕ*2). By *Rule* 4, � *Bi*(*a*1, *ϕ*1) ∧ *Bi*(*a*2, *ϕ*2) → *Bi*(*a*<sup>1</sup> + *a*2, *ϕ*<sup>1</sup> ∨ *ϕ*2)), therefore *μi*,Γ(*X*(*ϕ*<sup>1</sup> ∨ *ϕ*2)) ≥ *μi*,Γ(*X*(*ϕ*1)) + *μi*,Γ(*X*(*ϕ*2)). Since *X*(*ϕ*<sup>1</sup> ∨

, this contradicts *X*(*ϕ*) = *X*(*ψ*).

applying *Rule* 5 and *Rule* 6. This case has been discussed above.

there is a maximal consistent set Γ� such that {*ϕ*, ¬*ψ*} ⊆ Γ�

*PBLω*-consistent. Hence Γ is maximal.

*μi*,Γ(*X*(*ϕ*)) = *μi*,Γ(*X*(*ψ*)).

*sup*(*Proi*,Γ(*ϕ*)) ∈ *Proi*,Γ(*ϕ*) as desired.

*μi*,Γ(∅) holds, therefore *μi*,Γ(∅) = 0.

*μi*,Γ(*A*2).

*Proo f* . We only prove the following claim: if *A* ∈ *Xi*,Γ, then {Γ� |*μi*,Γ�(*A*) ≥ *a*} ∈ *Xi*,Γ. Other cases can be proved similarly. Since *A* ∈ *Xi*,Γ, so there is *ϕ* with *A* = *X*(*ϕ*). It is clear that *X*(*Bi*(*a*, *ϕ*)) ∈ *Xi*,Γ, so {Γ� |*μi*,Γ�(*A*) ≥ *a*} = {Γ� |*μi*,Γ�(*X*(*ϕ*)) ≥ *a*} = *X*(*Bi*(*a*, *ϕ*)) ∈ *Xi*,Γ.

In classical logic, it is easy to see that every consistent set of formulas can be extended to a maximal consistent set, but with the infinitary rules in *PBL<sup>ω</sup>* it cannot simply be proved in a naive fashion because the union of an increasing sequence of consistent sets need no longer be consistent. Therefore we give a detailed proof for this claim in the following lemma.

**Lemma 2.6** For any *PBLω*-consistent set of formulas Δ, there is a maximal *PBLω*-consistent set Γ such that Δ ⊆ Γ.

*Proo f* . To show that Δ can be extended to a maximal *PBLω*-consistent set, we construct a sequence Γ0, Γ1, ... of *PBLω*-consistent sets as follows. Let *ψ*1, *ψ*2, ... be a sequence of the formulas in *LPBL<sup>ω</sup>* . This sequence is not an enumeration sequence since the cardinal number of the set of real number is not enumerable, however, we can get a well-ordered sequence of the formulas by the choice axiom of set theory.

At first, we construct Γ<sup>0</sup> which satisfies the following conditions:

(1) Δ ⊆ Γ0;

(2) Γ<sup>0</sup> is consistent;

(3) For any agent *<sup>i</sup>* ∈ {1, ..., *<sup>n</sup>*} and every *<sup>ϕ</sup>* <sup>∈</sup> *<sup>L</sup>PBL<sup>ω</sup>* , there is some *ci*,*<sup>ϕ</sup>* <sup>∈</sup> [0, 1] such that {*Bi*(*a*, *ϕ*)|0 ≤ *a* ≤ *ci*,*ϕ*} ∪ {¬*Bi*(*b*, *ϕ*)|1 ≥ *b* > *ci*,*ϕ*} ⊆ Γ0.

Let <sup>Σ</sup><sup>0</sup> <sup>=</sup> <sup>Δ</sup>, then for agent 1 and every *<sup>ϕ</sup>* <sup>∈</sup> *<sup>L</sup>PBL<sup>ω</sup>* , there is some *<sup>c</sup>*1,*<sup>ϕ</sup>* <sup>∈</sup> [0, 1] such that <sup>Σ</sup><sup>0</sup> <sup>∪</sup> (∪*ϕ*∈*LPBL<sup>ω</sup>* ({*B*1(*a*, *<sup>ϕ</sup>*)|<sup>0</sup> <sup>≤</sup> *<sup>a</sup>* <sup>≤</sup> *<sup>c</sup>*1,*ϕ*} ∪ {¬*B*1(*b*, *<sup>ϕ</sup>*)|<sup>1</sup> <sup>≥</sup> *<sup>b</sup>* <sup>&</sup>gt; *<sup>c</sup>*1,*ϕ*})) is consistent, otherwise, for all *<sup>a</sup>*1,*<sup>ϕ</sup>* <sup>∈</sup> [0, 1], <sup>Σ</sup><sup>0</sup> <sup>∪</sup> (∪*ϕ*∈*LPBL<sup>ω</sup>* ({*B*1(*a*, *<sup>ϕ</sup>*)|<sup>0</sup> <sup>≤</sup> *<sup>a</sup>* <sup>≤</sup> *<sup>a</sup>*1,*ϕ*} ∪ {¬*B*1(*b*, *<sup>ϕ</sup>*)|<sup>1</sup> <sup>≥</sup> *<sup>b</sup>* <sup>&</sup>gt; *<sup>a</sup>*1,*ϕ*})) � *f alse*. By *Rule* 6, we have Σ<sup>0</sup> � *f alse*, and since Σ<sup>0</sup> = Δ is consistent, it is a contradiction. Let <sup>Σ</sup><sup>1</sup> <sup>=</sup> <sup>Σ</sup><sup>0</sup> <sup>∪</sup> (∪*ϕ*∈*LPBL<sup>ω</sup>* ({*B*1(*a*, *<sup>ϕ</sup>*)|<sup>0</sup> <sup>≤</sup> *<sup>a</sup>* <sup>≤</sup> *<sup>c</sup>*1,*ϕ*} ∪ {¬*B*1(*b*, *<sup>ϕ</sup>*)|<sup>1</sup> <sup>≥</sup> *<sup>b</sup>* <sup>&</sup>gt; *<sup>c</sup>*1,*ϕ*})), similarly, for agent *<sup>i</sup>* and for each *<sup>ϕ</sup>* <sup>∈</sup> *<sup>L</sup>PBL<sup>ω</sup>* there is *ci*,*<sup>ϕ</sup>* <sup>∈</sup> [0, 1] such that <sup>Σ</sup>*i*−<sup>1</sup> <sup>∪</sup> (∪*ϕ*∈*LPBL<sup>ω</sup>* ({*Bi*(*a*, *<sup>ϕ</sup>*)|<sup>0</sup> <sup>≤</sup> *<sup>a</sup>* <sup>≤</sup> *ci*,*ϕ*} ∪ {¬*Bi*(*b*, *<sup>ϕ</sup>*)|<sup>1</sup> <sup>≥</sup> *<sup>b</sup>* <sup>&</sup>gt; *ci*,*ϕ*})) is consistent. Let <sup>Σ</sup>*<sup>i</sup>* <sup>=</sup> <sup>Σ</sup>*i*−<sup>1</sup> <sup>∪</sup> (∪*ϕ*∈*LPBL<sup>ω</sup>* ({*Bi*(*a*, *<sup>ϕ</sup>*)|<sup>0</sup> <sup>≤</sup> *<sup>a</sup>* <sup>≤</sup> *ci*,*ϕ*} ∪ {¬*Bi*(*b*, *<sup>ϕ</sup>*)|<sup>1</sup> <sup>≥</sup> *<sup>b</sup>* <sup>&</sup>gt; *ci*,*ϕ*})) for *<sup>i</sup>* ∈ {1, ..., *<sup>n</sup>*} and <sup>Γ</sup><sup>0</sup> <sup>=</sup> <sup>∪</sup>*i*∈{1,...,*n*}Σ*<sup>i</sup>* <sup>=</sup> <sup>Σ</sup>*n*, here {1, ..., *n*} is the set of agent. Since Σ*<sup>n</sup>* is consistent, Γ<sup>0</sup> is also consistent.

Now we inductively construct the rest of the sequence according to *ψk*: (a) in the case of *k* = *n* + 1, take Γ*n*+<sup>1</sup> = Γ*<sup>n</sup>* ∪ {*ψn*+1} if the set is *PBLω*-consistent and otherwise take Γ*n*+<sup>1</sup> = Γ*n*. (b) in the case that *k* is a limit ordinal, take Γ*<sup>k</sup>* = ∪*n*<*k*Γ*<sup>n</sup>* ∪ {*ψk*} if the set is *PBLω*-consistent and otherwise take Γ*<sup>k</sup>* = ∪*n*<*k*Γ*n*. Let Γ = ∪Γ*k*. We will prove that Γ is a maximal *PBLω*-consistent set and Δ ⊆ Γ.

Firstly, we prove that Γ*<sup>k</sup>* is consistent by induction. We have already known that Γ<sup>0</sup> is consistent. Now we prove the claim when *k* > 0. In the case of (a), it is clear. In the case of (b), we only need to prove that ∪*n*<*k*Γ*<sup>n</sup>* is consistent. Suppose ∪*n*<*k*Γ*<sup>n</sup>* is not consistent, then there is a proof *C* of falsity from ∪*n*<*k*Γ*n*. If this proof does not apply *Rule* 5 and *Rule* 6, then one of Γ*<sup>n</sup>* contains the formulas in the proof, since Γ*<sup>n</sup>* is consistent, then there is a contradiction. If this proof does apply *Rule* 5, since our construction of Γ<sup>0</sup> ensures: for some *ci*,*<sup>ϕ</sup>* ∈ [0, 1], {*Bi*(*a*, *ϕ*)|0 ≤ *a* ≤ *ci*,*ϕ*} ∪ {¬*Bi*(*b*, *ϕ*)|1 ≥ *b* > *ci*,*ϕ*} ⊆ Γ0, hence if {*Bi*(*a*1, *ϕ*), *Bi*(*a*2, *ϕ*), ...} can be deduced from ∪*n*<*k*Γ*n*, then {*Bi*(*a*1, *ϕ*), *Bi*(*a*2, *ϕ*), ...}∪{*Bi*(*c*, *ϕ*)} ⊆ Γ0, here *c* = 10 Will-be-set-by-IN-TECH

cases can be proved similarly. Since *A* ∈ *Xi*,Γ, so there is *ϕ* with *A* = *X*(*ϕ*). It is clear that

In classical logic, it is easy to see that every consistent set of formulas can be extended to a maximal consistent set, but with the infinitary rules in *PBL<sup>ω</sup>* it cannot simply be proved in a naive fashion because the union of an increasing sequence of consistent sets need no longer be consistent. Therefore we give a detailed proof for this claim in the following lemma.

**Lemma 2.6** For any *PBLω*-consistent set of formulas Δ, there is a maximal *PBLω*-consistent

*Proo f* . To show that Δ can be extended to a maximal *PBLω*-consistent set, we construct a sequence Γ0, Γ1, ... of *PBLω*-consistent sets as follows. Let *ψ*1, *ψ*2, ... be a sequence of the formulas in *LPBL<sup>ω</sup>* . This sequence is not an enumeration sequence since the cardinal number of the set of real number is not enumerable, however, we can get a well-ordered sequence of

(3) For any agent *<sup>i</sup>* ∈ {1, ..., *<sup>n</sup>*} and every *<sup>ϕ</sup>* <sup>∈</sup> *<sup>L</sup>PBL<sup>ω</sup>* , there is some *ci*,*<sup>ϕ</sup>* <sup>∈</sup> [0, 1] such that

Let <sup>Σ</sup><sup>0</sup> <sup>=</sup> <sup>Δ</sup>, then for agent 1 and every *<sup>ϕ</sup>* <sup>∈</sup> *<sup>L</sup>PBL<sup>ω</sup>* , there is some *<sup>c</sup>*1,*<sup>ϕ</sup>* <sup>∈</sup> [0, 1] such that <sup>Σ</sup><sup>0</sup> <sup>∪</sup> (∪*ϕ*∈*LPBL<sup>ω</sup>* ({*B*1(*a*, *<sup>ϕ</sup>*)|<sup>0</sup> <sup>≤</sup> *<sup>a</sup>* <sup>≤</sup> *<sup>c</sup>*1,*ϕ*} ∪ {¬*B*1(*b*, *<sup>ϕ</sup>*)|<sup>1</sup> <sup>≥</sup> *<sup>b</sup>* <sup>&</sup>gt; *<sup>c</sup>*1,*ϕ*})) is consistent, otherwise, for all *<sup>a</sup>*1,*<sup>ϕ</sup>* <sup>∈</sup> [0, 1], <sup>Σ</sup><sup>0</sup> <sup>∪</sup> (∪*ϕ*∈*LPBL<sup>ω</sup>* ({*B*1(*a*, *<sup>ϕ</sup>*)|<sup>0</sup> <sup>≤</sup> *<sup>a</sup>* <sup>≤</sup> *<sup>a</sup>*1,*ϕ*} ∪ {¬*B*1(*b*, *<sup>ϕ</sup>*)|<sup>1</sup> <sup>≥</sup> *<sup>b</sup>* <sup>&</sup>gt; *<sup>a</sup>*1,*ϕ*})) � *f alse*. By *Rule* 6, we have Σ<sup>0</sup> � *f alse*, and since Σ<sup>0</sup> = Δ is consistent, it is a contradiction. Let <sup>Σ</sup><sup>1</sup> <sup>=</sup> <sup>Σ</sup><sup>0</sup> <sup>∪</sup> (∪*ϕ*∈*LPBL<sup>ω</sup>* ({*B*1(*a*, *<sup>ϕ</sup>*)|<sup>0</sup> <sup>≤</sup> *<sup>a</sup>* <sup>≤</sup> *<sup>c</sup>*1,*ϕ*} ∪ {¬*B*1(*b*, *<sup>ϕ</sup>*)|<sup>1</sup> <sup>≥</sup> *<sup>b</sup>* <sup>&</sup>gt; *<sup>c</sup>*1,*ϕ*})), similarly, for agent *<sup>i</sup>* and for each *<sup>ϕ</sup>* <sup>∈</sup> *<sup>L</sup>PBL<sup>ω</sup>* there is *ci*,*<sup>ϕ</sup>* <sup>∈</sup> [0, 1] such that <sup>Σ</sup>*i*−<sup>1</sup> <sup>∪</sup> (∪*ϕ*∈*LPBL<sup>ω</sup>* ({*Bi*(*a*, *<sup>ϕ</sup>*)|<sup>0</sup> <sup>≤</sup> *<sup>a</sup>* <sup>≤</sup> *ci*,*ϕ*} ∪ {¬*Bi*(*b*, *<sup>ϕ</sup>*)|<sup>1</sup> <sup>≥</sup> *<sup>b</sup>* <sup>&</sup>gt; *ci*,*ϕ*})) is consistent. Let <sup>Σ</sup>*<sup>i</sup>* <sup>=</sup> <sup>Σ</sup>*i*−<sup>1</sup> <sup>∪</sup> (∪*ϕ*∈*LPBL<sup>ω</sup>* ({*Bi*(*a*, *<sup>ϕ</sup>*)|<sup>0</sup> <sup>≤</sup> *<sup>a</sup>* <sup>≤</sup> *ci*,*ϕ*} ∪ {¬*Bi*(*b*, *<sup>ϕ</sup>*)|<sup>1</sup> <sup>≥</sup> *<sup>b</sup>* <sup>&</sup>gt; *ci*,*ϕ*})) for *<sup>i</sup>* ∈ {1, ..., *<sup>n</sup>*} and <sup>Γ</sup><sup>0</sup> <sup>=</sup> <sup>∪</sup>*i*∈{1,...,*n*}Σ*<sup>i</sup>* <sup>=</sup> <sup>Σ</sup>*n*, here

Now we inductively construct the rest of the sequence according to *ψk*: (a) in the case of *k* = *n* + 1, take Γ*n*+<sup>1</sup> = Γ*<sup>n</sup>* ∪ {*ψn*+1} if the set is *PBLω*-consistent and otherwise take Γ*n*+<sup>1</sup> = Γ*n*. (b) in the case that *k* is a limit ordinal, take Γ*<sup>k</sup>* = ∪*n*<*k*Γ*<sup>n</sup>* ∪ {*ψk*} if the set is *PBLω*-consistent and otherwise take Γ*<sup>k</sup>* = ∪*n*<*k*Γ*n*. Let Γ = ∪Γ*k*. We will prove that Γ is a maximal *PBLω*-consistent

Firstly, we prove that Γ*<sup>k</sup>* is consistent by induction. We have already known that Γ<sup>0</sup> is consistent. Now we prove the claim when *k* > 0. In the case of (a), it is clear. In the case of (b), we only need to prove that ∪*n*<*k*Γ*<sup>n</sup>* is consistent. Suppose ∪*n*<*k*Γ*<sup>n</sup>* is not consistent, then there is a proof *C* of falsity from ∪*n*<*k*Γ*n*. If this proof does not apply *Rule* 5 and *Rule* 6, then one of Γ*<sup>n</sup>* contains the formulas in the proof, since Γ*<sup>n</sup>* is consistent, then there is a contradiction. If this proof does apply *Rule* 5, since our construction of Γ<sup>0</sup> ensures: for some *ci*,*<sup>ϕ</sup>* ∈ [0, 1], {*Bi*(*a*, *ϕ*)|0 ≤ *a* ≤ *ci*,*ϕ*} ∪ {¬*Bi*(*b*, *ϕ*)|1 ≥ *b* > *ci*,*ϕ*} ⊆ Γ0, hence if {*Bi*(*a*1, *ϕ*), *Bi*(*a*2, *ϕ*), ...} can be deduced from ∪*n*<*k*Γ*n*, then {*Bi*(*a*1, *ϕ*), *Bi*(*a*2, *ϕ*), ...}∪{*Bi*(*c*, *ϕ*)} ⊆ Γ0, here *c* =



*Proo f* . We only prove the following claim: if *A* ∈ *Xi*,Γ, then {Γ�

At first, we construct Γ<sup>0</sup> which satisfies the following conditions:

{1, ..., *n*} is the set of agent. Since Σ*<sup>n</sup>* is consistent, Γ<sup>0</sup> is also consistent.

{*Bi*(*a*, *ϕ*)|0 ≤ *a* ≤ *ci*,*ϕ*} ∪ {¬*Bi*(*b*, *ϕ*)|1 ≥ *b* > *ci*,*ϕ*} ⊆ Γ0.


*X*(*Bi*(*a*, *ϕ*)) ∈ *Xi*,Γ, so {Γ�

set Γ such that Δ ⊆ Γ.

(1) Δ ⊆ Γ0;

(2) Γ<sup>0</sup> is consistent;

set and Δ ⊆ Γ.

the formulas by the choice axiom of set theory.

*sup*({*am*|*Bi*(*am*, *ϕ*)}), hence Γ<sup>0</sup> � *Bi*(*c*, *ϕ*). This proof can be transferred to a new proof *D* of falsity from ∪*n*<*k*Γ*<sup>n</sup>* which does not apply *Rule* 5, this reduces to the case that the proof does not apply *Rule* 5. If this proof does apply *Rule* 6, since our construction of Γ<sup>0</sup> ensures: for any *ϕ* ∈ Σ there is some *ci*,*<sup>ϕ</sup>* ∈ [0, 1], ∪*ϕ*∈Σ({*Bi*(*a*, *ϕ*)|0 ≤ *a* ≤ *ci*,*ϕ*} ∪ {¬*Bi*(*b*, *ϕ*)|1 ≥ *b* > *ci*,*ϕ*}) ⊆ Γ0, hence if (∪*n*<*k*Γ*n*) ∪ (∪*ϕ*∈Σ({*Bi*(*a*, *ϕ*)|0 ≤ *a* ≤ *ai*,*ϕ*} ∪ {¬*Bi*(*b*, *ϕ*)|1 ≥ *b* > *ai*,*ϕ*})) � *ψ* for any *ai*,*<sup>ϕ</sup>* ∈ [0, 1] then (∪*n*<*k*Γ*n*)=(∪*n*<*k*Γ*n*) ∪ Γ<sup>0</sup> � (∪*n*<*k*Γ*n*) ∪ (∪*ϕ*∈Σ({*Bi*(*a*, *ϕ*)|0 ≤ *a* ≤ *ci*,*ϕ*} ∪ {¬*Bi*(*b*, *ϕ*)|1 ≥ *b* > *ci*,*ϕ*})) � *ψ*. This proof can be transferred to a new proof *E* of falsity from ∪*n*<*k*Γ*<sup>n</sup>* which does not apply *Rule* 6, this reduces to the case that the proof does not apply *Rule* 6. Therefore a proof of falsity from ∪*n*<*k*Γ*<sup>n</sup>* can be transferred to a proof without applying *Rule* 5 and *Rule* 6. This case has been discussed above.

The proof of that Γ is consistent is similar to the above proof of that ∪*n*<*k*Γ*<sup>n</sup>* is consistent.

We claim that <sup>Γ</sup> is maximal, for suppose *<sup>ψ</sup>* <sup>∈</sup> *<sup>L</sup>PBL<sup>ω</sup>* and *<sup>ψ</sup>* <sup>∈</sup>/ <sup>Γ</sup>, since *<sup>ψ</sup>* must appear in our sequence, say as *ψk*, here we assume *k* is a successor ordinal, the case of limit ordinal *k* can be proved similarly. If Γ*<sup>k</sup>* ∪ {*ψk*} were *PBLω*-consistent, then our construction would guarantee that *ψ<sup>k</sup>* ∈ Γ*k*<sup>+</sup>1. Hence *ψ<sup>k</sup>* ∈ Γ. Because *ψ<sup>k</sup>* = *ψ* ∈/ Γ, it follows that Γ*<sup>k</sup>* ∪ {*ψ*} is not *PBLω*-consistent. Hence Γ is maximal.

By the above discussion, we have a maximal *PBLω*-consistent set Γ such that Δ ⊆ Γ.

**Lemma 2.7** For any Γ, *Pi*(Γ) is well defined, i.e., for any *S* ∈ *Xi*,Γ, the value of *μi*,Γ(*S*) is unique.

*Proo f* . It suffices to prove the following claim: if *S*1, *S*<sup>2</sup> ∈ *Xi*,<sup>Γ</sup> and *S*<sup>1</sup> = *S*2, then *μi*,Γ(*S*1) = *μi*,Γ(*S*2). Since *S*1, *S*<sup>2</sup> ∈ *Xi*,Γ, by the construction of *Xi*,Γ, there are *ϕ* and *ψ* such that *S*<sup>1</sup> = *X*(*ϕ*) and *S*<sup>2</sup> = *X*(*ψ*). Assume *S*<sup>1</sup> = *S*2, then *X*(*ϕ*) = *X*(*ψ*). It is clear that � *ϕ* ↔ *ψ*, suppose not, w.l.o.g, assume � *ϕ* → *ψ* does not hold, then by Lemma 2.6, {*ϕ*, ¬*ψ*} is consistent and there is a maximal consistent set Γ� such that {*ϕ*, ¬*ψ*} ⊆ Γ� , this contradicts *X*(*ϕ*) = *X*(*ψ*). Furthermore, by rule: � *ϕ* → *ψ* ⇒� *Bi*(*a*, *ϕ*) → *Bi*(*a*, *ψ*), we have � *Bi*(*a*, *ϕ*) ↔ *Bi*(*a*, *ψ*), so *μi*,Γ(*X*(*ϕ*)) = *μi*,Γ(*X*(*ψ*)).

**Lemma 2.8** Let *Proi*,Γ(*ϕ*) = {*a*|*Bi*(*a*, *ϕ*) is in Γ}, then *sup*(*Proi*,Γ(*ϕ*)) ∈ *Proi*,Γ(*ϕ*).

*Proo f* . Suppose *Proi*,Γ(*ϕ*) = {*a*|*Bi*(*a*, *ϕ*) is in Γ}, therefore Γ � *Bi*(*an*, *ϕ*) for all *an* ∈ *Proi*,Γ(*ϕ*), by *Rule* 5 of *PBLω*, Γ � *Bi*(*a*, *ϕ*), where *a* = *supn*∈*M*({*an*}) = *sup*(*Proi*,Γ(*ϕ*)), so we get *sup*(*Proi*,Γ(*ϕ*)) ∈ *Proi*,Γ(*ϕ*) as desired.

**Lemma 2.9** If *A* ∈ *Xi*,Γ, then 0 ≤ *μi*,Γ(*A*) ≤ 1. Furthermore, *μi*,Γ(∅) = 0, *μi*,Γ(*S*) = 1.

*Proo f* . By the construction of model, it is obvious that if *A* ∈ *Xi*,<sup>Γ</sup> then 0 ≤ *μi*,Γ(*A*) ≤ 1.

By rule: � *ϕ* ⇒� *Bi*(1, *ϕ*), therefore we have *μi*,Γ(*S*) = *μi*,Γ(*X*(*true*)) = 1 as desired. By axiom: *Bi*(0, *ϕ*), we get *Bi*(0, *f alse*), so *μi*,Γ(∅) = *μi*,Γ(*X*(*f alse*)) ≥ 0. By rule: � ¬(*ϕ* ∧ *ψ*) ⇒� *Bi*(*a*, *ϕ*) ∧ *Bi*(*b*, *ψ*) → *Bi*(*a* + *b*, *ϕ* ∨ *ψ*), where *a* + *b* ≤ 1, we have *μi*,Γ(*S*) = *μi*,Γ(*X*(*true* ∨ *f alse*)) ≥ *μi*,Γ(*X*(*true*)) + *μi*,Γ(*X*(*f alse*)) = *μi*,Γ(*S*) + *μi*,Γ(∅), since *μi*,Γ(*S*) = 1, 1 ≥ 1 + *μi*,Γ(∅) holds, therefore *μi*,Γ(∅) = 0.

**Lemma 2.10** If *A*<sup>1</sup> and *A*<sup>2</sup> are disjoint members of *Xi*,Γ, then *μi*,Γ(*A*<sup>1</sup> ∪ *A*2) = *μi*,Γ(*A*1) + *μi*,Γ(*A*2).

*Proo f* . Suppose *Ai* = *X*(*ϕi*), and � ¬(*ϕ*<sup>1</sup> ∧ *ϕ*2). By *Rule* 4, � *Bi*(*a*1, *ϕ*1) ∧ *Bi*(*a*2, *ϕ*2) → *Bi*(*a*<sup>1</sup> + *a*2, *ϕ*<sup>1</sup> ∨ *ϕ*2)), therefore *μi*,Γ(*X*(*ϕ*<sup>1</sup> ∨ *ϕ*2)) ≥ *μi*,Γ(*X*(*ϕ*1)) + *μi*,Γ(*X*(*ϕ*2)). Since *X*(*ϕ*<sup>1</sup> ∨

*Proo f* . Suppose not, then there is a *PBL<sup>ω</sup>* - consistent formulas set Φ = Γ ∪ {¬*ϕ*}, and there is no model *PM* such that Φ is satisfied in *PM*. For there is a *PBL<sup>ω</sup>* - maximal consistent formula set Σ such that Φ ⊆ Σ, by Lemma 2.15, Φ is satisfied in possible world Σ of *PM*. It is

Probabilistic Belief Logics for Uncertain Agents 29

Our proof of the above completeness is different from the proof in [3]. The main idea of our proof is to give a canonical model, which can be regarded as a generalization of canonical model method in Kripke semantics. In [3], Fagin and Halpern adopt another technique to get the completeness. Let *ϕ* be consistent with *AXMEAS*, they show firstly that an *i*-probability formula *<sup>ψ</sup>* <sup>∈</sup> *Sub*+(*ϕ*) is provably equivalent to a formula of the form <sup>Σ</sup>*s*∈*Scsμi*(*ϕs*) <sup>≥</sup> *<sup>b</sup>*, for some appropriate coefficients *cs*, where *S* consists of all maximal consistent subsets of *Sub*+(*ϕ*). Then for a fixed agent *i* and a fixed state *s*, they describe a set of linear equalities and inequalities corresponding to *i* and *s*, over variables of the form *xiss*� , for *s*� ∈ *S*. We can

distribution at state *<sup>s</sup>*. Assume that *<sup>ψ</sup>* is equivalent to <sup>Σ</sup>*s*∈*Scsμi*(*ϕs*) ≥ *<sup>b</sup>*. Observe that exactly one of *ψ* and ¬*ψ* is in *s*. If *ψ* ∈ *s*, then the corresponding inequality is Σ*s*�∈*Scs*� *xiss*� ≥ *b*. If ¬*<sup>ψ</sup>* ∈ *<sup>s</sup>*, then the corresponding inequality is <sup>Σ</sup>*s*�∈*Scs*� *xiss*� < *<sup>b</sup>*. Finally, we have the equality <sup>Σ</sup>*s*�∈*Sxiss*� = 1. As shown in Theorem 2.2 in Fagin et al. [4], since*ϕ<sup>s</sup>* is consistent, this set of

proof depends tightly on the axioms of linear equalities and inequalities, whereas there are no such axioms in our inference system. On the other hand, their proof cannot deal with the case of infinite set of formulas, because in this case, we will get an infinite set of linear equalities and inequalities, which contains infinite variables. But their axioms seem insufficient to describe

Proposition 2.1 and Proposition 2.2 show that the axioms and inference rules of *PBL<sup>ω</sup>* give us a sound and complete axiomatization for probabilistic belief. Moreover, we can prove the finite model property and decidability of the provability problem for some variants of *PBL<sup>ω</sup>*

It is not difficult to see that *Axioms* 1-6 and *Rules* 1-4 are not complete for our model. Because otherwise, the compactness property of *PBL<sup>ω</sup>* holds, but we can give the following example to show that the compactness property fails in *PBLω*: any finite sub set of {*Bi*(1/2, *ϕ*), *Bi*(2/3, *ϕ*), ..., *Bi*(*n*/*n* + 1, *ϕ*),...} ∪ {¬*Bi*(1, *ϕ*)} has model, whereas the whole set does not. But we do not know whether *Axioms* 1-6 and *Rules* 1-5 are complete for our model, i.e., whether *Rule* 6 is redundant in the inference system. Although we believe that *Rule* 6 is not redundant, we have

As is often the case in modal logics, the ideas in our completeness proof can be extended to get a finite model property. Therefore the question arises whether finite model property holds for *PBLω*, i.e., for every consistent formula *ϕ*, whether there is a finite sates model satisfies *ϕ*. Unfortunately, we cannot give a positive or negative answer here. Therefore, we seek for some weak variant of *PBL<sup>ω</sup>* whose finite model property can be proved. We call the variant *PBLf* , its reasoning system is the result of deleting *Axiom* 6 and *Rule* 6 from *PBLω*. In the semantics of *PBLf* , we assign an inner probability space to every possible world in the model, here "inner" means the measure does not obey the additivity condition, but obeys some weak

the existence of solutions of an infinite set of linear equalities and inequalities.

), i.e., the probability of state *s*� under agent *i*'s probability

*iss*� , *s*� ∈ *S*. From their idea, it is clear that their

a contradiction.

think of *xiss*� as representing *μi*,*s*(*s*�

in the following sections.

no proof up to now.

linear equalities and inequalities has a solution *x*∗

**3. PBL***<sup>f</sup>* **and its inner probabilistic semantics**

additivity conditions satisfied by inner probability measure.

*ϕ*2) = *A*<sup>1</sup> ∪ *A*2, we have *μi*,Γ(*A*<sup>1</sup> ∪ *A*2) = *μi*,Γ(*X*(*ϕ*<sup>1</sup> ∨ *ϕ*2)) ≥ *μi*,Γ(*X*(*ϕ*1)) + *μi*,Γ(*X*(*ϕ*2)) = *μi*,Γ(*A*1) + *μi*,Γ(*A*2).

Now, we prove *μi*,Γ(*A*1) + *μi*,Γ(*A*2) ≥ *μi*,Γ(*A*<sup>1</sup> ∪ *A*2). Suppose *μi*,Γ(*A*<sup>1</sup> ∪ *A*2) > *μi*,Γ(*A*1) + *μi*,Γ(*A*2). Let *μi*,Γ(*A*1) = *a*1, *μi*,Γ(*A*2) = *a*2, *μi*,Γ(*A*<sup>1</sup> ∪ *A*2) > *a*<sup>1</sup> + *a*2. Choose *e* > 0 such that 2*e* + *a*<sup>1</sup> + *a*<sup>2</sup> = *μi*,Γ(*A*<sup>1</sup> ∪ *A*2), then *μi*,Γ(*A*1) < *a*<sup>1</sup> + *e*, *μi*,Γ(*A*2) < *a*<sup>2</sup> + *e*.

Therefore we have Γ |= *Bi*(*a*<sup>1</sup> + *a*<sup>2</sup> + 2*e*, *ϕ*<sup>1</sup> ∨ *ϕ*2) ∧ ¬*Bi*(*a*<sup>1</sup> + *e*, *ϕ*1) ∧ ¬*Bi*(*a*<sup>2</sup> + *e*, *ϕ*2). Since Γ is a maximal consistent set, so Γ � *Bi*(*a*<sup>1</sup> + *a*<sup>2</sup> + 2*e*, *ϕ*<sup>1</sup> ∨ *ϕ*2) ∧ ¬*Bi*(*a*<sup>1</sup> + *e*, *ϕ*1) ∧ ¬*Bi*(*a*<sup>2</sup> + *e*, *ϕ*2), which contradicts *Axiom* 6 of *PBLω*.

**Lemma 2.11** For any *ϕ*, let Δ*<sup>a</sup> <sup>i</sup>* (*ϕ*) = {Γ� |*μi*,Γ�(*X*(*ϕ*)) ≥ *a*}. If *μi*,Γ(*X*(*ϕ*)) ≥ *a*, then *μi*,Γ(Δ*<sup>a</sup> <sup>i</sup>* (*ϕ*)) = 1.

*Proo f* . It is clear that *Bi*(*a*, *<sup>ϕ</sup>*) <sup>∈</sup> <sup>Γ</sup> <sup>⇒</sup> <sup>Γ</sup> <sup>∈</sup> <sup>Δ</sup>*<sup>a</sup> <sup>i</sup>* (*ϕ*), therefore *<sup>X</sup>*(*Bi*(*a*, *<sup>ϕ</sup>*)) <sup>⊆</sup> <sup>Δ</sup>*<sup>a</sup> <sup>i</sup>* (*ϕ*). Since *Bi*(*a*, *ϕ*) → *Bi*(1, *Bi*(*a*, *ϕ*)), so *Bi*(1, *Bi*(*a*, *ϕ*)) ∈ Γ, thereby *μi*,Γ(*X*(*Bi*(*a*, *ϕ*))) = *sup*({*b*|*Bi*(*b*, *Bi*(*a*, *<sup>ϕ</sup>*)) <sup>∈</sup> <sup>Γ</sup>}) = 1. For *<sup>X</sup>*(*Bi*(*a*, *<sup>ϕ</sup>*)) <sup>⊆</sup> <sup>Δ</sup>*<sup>a</sup> <sup>i</sup>* (*ϕ*), we have *<sup>μ</sup>i*,Γ(Δ*<sup>a</sup> <sup>i</sup>* (*ϕ*)) = 1 as desired.

**Lemma 2.12** For any *ϕ*, let Θ*<sup>a</sup> <sup>i</sup>* (*ϕ*) = {Γ� |*μi*,Γ�(*X*(*ϕ*)) < *a*}. If *μi*,Γ(*X*(*ϕ*)) < *a*, then *μi*,Γ(Θ*<sup>a</sup> <sup>i</sup>* (*ϕ*)) = 1.

*Proo f* . By the construction of the canonical model and Lemma 2.8, ¬*Bi*(*a*, *ϕ*) ∈ Γ ⇒ Γ ∈ Θ*a <sup>i</sup>* (*ϕ*), so *<sup>X</sup>*(¬*Bi*(*a*, *<sup>ϕ</sup>*)) <sup>⊆</sup> <sup>Θ</sup>*<sup>a</sup> <sup>i</sup>* (*ϕ*). Since ¬*Bi*(*a*, *ϕ*) → *Bi*(1, ¬*Bi*(*a*, *ϕ*)), so ¬*Bi*(*a*, *ϕ*) ∈ Γ ⇒ *Bi*(1, ¬*Bi*(*a*, *ϕ*)) ∈ Γ, thereby *μi*,Γ(*X*(¬*Bi*(*a*, *ϕ*))) = *sup*({*b*|*Bi*(*b*, ¬*Bi*(*a*, *ϕ*)) ∈ Γ}) = 1. For *<sup>X</sup>*(¬*Bi*(*a*, *<sup>ϕ</sup>*)) <sup>⊆</sup> <sup>Θ</sup>*<sup>a</sup> <sup>i</sup>* (*ϕ*), we have *<sup>μ</sup>i*,Γ(Θ*<sup>a</sup> <sup>i</sup>* (*ϕ*)) = 1 as desired.

**Lemma 2.13** For any Γ, *Pi*(Γ) is a *PBLω*-probability space.

*Proo f* . By Lemma 2.4-2.12, it is obvious.

**Lemma 2.14** The model *PM* is a *PBLω*-probabilistic model.

*Proo f* . It follows from Lemma 2.4 and Lemma 2.13.

The above lemmas state that the probability space *Pi*(Γ)=(*S*, *Xi*,Γ, *μi*,Γ) of the model satisfies all conditions in Definition 2.2, then as a consequence, the model *PM* is a *PBLω*-probabilistic model. In order to get the completeness, we further prove the following lemma, which states that *PM* is "canonical".

**Lemma 2.15** In the model *PM*, for any Γ and any *ϕ*, (*PM*, Γ) |= *ϕ* ⇔ *ϕ* ∈ Γ.

*Proo f* . We argue by the cases on the structure of *ϕ*, here we only give the proof in the case of *ϕ* ≡ *Bi*(*a*, *ψ*).

It suffices to prove that: (*PM*, Γ) |= *Bi*(*a*, *ψ*) ⇔ *Bi*(*a*, *ψ*) ∈ Γ.

If *Bi*(*a*, *ψ*) ∈ Γ, by the definition of *PM*, *μi*,Γ(*X*(*ψ*)) = *b* ≥ *a*, therefore (*PM*, Γ) |= *Bi*(*a*, *ψ*).

If *Bi*(*a*, *ψ*) ∈/ Γ, by Lemma 2.8, there exists *b* = *sup*({*c*|*Bi*(*c*, *ψ*) ∈ Γ}) such that *Bi*(*b*, *ψ*) ∈ Γ and *a* > *b*. By the definition of *PM*, *μi*,Γ(*X*(*ψ*)) = *b*, therefore (*PM*, Γ) �|= *Bi*(*a*, *ψ*).

Now it is ready to get the completeness of *PBLω*:

**Proposition 2.2** (Completeness of *PBLω*) If Γ |=*PBL<sup>ω</sup> ϕ*, then Γ �*PBL<sup>ω</sup> ϕ*.

12 Will-be-set-by-IN-TECH

*ϕ*2) = *A*<sup>1</sup> ∪ *A*2, we have *μi*,Γ(*A*<sup>1</sup> ∪ *A*2) = *μi*,Γ(*X*(*ϕ*<sup>1</sup> ∨ *ϕ*2)) ≥ *μi*,Γ(*X*(*ϕ*1)) + *μi*,Γ(*X*(*ϕ*2)) =

Now, we prove *μi*,Γ(*A*1) + *μi*,Γ(*A*2) ≥ *μi*,Γ(*A*<sup>1</sup> ∪ *A*2). Suppose *μi*,Γ(*A*<sup>1</sup> ∪ *A*2) > *μi*,Γ(*A*1) + *μi*,Γ(*A*2). Let *μi*,Γ(*A*1) = *a*1, *μi*,Γ(*A*2) = *a*2, *μi*,Γ(*A*<sup>1</sup> ∪ *A*2) > *a*<sup>1</sup> + *a*2. Choose *e* > 0 such that

Therefore we have Γ |= *Bi*(*a*<sup>1</sup> + *a*<sup>2</sup> + 2*e*, *ϕ*<sup>1</sup> ∨ *ϕ*2) ∧ ¬*Bi*(*a*<sup>1</sup> + *e*, *ϕ*1) ∧ ¬*Bi*(*a*<sup>2</sup> + *e*, *ϕ*2). Since Γ is a maximal consistent set, so Γ � *Bi*(*a*<sup>1</sup> + *a*<sup>2</sup> + 2*e*, *ϕ*<sup>1</sup> ∨ *ϕ*2) ∧ ¬*Bi*(*a*<sup>1</sup> + *e*, *ϕ*1) ∧ ¬*Bi*(*a*<sup>2</sup> + *e*, *ϕ*2),

Since *Bi*(*a*, *ϕ*) → *Bi*(1, *Bi*(*a*, *ϕ*)), so *Bi*(1, *Bi*(*a*, *ϕ*)) ∈ Γ, thereby *μi*,Γ(*X*(*Bi*(*a*, *ϕ*))) =

*Proo f* . By the construction of the canonical model and Lemma 2.8, ¬*Bi*(*a*, *ϕ*) ∈ Γ ⇒ Γ ∈

*Bi*(1, ¬*Bi*(*a*, *ϕ*)) ∈ Γ, thereby *μi*,Γ(*X*(¬*Bi*(*a*, *ϕ*))) = *sup*({*b*|*Bi*(*b*, ¬*Bi*(*a*, *ϕ*)) ∈ Γ}) = 1. For

The above lemmas state that the probability space *Pi*(Γ)=(*S*, *Xi*,Γ, *μi*,Γ) of the model satisfies all conditions in Definition 2.2, then as a consequence, the model *PM* is a *PBLω*-probabilistic model. In order to get the completeness, we further prove the following lemma, which states

*Proo f* . We argue by the cases on the structure of *ϕ*, here we only give the proof in the case of

If *Bi*(*a*, *ψ*) ∈ Γ, by the definition of *PM*, *μi*,Γ(*X*(*ψ*)) = *b* ≥ *a*, therefore (*PM*, Γ) |= *Bi*(*a*, *ψ*).

and *a* > *b*. By the definition of *PM*, *μi*,Γ(*X*(*ψ*)) = *b*, therefore (*PM*, Γ) �|= *Bi*(*a*, *ψ*).

**Proposition 2.2** (Completeness of *PBLω*) If Γ |=*PBL<sup>ω</sup> ϕ*, then Γ �*PBL<sup>ω</sup> ϕ*.

If *Bi*(*a*, *ψ*) ∈/ Γ, by Lemma 2.8, there exists *b* = *sup*({*c*|*Bi*(*c*, *ψ*) ∈ Γ}) such that *Bi*(*b*, *ψ*) ∈ Γ

*<sup>i</sup>* (*ϕ*)) = 1 as desired.


*<sup>i</sup>* (*ϕ*), we have *<sup>μ</sup>i*,Γ(Δ*<sup>a</sup>*


*<sup>i</sup>* (*ϕ*). Since ¬*Bi*(*a*, *ϕ*) → *Bi*(1, ¬*Bi*(*a*, *ϕ*)), so ¬*Bi*(*a*, *ϕ*) ∈ Γ ⇒

*<sup>i</sup>* (*ϕ*), therefore *<sup>X</sup>*(*Bi*(*a*, *<sup>ϕ</sup>*)) <sup>⊆</sup> <sup>Δ</sup>*<sup>a</sup>*

*<sup>i</sup>* (*ϕ*).

*<sup>i</sup>* (*ϕ*)) = 1 as

2*e* + *a*<sup>1</sup> + *a*<sup>2</sup> = *μi*,Γ(*A*<sup>1</sup> ∪ *A*2), then *μi*,Γ(*A*1) < *a*<sup>1</sup> + *e*, *μi*,Γ(*A*2) < *a*<sup>2</sup> + *e*.

*<sup>i</sup>* (*ϕ*) = {Γ�

*<sup>i</sup>* (*ϕ*) = {Γ�

**Lemma 2.15** In the model *PM*, for any Γ and any *ϕ*, (*PM*, Γ) |= *ϕ* ⇔ *ϕ* ∈ Γ.

*μi*,Γ(*A*1) + *μi*,Γ(*A*2).

*<sup>i</sup>* (*ϕ*)) = 1.

*<sup>i</sup>* (*ϕ*)) = 1.

*<sup>X</sup>*(¬*Bi*(*a*, *<sup>ϕ</sup>*)) <sup>⊆</sup> <sup>Θ</sup>*<sup>a</sup>*

that *PM* is "canonical".

*ϕ* ≡ *Bi*(*a*, *ψ*).

*μi*,Γ(Δ*<sup>a</sup>*

desired.

*μi*,Γ(Θ*<sup>a</sup>*

Θ*a*

which contradicts *Axiom* 6 of *PBLω*. **Lemma 2.11** For any *ϕ*, let Δ*<sup>a</sup>*

**Lemma 2.12** For any *ϕ*, let Θ*<sup>a</sup>*

*<sup>i</sup>* (*ϕ*), so *<sup>X</sup>*(¬*Bi*(*a*, *<sup>ϕ</sup>*)) <sup>⊆</sup> <sup>Θ</sup>*<sup>a</sup>*

*Proo f* . By Lemma 2.4-2.12, it is obvious.

*Proo f* . It is clear that *Bi*(*a*, *<sup>ϕ</sup>*) <sup>∈</sup> <sup>Γ</sup> <sup>⇒</sup> <sup>Γ</sup> <sup>∈</sup> <sup>Δ</sup>*<sup>a</sup>*

*sup*({*b*|*Bi*(*b*, *Bi*(*a*, *<sup>ϕ</sup>*)) <sup>∈</sup> <sup>Γ</sup>}) = 1. For *<sup>X</sup>*(*Bi*(*a*, *<sup>ϕ</sup>*)) <sup>⊆</sup> <sup>Δ</sup>*<sup>a</sup>*

*<sup>i</sup>* (*ϕ*), we have *<sup>μ</sup>i*,Γ(Θ*<sup>a</sup>*

**Lemma 2.13** For any Γ, *Pi*(Γ) is a *PBLω*-probability space.

**Lemma 2.14** The model *PM* is a *PBLω*-probabilistic model.

It suffices to prove that: (*PM*, Γ) |= *Bi*(*a*, *ψ*) ⇔ *Bi*(*a*, *ψ*) ∈ Γ.

Now it is ready to get the completeness of *PBLω*:

*Proo f* . It follows from Lemma 2.4 and Lemma 2.13.

*Proo f* . Suppose not, then there is a *PBL<sup>ω</sup>* - consistent formulas set Φ = Γ ∪ {¬*ϕ*}, and there is no model *PM* such that Φ is satisfied in *PM*. For there is a *PBL<sup>ω</sup>* - maximal consistent formula set Σ such that Φ ⊆ Σ, by Lemma 2.15, Φ is satisfied in possible world Σ of *PM*. It is a contradiction.

Our proof of the above completeness is different from the proof in [3]. The main idea of our proof is to give a canonical model, which can be regarded as a generalization of canonical model method in Kripke semantics. In [3], Fagin and Halpern adopt another technique to get the completeness. Let *ϕ* be consistent with *AXMEAS*, they show firstly that an *i*-probability formula *<sup>ψ</sup>* <sup>∈</sup> *Sub*+(*ϕ*) is provably equivalent to a formula of the form <sup>Σ</sup>*s*∈*Scsμi*(*ϕs*) <sup>≥</sup> *<sup>b</sup>*, for some appropriate coefficients *cs*, where *S* consists of all maximal consistent subsets of *Sub*+(*ϕ*). Then for a fixed agent *i* and a fixed state *s*, they describe a set of linear equalities and inequalities corresponding to *i* and *s*, over variables of the form *xiss*� , for *s*� ∈ *S*. We can think of *xiss*� as representing *μi*,*s*(*s*� ), i.e., the probability of state *s*� under agent *i*'s probability distribution at state *<sup>s</sup>*. Assume that *<sup>ψ</sup>* is equivalent to <sup>Σ</sup>*s*∈*Scsμi*(*ϕs*) ≥ *<sup>b</sup>*. Observe that exactly one of *ψ* and ¬*ψ* is in *s*. If *ψ* ∈ *s*, then the corresponding inequality is Σ*s*�∈*Scs*� *xiss*� ≥ *b*. If ¬*<sup>ψ</sup>* ∈ *<sup>s</sup>*, then the corresponding inequality is <sup>Σ</sup>*s*�∈*Scs*� *xiss*� < *<sup>b</sup>*. Finally, we have the equality <sup>Σ</sup>*s*�∈*Sxiss*� = 1. As shown in Theorem 2.2 in Fagin et al. [4], since*ϕ<sup>s</sup>* is consistent, this set of linear equalities and inequalities has a solution *x*∗ *iss*� , *s*� ∈ *S*. From their idea, it is clear that their proof depends tightly on the axioms of linear equalities and inequalities, whereas there are no such axioms in our inference system. On the other hand, their proof cannot deal with the case of infinite set of formulas, because in this case, we will get an infinite set of linear equalities and inequalities, which contains infinite variables. But their axioms seem insufficient to describe the existence of solutions of an infinite set of linear equalities and inequalities.

Proposition 2.1 and Proposition 2.2 show that the axioms and inference rules of *PBL<sup>ω</sup>* give us a sound and complete axiomatization for probabilistic belief. Moreover, we can prove the finite model property and decidability of the provability problem for some variants of *PBL<sup>ω</sup>* in the following sections.

It is not difficult to see that *Axioms* 1-6 and *Rules* 1-4 are not complete for our model. Because otherwise, the compactness property of *PBL<sup>ω</sup>* holds, but we can give the following example to show that the compactness property fails in *PBLω*: any finite sub set of {*Bi*(1/2, *ϕ*), *Bi*(2/3, *ϕ*), ..., *Bi*(*n*/*n* + 1, *ϕ*),...} ∪ {¬*Bi*(1, *ϕ*)} has model, whereas the whole set does not. But we do not know whether *Axioms* 1-6 and *Rules* 1-5 are complete for our model, i.e., whether *Rule* 6 is redundant in the inference system. Although we believe that *Rule* 6 is not redundant, we have no proof up to now.
