**3.4 Finite model property of** *PBLf*

We now turn our attention to the finite model property of *PBLf* . It needs to show that if a formula is *PBLf* -consistent, then it is satisfiable in a finite structure. The idea is that rather than considering maximal consistent formulas set when trying to construct a structure satisfying a formula *ϕ*, we restrict our attention to sets of subformulas of *ϕ*.

**Definition 3.3** Suppose *ζ* is a consistent formula with respect to *PBLf* , *Sub*∗(*ζ*) is a set of formulas defined as follows: let *<sup>ζ</sup>* <sup>∈</sup> *<sup>L</sup>PBLf* , *Sub*(*ζ*) is the set of subformulas of *<sup>ζ</sup>*, then *Sub*∗(*ζ*) = *Sub*(*ζ*) ∪ {¬*ψ*|*ψ* ∈ *Sub*(*ζ*)}. It is clear that *Sub*∗(*ζ*) is finite.

**Definition 3.4** The inner probabilistic model *PM<sup>ζ</sup>* with respect to formula *ζ* is (*S<sup>ζ</sup>* , *P*1,*<sup>ζ</sup>* , ..., *Pn*,*<sup>ζ</sup>* , *πζ* ).

(1) Here *S<sup>ζ</sup>* = {Γ|Γ is a maximal consistent formulas set with respect to *PBLf* and Γ ⊆ *Sub*∗(*ζ*)}.

(2) For any Γ ∈ *S<sup>ζ</sup>* , *Pi*,*<sup>ζ</sup>* (Γ)=(*S<sup>ζ</sup>* , *X<sup>ζ</sup>* , *μζ*,*i*,Γ), where *X<sup>ζ</sup>* = {*X*(*ϕ*)| *X*(*ϕ*) = {Γ� | *ϕ* is a Boolean combination of formulas in *Sub*∗(*ζ*) and Γ �*PBLf ϕ*}}; *μζ*,*i*,<sup>Γ</sup> is an inner probability assignment: *X<sup>ζ</sup>* → [0, 1], and *μζ*,*i*,Γ(*X*(*ϕ*)) = *sup*({*a*|*Bi*(*a*, *ϕ*) is provable from Γ in *PBLf* }).

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

hence *μζ*,*i*,Γ(*S<sup>ζ</sup>* ) ≥ *μζ*,*i*,Γ(*S<sup>ζ</sup>* ) + *μζ*,*i*,Γ(∅). Since *μζ*,*i*,Γ(*S<sup>ζ</sup>* ) = 1, so 1 ≥ 1 + *μζ*,*i*,Γ(∅), therefore

Probabilistic Belief Logics for Uncertain Agents 33

*Proo f* . Since *A*1, *A*<sup>2</sup> ∈ *X<sup>ζ</sup>* , assume *A*<sup>1</sup> = *X*(*ϕ*), *A*<sup>2</sup> = *X*(*ψ*). If *X*(*ϕ*) ⊆ *X*(*ψ*), by rule: � *ϕ* → *ψ* ⇒ � *Bi*(*a*, *ϕ*) → *Bi*(*a*, *ψ*), we have *μζ*,*i*,Γ(*X*(*ϕ*)) ≤ *μζ*,*i*,Γ(*X*(*ψ*)). Therefore if *A*1, *A*<sup>2</sup> ∈ *X<sup>ζ</sup>* and

**Lemma 3.8** If *A*1, *A*<sup>2</sup> ∈ *X<sup>ζ</sup>* and *A*<sup>1</sup> ∩ *A*<sup>2</sup> = ∅, then *μζ*,*i*,Γ(*A*<sup>1</sup> ∪ *A*2) ≥ *μζ*,*i*,Γ(*A*1) + *μζ*,*i*,Γ(*A*2). *Proo f* . Since *A*1, *A*<sup>2</sup> ∈ *X<sup>ζ</sup>* , assume *A*<sup>1</sup> = *X*(*ϕ*), *A*<sup>2</sup> = *X*(*ψ*), by rule: � ¬(*ϕ* ∧ *ψ*) ⇒ � *Bi*(*a*, *ϕ*) ∧ *Bi*(*b*, *ψ*) → *Bi*(*a* + *b*, *ϕ* ∨ *ψ*), where *a*<sup>1</sup> + *a*<sup>2</sup> ≤ 1, we have *μζ*,*i*,Γ(*X*(*ϕ*) ∪ *X*(*ψ*)) ≥ *μζ*,*i*,Γ(*X*(*ϕ*)) + *μζ*,*i*,Γ(*X*(*ψ*)). Therefore if *A*1, *A*<sup>2</sup> ∈ *X<sup>ζ</sup>* and *A*<sup>1</sup> ∩ *A*<sup>2</sup> = ∅, then *μζ*,*i*,Γ(*A*<sup>1</sup> ∪ *A*2) ≥ *μζ*,*i*,Γ(*A*1) +

*Proo f* . Since *C*, *D* ∈ *X<sup>ζ</sup>* , assume *C* = *X*(*ϕ*), *D* = *X*(*ψ*), by axiom: *Bi*(*a*, *ϕ*) ∧ *Bi*(*b*, *ψ*) → *Bi*(*max*(*a* + *b* − 1, 0), *ϕ* ∧ *ψ*), we get *μζ*,*i*,Γ(*X*(*ϕ*) ∩ *X*(*ψ*)) ≥ *μζ*,*i*,Γ(*X*(*ϕ*)) + *μζ*,*i*,Γ(*X*(*ψ*)) − 1.

*Bi*(*a*, *ϕ*) ∧ *Bi*(*b*, *ψ*) → *Bi*(*max*(*a* + *b* − 1, 0), *ϕ* ∧ *ψ*), we have that ∧*Bi*(1, *ϕn*) → *Bi*(1, ∧*ϕn*),

*Proo f* . By the definition of *S<sup>ζ</sup>* , the cardinality of *S<sup>ζ</sup>* is no more than the cardinality of

Similar to the proof of completeness of *PBLω*, the above lemmas show that *PM<sup>ζ</sup>* is a finite

**Lemma 3.14** For the finite canonical model *PM<sup>ζ</sup>* , for any Γ ∈ *S<sup>ζ</sup>* and any *ϕ* ∈ *Sub*∗(*ζ*),

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

.

}, then *μζ*,*i*,Γ(*B*<sup>−</sup>

)}, then *μζ*,*i*,Γ(Λ*i*,Γ) = 1.

*<sup>i</sup>* (Γ). If *Bi*(*a*, *ϕ*) ∈ Γ, by rule: *Bi*(*a*, *ϕ*) → *Bi*(1, *Bi*(*a*, *ϕ*)), we get

, which means for any *A* ∈ *X<sup>ζ</sup>* , *μζ*,*i*,Γ(*A*) = *μζ*,*i*,Γ�(*A*), so Γ� ∈ Λ*i*,Γ,

*<sup>i</sup>* (Γ)) = 1.

*<sup>i</sup>* (Γ) = *<sup>X</sup>*(∧*Bi*(1,*ϕn*)∈Γ*ϕn*), by axiom:

. If ¬*Bi*(*a*, *ϕ*) ∈ Γ, by rule: ¬*Bi*(*a*, *ϕ*) →

*<sup>i</sup>* (Γ)) = 1, we get *μζ*,*i*,Γ(Λ*i*,Γ) = 1 as

. Therefore

*<sup>i</sup>* (Γ)) = 1.

*<sup>i</sup>* (Γ), hence ¬*Bi*(*a*, *ϕ*) ∈ Γ�


*<sup>i</sup>* (Γ), hence *Bi*(*a*, *ϕ*) ∈ Γ�

*<sup>i</sup>* (Γ) ⊆ Λ*i*,Γ. By Lemma 3.10, *μζ*,*i*,Γ(*B*<sup>−</sup>

**Lemma 3.12** For any Γ ∈ *S<sup>ζ</sup>* , *Pi*,*<sup>ζ</sup>* (Γ) is a *PBLf* -inner probability space. *Proo f* . By Lemma 3.6 to Lemma 3.11, we can get the claim immediately.

**Lemma 3.13** The inner probabilistic model *PM<sup>ζ</sup>* is a finite model.

*PBLf* -model and the following lemma states that *PM<sup>ζ</sup>* is canonical.

It suffices to prove: (*PM<sup>ζ</sup>* , Γ) |= *Bi*(*a*, *ψ*) ⇔ *Bi*(*a*, *ψ*) ∈ Γ.

**Lemma 3.7** If *A*1, *A*<sup>2</sup> ∈ *X<sup>ζ</sup>* and *A*<sup>1</sup> ⊆ *A*2, then *μζ*,*i*,Γ(*A*1) ≤ *μζ*,*i*,Γ(*A*2).

**Lemma 3.9** For any *C*, *D* ∈ *X<sup>ζ</sup>* , *μζ*,*i*,Γ(*C* ∩ *D*) ≥ *μζ*,*i*,Γ(*C*) + *μζ*,*i*,Γ(*D*) − 1.

*μζ*,*i*,Γ(∅) = 0 as desired.

*μζ*,*i*,Γ(*A*2).

**Lemma 3.10** Let *B*−

**Lemma 3.11** Let Λ*i*,<sup>Γ</sup> = {Γ�

*Proo f* . Suppose Γ� ∈ *B*<sup>−</sup>

*Bi*(*a*, *ϕ*) ∈ Γ iff *Bi*(*a*, *ϕ*) ∈ Γ�

and furthermore *B*−

(*PM<sup>ζ</sup>* , Γ) |= *ϕ* ⇔ *ϕ* ∈ Γ.

*ϕ* ≡ *Bi*(*a*, *ψ*):

desired.

*Bi*(1, *Bi*(*a*, *ϕ*)) ∈ Γ, for Γ� ∈ *B*<sup>−</sup>

<sup>℘</sup>(*Sub*∗(*ζ*)), which means <sup>|</sup>*S<sup>ζ</sup>* | ≤ <sup>2</sup>|*Sub*∗(*ζ*)<sup>|</sup>

*A*<sup>1</sup> ⊆ *A*2, then *μζ*,*i*,Γ(*A*1) ≤ *μζ*,*i*,Γ(*A*2).

*<sup>i</sup>* (Γ) = {Γ�

*Proo f* . For Γ is a finite formulas set, therefore *B*−

*Bi*(1, ¬*Bi*(*a*, *ϕ*)), we get *Bi*(1, ¬*Bi*(*a*, *ϕ*)) ∈ Γ, for Γ� ∈ *B*<sup>−</sup>

so *Bi*(1, <sup>∧</sup>*Bi*(1,*ϕn*)∈Γ*ϕn*) can be proved from <sup>Γ</sup> in *PBLf* , so *μζ*,*i*,Γ(*B*<sup>−</sup>


Similar to the proof of completeness of *PBLω*, we mainly need to show that the above canonical model *PM<sup>ζ</sup>* is a *PBLf* -inner probabilistic model. The following lemmas from Lemma 3.1 to Lemma 3.13 contribute to this purpose. Furthermore, Lemma 3.14 states that *PM<sup>ζ</sup>* is "canonical", i.e., for any consistent formula *ϕ* ∈ *Sub*∗(*ζ*), there is a state *s*, such that (*PM<sup>ζ</sup>* ,*s*) |= *ϕ*. Since we can prove that *PM<sup>ζ</sup>* is a finite model, these lemmas imply the finite model property of *PBLf* .

**Lemma 3.1** *S<sup>ζ</sup>* is a nonempty finite set.

*Proo f* . Since the rules and axioms of *PBLf* are consistent, *S<sup>ζ</sup>* is nonempty. For *Sub*∗(*ζ*) is a finite set, by the definition of *S<sup>ζ</sup>* , the cardinality of *S<sup>ζ</sup>* is no more than the cardinality of ℘(*Sub*∗(*ζ*)).

**Lemma 3.2** *X<sup>ζ</sup>* is the power set of *S<sup>ζ</sup>* .

*Proo f* . Firstly, since *Sub*∗(*ζ*) is finite, so if Γ ∈ *S<sup>ζ</sup>* then Γ is finite. We can let *ϕ*<sup>Γ</sup> be the conjunction of the formulas in <sup>Γ</sup>. Secondly, if *<sup>A</sup>* ⊆ *<sup>S</sup><sup>ζ</sup>* , then *<sup>A</sup>* = *<sup>X</sup>*(∨Γ∈*<sup>A</sup> <sup>ϕ</sup>*Γ). By the above argument, we have that *X<sup>ζ</sup>* is the power set of *S<sup>ζ</sup>* .

**Lemma 3.3** If *ϕ* is consistent (here *ϕ* is a Boolean combination of formulas in *Sub*∗(*ζ*)), then there exists Γ such that *ϕ* can be proved from Γ, here Γ is a maximal consistent set with respect to *PBLf* and Γ ⊆ *Sub*∗(*ζ*).

*Proo f* . For *ϕ* is a Boolean combination of formulas in *Sub*∗(*ζ*), therefore by regarding the formulas in *Sub*∗(*ζ*) as atomic formulas, *ϕ* can be represented as disjunctive normal form. Since *ϕ* is consistent, so there is a consistent disjunctive term in disjunctive normal form expression of *ϕ*, let such term be *ψ*1∧...∧*ψn*, then *ϕ* can be derived from the maximal consistent set Γ which contains {*ψ*1, ..., *ψn*}.

**Lemma 3.4** For any Γ ∈ *S<sup>ζ</sup>* , *Pi*,*<sup>ζ</sup>* (Γ) is well defined.

*Proo f* . It suffices to prove the following claim: if *X*(*ϕ*) = *X*(*ψ*), then *μζ*,*i*,Γ(*X*(*ϕ*)) = *μζ*,*i*,Γ(*X*(*ψ*)). If *X*(*ϕ*) = *X*(*ψ*), it is clear that � *ϕ* ↔ *ψ*. For suppose not, then *ϕ* ∧ ¬*ψ* is consistent, by Lemma 3.3, there is Γ� such that *ϕ* ∧ ¬*ψ* can be proved from Γ� , therefore Γ� ∈ *X*(*ϕ*) and Γ� ∈/ *X*(*ψ*), it is a contradiction. Thus � *ϕ* ↔ *ψ*. By rule: � *ϕ* → *ψ* ⇒ � *Bi*(*a*, *ϕ*) → *Bi*(*a*, *ψ*), we get � *Bi*(*a*, *ϕ*) ↔ *Bi*(*a*, *ψ*), which means *μζ*,*i*,Γ(*X*(*ϕ*)) = *μζ*,*i*,Γ(*X*(*ψ*)).

**Lemma 3.5** Let *Proζ*,*i*,Γ(*ϕ*) = {*a*|*Bi*(*a*, *ϕ*) can be proved from Γ in *PBLf* }, then *sup*(*Proζ*,*i*,Γ(*ϕ*)) ∈ *Proζ*,*i*,Γ(*ϕ*).

*Proo f* . Suppose *Proζ*,*i*,Γ(*ϕ*) = {*a*|*Bi*(*a*, *ϕ*) can be proved from Γ in *PBLf* }, therefore Γ � *Bi*(*an*, *ϕ*) for all *an* ∈ *Proζ*,*i*,Γ(*ϕ*), by *Rule* 5 of *PBLf* , Γ � *Bi*(*a*, *ϕ*), where *a* = *supn*∈*M*({*an*}) = *sup*(*Proζ*,*i*,Γ(*ϕ*)), so we get *sup*(*Proζ*,*i*,Γ(*ϕ*)) ∈ *Proζ*,*i*,Γ(*ϕ*) as desired.

**Lemma 3.6** If *A* ∈ *X<sup>ζ</sup>* , then 0 ≤ *μζ*,*i*,Γ(*A*) ≤ 1. Furthermore, *μζ*,*i*,Γ(∅) = 0, *μζ*,*i*,Γ(*S<sup>ζ</sup>* ) = 1.

*Proo f* . By the definition, if *Bi*(*a*, *ϕ*) is a well formed formula, then 0 ≤ *a* ≤ 1; furthermore, check the axioms and rules of *PBLf* , any formula derived from well formed formulas is also a well formed formula, so 0 ≤ *μζ*,*i*,Γ(*X*(*ϕ*)) ≤ 1. Therefore, if *A* ∈ *X<sup>ζ</sup>* , then 0 ≤ *μζ*,*i*,Γ(*A*) ≤ 1.

By rule: � *ϕ* ⇒ � *Bi*(1, *ϕ*), therefore we have *μζ*,*i*,Γ(*S<sup>ζ</sup>* ) = 1 as desired. By axiom: *Bi*(0, *ϕ*), we get *Bi*(0, *f alse*), so *μζ*,*i*,Γ(∅) ≥ 0. By rule: � ¬(*ϕ* ∧ *ψ*) ⇒ � *Bi*(*a*, *ϕ*) ∧ *Bi*(*b*, *ψ*) → *Bi*(*a* + *b*, *ϕ* ∨ *ψ*), where *a* + *b* ≤ 1, we have � *Bi*(*a*, *f alse*) ∧ *Bi*(*b*, *true*) → *Bi*(*a* + *b*, *f alse* ∨ *true*), 16 Will-be-set-by-IN-TECH

Similar to the proof of completeness of *PBLω*, we mainly need to show that the above canonical model *PM<sup>ζ</sup>* is a *PBLf* -inner probabilistic model. The following lemmas from Lemma 3.1 to Lemma 3.13 contribute to this purpose. Furthermore, Lemma 3.14 states that *PM<sup>ζ</sup>* is "canonical", i.e., for any consistent formula *ϕ* ∈ *Sub*∗(*ζ*), there is a state *s*, such that (*PM<sup>ζ</sup>* ,*s*) |= *ϕ*. Since we can prove that *PM<sup>ζ</sup>* is a finite model, these lemmas imply the finite

*Proo f* . Since the rules and axioms of *PBLf* are consistent, *S<sup>ζ</sup>* is nonempty. For *Sub*∗(*ζ*) is a finite set, by the definition of *S<sup>ζ</sup>* , the cardinality of *S<sup>ζ</sup>* is no more than the cardinality of

*Proo f* . Firstly, since *Sub*∗(*ζ*) is finite, so if Γ ∈ *S<sup>ζ</sup>* then Γ is finite. We can let *ϕ*<sup>Γ</sup> be the conjunction of the formulas in <sup>Γ</sup>. Secondly, if *<sup>A</sup>* ⊆ *<sup>S</sup><sup>ζ</sup>* , then *<sup>A</sup>* = *<sup>X</sup>*(∨Γ∈*<sup>A</sup> <sup>ϕ</sup>*Γ). By the above

**Lemma 3.3** If *ϕ* is consistent (here *ϕ* is a Boolean combination of formulas in *Sub*∗(*ζ*)), then there exists Γ such that *ϕ* can be proved from Γ, here Γ is a maximal consistent set with respect

*Proo f* . For *ϕ* is a Boolean combination of formulas in *Sub*∗(*ζ*), therefore by regarding the formulas in *Sub*∗(*ζ*) as atomic formulas, *ϕ* can be represented as disjunctive normal form. Since *ϕ* is consistent, so there is a consistent disjunctive term in disjunctive normal form expression of *ϕ*, let such term be *ψ*1∧...∧*ψn*, then *ϕ* can be derived from the maximal consistent

*Proo f* . It suffices to prove the following claim: if *X*(*ϕ*) = *X*(*ψ*), then *μζ*,*i*,Γ(*X*(*ϕ*)) = *μζ*,*i*,Γ(*X*(*ψ*)). If *X*(*ϕ*) = *X*(*ψ*), it is clear that � *ϕ* ↔ *ψ*. For suppose not, then *ϕ* ∧ ¬*ψ*

Γ� ∈ *X*(*ϕ*) and Γ� ∈/ *X*(*ψ*), it is a contradiction. Thus � *ϕ* ↔ *ψ*. By rule: � *ϕ* → *ψ* ⇒ � *Bi*(*a*, *ϕ*) → *Bi*(*a*, *ψ*), we get � *Bi*(*a*, *ϕ*) ↔ *Bi*(*a*, *ψ*), which means *μζ*,*i*,Γ(*X*(*ϕ*)) = *μζ*,*i*,Γ(*X*(*ψ*)). **Lemma 3.5** Let *Proζ*,*i*,Γ(*ϕ*) = {*a*|*Bi*(*a*, *ϕ*) can be proved from Γ in *PBLf* }, then

*Proo f* . Suppose *Proζ*,*i*,Γ(*ϕ*) = {*a*|*Bi*(*a*, *ϕ*) can be proved from Γ in *PBLf* }, therefore Γ � *Bi*(*an*, *ϕ*) for all *an* ∈ *Proζ*,*i*,Γ(*ϕ*), by *Rule* 5 of *PBLf* , Γ � *Bi*(*a*, *ϕ*), where *a* = *supn*∈*M*({*an*}) =

*Proo f* . By the definition, if *Bi*(*a*, *ϕ*) is a well formed formula, then 0 ≤ *a* ≤ 1; furthermore, check the axioms and rules of *PBLf* , any formula derived from well formed formulas is also a well formed formula, so 0 ≤ *μζ*,*i*,Γ(*X*(*ϕ*)) ≤ 1. Therefore, if *A* ∈ *X<sup>ζ</sup>* , then 0 ≤ *μζ*,*i*,Γ(*A*) ≤ 1. By rule: � *ϕ* ⇒ � *Bi*(1, *ϕ*), therefore we have *μζ*,*i*,Γ(*S<sup>ζ</sup>* ) = 1 as desired. By axiom: *Bi*(0, *ϕ*), we get *Bi*(0, *f alse*), so *μζ*,*i*,Γ(∅) ≥ 0. By rule: � ¬(*ϕ* ∧ *ψ*) ⇒ � *Bi*(*a*, *ϕ*) ∧ *Bi*(*b*, *ψ*) → *Bi*(*a* + *b*, *ϕ* ∨ *ψ*), where *a* + *b* ≤ 1, we have � *Bi*(*a*, *f alse*) ∧ *Bi*(*b*, *true*) → *Bi*(*a* + *b*, *f alse* ∨ *true*),

**Lemma 3.6** If *A* ∈ *X<sup>ζ</sup>* , then 0 ≤ *μζ*,*i*,Γ(*A*) ≤ 1. Furthermore, *μζ*,*i*,Γ(∅) = 0, *μζ*,*i*,Γ(*S<sup>ζ</sup>* ) = 1.

, therefore

is consistent, by Lemma 3.3, there is Γ� such that *ϕ* ∧ ¬*ψ* can be proved from Γ�

*sup*(*Proζ*,*i*,Γ(*ϕ*)), so we get *sup*(*Proζ*,*i*,Γ(*ϕ*)) ∈ *Proζ*,*i*,Γ(*ϕ*) as desired.

model property of *PBLf* .

to *PBLf* and Γ ⊆ *Sub*∗(*ζ*).

set Γ which contains {*ψ*1, ..., *ψn*}.

*sup*(*Proζ*,*i*,Γ(*ϕ*)) ∈ *Proζ*,*i*,Γ(*ϕ*).

℘(*Sub*∗(*ζ*)).

**Lemma 3.1** *S<sup>ζ</sup>* is a nonempty finite set.

**Lemma 3.2** *X<sup>ζ</sup>* is the power set of *S<sup>ζ</sup>* .

argument, we have that *X<sup>ζ</sup>* is the power set of *S<sup>ζ</sup>* .

**Lemma 3.4** For any Γ ∈ *S<sup>ζ</sup>* , *Pi*,*<sup>ζ</sup>* (Γ) is well defined.

hence *μζ*,*i*,Γ(*S<sup>ζ</sup>* ) ≥ *μζ*,*i*,Γ(*S<sup>ζ</sup>* ) + *μζ*,*i*,Γ(∅). Since *μζ*,*i*,Γ(*S<sup>ζ</sup>* ) = 1, so 1 ≥ 1 + *μζ*,*i*,Γ(∅), therefore *μζ*,*i*,Γ(∅) = 0 as desired.

**Lemma 3.7** If *A*1, *A*<sup>2</sup> ∈ *X<sup>ζ</sup>* and *A*<sup>1</sup> ⊆ *A*2, then *μζ*,*i*,Γ(*A*1) ≤ *μζ*,*i*,Γ(*A*2).

*Proo f* . Since *A*1, *A*<sup>2</sup> ∈ *X<sup>ζ</sup>* , assume *A*<sup>1</sup> = *X*(*ϕ*), *A*<sup>2</sup> = *X*(*ψ*). If *X*(*ϕ*) ⊆ *X*(*ψ*), by rule: � *ϕ* → *ψ* ⇒ � *Bi*(*a*, *ϕ*) → *Bi*(*a*, *ψ*), we have *μζ*,*i*,Γ(*X*(*ϕ*)) ≤ *μζ*,*i*,Γ(*X*(*ψ*)). Therefore if *A*1, *A*<sup>2</sup> ∈ *X<sup>ζ</sup>* and *A*<sup>1</sup> ⊆ *A*2, then *μζ*,*i*,Γ(*A*1) ≤ *μζ*,*i*,Γ(*A*2).

**Lemma 3.8** If *A*1, *A*<sup>2</sup> ∈ *X<sup>ζ</sup>* and *A*<sup>1</sup> ∩ *A*<sup>2</sup> = ∅, then *μζ*,*i*,Γ(*A*<sup>1</sup> ∪ *A*2) ≥ *μζ*,*i*,Γ(*A*1) + *μζ*,*i*,Γ(*A*2).

*Proo f* . Since *A*1, *A*<sup>2</sup> ∈ *X<sup>ζ</sup>* , assume *A*<sup>1</sup> = *X*(*ϕ*), *A*<sup>2</sup> = *X*(*ψ*), by rule: � ¬(*ϕ* ∧ *ψ*) ⇒ � *Bi*(*a*, *ϕ*) ∧ *Bi*(*b*, *ψ*) → *Bi*(*a* + *b*, *ϕ* ∨ *ψ*), where *a*<sup>1</sup> + *a*<sup>2</sup> ≤ 1, we have *μζ*,*i*,Γ(*X*(*ϕ*) ∪ *X*(*ψ*)) ≥ *μζ*,*i*,Γ(*X*(*ϕ*)) + *μζ*,*i*,Γ(*X*(*ψ*)). Therefore if *A*1, *A*<sup>2</sup> ∈ *X<sup>ζ</sup>* and *A*<sup>1</sup> ∩ *A*<sup>2</sup> = ∅, then *μζ*,*i*,Γ(*A*<sup>1</sup> ∪ *A*2) ≥ *μζ*,*i*,Γ(*A*1) + *μζ*,*i*,Γ(*A*2).

**Lemma 3.9** For any *C*, *D* ∈ *X<sup>ζ</sup>* , *μζ*,*i*,Γ(*C* ∩ *D*) ≥ *μζ*,*i*,Γ(*C*) + *μζ*,*i*,Γ(*D*) − 1.

*Proo f* . Since *C*, *D* ∈ *X<sup>ζ</sup>* , assume *C* = *X*(*ϕ*), *D* = *X*(*ψ*), by axiom: *Bi*(*a*, *ϕ*) ∧ *Bi*(*b*, *ψ*) → *Bi*(*max*(*a* + *b* − 1, 0), *ϕ* ∧ *ψ*), we get *μζ*,*i*,Γ(*X*(*ϕ*) ∩ *X*(*ψ*)) ≥ *μζ*,*i*,Γ(*X*(*ϕ*)) + *μζ*,*i*,Γ(*X*(*ψ*)) − 1.

**Lemma 3.10** Let *B*− *<sup>i</sup>* (Γ) = {Γ� |{*ϕ* : *Bi*(1, *ϕ*) ∈ Γ} ⊆ Γ� }, then *μζ*,*i*,Γ(*B*<sup>−</sup> *<sup>i</sup>* (Γ)) = 1.

*Proo f* . For Γ is a finite formulas set, therefore *B*− *<sup>i</sup>* (Γ) = *<sup>X</sup>*(∧*Bi*(1,*ϕn*)∈Γ*ϕn*), by axiom: *Bi*(*a*, *ϕ*) ∧ *Bi*(*b*, *ψ*) → *Bi*(*max*(*a* + *b* − 1, 0), *ϕ* ∧ *ψ*), we have that ∧*Bi*(1, *ϕn*) → *Bi*(1, ∧*ϕn*), so *Bi*(1, <sup>∧</sup>*Bi*(1,*ϕn*)∈Γ*ϕn*) can be proved from <sup>Γ</sup> in *PBLf* , so *μζ*,*i*,Γ(*B*<sup>−</sup> *<sup>i</sup>* (Γ)) = 1.

**Lemma 3.11** Let Λ*i*,<sup>Γ</sup> = {Γ� |*Pi*,*<sup>ζ</sup>* (Γ) = *Pi*,*<sup>ζ</sup>* (Γ� )}, then *μζ*,*i*,Γ(Λ*i*,Γ) = 1.

*Proo f* . Suppose Γ� ∈ *B*<sup>−</sup> *<sup>i</sup>* (Γ). If *Bi*(*a*, *ϕ*) ∈ Γ, by rule: *Bi*(*a*, *ϕ*) → *Bi*(1, *Bi*(*a*, *ϕ*)), we get *Bi*(1, *Bi*(*a*, *ϕ*)) ∈ Γ, for Γ� ∈ *B*<sup>−</sup> *<sup>i</sup>* (Γ), hence *Bi*(*a*, *ϕ*) ∈ Γ� . If ¬*Bi*(*a*, *ϕ*) ∈ Γ, by rule: ¬*Bi*(*a*, *ϕ*) → *Bi*(1, ¬*Bi*(*a*, *ϕ*)), we get *Bi*(1, ¬*Bi*(*a*, *ϕ*)) ∈ Γ, for Γ� ∈ *B*<sup>−</sup> *<sup>i</sup>* (Γ), hence ¬*Bi*(*a*, *ϕ*) ∈ Γ� . Therefore *Bi*(*a*, *ϕ*) ∈ Γ iff *Bi*(*a*, *ϕ*) ∈ Γ� , which means for any *A* ∈ *X<sup>ζ</sup>* , *μζ*,*i*,Γ(*A*) = *μζ*,*i*,Γ�(*A*), so Γ� ∈ Λ*i*,Γ, and furthermore *B*− *<sup>i</sup>* (Γ) ⊆ Λ*i*,Γ. By Lemma 3.10, *μζ*,*i*,Γ(*B*<sup>−</sup> *<sup>i</sup>* (Γ)) = 1, we get *μζ*,*i*,Γ(Λ*i*,Γ) = 1 as desired.

**Lemma 3.12** For any Γ ∈ *S<sup>ζ</sup>* , *Pi*,*<sup>ζ</sup>* (Γ) is a *PBLf* -inner probability space.

*Proo f* . By Lemma 3.6 to Lemma 3.11, we can get the claim immediately.

**Lemma 3.13** The inner probabilistic model *PM<sup>ζ</sup>* is a finite model.

*Proo f* . By the definition of *S<sup>ζ</sup>* , the cardinality of *S<sup>ζ</sup>* is no more than the cardinality of <sup>℘</sup>(*Sub*∗(*ζ*)), which means <sup>|</sup>*S<sup>ζ</sup>* | ≤ <sup>2</sup>|*Sub*∗(*ζ*)<sup>|</sup> .

Similar to the proof of completeness of *PBLω*, the above lemmas show that *PM<sup>ζ</sup>* is a finite *PBLf* -model and the following lemma states that *PM<sup>ζ</sup>* is canonical.

**Lemma 3.14** For the finite canonical model *PM<sup>ζ</sup>* , for any Γ ∈ *S<sup>ζ</sup>* and any *ϕ* ∈ *Sub*∗(*ζ*), (*PM<sup>ζ</sup>* , Γ) |= *ϕ* ⇔ *ϕ* ∈ Γ.

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

It suffices to prove: (*PM<sup>ζ</sup>* , Γ) |= *Bi*(*a*, *ψ*) ⇔ *Bi*(*a*, *ψ*) ∈ Γ.

problem of *PBLf* . Roughly speaking, let Γ be a finite set of formulas, weak completeness means Γ |= *ϕ* ⇒ Γ � *ϕ*, and decidability of the provability problem of *PBLf* means there is an algorithm that, given as input a formula *ϕ*, will decide whether *ϕ* is provable in *PBLf* .

Probabilistic Belief Logics for Uncertain Agents 35

**Definition 4.1** The set of well formed formulas set of *PBLr*, called *LPBLr* , is given by the

Remark: A significant difference between *PBLr* and *PBL�* (*PBLf* ) is that in the definition of

The inner probabilistic model of *PBLr* is the same as the inner probabilistic model of *PBLf* ,

The inference system of *PBLr* consists of axioms and inference rules of proposition logic and the *Axioms* 1-5 and *Rules* 1-4 of *PBL�*. But it is necessary to note that by the definition of well formed formulas of *PBLr*, all the probabilities in the axioms and inference rules of *PBLr* should be modified to be rational numbers. For example, *Axiom* 5 of *PBLω*: "*Bi*(*a*, *ϕ*) → *Bi*(*b*, *ϕ*), where 1 ≥ *a* ≥ *b* ≥ 0" should be modified as "*Bi*(*a*, *ϕ*) → *Bi*(*b*, *ϕ*), where 1 ≥ *a* ≥ *b* ≥ 0 and *a*, *b* are rational numbers" in *PBLr*. Since the probabilities *a* and *b* in the formulas *Bi*(*a*, *ϕ*) and *Bi*(*b*, *ψ*) are rational numbers, so the probability *max*(*a* + *b* − 1, 0) in the scope of *Bi*(*max*(*a* + *b* − 1, 0), *ϕ* ∧ *ψ*) in *Axiom* 2 and the probability *a* + *b* in the scope of *Bi*(*a* + *b*, *ϕ* ∨ *ψ*)

The proof of the soundness of *PBLr* is similar to the soundness of *PBLf* , and we do not give

In order to prove the weak completeness of *PBLr*, we first present a probabilistic belief logic - *PBLr*(*N*), where *N* is a given natural number. The finite model property of *PBLr*(*N*) is then proved. From this property, we get the weak completeness and the decidability of *PBLr*.

The syntax of *PBLr*(*N*) is the same as the syntax of *PBLr* except that the probabilities in formulas should be rational numbers like *k*/*N*. For example, every probability in formulas of *PBLr*(3) should be one of 0/3, 1/3, 2/3 or 3/3. Therefore, *Bi*(1/3, *ϕ*) and *Bi*(2/3, *Bj*(1/3, *ϕ*)) are well formed formulas in *PBLr*(3), but *Bi*(1/2, *ϕ*) is not a well formed formula in *PBLr*(3). The inner probabilistic model of *PBLr*(*N*) is also the same as *PBLr* except that the measure assigned to every possible world should be the form of *k*/*N* respectively. Therefore, in an inner probabilistic model of *PBLr*(3), the measure in a possible world may be 1/3, 2/3 and

The inference system of *PBLr*(*N*) is also similar to *PBLr* but all the probabilities in the axioms and inference rules should be the form of *k*/*N* respectively. For example, *Axiom* 5 of *PBLω*:

(4) If *<sup>ϕ</sup>* <sup>∈</sup> *<sup>L</sup>PBLr* and *<sup>a</sup>* is a rational number in [0,1], then *Bi*(*a*, *<sup>ϕ</sup>*) <sup>∈</sup> *<sup>L</sup>PBLr* .

syntax, the probability in the scope of *Bi*(*a*, *ϕ*) in the former is a rational number.

except that the value of *PBLr*-inner probability measure is a rational number.

**Proposition 4.1** (Soundness of *PBLr*) If Γ �*PBLr ϕ* then Γ |=*PBLr ϕ*.

**4.1 Finite model property and decidability of** *PBLr*

following grammar:

(1) If *<sup>ϕ</sup>* <sup>∈</sup>Atomic formulas set, then *<sup>ϕ</sup>* <sup>∈</sup> *<sup>L</sup>PBLr* ;

(3) If *<sup>ϕ</sup>*1,*ϕ*<sup>2</sup> <sup>∈</sup> *<sup>L</sup>PBLr* , then *<sup>ϕ</sup>*<sup>1</sup> <sup>∧</sup> *<sup>ϕ</sup>*<sup>2</sup> <sup>∈</sup> *<sup>L</sup>PBLr* ;

(2) If *<sup>ϕ</sup>* <sup>∈</sup> *<sup>L</sup>PBLr* , then <sup>¬</sup>*<sup>ϕ</sup>* <sup>∈</sup> *<sup>L</sup>PBLr* ;

in *Rule* 4 are also rational numbers.

etc, but can not be 1/2 or 1/4.

the details.

If *Bi*(*a*, *ψ*) ∈ Γ, by the definition of *PM<sup>ζ</sup>* , *μζ*,*i*,Γ(*X*(*ψ*)) = *b* ≥ *a*, therefore (*PM<sup>ζ</sup>* , Γ) |= *Bi*(*a*, *ψ*).

If *Bi*(*a*, *ψ*) ∈/ Γ, by Lemma 3.5, there exists *b* = *sup*({*c*|*Bi*(*c*, *ψ*) ∈ Γ}) such that *Bi*(*b*, *ψ*) ∈ Γ and *a* > *b*. By the definition of *PM<sup>ζ</sup>* , *μζ*,*i*,Γ(*X*(*ψ*)) = *b*, therefore (*PM<sup>ζ</sup>* , Γ) �|= *Bi*(*a*, *ψ*).

From the above lemmas, we know that *PM<sup>ζ</sup>* is a finite *PBLf* -model that is canonical. Now it is no difficult to get the following proposition.

**Proposition 3.2** (Finite model property of *PBLf* ) If Γ is a finite set of consistent formulas, then there is a finite *PBLf* -model *PM* such that *PM* |=*PBLf* Γ.

*Proo f* . By Lemma 3.14, there exists a finite *PBLf* -model *PM*∧<sup>Γ</sup> such that Γ is satisfied in *PM*∧Γ.

**Proposition 3.3** (Weak completeness of *PBLf* ) If Γ is a finite set of formulas, *ϕ* is a formula, and Γ |= *PBLf ϕ*, then Γ �*PBLf ϕ*.

*Proo f* . Suppose not, then (∧Γ) ∧ ¬*ϕ* is consistent with respect to *PBLf* , by Proposition 3.2, there exists an inner probabilistic model *PM*(∧Γ)∧¬*<sup>ϕ</sup>* such that (∧Γ) ∧ ¬*<sup>ϕ</sup>* is satisfied in *PM*(∧Γ)∧¬*ϕ*, but this contradicts our assumption that <sup>Γ</sup> <sup>|</sup><sup>=</sup> *PBLf <sup>ϕ</sup>*, thus the proposition holds.

As to *PBL<sup>ω</sup>* case, the construction of canonical model like Definition 3.4 fails to get the finite model property. The main problem lies in how to define measure assignment *μζ*,*i*,Γ, in Definition 3.4, *μζ*,*i*,Γ(*X*(*ϕ*)) = *sup*({*a*|*Bi*(*a*, *ϕ*) is provable from Γ in *PBLf* }), but *Rule* 6 fails under this definition. Thus there is an unsolved problem about how to construct a finite model with respect to a *PBLω*-consistent formula.

Usually, in the case of modal logics, one can get decidability of the provability problem from finite model property. At first, one can simply construct every model with finite (for example, say 2|*Sub*∗(*ϕ*)<sup>|</sup> ) states. One then check if *ϕ* is true at some state of one of these models (note that the number of models that have 2|*Sub*∗(*ϕ*)<sup>|</sup> states is finite). By finite model property, if a formula *ϕ* is consistent, then *ϕ* is satisfiable with respect to some models. Conversely, if *ϕ* is satisfiable with respect to some models, then *ϕ* is consistent.

But it becomes different for *PBLf* . Because there may be infinitely many *PBLf* -inner probability measure assigned to the set *X* (since real number in [0,1] is infinite), there are infinitely many probabilistic models associated to a given number of states (for example, say 2|*Sub*∗(*ϕ*)<sup>|</sup> ). Therefore the above argument fails. On the contrary, in the next section, we will present another variant-*PBLr*, and prove that the decidability of the provable problem holds for *PBLr*.
