This section proves the strong completeness of ΛREL with respect to the class of REL
models. The proof is based on the the completeness-via-canonicity approach. In particular, we construct of a canonicalRELmodel for each LREL-consistent theoryT0.
Definition 69 (Canonical model for T0). A canonical model for T0 is a structure Mc =
hWc,hRc,ci, Vc, Agci with:
• Wc:={T |T is a maximally consistent theory andR∀T
0T}
• Rc:={R0ϕ |ϕ∈L
REL and (∃0ϕ)∈T0}
• c is a priority order on RcwithR0 ≺R0> for allR0 ∈Rc\ {R0>} • Vcis a valuation function given by Vc(p) :=kpk
• Agc is an aggregator forWcwith
Agc(hR,i) =
(
R~π if hR,i=hRcupJ~πKM c,cupJ~πKM ci
Wc×Wc otherwise where:
• R∀is the relation onWcgiven by: R∀T Siff for allϕ∈L
REL: (∀ϕ)∈T ⇒(∀ϕ)∈S.
• for each ϕ∈ LREL, R0ϕ is the relation onWc given by: R0ϕT S iff 0ϕ∈ T ⇒
0ϕ∈S.
• for each ϕ∈LREL,kϕk:={T ∈Wc|ϕ∈T}
• for each~π ∈S0(Π∗),R~π is the relation onWcgiven by: R~πT S iff for allϕ∈LREL: ~πϕ∈T ⇒ϕ∈S.
/
4.2. REL+: dynamics of prioritized addition 91
Proposition 20. Mc is aREL model.
Proof. In order to show that Mc is anREL model, we have to show that: 1. Rc is a family of evidence, i.e., everyR∈R is a preorder.
2. R0>=Wc×Wc, i.e., the trivial evidence order is an element of Rc, as required. 3. R~π is a preorder for each~π, and thus Agc is well-defined.
The rest of the model meets the conditions of a REL model, so let’s turn to the three points just indicated.
For item 1, let ϕ∈ L be arbitrary. Let R ∈ R be arbitrary. Then R = R0ϕ for some
ϕ. The reflexivity of R is immediate from the definition of R0ϕ. For the transitivity, let
T, S, U ∈Mcand suppose thatR0ϕT S andR0ϕSU. Either 0ϕ6∈T or0ϕ∈T. Note that, by definition of R0ϕ, if0ϕ6∈T, then R0ϕ[T] =Wc and thus R0ϕT U. Suppose
now that 0ϕ ∈ T. Then by definition of R0ϕ, given R0ϕT S we have 0ϕ ∈ S, and
thus asR0ϕSU we get0ϕ∈U, which implies R0ϕT U.
For item 2, observe thatN, i.e.,0>, is an axiom of our system. Thus it is a member of
any maximal consistent set, which implies thatR0> =Wc×Wc.
For item 3, take anyR~π. For reflexivity, suppose that(~πϕ)∈T for someT ∈Mc. AsT~π
is an axiom andT is maximal consistent,(~πϕ→ϕ)∈T. As (~πϕ)∈T and T is closed under modus ponens, we have ϕ ∈ T. Thus R~πT T. For transitivity, let T, S, U ∈ Mc
and suppose that R~πT S and R~πSU. Suppose (~πϕ) ∈ T. As 4~π is an axiom and T is maximally consistent,(~πϕ→~π~πϕ)∈T. As (~πϕ)∈T andT is closed under modus
ponens, we have ~π~πϕ ∈ T. As R~πT S, we then have ~πϕ ∈ S. Hence, as R~πSU, we
haveϕ∈U. As ϕwas arbitrary, this holds for eachϕand hence we haveR~πT U.
Having established thatMcis a REL model, we prove now the standard lemmas to show that the canonical model works as expected.
Lemma 13 (Existence Lemma for∀). k∃ϕk 6=∅ iff kϕk 6=∅. Proof.
(⇒). Assume T ∈ k∃ϕk, i.e.,(∃ϕ)∈T ∈Wc. We first prove the following:
Claim. The set Γ :={∀ψ|(∀ψ)∈T} ∪ {ϕ} is consistent.
Proof. Suppose that Γ is inconsistent, i.e., Γ `LREL ⊥. Then there are finitely many sentences∀ψ1, . . . ,∀ψn∈T such that`LREL ∀ψ1∧ · · · ∧ ∀ψn→ ¬ϕ. By Necessitation for
∀we have`LREL ∀(∀ψ∧ · · · ∧ ∀ψn→ ¬ϕ)and from this, by K∀ and modus ponens we get
`LREL ∀(∀ψ1∧ · · · ∧ ∀ψn) → ∀¬ϕ. The system S5 has the theorem `LREL (∀∀ψ1∧ · · · ∧
∀∀ψn)→ ∀(∀ψ1∧ · · · ∧ ∀ψn) (see, e.g., [29, p. 20]). Hence by propositional logic we have `LREL (∀∀ψ1∧· · ·∧∀∀ψn)→ ∀¬ϕ. Given4∀we have`LREL ∀ψ1→ ∀∀ψ1, . . . ,`LREL∀ψn→
∀∀ψn, which by propositional logic implies `LREL (∀ψ1∧ · · · ∧ ∀ψn)→ ∀∀ψ1∧ · · · ∧ ∀∀ψn. Thus we have`LREL(∀ψ1∧· · ·∧∀ψn)→ ∀¬ϕ. Hence asT is maximal consistent and closed
under modus ponens, we get(∀¬ϕ) ∈T. But we also have (∃ϕ) ∈T, i.e., (¬∀¬ϕ) ∈ T, and sinceT is maximal consistent, this means that (∀¬ϕ)6∈T. Contradiction.
Given the Claim, by Lindenbaum’s Lemma, there is some maximally consistent the- ory S such that Γ ⊆ S. As ϕ ∈ Γ we have ϕ ∈ S. Moreover, as {∀ψ | (∀ψ) ∈
T} ⊆ {∀χ | (∀χ) ∈ S} we have R∀T S. As T ∈ Wc, we also have R∀T
0T. That is,
{∀θ|(∀θ)∈ T0} ⊆ {∀ψ |(∀ψ) ∈T}. Thus {∀θ |(∀θ) ∈T0} ⊆ {∀χ|(∀χ) ∈S} and thus
R∀T0S. HenceS∈Wc, which together withϕ∈S gives us S ∈ kϕk.
(⇐) AssumeT ∈ kϕk, i.e., ϕ∈T. Given T∀ we have`LREL ∀¬ϕ→ ¬ϕ, and by contrapo- sition we get `LREL ¬¬ϕ→ ¬∀¬ϕ, i.e., `LRELϕ→ ∃ϕ. Hence (ϕ→ ∃ϕ)∈T and as T is closed under modus ponens, given alsoϕ∈T we get(∃ϕ)∈T, i.e.,T ∈ k∃ϕk.
Lemma 14 (Existence Lemma for ~π). Let ~π =hπ1, . . . , πni be arbitrary. T ∈ k♦~πϕk iff
there is an S∈ kϕk such that R~πT S.
Proof.
(⇒). Assume T ∈ k♦~πϕk, i.e., ♦~πϕ∈T ∈Wc. We first prove the following:
Claim. The set Γ :={ψ|(~πψ)∈T} ∪ {∀θ| ∀θ∈T0} ∪ {ϕ}is consistent.
Proof. Suppose that Γ is inconsistent. Then there is a finite Γ0 ⊆ Γ such that Γ0 `LREL
⊥. By the theorems `LREL ~π(ψ
i1 ∧ · · · ∧ψin) ↔ ( ~πψ i1 ∧ · · · ∧ ~ πψ in) and `LREL
∀(θj1 ∧ · · · ∧θjn)↔ (∀θj1 ∧ · · · ∧ ∀θjn) we can assume that Γ0 ={
~
πψ,∀θ,¬ϕ} for some
~πψ,∀θ∈T. That is, we have`LREL
~
πψ∧ ∀θ→ ¬ϕ. By Necessitation for~π we obtain `LREL ~π(π~ψ∧ ∀θ → ¬ϕ). From this, by K
~π we get `LREL ~π(~πψ∧ ∀θ) → ~π¬ϕ.
By the theorem `LREL ~π(~πψ∧ ∀θ) ↔ (~π~πψ∧~π∀θ), from propositional logic we
get `LREL (
~
π~πψ∧~π∀θ) → ~π¬ϕ. Given the axioms in our system we have `
LREL
~πψ→~π~πψand`
LREL∀(∀θ)→
~π(∀θ). Using these, by propositional logic we obtain `LREL (~πψ∧ ∀∀θ)→~π¬ϕ. Given our axioms, we also have`
LREL∀θ→ ∀∀θ. Hence by propositional logic we get `LREL (
~πψ∧ ∀θ) → ~π¬ϕ. As ~πψ,∀θ ∈T and T is closed
under modus ponens, we get(~π¬ϕ)∈T. But we also have(♦~πϕ)∈T, i.e.,(¬~π¬ϕ)∈T,
and sinceT is maximal consistent, this means that (~π¬ϕ)6∈T. Contradiction.
Given the Claim, by Lindenbaum’s Lemma, there is some maximally consistent theoryS
such thatΓ ⊆S. As ϕ∈Γ we have ϕ∈ S. Moreover, as{ψ |(~πψ) ∈T} ⊆ S, we have
R~πT S. Additionally, we have {∀θ | (∀θ) ∈ T0} ⊆ S and thus R∀T0S. Hence S ∈ Wc,
which together withϕ∈S gives us S ∈ kϕk.
(⇐) Assume T ∈ kϕk, i.e., ϕ ∈ T. Given T~π we have `LREL ~π¬ϕ → ¬ϕ, and by
contraposition we get`LREL ¬¬ϕ→ ¬~π¬ϕ, i.e.,`
LREL ϕ→ ♦
~
πϕ. Hence,(ϕ→♦~πϕ)∈
T and as T is closed under modus ponens, given also ϕ ∈ T we get (♦~πϕ) ∈ T, i.e.,
T ∈ k♦~πϕk.
Lemma 15 (Existence Lemma for 0). T ∈ k0ϕk iff there is an R ∈ Rc such that
R[T]⊆ kϕk.
Proof. (⇒). Assume T ∈ k0ϕk, i.e., (0ϕ)∈T ∈Wc. We first prove the following:
Claim. ∃0ϕ∈T0.
Proof. Suppose not. As T0 is maximal consistent, we have¬∃0ϕ∈T0, i.e., ∀¬0ϕ∈T0.
AsT ∈Wc, we have R∀T0T. So given ∀¬0ϕ∈T0 we have ∀¬0ϕ∈T. ByT∀ we have
`LREL ∀¬0ϕ→ ¬0ϕ, i.e.,(∀¬0ϕ→ ¬0ϕ)∈T. As T is closed under modus ponens, given(∀¬0ϕ)∈T we get(¬0ϕ)∈T. But we also have(0ϕ)∈T ∈Wc and thusT is
4.2. REL+: dynamics of prioritized addition 93
Hence R0ϕ ∈ Rc. We will show that R0ϕ[T] ⊆ kϕk. Let S ∈ Wc be arbitrary and
suppose that R0ϕT S. By definition of R0ϕ, we have (0ϕ) ∈ T implies (0ϕ) ∈ S.
As (0ϕ) ∈ T we get (0ϕ) ∈ S. Given T0 we have `LREL 0ϕ → ϕ and thus
(0ϕ→ ϕ) ∈S. SinceS is closed under modus ponens we thus getϕ∈S, i.e.,S ∈ kϕk.
AsS was picked arbitrarily, we haveR0ϕ[T]⊆ kϕk.
(⇐) LetT ∈Wc and suppose there is an R ∈Rc such thatR[T]⊆ kϕk. By definition of
Rc, R = R0θ for some θ ∈ LREL such that (∃0θ) ∈ T0. Either 0θ∈ T or 0θ6∈ T.
We consider both cases.
Case 1: Suppose that0θ∈T ∈Wc. We first prove the following:
Claim. The set Γ :={0θ} ∪ {∀ψ|(∀ψ)∈T} ∪ {¬ϕ}is inconsistent.
Proof. Suppose that Γ is consistent. By Lindenbaum’s Lemma there is some maximal consistent theoryS such thatΓ⊆S. Moreover, as{∀ψ|(∀ψ)∈T0} ⊆ {∀ψ|(∀ψ)∈T} ⊆
S, we have R∀T0S and thusS ∈ Wc. As ¬ϕ∈Γ we have ¬ϕ∈S. Since S is consistent
we haveϕ6∈S, i.e., S 6∈ kϕk. From 0θ∈Γ we have 0θ∈S. By definition of R0θ, we getR0θT S. But then, givenS6∈ kϕk, we haveR0θ[T]6⊆ kϕk. Contradiction.
Given the Claim, there is a finite Γ0 ⊆ Γ such that Γ0 `LREL ⊥. By the theorem
`LREL ∀(ψ1 ∧ · · · ∧ψn) ↔ (∀ψ1 ∧ · · · ∧ ∀ψn) we can assume that Γ0 = {0θ,∀ψ,¬ϕ} for some ψ ∈ T. Since Γ0 `LREL ⊥ we have `LREL (0θ ∧ ∀ψ∧ ¬ϕ) → ⊥, so by propositional logic `LREL (0θ∧ ∀ψ) → (¬ϕ → ⊥), i.e., `LREL (0θ∧ ∀ψ) → (¬¬ϕ), i.e., `LREL (0θ∧ ∀ψ) → ϕ. Given the Pullout axiom, we have `LREL 0(θ∧ ∀ψ) →
(0θ∧ ∀ψ) and thus `LREL 0(θ∧ ∀ψ) → ϕ. By the Monotonicity Rule for 0, we get
`LREL 00(θ∧ ∀ψ)→0ϕ. By40, we have`LREL0(θ∧ ∀ψ)→00(θ∧ ∀ψ)and thus
`LREL 0(θ∧ ∀ψ)→0ϕ. By the Pullout axiom, we have`LREL(0θ∧ ∀ψ)→0(θ∧ ∀ψ).
Hence`LREL (0θ∧ ∀ψ)→0ϕ. Therefore ((0θ∧ ∀ψ)→0ϕ)∈T. As(0θ)∈T and
(∀ψ)∈T, by closure under modus ponens, we have 0ϕ∈T. That is,T ∈ k0ϕk.
Case 2: Suppose that0θ6∈T. Note that0θ6∈T implies thatR0θ[T] =Wc, and since
we haveR =R0θ and R[T]⊆ kϕk, all this gives us that Wc ⊆ kϕkc, i.e. all theories in the canonical model containϕ. We now prove the following:
Claim. The set Γ :={∀ψ|(∀ψ)∈T} ∪ {¬ϕ} is inconsistent.
Proof. Suppose that Γ is consistent. By Lindenbaum’s Lemma there is some maximal consistent theoryS such thatΓ⊆S. Moreover, as{∀ψ|(∀ψ)∈T0} ⊆ {∀ψ|(∀ψ)∈T} ⊆
S, we have R∀T0S and thus S ∈ Wc. As ¬ϕ ∈ Γ we have ¬ϕ ∈S and thus S ∈ k¬ϕk.
ThereforeWc6⊆ kϕk (contradiction).
Given the Claim, there is a finiteΓ0 ⊆Γ such that Γ0 `LREL ⊥. By the theorem `LREL
∀(ψ1∧ · · · ∧ψn)↔(∀ψ1∧ · · · ∧ ∀ψn) we can assume that Γ0 ={∀ψ,¬ϕ} for some ψ∈T.
SinceΓ0 `LREL⊥we have`LREL (∀ψ∧ ¬ϕ)→ ⊥, so by propositional logic `LREL(∀ψ)→
(¬ϕ→ ⊥), i.e., `LREL (∀ψ)→(¬¬ϕ), i.e., `LREL (∀ψ)→ϕ. By propositional logic, given
`LREL (∀ψ)→ϕwe can strengthen the antecedent getting`LREL (0> ∧ ∀ψ)→ϕ. Given the Pullout axiom↔, we have `LREL 0(> ∧ ∀ψ) ↔ (0> ∧ ∀ψ) and thus`LREL 0(> ∧
∀ψ)→ϕ. By the Monotonicity Rule for 0, we get`LREL00(> ∧ ∀ψ)→0ϕ. By40, we have`LREL 0(> ∧ ∀ψ)→00(> ∧ ∀ψ)and thus`LREL 0(> ∧ ∀ψ)→0ϕ. By the Pullout axiom↔, we have`LREL (0> ∧ ∀ψ)↔0(> ∧ ∀ψ). Hence`LREL (0> ∧ ∀ψ)→
(0>)∈T and(∀ψ)∈T. Hence by closure under modus ponens, we have0ϕ∈T. That
is,T ∈ k0ϕk.
Lemma 16 (Truth Lemma). For every formula ϕ∈LREL, we have: JϕKMc =kϕk.
Proof. The proof is by induction on the complexity of ϕ. The base case follows from the definition of Vc. For the inductive case, suppose that for all T ∈ Wc and all formulas
ψ of lower complexity than ϕ, we have JψKMc = kψk. The Boolean cases where ϕ =
¬ψ and ϕ = ψ1 ∧ψ2 follow from the induction hypothesis together with the standard
facts about maximal consistent theories included in Proposition 12. Only the modalities remain. Let ϕ = ∃ψ and consider any T ∈ Mc. We have T ∈ k∃ψk iff (Proposition 13) kϕk 6= ∅ iff (induction hypothesis) JψKMc iff
J∃ψKMc = W
c iff T ∈
J∃ψKMc. Now
let ϕ = 0ψ and consider any T ∈ Mc. We have T ∈ k0ψk iff (Proposition 15) there
is an R ∈ Rc such that R[T] ⊆ kψk iff (induction hypothesis) there is an R ∈ Rc such
that R[T] ⊆ JψKMc iff T ∈
J0ψKMc. Finally, let ~π be arbitrary and let ϕ = ♦
~
πψ. We
have T ∈ k♦~πψk iff (Proposition 14) there is an S ∈ kψk such that R~πT S iff (induction
hypothesis) there is an S ∈ JψKMc such that R~πT S iff there is an S ∈
JψKMc such that (T, S)∈Agc(hRcupJ~πKM c,cupJ~πKM ci) iffT ∈ J♦ ~ πψ KMc.
Lemma 17. ΛREL is strongly complete with respect to the class of REL models.
Proof. By Proposition 6, it suffices to show that every ΛREL-consistent set of formulas
is satisfiable on some REL model. Let Γ be an ΛREL-consistent set of formulas. By
Lindenbaum’s Lemma, there is a maximally consistent set T0 such that Γ ⊆ T0. Choose
any canonical modelMcfor T0. By Lemma 16, Mc, T0 |=ϕfor all ϕ∈T0.
Now that we have shown soundness and completeness for the systemLREL, we study next
the dynamic logic of iterated prioritized evidence addition,REL+.