Using the reduction equivalences from the previous sections, we will show that any formula of AMLQ can be reduced to a formula of the static language of IEL. The proof style we will use is different from that ofAMLand IDEL: while the reduction proofs for these logics proceed by a straightforward induction on the structure of formulas, we cannot do the same, due to the lack of a reduction equivalence for formulas of the form [s](ϕ→ψ) (in particular, the case whensis a set of more than one action andϕ→ψ is a question).
To show that we can reduce formulas even when they have this form, we need to use the fact that every formula is equivalent to the inquisitive disjunction of its resolutions (Propo- sition 3.4.5). We can then perform an induction on the structure of these resolutions. As these are declaratives by definition, we will have a way of reducing them.
However, another complication is that a subformula of a declarative is not necessarily a declarative itself (e.g. a declarative can be of the form Kaµ with µ a question). As a
consequence, we cannot perform a straightforward induction on the structure of declaratives. Instead, we do our induction inside an induction on the modal depth of formulas, which we define as follows.
Definition3.8.1. Modal depth
For any ϕ ∈ LAMLQ, its modal depth md(ϕ) is the maximum number of nested epistemic modalitiesKa andEa occurring inϕ, defined inductively as follows:
• md(p) = 0 • md(⊥) = 0
• md(ϕ◦ψ) =max(md(ϕ),md(ψ)) for◦ ∈ {∧,→,> } • md(ϕ) =md(ϕ) + 1 for∈ {Ka, Ea}
• md([s]ϕ) =md(ϕ)
In our reduction proof, we will use this notion together with the following lemma.
Lemma3.8.1. Resolutions never have a higher modal depth
Ifα∈ R(ϕ), thenmd(α)≤md(ϕ).
Proof: A straightforward induction on the complexity ofϕ, using the definitions of resolu- tions and modal depth, suffices to show this.
We are now ready to prove that we can reduce every formula ofLAMLQ to one ofLIEL.
Theorem 3.8.1. Every formula ofAMLQ is equivalent to some formula of IEL
For anyϕ∈ LAMLQ, there is someϕ∗∈ LIEL such thatϕ≡ϕ∗.
Proof: We start by proving the claim for the most basic fragment of our logic, namely LAMLQ0. Take any ϕ ∈ LAMLQ0. We perform an induction on the complexity of
ϕ. All steps are immediate, except for the one where ϕis [s]ψ. By the induction hypothesis we have aψ∗∈ LIELsuch thatψ≡ψ∗, soϕ≡[s]ψ∗. Therefore, what we
need to show is that [s]ψ∗≡ϕ∗ for someϕ∗∈ LIEL. We show this by induction on the modal depth of [s]ψ∗.
• Base case. Let md([s]ψ∗) = 0. Take any resolution α ∈ R([s]ψ∗). Then by
Definition 3.4.2,α= [s]β∗ withβ∗∈ R(ψ∗). This means thatβ∗∈ LIEL. We will show that we can find a formulaα∗ ∈ LIEL that is equivalent to [s]β∗.
We do this by another induction, on the complexity ofβ∗.
• Base case. Supposeβ∗is an atompor⊥. Take any set of actionst. Then byCorollary 3.8.1or3.8.2we have:
[t]β∗≡V
x∈t(pre(x)→β∗)
Which means that in case of [s]β∗ we can letα∗:=V
x∈spre(x)→β∗. • Inductive step. Induction hypothesis: for allγ less complex thanβ∗, for
all sets of actionst, there is someγ∗∈ LIEL such that [t]γ≡γ∗.
Notice that because β∗ is a declarative with modal depth 0, the only con- nectives it can contain are conjunction and implication.
(∧) Suppose β∗ is γ∧γ0. Then by Proposition 3.8.3, [s]β∗ ≡[s]γ∧[s]γ0. By the induction hypothesis, we have some γ∗, γ0∗ ∈ LIEL which are equivalent to [s]γ and [s]γ0 respectively. So we can letα∗:=γ∗∧γ0∗. (→) Supposeβ∗ isγ→γ0.
Then by Proposition 3.6.2and3.8.6, [s]β∗ ≡V
x∈s([x]γ →[x]γ
0). Take
any x ∈ s. By the induction hypothesis, we have someγ∗, γ0∗ ∈ LIEL which are equivalent to [x]γand [x]γ0 respectively. Letθx :=γ∗ →γ0∗. Thenθx≡[x]γ→[x]γ0. As we can define such aθx for allx∈s, we can letα∗:=V
x∈sθx.
This concludes the proof of the claim that we have someα∗ ∈ LIEL equivalent to everyα∈ R([s]ψ∗). Let Γ be the set of these formulas. Then
>
Γ is alsoa formula ofLIEL, which by the support condition of inquisitive disjunction is equivalent to
>
R([s]ψ∗). By Proposition 3.4.5, it is also equivalent to [s]ψ∗. This means we can letϕ∗:=>
Γ.What we have shown is that ifmd([s]ψ∗) = 0, then there is someϕ∗∈ LIELsuch that [s]ψ∗≡ϕ∗.
• Inductive step. By the induction hypothesis, we have for any [s]χ∗ of modal depth< n, some equivalent formula inLIEL.
Letmd([s]ψ∗) =n.
Like in the base case, we take an arbitrary resolution α ∈ R([s]ψ∗). Then
α= [s]β∗ withβ∗∈ R(ψ∗). This means thatβ∗∈ LIEL.
We will show that we can find a formulaα∗ ∈ LIEL that is equivalent to [s]β∗.
Like in the base case, we perform an induction on the complexity ofβ∗. The base case and the inductive steps for conjunction and implication carry over. We only need to show the cases for the modalitiesKa andEa.
(K) Supposeβ∗isKaχ. Then byCorollary 3.8.3we have [s]β∗≡Vx∈s(pre(x)→
Ka[δa(x)]χ). Asmd(χ)< n, by the induction hypothesis we have someχ∗x ∈ LIELequivalent to each [δ
a(x)]χ. So we can letα∗:=Vx∈s(pre(x)→Kaχ∗x). (E) Supposeβ∗isEaχ. Then byCorollary 3.8.4we have [s]β∗≡Vx∈s(pre(x)→
V
y∼axEa(cont(y)→[y]χ)). Asmd(χ)< n, by the induction hypothesis we
have some χ∗y ∈ LIEL equivalent to each [y]χ. So we can let α∗ be defined
as: V x∈s(pre(x)→ V y∼axEa(cont(y)→χ ∗ y))
Like in the base case, we have now shown that we have someα∗∈ LIELequivalent to everyα∈ R([s]ψ∗). Again, we let Γ be the set of these formulas. Then
>
Γis a formula ofLIEL equivalent to [s]ψ∗. This means we can letϕ∗:=
>
Γ.This concludes the inductive proof that no matter what the modal depth of [s]ψ∗ is, we can find an equivalent formula ϕ∗ ∈ LIEL. As this was the only step we needed to prove, we have thereby shown that for everyϕ∈ LAMLQ0, there is someϕ∗∈ LIEL
such thatϕ≡ϕ∗.
We can now generalize this claim to all formulas ofLAMLQ. Recall that by definition, all formulas ofLAMLQ are inLAMLQi for some i≥0. This means we can perform an
induction oni.
• Base case. We have already shown that the claim holds for allϕ∈ LAMLQ0. • Inductive step. By the induction hypothesis, for any ϕ ∈ LAMLQi, there is
some ϕ∗ ∈ LIEL such thatϕ≡ϕ∗. We need to show that this is also the case
for anyϕ∈ LAMLQ(i+1).
The proof for the base case can be repeated. However, this time when we want to usepre(x) orcont(x) of some actionxin our reduction, it might be the case that these are not formulas ofIEL. But by Definition 3.5.1 they are formulas of LAMLQi, which means that by the induction hypothesis we can obtain an
equivalent formula fromIELthat we can use instead.