Like inAML, we want it to be possible inAMLQto reduce a formula of the form [s][t]ϕto [s;t]ϕ, where the set of actionss;tstands for the information that one action from the sets and one action from the sett are executed in sequence.
We can do this by taking the two action models from the two modalities and combining them into one action model that has, as actions, pairs of which the first element is an action from the first action model and the second element an action from the second action model. In this way, we construct an action model that encodes all sequences of epistemic actions from the first and the second action model.
Definition3.7.1. Composition of AMLQ action models
LetM=hS,{∼a | a∈ A},contiand M0 =hS0,{∼0a |a∈ A},cont0ibe two AMLQaction
models. Their compositionM;M0 is the action modelhS00,{∼00
a | a∈ A},cont00isuch that:
• S00=S×S0 • hx,x0i ∼00
a hy,y0iiffx∼ayandx0∼0ay0
• cont00(hx,x0i) =cont(x)∧[M,x]cont0(x0)
The composition of two pointed action models (M,s) and (M0,s0) is the pointed action model (M00,s×s0) withM00=M;M0 defined as above.
It is easy to show that∼00
a is an equivalence relation if∼aand∼0aare. This guarantees that
ifMandM0 are action models, then so is M;M0. We also need to show that updating with Mfirst and then withM0 gives the same result as updating withM;M0 at once.
Strictly speaking, any (M⊗M)⊗M0andM⊗(M;M0) will never be the same model, as their worlds are pairs with a different internal structure. However, this internal structure only shows the history of actions, which is not relevant for describing the knowledge and issues of agents. Therefore, for our purposes we can regardhhw,xi,x0iandhw,hx,x0iias one world under two different names. We prove this in the following proposition.
Proposition 3.7.1. Isomorphism between updated models
For every inquisitive epistemic modelM and every two action modelsMandM0, (M⊗M)⊗M0 is isomorphic toM⊗(M;M0).
Proof: We define three updated modelsM0,M00 andM000: • M0=M⊗M=hW0,{Σ0 a | a∈ A}, V0i • M00=M0⊗M0= hW00,{Σ00 a |a∈ A}, V00i • M000=M⊗(M;M0) =hW000,{Σ000a |a∈ A}, V000i
We want to show that M00 and M000 are isomorphic. That is, there is a bijective functionf that maps the worlds in M00 to the worlds in M000, which preserves the exact structure of the models. We show this by mapping eachhhw,xi,x0i ∈W00 to hw,hx,x0ii ∈W000.
We start by showing thathhw,xi,x0i ∈W00 just in casehw,hx,x0ii ∈W000. (⇒) Assumehhw,xi,x0i ∈W00
Then by definition ofW00, we havehw,xi ∈W0,x0∈S0andM0,hw,xi |=pre0(x0). By definition ofW0,hw,xi ∈W0 impliesw∈W,x∈S andM, w|=pre(x).
ByProposition 3.2.1we obtainM0,hw,xi |=cont0(x0) andM, w|=cont(x). Then as hw,xi =w[x], by the support condition of the dynamic modality, M, w |= [M,x]cont0(x0). This means thatM, w|=cont(x)∧[M,x]cont0(x0). It follows that
M, w|=cont00(hx,x0i) and thus thatM, w|=pre00(hx,x0i). Byx∈Sandx0 ∈S0 we also havehx,x0i ∈S×S0. Then by definition ofW000,hw,hx,x0ii ∈W000. (⇐) Assumehw,hx,x0ii ∈W000.
Then by definition ofW000, we havew∈W,hx,x0i ∈S00andM, w|=pre00(hx,x0i). By Proposition 3.2.1, we obtain M, w |= cont00(hx,x0i), which implies that
M, w|=cont(x)∧[M,x]cont0(x0). AsS00=S×S0,x∈Sandx0∈S0.
We can useProposition 3.2.1and the support condition of the dynamic modality to obtain fromM, w|=cont(x) andM, w|= [M,x]cont0(x0) that M, w|=pre(x), andM,hw,xi |=pre0(x0).
This also means it must be the case thathw,xi ∈W0. This fact together with
M,hw,xi |=pre0(x0) andx0 ∈S0, implieshhw,xi,x0i ∈W00.
From this we can conclude that if we let f(hhw,xi,x0i) = hw,hx,x0ii, then f is a bijection betweenW00 andW000. Now let us show that it is indeed an isomorphism. For this, we need to show two things:
(i) The mapping preserves the structure of the state maps. That is, if we let f(s) be{f(w)|w∈s}, then we have:
s∈Σ00a(hhw,xi,x0i) ⇐⇒ f(s)∈Σ000a(f(hhw,xi,x0i)
(ii) The mapping preserves the valuation: V00(hhw,xi,x0i) =V000(f(hhw,xi,x0i)) We start by showing (i). Take any worldhhw,xi,x0i ∈W00 and any states00 ⊆W00.
Let s000 = f(s00). Then we need to show that s00 ∈ Σ00a(hhw,xi,x0i) ⇐⇒ s000 ∈ Σ000a (hw,hx,x0ii).
(⇒) Assumes00∈Σ00a(hhw,xi,x0i).
If s00 is empty, we are done, so assume it is not. We know that it satisfies conditions (i-iv) ofDefinition 3.2.4.
By condition (iii), π2(s00) contains one action y0. Let s0 = π1(s00). Then by condition (iv),M0, s0 |=cont0(y0). By condition (ii), x0 ∼a y0 and by condition
(i), s0 ∈ Σ0a(hw,xi). That means s0 in turn satisfies conditions (i-iv) to be in Σ0a(hw,xi).
By condition (iii),π2(s0) contains one actiony. Then by condition (iv), we have
M, π1(s0) |= cont(y). By condition (ii) we have x ∼a y and by condition (i),
π1(s0)∈Σa(w).
We can now show thats000satisfies the four conditions to be in Σ000a(hw,hx,x0ii). Because there is only oney0∈π2(s00) and oney∈π2(s0),π2(s000) ={hy,y0i}. So condition (iii) is satisfied.
Notice thatπ1(s000) =π1(s0). Asπ2(s0) ={y},s0 =π1(s0)[y] =π1(s000)[y]. There- fore, we haveM, π1(s000)|=cont(y) byM, π1(s0)|=cont(y) andM0, π1(s000)[y]|= cont0(y0) byM0, s0|=cont0(y0). By the support condition of the dynamic modal- ity, we haveM, π1(s000)|= [y]cont0(y0). It follows thatM, π1(s000)|=cont00(hy,y0i), which means condition (iv) is satisfied as well.
As for condition (ii), we have x ∼a y and x0 ∼0a y0, so by the definition of
composition we have hx,x0i ∼00
a hy,y0i. Finally, we have condition (i) because
(⇐) Assumes000∈Σ000a(hw,hx,x0ii).
If s000 is empty, we are done, so assume it is not. We know that it satisfies conditions (i-iv) ofDefinition 3.2.4.
By condition (iii),π2(s000) contains one actionhy,y0i. By condition (iv), we have
M0, π1(s000)|=cont00(hy,y0i). By condition (ii),hx,x0i ∼00
a hy,y0iand by condition
(i),π1(s000)∈Σa(w).
By the definition of composition, byM0, π1(s000) |=cont00(hy,y0i) we have that
M0, π1(s000)|=cont(y)∧[M,y]cont0(y0) andhx,x0i ∼00
a hy,y0i impliesx∼a y and
x0 ∼0 a y0.
Again, lets0 =π1(s00). Then as π1(s0) = π1(s000) and π2(s0) = {y}, it follows from the above thats0 satisfies conditions (i-iv) to be in Σ0a(hw,xi). This, in
turn, is condition (i) for s00 to be in Σ00a(hhw,xi,x0i). As π2(s00) = {y0}, and
we already havex0 ∼0
a y0, condition (iii) and (ii) are also satisfied. Then from
M0, π
1(s0)|= [M,y]cont0(y0) we obtain M, π1(s0)[y] |=cont0(y0). As π1(s0)[y] =
π1(s00), this establishes condition (iv).
Finally, by the definition of valuation in updated models, hhw,xi,x0i ∈ V00(p) iff hw,xi ∈V0(p) iffw∈V(p) iffhw,hx,x0ii ∈V000(p), which settles condition (ii) forf
to be an isomorphism. This concludes the proof thatM00andM000are isomorphic.
We continue by showing that updating a state in an original model twice, first with a set of actionssfrom one action model and then with a set of actionstfrom another action model, gives the same result as updating it once with the corresponding set of actions from the composed action model.
Proposition 3.7.2. s[s;t] is the image ofs[s][t] under isomorphism f
For every information statesand every two sets of actionssandt: hhw,xi,x0i ∈s[s][t] ⇐⇒ hw,hx,x0ii ∈s[s;t] Proof: (⇒) Assumehhw,xi,x0i ∈s[s][t].
Then by Definition 3.2.6, M0,hw,xi |= pre0(x0), hw,xi ∈ s[s] and x0 ∈ t. By
Proposition 3.2.1we obtain M0,hw,xi |=cont0(x0). By the support condition of the dynamic modality,M, w|= [x]cont0(x0).
As hw,xi ∈ s[s], by Definition 3.2.6 we also have M, w |= pre(x), w ∈ s and x∈ s. This means that M, w |= cont(x). As x∈ s and x0 ∈ t, hx,x0i ∈s×t. FromM, w|=cont(x)∧[M,x]cont0(x0) it follows thatM, w|=cont00(hx,x0i) and thus thatM, w|=pre00(hx,x0i).
Then byDefinition 3.2.6,hw,hx,x0ii ∈s[s;t]. (⇐) Assumehw,hx,x0ii ∈s[s;t].
Then byDefinition 3.2.6,M, w|=pre00(hx,x0i),w∈sandhx,x0i ∈s×t, which means that x ∈ s and x0 ∈ t. We obtain M, w |= cont00(hx,x0i) by Proposi- tion 3.2.1, which means by the definition of composition thatM, w|=cont(x)∧ [M,x]cont0(x0). This in turn means thatM, w|=pre(x) andM0,hw,xi |=pre0(x0). AsM, w |=pre(x), w∈ sand x∈ s, we have by Definition 3.2.6that hw,xi ∈
s[s]. Similarly, because M0,hw,xi |=pre0(x0), hw,xi ∈s[s] and x0 ∈t, we have hhw,xi,x0i ∈s[s][t].
From the previous two propositions we can conclude that these states support the same formulas. A straightforward induction on the structure of formulas would suffice to show
this. That means we can show that for any statesand formulaϕ, any two formulas of the form [s][t]ϕand [s;t]ϕare indeed equivalent.
Proposition 3.7.3. [s][t]ϕ≡[s;t]ϕ
Proof: From the support condition of the dynamic modality and the previous propositions, we obtain the following proof:
M, s|= [s][t]ϕ ⇐⇒ M0, s[s]|= [t]ϕ
⇐⇒ M00, s[s][t]|=ϕ
⇐⇒ M000, s[s;t]|=ϕ
⇐⇒ M, s|= [s;t]ϕ
This proposition shows that any formula of the form [s][t]ϕcan be reduced to [s;t]ϕ.