Step 2 unravelled the canonical pre-model from step 1. Using the structure from the unrav- elled tree, we now define aRELmodelM˜ from it. We then show that this model is in fact alex-model. Finally, we define a variant of a bounded morphism defined for RELmodels, which we call bounded aggregation-morphism. Bounded aggregation-morphisms work on
REL models in the same way as standard bounded-morphisms do on Kripke models: for
REL models, modal satisfaction is invariant under bounded aggregation-morphisms. We then show thatMc is a bounded-morphic image ofM˜, which gives us completeness. We first define the modelM˜.
Definition 57 (M˜). Let K˜ be the unravelling of Mc around T0. The structure M˜ =
hW ,˜ hR˜,i˜ ,V ,˜ Ag˜ i has: ˜ R :={R˜0ϕ|(∃0ϕ)∈T0} ∪ {R0l, R 0 r} ∪ {W˜ ×W˜} where: ˜ R0ϕ := (→0ϕ)∗ R0l= (→ ∪[{→(ϕ,l)|ϕ∈L})∗
3.4. Completeness ofLlex 73
R0r= (→ ∪[{→(ϕ,r)|ϕ∈L})∗
Moreover, the priority order˜ is the preorder onR˜with:
R≺˜Rl0, R0r for all R∈R˜\ {R0l, Rr0,W˜ ×W˜}
and
R0l, R0r≺˜W˜ ×W˜
Finally, the aggregatorAg˜ is given by:
˜ Ag(hR,i) = ( ˜ R := (→)∗ if hR,i=hR˜,i˜ lex(hR,i) otherwise /
Proposition 14. All the evidence relations in R˜\ {W˜ ×W˜} are reflexive, transitive and anti-symmetric.
Proof. Reflexivity and transitivity follow immediately from the fact that each R ∈ R˜\ {W˜ ×W˜}is the reflexive transitive closure of some other relation, andW˜ ×W˜ is reflexive and transitive. Hence we just need to show the anti-symmetry of the relations. Let
R∈R˜\ {W˜ ×W˜}and suppose Rwv andRvw. First, we consider the caseR= ˜R0ϕ for some ϕ such that ∃0ϕ∈ T0. I.e. R = (→0ϕ)∗. Given Rwv there is some n ≥ 0 such
that:
w=w0→0ϕ w1 →0ϕ· · · →0ϕ wn=v
Similarly, givenRvw there is some m≥0 such that:
v=w00 →0ϕw01 →0ϕ · · · →0ϕw0m=w
By definition of→0ϕ, we havew
1= (w,0ϕ, T1)for someT1 ∈Wc,w2= (w,0ϕ, T1,0ϕ, T2)
for someT2∈Wc, and proceeding in this way we get
wn=v= (w,0ϕ, T1,0ϕ, T2, . . . ,0ϕ, Tn) whereTi∈Wc,for i≤n (3.1)
Similarly, we have w10 = (v,0ϕ, T10) for some T10 ∈ Wc, w20 = (v,0ϕ, T10,0ϕ, T20) for
someT20 ∈Wc, and proceeding in this way we get
w0m=w= (v,0ϕ, T10,0ϕ, T20, . . . ,0ϕ, Tn0) whereT 0
i ∈Wc,for i≤m (3.2)
Hence we must haven=m = 0. For otherwise, substituting v in 3.2 with the expression in 3.1 we get
w= (w,0ϕ, T1,0ϕ, T2, . . . ,0ϕ, Tn, v,0ϕ, T10,0ϕ, T20, . . . ,0ϕ, Tm0 )for n >0 orm >0
which is impossible. Thereforew=w0 =wn=v, as required.
Now we consider the caseR=R0l, i.e. R= (→ ∪S
{→(ϕ,l)|ϕ∈L})∗. Given Rwvthere
is somen≥0 such that:
w=w0 →i0 w1→i2 · · · →in−1 wn=v whereik∈ {}∪{(ϕ, l)|ϕ∈L}, for k= 1, . . . , n−1
Similarly, givenRvw there is some m≥0 such that:
v=w00→j0 w0
1 →j2 · · · →jm−1 w
0
Reasoning as we did in the case of R = ˜R0ϕ, we conclude that m = n = 0 and hence
w=v. The case ofR=R0r is analogous to the one just discussed, and we are done.
Proposition 15. In M˜ we have:
R0l∩Rd0 = ˜R
Proof.
(⊆) Let(w, v)∈Rl0∩R0d. Then we have (w, v)∈R0l, i.e.,
(w, v)∈(→ ∪[{→(ϕ,l)|ϕ∈L})∗
Hence, there is somen≥0 such that:
w=w0 →i0 w1→i2 · · · →in−1 wn=v where ik∈ {}∪{(ϕ, l)|ϕ∈L}, for k= 1, . . . , n−1
Similarly, we have(w, v)∈R0r, i.e.,
(w, v)∈(→ ∪[{→(ϕ,r)|ϕ∈L})∗
Hence, there is somem≥0 such that:
w=w00 →j0 w0 1→j2 · · · →jm−1 w 0 m =v where jk∈ {}∪{(ϕ, r)|ϕ∈L}, for k= 1, . . . , m−1 By definition of →ik, we have w 1 = (w, i0, T1) for some T1 ∈Wc, w2 = (w, i0, T1, i1, T2) for someT2∈Wc, and proceeding in this way we get
wn=v= (w, i0, T1, i1, T2, . . . , in−1, Tn) (3.3)
where Ti ∈ Wc and ik ∈ {} ∪ {(ϕ, l) | ϕ ∈ L}, for k = 1, . . . , n−1. Reasoning in a
similar way, we get
w0m=v = (w, j0, T10, j1, T20, . . . , jm−1, Tm0 ) (3.4)
whereTi0 ∈Wc and jk ∈ {} ∪ {(ϕ, r)|ϕ∈L}, for k= 1, . . . , m−1.
Given the expressions 3.3 and 3.4, we havewn=v=w0m. Hencen=mand for all k < n,
ik=jk. Hence we must have ik==jk for all k < n. This means that the path
w=w0→i0 w1 →i2 · · · →in−1 wn=v
can be rewritten as
w=w0 → w1→· · · → wn=v
which is an {}-path fromw to v. Hence (w, v)∈R˜= (→)∗. (⊇) Let (w, v)∈R˜ = (→)∗. Then there is some n≥0 such that:
w=w0 → w1→· · · → wn=v
The {}-path described above is also an {} ∪S
{(ϕ, l) | ϕ ∈ L}-path and an {} ∪
S{(
ϕ, r) | ϕ ∈ L}-path. Hence we have (w, v) ∈ Rl0 and (w, v) ∈ R0r. Thus (w, v) ∈
3.4. Completeness ofLlex 75
Proposition 16. M˜ is a lexmodel.
Proof. To establish thatM˜ is a lexmodel, we need to show that it meets the condition of aREL model and thatAg˜ =lex. That is, we have to show:
1. R˜ is a family of evidence, i.e., everyR∈R˜ is a preorder.
2. W˜ ×W˜ ∈R˜, i.e., the trivial evidence order is a piece of available evidence.
3. R˜ =lex(hR˜,i)˜ (which given the definition of Ag˜ , gives Ag˜ =lexas required)
Item 1 follows from 14, and Item 2 follows from the definition ofM˜. Hence Item 3 remains to be shown. Note first that by 15, we haveR0l∩Rd0 = ˜R.
(⊆) Suppose that(w, v)∈lex(hR˜,)˜ . Note thatlexis given here by
(w, v)∈lex(hR˜,i)˜ iff∀R0 ∈R (R0wv ∨ ∃R∈R˜(R0≺˜R∧R<wv)) (3.5)
Suppose for reductio that(w, v)6∈R0l∩R0d. Then(w, v)6∈R0l or(w, v)6∈R0r. Without loss of generality, suppose(w, v)6∈R0l. Given 3.5, we have in particular:
(R0lwv ∨ ∃R∈R˜(R0l≺˜R∧R<wv)) (3.6) Note that the definition of˜ is such that R0l has no relation strictly above it other than
˜
W×W˜. AndW˜ ×W˜ is symmetric and thus it is not the case that( ˜W×W˜)<wv. Hence the right disjunct in 3.6 is false. Therefore we must haveRl0wv, contradicting our assumption to the contrary.
Suppose that (w, v) ∈ R0l∩R0d. Then R0lwv and Rr0wv. Suppose first that w = v. As
lex(hR˜,i)˜ is a preorder, we have(w, v)∈lex(hR˜,i)˜ and we are done. Suppose now that
w 6= v. By Proposition 14, R0l and Rr0 are antisymmetric. Thus from w 6= v, Rl0wv and
R0rwv, we get (R0l)<wv and (Rr0)<wv. Hence, as we have
R≺˜R0l, R0r for all R∈R˜\ {Rl0, R0r}
from the definition oflexwe get(w, v)∈lex(hR˜,)˜ as required.
We now introduce the notion of a bounded aggregation-morphism. This is, as we will show, a truth-preserving map betweenRELmodels, which works similarly to standard bounded morphisms for Kripke models.
Definition 58 (Bounded aggregation-morphism). LetM =hW,hR,i, V, Agi andM0 = hW0,hR0,0i, V0, Ag0i be two REL models. A mapping f : W → W0 is a bounded
aggregation-morphism if the following hold:
1. Valuation condition: for allw∈W,w∈V(p)iff f(w)∈V0(p)
2. Forth conditions:
(a) for all R ∈R, for all w∈W, there exists some R0 ∈R0 such that R0[f(w)] ⊆ {f(v)|Rwv}
(b) for allw, v∈W, ifAg(hR,i)wv thenAg0(hR0,0i)f(w)f(v)
3. Back conditions:
(a) for all R0 ∈ R0 and all w ∈ W there exists some R ∈ R such that {f(v) |
(b) for all w ∈ W, v0 ∈ W0, if Ag0(hR0,0i)f(w)v0 then there exists some world
v ∈W such thatAg(hR,i)wv and f(v) =v0.
/
Proposition 17. Let M = hW,hR,i, V, Agi and M0 = hW0,hR0,0i, V0, Ag0i be two REL models. Let f : W → W0 be a surjective bounded aggregation-morphism. Then for all w ∈ W and ϕ ∈ L: M, w |= ϕ iff M0, f(w) |= ϕ. That is: modal satisfaction is invariant under surjective bounded aggregation-morphisms.
Proof. By induction on the structure of ϕ. The base case holds by the valuation condi- tion. The boolean cases are shown by unfolding the definitions, so we consider the cases involving modalities.
Suppose M, w |= ♦0ψ. Then for all R ∈ R there is some v ∈ W such that Rwv and
M, v|=ψ. Now we want to show: M, f(w)|=♦0ψ. That is, for all R0 ∈R0 there is some
v0 ∈W0 such thatR0f(w)v0 andM0, v0 |=ψ. LetR0 ∈R0 be arbitrary. By the back condi-
tion 3(a), there is someR∈R such that{f(v)|Rwv} ⊆R0[f(w)]. Hence given Rwv, we havef(v)∈R0[f(w)]. That is,R0f(w)f(v). By induction hypothesis, givenM, v|=ψ we haveM0, f(v)|=ψ. AsR0 was arbitrarily picked, this holds for all relations inR0. Hence
M0, f(w)|=♦0ψ.
Suppose now thatM0, f(w)|=♦0ψ. Then for all R0 ∈R0 there is some v0 ∈W0 such that
R0f(w)v0 and M0, v0 |= ψ. Now we want to show: M, w |= ♦0ψ. That is, for all R ∈R
there is some v ∈ W such that Rwv and M, v |= ψ. Let R ∈ R be arbitrary. By the forth condition 2(a), there exists some R0 ∈R0 such that R0[f(w)] ⊆ {f(v) | Rwv}. We
have R0f(w)v0 and M0, v0 |= ψ for some v0 ∈ W0. As f is surjective, we have v0 =f(u)
for some u ∈ W. Hence given R0[f(w)] ⊆ {f(v) | Rwv} and f(u) ∈ R0[f(w)], we get
f(u) ∈ {f(v) |Rwv}. Hence Rwu. By induction hypothesis, given M0, f(u) |=ψ we get
M, u |= ψ. As R was arbitrarily picked, this holds for all relations inR. Hence we have
M, w|=0ψ.
Now suppose M, w |= ♦ψ. Then there is some v ∈ W such that Ag(iR, h)wv and
M, v |= ψ. By the forth condition 2(b), we have Ag0(iR0,0 h)f(w)f(v). By induction hypothesis,M0, f(v)|=ψ. HenceM0, f(w)|=ψ.
Lastly, suppose M, f(w) |=♦ψ. Then there is some v0 ∈ W0 such that Ag0(iR0,0 h)wv
and M0, v0 |= ψ. Hence by the back condition 3(b), there is some world v ∈ W such that Ag(iR, h)wv and f(v) = v0. By induction hypothesis, we get M, v |= ψ. Hence
M, w|=♦ψ.
Proposition 18. The map β: ˜W →Wc is a surjective bounded aggregation-morphism. Proof. We need to check thatβsatisfies the conditions of a surjective bounded aggregation- morphism.
1. Surjectivity: Let T ∈Wc be arbitrary. We need to show that there is some h ∈W˜
such that β(h) = T. Recall that we showed in 13.2. that Wc×Wc = R0> ∈ R. HenceR0>T
0T. Thus the historyh= (T0,0>, T) is an element ofW˜ withβ(h) =
3.4. Completeness ofLlex 77
2. Valuation condition. This follows from the definition ofV˜, i.e.,
˜
V(p) :={h∈W˜ |β(h)∈Vc(p)}
3. Forth conditions:
(a) We need to show that for all R ∈ R˜, for all w ∈ W˜, there exists some
R00 ∈ Rc such that R00[β(w)] ⊆ {β(v) | Rwv}. Let R ∈ R˜ and w ∈ W˜ be
arbitrary. Suppose first that R = ˜R0ϕ for some ϕ with ∃
0ϕ ∈ T0. Con-
sider R0ϕ ∈ Rc. Take any T ∈ R0ϕ[β(w)], i.e., R0ϕβ(w)T. We will
show that T ∈ {β(v) | R˜0ϕwv}. Note that, given R0ϕβ(w)T, the history
w0 = (w,0ϕ, T) is in W˜. This means that w →0ϕ w0. Hence (→0ϕ)∗ww0,
i.e., R˜0ϕww0. Given β(w0) =T, we getT ∈ {β(v)|R˜0ϕwv}, as required. Suppose now that R = R0l = (→ ∪S{→(ϕ,l)| ϕ ∈ L})∗. Consider R0 =
(R∪S
{R(ϕ,l)})∗ ∈Rc. Take anyT ∈R0[β(w)], i.e., R0β(w)T. We will show
that T ∈ {β(v)|R0lwv}. Given R0β(w)T, for some n≥0, there is a path:
β(w) =S0R0S1R2, . . . , Rn−1Sn=T
whereSi ∈Wc,k∈ {}∪{(ϕ, l)|ϕ∈L}, fork < n. Hence there are histories
w1 = (w, 0, S1),w2= (w, 0, S1, 1, S2), up town= (w, 0, S1, 1, S2, . . . , n−1, T).
Hence, by definition of→k, for eachk < n, we havew
k→k wk+1. Hence there
is a path
w=w0 →0 w1 →2, . . . ,→n−1 wn
And hence R0lwwn. Given β(wn) =T, we getT ∈ {β(v)|R0lwv}, as required.
The case of R=R0r is analogous to the one above. Hence we have left the case
R = ˜W ×W˜. Consider R0> =Wc×Wc ∈ Rc. Take any T ∈ R0>[β(w)],
i.e., R0>β(w)T. This just means that T ∈Wc. We will show thatT ∈ {β(v)| (w, v)∈W˜ ×W˜}={β(v) |v ∈W˜}. As β is surjective, we know that there is some u∈W˜ such thatβ(u) =T, and we are done.
(b) We need to show that for all w, v ∈ W˜, if Ag˜ (hR˜,i)˜ wv then Agc(hRc,c i)β(w)β(v). Let w, v ∈ W˜ be arbitrary and suppose that Ag˜ (hR˜,i)˜ wv. By Proposition 16.3, given Ag˜ (hR˜,i)˜ wv we have R˜wv, i.e., (→)∗wv. Hence,
for some n≥0, there is a path:
w=w0→w1 → · · · →wn=v
Hence there are histories w1 = (w,, S1), w2 = (w,, S1,, S2), up to wn =
v = (w,, S1,, S2, . . . ,, Sn). Hence by definition of wn we have
β(w)RS1RS2, . . . , RSn
And since R is transitive, we getRβ(w)S
n, i.e.,Rβ(w)β(v), as required.
4. Back conditions:
(a) We need to show that for all R00∈Rc and allw∈W˜ there exists some R∈R˜
such that{β(v)|Rwv} ⊆R00[β(w)]. LetR00∈Rcandw∈W˜ be arbitrary. We
reason by cases. First, suppose that R00=R0ϕ. ConsiderR˜0ϕ ∈R˜. We will show that {β(v)|R˜0ϕwv} ⊆R0ϕ[β(w)]. Take any β(u)∈ {β(v)|R˜0ϕwv}.
We have R˜0ϕwu, i.e.,(→0ϕ)∗wu. Hence for somen≥0, there is a path:
w=w0 →0ϕ w1 →0ϕ, . . . ,→0ϕ wn=u
where wi ∈ W˜, for k ≤ n. Hence there are histories w1 = (w,0ϕ, S1),
w2 = (w,0ϕ, S1,0ϕ, S2), up town=u= (w,0ϕ, S1,0ϕ, S2, . . . ,0ϕ, Sn).
Hence, by definition of wn, we have
β(w)R0S1R0, . . . , R0β(u)
And since R0 is transitive, we get R0β(w)β(u), as required.
Suppose now that R00 = R0 ∈ Wc. Consider R0l ∈ R˜. We will show that
{β(v) | R0lwv} ⊆ R0[β(w)]. Take any β(u) ∈ {β(v) | R0lwv}. We have R0lwu, i.e., (→ ∪S
{→(ϕ,l)|ϕ∈L})∗wu. Hence for some n≥0, there is a path:
w=w0 →0 w1→1, . . . ,→n−1 wn=u
where wi ∈W˜,k ∈ {} ∪S{(ϕ, l)|ϕ∈L} for k < n. By definition of →k,
w1 = (w, 0, S1) for some S1 ∈Wc, w2 = (w, 0, S1, 1ϕ, S2) for some S2 ∈Wc,
up to wn=u= (w, 0, S1, 1, S2, . . . , n−1, Sn)whereSi∈Wc fori≤n. Hence,
by definition of wn, we have a path
β(w)R0S
1R1S2, . . . , Rn−1β(u)
SinceR0= (R∪S
{R(ϕ,l)})∗, the path above is a path fromβ(w)toβ(u)along
R0, i.e., we haveR0β(w)β(u), as required.
(b) We need to show that for all w ∈ W˜, T ∈ Wc, if Agc(hRc,ci)β(w)T then
there exists some history v ∈ W˜ such that Ag˜ (hR˜,i)˜ wv and β(v) = T. Let
w ∈W˜ and T ∈ Wc be arbitrary, and suppose that Agc(hRc,ci)β(w)T, i.e.,
Rβ(w)T. Then the historyw0 = (w,, T)is inW˜ andβ(w0) =T, as required.
Theorem 10. Λlex is complete with respect to the class of lexmodels.
Proof. Let Γ be a Λlex-consistent set of formulas. By Lindenbaum’s Lemma, Γ can be
extended to a maximal consistent set T0. Choose any canonical pre-model Mc for T0.
By Lemma 14, Mc, T0 |= ϕ for all ϕ ∈ T0. Let K˜ be the unraveling of Mc around T0
and let M˜ be the lex model generated from K˜. Note that the history (T0) ∈ W˜. Let
β : ˜W → Wc be the map defined above. By Proposition 18, β is a surjective bounded aggregation-morphism. By Proposition 17, we haveMc, T0 |=ψ iff M , β˜ (T0)|=ψ. Hence,
in particular,M , β˜ (T0)|=ϕfor all ϕ∈T0.