Tools and Techniques in Modal Logic Marcus Kracht II ...

7.9. The Lattice of Tense Logics 369
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Figure 7.16. hk
�✒ ❅■
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· · ·
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❅ ❝
E K4.t even these cases are ruled out. Look at the sequence 〈gl ◦ n : n ∈ ω〉 where gl ◦ n
differs from gl n in that n ⋪ n. The maps π and ρ defined by π : gl ◦ n � gl 1 : j ↦→ j
(mod 2) and ρ : gl ◦ n � gl 0 : j ↦→ 0 are n–localic with respect to 0. Thus this sequence
subreduces both frames in E K4.t and in E K.t.
Lemma 7.9.11. For no n, Th gl n splits E K.t, E K4.t.
It now remains to treat the clusters. Here the situation is quite similar to the
situation of the garlands.
Lemma 7.9.12. Suppose g is a cluster. Then Th g splits E K.t and E K4.t only if
g � • and E S4.t only if g � ◦ .
Proof. Let n := ♯g > 1 and hk = 〈hk, ⊳〉 with
hk = {0, . . . , k} × {1, . . . , n} − {〈k, n〉}
and 〈i, j〉 ⊳ 〈i ′ , j ′ 〉 iff (i) i is odd, i ′ = i + 1 or i − 1 or (ii) i is even and i ′ = i. This can
be visualized by Here, denotes a cluster with n points and ◦ a cluster with 1 point.
There is no p–morphism from hk into g as there is no way to map a point belonging
to an n − 1–point cluster onto a n–point cluster.
Now look at the k–transit of 〈0, 0〉 in hk; call it e. Let e be its underlying set.
Every point in e is contained in an n–point cluster since 〈i, j〉 ∈ e iff i < k. Thus
there is a p–morphism π : e → g. Extend π to a map p + : hk � g. p + is k–localic
with respect to 〈0, i〉 for every i. Hence hk is k–consistent with g. It follows that
〈Th hk : k ∈ ω〉 is a subreduction of Th g. �
Now we have collected all the material we need to prove the splitting theorems.
Notice that a splitting frame for any of these logics can only be one–point cluster or
a two–point garland. We will now show that the frames not excluded by the above
lemmata are indeed splitting frames.
Theorem 7.9.13. Λ splits the lattice ES4.t iff Λ = Th gl1 = Th ◦
Λ = Th gl0 = Th ◦ .
✲◦ or
Proof. (⇐) The nontrivial part is gl 1. We will show E S4.t/gl 1 = S5.t by proving
that (†) of the Splitting Theorem holds for m = 1. Therefore let A be an algebra
satisfying Th A � S5t. Then there is a set c ∈ A of A such that 0 < c ∩ � � − c.
Consequently, in the underlying Kripke–frame there are two points s ⊳ t such that