Weakly Aggregative Modal Logic: Characterization and Interpolation

arXiv:1803.10953v1 [cs.LO] 29 Mar 2018

Weakly Aggregative Modal Logic: Characterization and Interpolation Jixin Liu Department of Philosophy, Peking University & UC Berkeley

Yanjing Wang Department of Philosophy, Peking University

Abstract In this paper, we study the model theoretical aspects of Weakly Aggregative Modal Logic (WAML), which is a collection of disguised polyadic modal logics with n-ary modalities whose arguments are all the same. We give a van-Benthem-Rosen characterization theorem of WAML based on an intuitive notion of bisimulation, and show that WAML has Craig Interpolation. Keywords: weakly aggregative modal logic, Craig interpolation, bisimulation.

1

Introduction

You are invited to a dinner party for married couples after a logic conference in China. The host tells you the following facts: •

At least one person of each couple is a logician,

•

At least one person of each couple is Chinese.

Given these two facts, can you infer that at least one person of each couple is a Chinese logician? The answer is clearly negative, since there might be a couple consisting of a foreign logician and a Chinese spouse who is not a logician. Now, suppose that the host adds another fact: •

At least one person of each couple likes spicy food.

What do you know now? Actually, you can infer that for each couple, one of the two people must be either: •

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a Chinese logician, or

We thank Yifeng Ding for suggesting the construction of the n-ary unraveling. We are grateful to Emil Jeˇr´abek for the proof idea of a special case of Lemma 4.5, and thanks Guozhen Shen for posting the corresponding question on mathoverflow.

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Weakly Aggregative Modal Logic:Characterization and Interpolation

•

a logician who likes spicy food, or

•

a Chinese who likes spicy food.

This can be verified by the Pigeonhole Principle: for each couple, there are a logician, a Chinese, and a fan for spicy food, thus there must be at least one person of the couple who has two of those three properties. This can clearly be generalized to n-tuples of things w.r.t. n + 1 properties. Now, going back to logic, if we express “at least one person of each couple has property ϕ” by ✷ϕ then the above reasoning shows that the following is not valid: C : ✷p ∧ ✷q → ✷(p ∧ q). On the other hand, the following should be valid: K2 : ✷p ∧ ✷q ∧ ✷r → ✷((p ∧ q) ∨ (p ∧ r) ∨ (q ∧ r)). In general, if ✷ϕ expresses “at least one thing of each (relevant) n-tuple of things has property ϕ” then the following is intuitively valid: _ (pi ∧ pj ). Kn : ✷p0 ∧ · · · ∧ ✷pn → ✷ (0≤i 1. 2 The 2

Please do not confuse it with the non-contingency operator, which is also denoted as ∇ in noncontingency or knowing whether logics [10].

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semantics of ∇(ϕ1 , . . . , ϕn ) is based on Kripke models with n + 1-ary relations R [15,7]: ∇(ϕ1 , . . . , ϕn ) holds at s iff for all s1 , ..., sn such that Rss1 . . . sn there exists some i ∈ [1, n] such that ϕi holds at si . Note that the quantifier alternation pattern ∀∃ in the above (informal) semantics for ∇. Actually, the reading we mentioned for ✷ϕ in our motivating story is simply the semantics for ∇(ϕ1 , . . . , ϕn ) where ϕ1 = · · · = ϕn . Essentially, the formulas ✷ϕ under the new reading can be viewed as special cases of the modal formulas in polyadic modal languages. Due to the fact that the arguments are the same in ∇(ϕ, . . . , ϕ), we can also call the ✷ under the new reading the diagonal n-modalities. 3 In this light, we may call the new semantics for ✷ϕ the diagonal n-semantics (given frames with n-ary relations). Diagonal modalities also arise in other settings in disguise. For example, in epistemic logic of knowing value [12], the formula Kv(ϕ, c) says that the agent knows the value of c given ϕ, which semantically amounts to that for all the pairs of ϕ worlds that the agent cannot distinguish from the actual worlds, c has the same value. In other words, in every pair of the indistinguishable worlds where c has different values, there is a ¬ϕ world, which can be expressed by ✷c ¬ϕ with the diagonal 2-modality (✷c ) based on intuitive ternary relations (see details in [12]). As another example in epistemic logic, [9] proposed a local reasoning operator based on models where each agent on each world may have different frames of mind (sets of indistinguishable worlds). One agent believes ϕ then means that in one of his current frame of mind, ϕ is true everywhere. This belief modality can also be viewed as the dual of a diagonal 2-modality (noticing the quantifier alternation ∃∀ in the informal semantics). Yet another important reason to study diagonal modalities comes from the connection with paraconsistent reasoning established by Schotch and Jennings [24]. In a nutshell, [24] introduces a notion of n-forcing where a set of formulas Γ n-forces ϕ (Γ ⊢n ϕ) if for each n-partition of Γ there is a cell ∆ such that ϕ follows from ∆ classically w.r.t. some given logic (Γ ⊢ ϕ). This leads to a notion of n-coherence relaxing the notion of consistency: Γ 0n ⊥ (Γ is n-coherent) iff there exists an n-partition of Γ such that all the cells are classically consistent. These notions led the authors of [24] to the discovery of the diagonal semantics for ✷ based on frames with n + 1-ary relations, by requiring ✷(u) = {ϕ | u ✷ϕ} to be an n-theory based on the closure over n-forcing, under some other minor conditions. Since the derivation relation of basic normal modal logic K can be characterized by a proof system extending the propositional one with the rule Γ ⊢ ϕ/✷(Γ) ⊢ ϕ where ✷(Γ) = {✷ϕ | ϕ ∈ Γ}, it is interesting to ask whether adding Γ ⊢n ϕ/✷(Γ) ⊢ ✷ϕ characterizes exactly the valid consequences for modal logic under the diagonal semantics based on frames with n-ary relations. Apostoli and Brown answered this question positively in [4] 15 years later, and they characterize ⊢n by a Gentzen-style sequent 3

Name mentioned by Yde Venema via personal communications.

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calculus based on the compactness of ⊢n proved by using a compact result for coloring hypergraphs. 4 Moreover, they show that the WAML proof systems with Kn are also complete w.r.t. the class of all frames with n-ary relations respectively. The latter proof is then simplified in [21] without using the graph theoretical compactness result. This completeness result is further generalized to the extensions of WAML with extra one-degree axioms in [3]. The computational complexity issues of such logics are discussed in [1], and this concludes our relatively long introduction to WAML, which might not be that well-known to many modal logicians. In this paper, we continue the line of work on WAML by looking at the model theoretical aspects. In particular, we mainly focus on the following two questions: •

How to characterize the expressive power of WAML structurally within firstorder logic over (finite) pointed models?

•

Whether WAML has Craig Interpolation?

For the first question, we propose a notion of bisimulation to characterize WAML within the corresponding first-order logic. The second question is much harder to answer. We are inspired by the connection between the diagonal semantics and neighborhood semantics established in [2,5], which leads us to the model theoretical techniques for proving the interpolation theorem for monotonic neighborhood logic [13]. The last piece of the puzzle comes from a combinatorial result in lattice theory, which plays an important role in our proof. In the rest of the paper, we lay out the basics of WAML in Section 2, prove the characterization theorem based on a bisimulation notion in Section 3, and finally prove the interpolation theorem in Section 4 before concluding with future work in 5.

2

Preliminaries

In this section we review some basic definitions and existing results in the literature. 2.1 Weakly Aggregative Modal Logic The language for WAML is the same as the language for basic (monadic) modal logic. Definition 2.1 Given a set of propositional letters Φ and a single unary modality ✷, the language of WAML is defined by: ϕ := p | ¬ϕ | (ϕ ∧ ϕ) | ✷ϕ where p ∈ Φ. We define ⊤, ϕ ∨ ψ, ϕ → ψ, and ✸ϕ as usual. 4 Other connections between WAML and graph coloring problems can be found in [20] where the four-color problem is coded by the validity of some formulas in the WAML language.

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However, given n, WAML can be viewed as a fragment of poliadic modal logic with a n-ary modality, since ✷ϕ is essentially (ϕ, . . . , ϕ). Notation: in the sequel, we use WAMLn , where n > 1, to denote the logical framework with the semantics based on n-models defined below: Definition 2.2 (n-Semantics) An n-frame is a pair hW, Ri where W is an nonempty set and R is an n + 1-ary relation over W . A n-model M is a pair hF , V i where the valuation function V assigns each w ∈ W a subset of Φ. We say M is an image-finite model if there are only finitely many n-ary successors of each point. The semantics for ✷ϕ (and ✸ϕ) is defined by: M, w |= ✷ϕ iff for all v1 , . . . vn ∈ W with Rwv1 . . . , vn , M, vi |= ϕ for some i ≤ n. M, w |= ✸ϕ iff there are v1 , . . . vn ∈ W st. Rwv1 . . . , vn and M, vi |= ϕ for all i ≤ n.

According to the above semantics, it is not hard to see that the aggregation axiom ✷ϕ ∧ ✷ψ → ✷(ϕ ∧ ψ) in basic normal modal logic is not valid on nframes for any given n. [24] proposed the following proof systems Kn for each k. Definition 2.3 (Weakly aggregative modal logic) The logic Kn is a modal logic including propositional tautologies, the axiom Kn and closed under the rules N and RM: W Kn ✷p0 ∧ · · · ∧ ✷pn → ✷ (pi ∧ pj ) (0≤i q, even l = 1, the Duplicator could have a winning strategy, since any bijection will be a partial isomorphism. So the distance we define here is not a appropriate one for us to find the minimal bound l. Here we conjecture that the bound should be the least integer l s.t. l ≥ (2q − 1)/n.

4

Interpolation

In this section, we will show that WAMLn has the Craig Interpolation. The main idea is to use the alternative neighborhood semantics of WAMLn , and then follow the model theoretical proof strategy for interpolation in monotonic modal logic as in [13]. The major technical difficulty lies in proving a crucial combinatorial fact that we need in the proof of the interpolation theorem. Let us start with the neighborhood semantics perspective of WAMLn . Apostoli has shown that for each n ∈ N, Kn is complete w.r.t. all its neighborhood frames in [2] by a canonical model argument. It is also shown that the frames of Kn are those with neighborhoods which are so-called n-filters: Definition 4.1 (n-filter [2]) An n-filter F on a non-empty set I is a collection of subsets of I which satisfies (i) I ∈ F ; (ii) If X ∈ F and I ⊇ Y ⊇ X, then Y ∈ F ; S {Xi ∩ Xj : 0 ≤ i < j ≤ n} ∈ F .

(iii) If X0 , . . . , Xn ∈ F , then

A filter is proper if ∅ ∈ / F . Obviously, an 1-filter is just an ordinary filter. It is not hard to see that the three properties correspond to N, RM, and Kn in Kn w.r.t. the neighborhood semantics. We call a neighborhood frame (model) a

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Kn -frames (models) if the neighborhoods of all its points are n-filters. Theorem 4.2 (Apostoli [2]) Kn is strongly complete w.r.t. Kn -frames. Now we can define a closure operator on sets to obtain an n-filter. For each nonempty set X ⊆ P(I), Let CLIn (X ) be the closure of X under the following two operations: (i) Taking supersets on I; S (ii) {Xi ∩ Xj : 0 ≤ i < j ≤ n} for any X0 . . . Xn already in the set.

Definition 4.3 Formally, we can construct CLIn (X ) as follows by recursion. A0 = X ; S Am+1 = {Y | ∃X0 , . . . , Xn ∈ Am (Y = {Xi ∩ Xj : 0 ≤ i < j ≤ n})}. It is easy to verify that Am ⊆ Am+1 for each m ∈ N, so we define A = ∪m∈N Am and CLIn (X ) = {B ⊆ I | B ⊇ A ∈ A}. Proposition 4.4 CLIn (X ) is an n-filter. Proof. Since X is nonempty, there must be some X ∈ X , which means X ∈ A0 ⊆ A and hence I ∈ CLIn (X ) by I ⊇ X. If I ⊇ Y ⊇ X ∈ CLIn (X ), then Y ⊇ X ⊇ A ∈ A which means Y ∈ CLIn (X ). Suppose that X0 , ..., Xn ∈ CLIn (X ). By definition there are Y0 , ..., Yn ∈ A s.t. Xi ⊇ Yi , which follows that there is a minimalSk s.t. Y0 , ..., Yn ∈ Ak . Thus, by our definitionSof the sequence {Am | m ∈ N}, S{Yi ∩ Yj : 0 ≤ i < j ≤ n} ∈ Ak+1 ⊆ A. But {Xi ∩ Xj : 0 ≤ i < j ≤ n} ⊇ {Yi ∩ Yj : 0 ≤ i < j ≤ n}, S which means {Xi ∩ Xj : 0 ≤ i < j ≤ n} ∈ CLIn (X ). ✷

The following fact plays a key role in our later proof for the interpolation theorem. We first state it below for now, then prove it after giving the proof of the interpolation theorem assuming it, since its proof is highly non-trivial. Lemma 4.5 (Combinatorial Fact) Let A, B be two proper n-filters on W . Let X = {A×W | A ∈ A} and Y = {W ×B | B ∈ B}. For any X ∈ CLnW ×W (X ∪Y), for any (A, B) ∈ P(W ) × P(W )(X = A × B implies A ∈ A and B ∈ B).

Given this lemma, we can show that WAMLn has Craig interpolation. The general strategy is similar to the proof of Theorem 5.11 in [13], but instead of ultrafilter extensions used in [13] in order to obtain saturated models for suitable bisimilarity, we make a crucial use of the combinatorial fact of Lemma 4.5 to obtain the right models using elementary constructions only, and show an amalgamation-like property. Theorem 4.6 Each Kn has the Craig Interpolation, which means for every modal formula ϕ and ψ, if ϕ ⊢Kn ψ, then there is a χ satisfies the following: •

atom(χ) ⊆ atom(ϕ) ∩ atom(ψ).

•

ϕ ⊢Kn χ and χ ⊢Kn ψ.

where atom(ϕ) is the set of proposition letters in ϕ.

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Proof. Here we fix some n and just use the notions ⊢ and |=. For convenience, let atom(ϕ) = atom1 and atom(ψ) = atom2 . Let atom0 = atom1 ∩ atom2 . Let Cons0 (ϕ) = {α | atom(α) ⊆ atom0 and ϕ ⊢ α}. We need to prove that Cons0 (ϕ) ⊢ ψ, which by compactness means there are χ1 , . . . χm ∈ Cons0 (ϕ) s.t. χ1 , . . . χm ⊢ ψ. Thus we find the interpolation χ1 ∧ · · · ∧ χm . Since we know our logic is strongly complete w.r.t. all its neighborhood frames by Theorem 4.2, we can freely switch between ⊢ and |= here. Now suppose that there is a model N , t |= Cons0 (ϕ). We will show that N , t |= ψ. First note that Th0 (N , t) = {α | N , t |= α and atom(α) ⊆ atom0 } is consistent with ϕ: if not, we will find α1 , . . . , αn ∈ Th0 (N , t) s.t. ⊢ α1 ∧· · ·∧αm → ¬ϕ, which means ϕ ⊢ ¬(α1 ∧ · · · ∧ αm ). It follows that ¬(α1 ∧ · · · ∧ αm ) ∈ Cons0 (ϕ) and hence N , t |= ¬(α1 ∧ · · · ∧ αm ), a contradiction. So we could find a model M, s |= Th0 (N , t) ∪ {ϕ}. Let M + N = (Wd , ν, V ) be the disjoint union of M and N . Obviously M+N , s ≡ M, s and M+N , t ≡ N , t. In the following we use ≡i to denote the logical equivalence of WAML w.r.t. atomi . The later proof can be summarized by the following diagram. /M+N o N , t Cons0 (ϕ) ϕ M, s O π0 π1

(M + N )2

⊇

Z 2

We first build a product-like model (M + N ) and then select the pairs which are matched by ≡0 to form a submodel Z, and finally show that there are natural neighborhood bounded morphisms from Z to (M + N ). In the proof we only need to use the frame F of (M + N )2 . We define F = (W, ν ∗ ) as below: •

W = Wd × Wd ;

•

− − For each a = (x, y) ∈ W , ν ∗ (a) = CLW n ({π1 [X] | X ∈ ν(x)} ∪ {π2 [Y ] | Y ∈ ν(y)}) where πi is the projective function for the i-th coordinate.

By the construction of M, M, s ≡0 N , t, so let Z = {(x, y) | M + N , x ≡0 M + N , y}, and hence (s, t) ∈ Z ⊆ W . For each a ∈ Z, let νZ∗ (a) = {X ∩ Z | X ∈ ν ∗ (a)}, and let the frame F′ of Z be defined as F′ = (Z, νZ∗ ). Notice that M, N are Kn models, which means M + N is also a Kn model. We claim the following: Claim 1 F and F′ are Kn -frames. Claim 2 πi ◦ f are frame bounded morphisms, where f is the identity from F′ to F. If the above two claims are true, then we can define a valuation V ′ on F′ as follows: For each (x, y) ∈ Z,   x ∈ V (p) if p ∈ atom1 y ∈ V (p) if p ∈ atom2 (x, y) ∈ V ′ (p) ⇐⇒  never if otherwise

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where V is the valuation on M + N . Note that the definition here is welldefined, since M + N , x ≡0 M + N , y. Let Z = (F′ , V ′ ). Thus, we will have πi ◦ f is a (model) bounded morphism on atomi from Z to M + N , which means Z, (s, t) ≡1 M + N , s and M′ , (s, t) ≡2 M + N , t. Since M, s ϕ, Z, (s, t) |= ϕ and by ϕ ψ, Z, (s, t) |= ψ. Thus, N , t |= ψ. Below we prove the two claims: Proof of Claim 1: It is easy to show that W ∈ {π1− [X] | X ∈ ν(x)}, since Wd ∈ ν(x), and hence by our definition of ν ∗ (a), F must be a Kn frame. Because W ∈ ν ∗ (a) for each a ∈ W , we know that Z = W ∩ Z ∈ νZ∗ (b) for each b ∈ Z. For X ′ ∈ νZ∗ (a) and Z ⊇ Y ′ ⊇ X ′ , we know that X ′ = X ∩ Z for some X ∈ ν ∗ (a). Thus, Y ′ = (X ∪ Y ′ ) ∩ Z and since ν ∗ (a) is closed under taking supersets, Y ′ ∈ νZ∗ (a). For S X0 , . . . , Xn ∈ νZ∗ (a), we knowSthere are Y0 , . . . , Yn ∈ ν ∗ (a) s.t. Xi = Yi ∩Z. S So {Xi ∩Xj : 0 ≤ i < j ≤ n} = {Yi ∩Yj : 0 ≤ i < j ≤ n} ∩ Z, which means {Xi ∩ Xj : 0 ≤ i < j ≤ n} ∈ νZ∗ (a) by the closure property of ν ∗ (a). Proof of Claim 2: Here we only consider the case for π1 , the case for π2 is similar. First note that for each a = (x, y) ∈ Z and each X ⊆ Wd , by the definitions and the fact that f is an identity function, we have: (π1 ◦f )− [X] ∈ νZ∗ (a) iff f − [π1− [X]] ∈ νZ∗ (a) iff π1− [X]∩Z ∈ νZ∗ (a) iff π1− [X] ∈ ν ∗ (a) and X ∈ ν(π1 ◦ f (a)) iff X ∈ ν(x) According to the definition of neighborhood bounded morphism, we need to show that (π1 ◦ f )− [X] ∈ νZ∗ (a) iff X ∈ ν(π1 ◦ f (a)). Based on the above, the only missing part is to prove the following: for each (x, y) ∈ Z and each X ⊆ Wd , π1− [X] ∈ ν ∗ ((x, y)) iff X ∈ ν(x). The ⇐= part is clear by our definition. For =⇒ , let π1− [X] ∈ ν ∗ ((x, y)). We may assume that ∅ ∈ / ν(x), since otherwise X ∈ ν(x) will be trivially true by monotonicity of M + N , and note that ∅ ∈ / ν(y) since M + N , x ≡0 M + N , y. Obviously, π1− [X] = X × Wd , and we need to show that X ∈ ν(x), which follows from Lemma 4.5, since − − π1− [X] ∈ ν ∗ ((x, y)) = CLW n ({π1 [X] | X ∈ ν(x)} ∪ {π2 [Y ] | Y ∈ ν(y)})

Hence we can conclude that π1 ◦ f : F′ → M + N are frame bounded morphisms, and this completes the proof. ✷ 4.1 Proof of Lemma 4.5 It is quite non-trivial to prove Lemma 4.5. Here we present a proof generalizing the proof strategy sketched by Emil Jeˇra ´bek for a special case of Lemma 4.5 [17] 6 . The basic idea is that we can translate the combinatorial statement into facts about Boolean functions and clones to be defined below. 6

The question is proposed by Guozhen Shen [14].

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In order to prove this lemma, we need to use some tools in Post’s lattice theory [19]. We first present the relevant definitions. Definition 4.7 A Boolean function is an l-ary operation f : 2l → 2 for some l ≥ 1, where 2 denotes the two-element set {0, 1}. Particular Boolean functions are the projections πkl (x1 , . . . , xl ) = xk ; and given an m-ary function f , and l-ary functions g1 , . . . , gm , we can construct another l-ary function as their composition: f ◦ (g1 . . . gm )(x1 , . . . , xl ) = f (g1 (x1 , . . . , xl ), . . . , gm (x1 , . . . , xl )). A set of functions closed under composition, and containing all projections, is called a clone. There are some important W functions we will use later: 1. thn2 (x1 , . . . , xn ) = (xi ∧ xj ). 0 2q−m ) from any other previously selected worlds in T , then Duplicator just selects a corresponding world in a “new” tree which is isomorphic to T in another model (there are enough new trees). We will show that: After m rounds (0 ≤ m ≤ q), the following two hold: (Let (ai , bi ) be the pair selected at i round where each ai ∈ M∗ and bi ∈ N ∗ , especially a0 = w∗ and b0 = v ∗ . Let S(m) = {ai | i ≤ m}, Ni (m) be the neighborhood of ai within distance ′ of 2q−m −1, and Ni (m) be the neighborhood of ai within distance of 2q−(m+1) .) (1). the selected points form a partial isomorphism I: M∗ → N ∗ . (2). if m < q then there is a sequence (I0 , . . . , Im ) s.t. for each i ≤ m, a). Ii ⊇ I is a partial isomorphism with Dom(Ii ) = Ni (m) ∪ S(m); ′ ′ b). ∀h, j ≤ m∀x ∈ Nh (m) ∩ Nj (m)(Ih (x) = Ij (x)). Notice that Ii [Ni (m)] must be a similar neighborhood of bi and if y is within distance of 2q−m − 1 of bi , Ii−1 (y) ∈ Ni (m). We prove this by induction on m. When m = 0, (w∗ , v ∗ ) clearly forms a partial isomorphism, and the corresponding points in the neighborhood of M ∗ , w∗ and N ∗ , v ∗ within distance 2q − 1 respectively are isomorphic thus the partial isomorphism can be extended to them. Suppose that (1) and (2) hold for m = k < q. Now when m = k + 1 ≤ q, we need to consider two cases according to the strategy of Duplicator at the k + 1 round: (i) Spoiler selects some point outside each “2q−(k+1) −neighborhood” and Duplicator should choose a corresponding point in a new isomorphic copy in the other model (clearly there are enough new copies); (ii) Spoiler selects a point s in the ”2q−(k+1) −neighborhood” of some selected point, and Duplicator certainly has some nice point to respond in order to preserve an extended partial isomorphism due to claim (2) in the I.H. for m = k (it should be noticed that if s is in more than 1-neighborhood, the condition b) will ensure that Duplicator has a unique point to play.) Now we verify that the resulting selection satisfies the two claims. For claim (1), it’s trivial in the case (ii) due to the I.H. and our selection. For the case (i), since k < q, 2q−(k+1) ≥ 20 = 1, which means the point s that Spoiler selects is not directly linked to any selected points. Therefore the point in a new copy is safe w.r.t. s to extend the previous partial isomorphism. Now we consider claim (2) where we can assume that m = k + 1 < q and let s be the point Spoiler chose in this round. Suppose that when m = k, the previous partial isomorphism is I and the extended sequence is (I0 , . . . , Ik ). For case (i),let Ik+1 be defined as follows: ′ ′ x x ∈ Nk+1 (k + 1) Ik+1 (x) = , where x is the corresponding point I(x) x ∈ S(k) ′ of x in N ∗ . For each i ≤ k, Let Ii = Ii |Ni (k+1)∪S(k) ∪(ak+1 , bk+1 ), which is ′ well-defined since Dom(Ii ) = Ni (k) ∪ S(k) ⊇ Ni (k + 1) ∪ S(k). One can also

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see s ∈ {ak+1 , bk+1 }. ′ ′ We now check that (I0 , . . . , Ik , Ik+1 ) is a required sequence. First for condition a) we see that s is more than 2q−(k+1) − (2q−(k+1) − 1) = 1 far from any point in the ”2q−(k+1) − 1−neighborhood” of any previous selected points ′ (Remark 3.14), which means s is not linked to any such points, so Ii is still a partial isomorphism since Duplicator play a point in a new copy. Similarly, any previous selected points is more than 2q−(k+1) − (2q−(k+1) − 1) = 1 far from any point in the ”2q−(k+1) − 1−neighborhood” of s (Remark 3.14), hence Ik+1 is also a partial isomorphism. Then for condition b), we only need to ′ ′ ′ show that Nk+1 (k + 1) ∩ Ni (k + 1) = ∅ for each i ≤ k: each Ii is just a union of {(ak+1 , bk+1 )} and the restriction of Ii , so their comparability is followed by I.H. But s is more than 2q−(k+1) far from any previous selected points, which means the 2q−((k+1)+1) −neighborhoods of s is more than 2q−(k+1) − (2q−((k+1)+1) + 2q−((k+1)+1) ) = 0 far from the 2q−((k+1)+1) −neighborhoods of ′ ′ any previous selected points (Remark 3.14), so Nk+1 (k + 1) ∩ Ni (k + 1) = ∅. For case (ii), suppose that s is in the ”2q−(k+1) −neighborhood” of aj or bj , and the corresponding point that Duplicator picks is t. It is easy to see that t is in the ”2q−(k+1) −neighborhood” of aj (bj ) iff s is in the ”2q−(k+1) −neighborhood” of bj (aj ). (#) Actually an easy induction will show that the distance between t and aj must be the same with the distance between s and bj . We may assume s is in N ∗ , which seems to be the harder case, and the other situation is similar. Thus, s is a neighborhood of bj and t is a neighborhood of aj . Let Ik+1 be defined as follows: Ij (x) x ∈ Nk+1 (k + 1) Ik+1 (x) = ; I(x) x ∈ S(k) Notice that this is well-defined: if x ∈ Nk+1 (k + 1), the distance between x and t is at most 2q−(k+1) − 1, which means the distance between x and aj is at most 2q−(k+1) + 2q−(k+1) − 1 = 2q−k − 1 (Remark 3.14). So x ∈ ′ Nj (k) and hence Dom(Ij ) ⊇ Nj (k) ⊇ Nk+1 (k + 1). For each i ≤ k, Let Ii = ′ ′ Ii |Ni (k+1)∪S(k) ∪{(t, s)}. Next we show that (I0 , . . . , Ik , Ik+1 ) satisfies condition a) and b). b) is trivial since all of the new isomorphism are basically restrictions of some previous one, and the only new pair is the same: (t, s). For a), ∀h ≤ k, ′ there are two cases: 1). if ah is within 2q−(k+1) distance of t, then Ih must be a partial isomorphism by I.H; 2). if ah is far away from t by distance 2q−(k+1) , then since 2q−(k+1) −(2q−(k+1) −1) ≥ 1, t is not linked to any point in Nh (k +1) (Remark 3.14). A similar argument will show that s is not linked to any point ′ ′ in the neighborhood of Ih (ah ) by (#). So Ih is a partial isomorphism. Ik+1 is clearly a partial isomorphism since it’s a restriction of Ij .

B Proof of Proposition 4.9 Proof. Let the right-hand-side set be Θ. First we consider the ⊇ direction. We know that CLW n (X ) is closed under taking supersets. It is not hard to show that CLW (X ) is closed under Sf for all f ∈ T1n : obviously it is closed under n projections and thn+1 ; it is closed under → since for any x and y, x → y ≥ y; for 2 W any f, g1 , . . . gk ∈ T1n , if CLW n (X ) is closed under Sf and Sgi , clearly CLn (X ) is

20


closed under Sf ◦(g1 ,...gk ) . For the ⊆ direction, recall definition 4.3. Suppose that X ∈ CLW n (X ), which means X ⊇ A ∈ A and hence A ∈ Am for some m. We will show that each Am ⊆ S by an induction on m, and therefore the consequence is directly from that S is closed under taking supersets. The case for m = 0 is followed from the fact that T1n contains projections. Assume m = k + 1 and that B ∈ Am . Thus B = Sthn+1 (X0 , . . . , Xn ) where each Xi ∈ Ak . We only need to show that 2 Θ is closed under Sthn+1 , and then B ∈ Θ by the induction hypothesis. 2 Suppose that B0 , . . . , Bn ∈ Θ. By the definition of Θ, for each i ≤ n, and there are C1i , . . . , Clii ∈ X s.t. Bi ⊇ Sfi (C1i , . . . , Clii ) for some fi ∈ T1n . But by the monotonicity of thn+1 , 2 Sthn+1 (B0 , . . . , Bn ) ⊇ Sthn+1 (Sf0 (C10 , . . . , Cl00 ), . . . , Sfn (C1n , . . . , Clnn )). 2 2 Now we come to show that Sthn+1 (Sf0 (C10 , . . . , Cl00 ), . . . , Sfn (C1n , . . . , Clnn )) 2

∈ Θ, which is enough for our proof. First we reorder all these Cij as C1 , . . . , Ch where Ci = Ci0 for 0 < i ≤ l0 ; 1 Ci = Ci−l for l0 < i ≤ l0 + l1 ;... 0 n−1 n Ci = Ci−(Σn−1 l ) for Σj=0 lj < i ≤ Σnj=0 lj . j=0 j

One can check the following two principles: 1.Sπih (X1 , . . . , Xh ) = Xi . 2.Sf ◦(g0 ,...,gk ) (X1 , . . . , Xh ) = Sf (Sg0 (X1 , . . . , Xh ), . . . , Sgk (X1 , . . . , Xh )). So we have Sπih (C1 , . . . , Ch ) = Ci , and hence: Sthn+1 (Sf0 (C10 , . . . , Cl00 ), . . . , Sfn (C1n , . . . , Clnn )) 2

=Sthn+1 (Sf0 (Sπ1h (C1 , .., Ch ), .., Sπlh (C1 , .., Ch )), .., Sfn (Sπh n−1 2

0

Σ

=Sthn+1 (Sf0 ◦(π1h ,...,πlh ) (C1 , .., Ch ), .., Sfn ◦(πh n−1 2

0

Σ

=Sthn+1 ◦(f0 ◦(πh ,...,πh ),...,fn ◦(πh 2

1

l0

Σ

h )) ,...,πh n−1 l +1 j=0 j

l +1 j=0 j

l +1 j=0 j

h ) (C1 , .., Ch )) ,...,πh

(C1 , .., Ch ), which is in Θ.

h (since T1n is a clone, thn+1 ◦ (f0 ◦ (π1h , . . . , πlh0 ), . . . , fn ◦ (πΣ n−1 2 l

T1n )

(C1 , .., Ch ), .., Sπhh (C1 , .., Ch )))

j=0 j +1

, . . . , πhh )) ∈ ✷

C Proof of Lemma 4.10 Proof. Define a function f : 2k → 2 by f (a) = 0 iff ∃u ∈ W − Y s.t. χA1 ,...,Ak (u) = a. For each x ∈ W , by definition, x ∈ Sf (A1 , . . . , Ak ) iff f (χA1 ,...,Ak (x)) = 1. So if x ∈ / Y , then f (χA1 ,...,Ak (x)) = 0, which means x ∈ / Sf (A1 , . . . , Ak ). Hence Sf (A1 , . . . , Ak ) ⊆ Y ∈ / F , which means Sf (A1 , . . . , Ak ) ∈ / F . Note that since F is a proper n-filter, it is clear that CLW (F ) = F . Now by Proposin / T1n . Thus, there are b1 , . . . , bn s.t. b1 ∨ · · · ∨ bn = 1 tion 4.9 we know that f ∈ and f (b1 ) ∨ · · · ∨ f (bn ) = 0. It follows that there are u1 , . . . , un ∈ W − Y s.t. bi = χA1 ,...,Ak (ui ). ✷