The logic of subset spaces, topologic and the local difference modality K

Transcription

1 The logic of subset spaces, topologic and the local difference modality K Isabel Bevort July 18, 2013 Bachelor Thesis in Mathematics Supervisor: dr. Alexandru Baltag Korteweg-De Vries Instituut voor Wiskunde Faculteit der Natuurwetenschappen, Wiskunde en Informatica Universiteit van Amsterdam

2 Abstract In this thesis we will study the logic of subset spaces and the logical system topologic. These are logics that are strong enough for elementary topological reasoning. The first chapter will provide a quick introduction to basic modal logic on relational models with syntax, semantics, some correspondence theory, the canonical model and a description of the standard modal completeness proof. In the second chapter we will give the syntax and semantics of the logic of subset spaces and look at the soundness of the base axioms. Also, we will prove completeness of the base axioms for subset spaces in this second chapter. The third and last chapter is dedicated to the local difference modality K, with the appropriate syntax, semantics and the K -axioms. We shall finish this thesis with the open question of completeness of the K -axioms on subset spaces, with an explicit explanation as to why this remains an open question. Title: The logic of subset spaces, topologic and the local difference modality K Authors: Isabel Bevort, isabel.bevort@student.uva.nl, Supervisor: dr. Alexandru Baltag Second grader: Prof. dr. Dick de Jongh Date: July 18, 2013 Korteweg-De Vries Instituut voor Wiskunde Universiteit van Amsterdam Science Park 904, 1098 XH Amsterdam 2

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4 Contents Introduction 5 1. Modal ogic on Relational Models Syntax & Semantics Correspondence theory Canonical model ogic of subset spaces & topologic Syntax Semantics Axioms, inference rules & soundness Completeness for subset spaces Properties of theories The proof Construction Details of the construction The system topologic The local difference modality K Elementary topological reasoning The local difference modality K Syntax Semantics Axioms, inference rules & soundness Topology with the K -modality Open problem: completeness Properties of theories The proof Construction Details of the construction Popular summary in Dutch 43 Bibliography 45 A. Notes on Topological reasoning and the logic of knowledge from Dabrowski, Moss and Parikh [3] 46 4

5 Introduction Half a year away from completing my bachelor in mathematics, I realized that the courses in mathematical logic were the ones that attracted most of my attention and enthusiasm. This is why I decided to ask dr. Alexandru Baltag from the Institute of ogic, anguage and Computation, who taught the course Introduction to Modal ogic in the first semester of this schoolyear, to be the supervisor of my bachelorthesis. He accepted and soon came with a subject that I was very excited to start working on, Modal ogics of Space. After having read the chapters Modal ogics of Space and Topology and Epistemic ogic from the Handbook of Spatial ogics [1], my attention was drawn to the latter, and in particular the logic of subset spaces and the logical system of topologic. This is an area of research that considers the connection between topology and modal logic, in particular epistemic logic. This area of research is relatively young as it started with the work of Alfred Tarski and J.C.C. McKinsey in the 1940 s, who reviewed the connection between the modal logic S4 and topology. When stated in the modern truthconditional format, the modality can be interpreted as various topological concepts, such as the interior operator or the derivative. In this thesis we shall mostly work in a modal language with two modalities, the and K. In some way, the can be interpreted as effort because it shrinks an open, and K as knowledge, because it tells us exactly where the point is. We will give an introduction to basic modal logic on relational models in the first chapter, which will be a quick overview of some standard definitions and results. The second chapter is dedicated to the logic of subset spaces and topologic. First we will define our language and give the semantics on this language. Then we will define the logic of subset spaces by giving the base axioms and inference rules. The soundness of the axioms will also be proven. We will spend an entire section on the completeness proof of the base axioms for subset spaces, which is an exciting adaptation of the standard modal completeness proof. We have followed the proof given by Dabrowski, Moss and Parikh given partly in [8] and fully in [3]. astly, we will define the logical system topologic, which is an extension of the logic of subset spaces with the addition of two more axioms, (Un) and (WD). In the third and last chapter we will consider a new modality, the local difference modality K. Before defining a new language, we will give a connection between topology and the logic of subset spaces by expressing some topological notions in the language of the subset space logic. This will make clear why the new modality is needed, since the possibilities of expressing topological notions are very limited. The modality K will allow us to express difference between points, and hence enlarge our capability of expression. We will give the accompanying axioms and inference rules and define a system called the K -system. After proving that the axioms are sound, we will show 5

6 what new topological notions can then be expressed. The completeness of the K axioms will be given as an open problem. We will follow the pattern of the completeness proof for the subset space logic, adapt it to this new logic, and show exactly at what point the adapted proof fails. A section with notes on the article by Dabrowski, Moss and Parikh [3] is added as an appendix. I have very much enjoyed working on this project and would like to conclude by thanking my supervisor, dr. Alexandru Baltag, for his kind and enthusiastic guidance. His patient assistance in helping me overcome the problems I encountered, made writing this bachelor thesis a rewarding and thought-provoking experience. 6

7 1. Modal ogic on Relational Models In this chapter we will introduce basic modal logic on relational models. The definitions and results given in this chapter can be found here [2]. The first definition we need is the following: Definition 1.1. A relational structure is a tuple F whose first component is a nonempty set W called the universe (or domain) of F, and whose remaining components are relations on W. The elements of W are called states or points Syntax & Semantics In this section we will define the syntax and semantics of basic modal logic on relational models. et us begin with the syntax. Definition 1.2. The basic modal language is defined using a set of proposition letters Φ whose elements are usually denoted by p, q, r,..., and a unary modal operator ( box ). The well-formed formulas φ of the basic modal language are given by the rule: φ ::= p φ ψ ψ φ, where p ranges over elements of Φ. This definition means that a formula is either a propositional letter, the propositional constant falsum ( bottom ), a negated formula, a disjunction of formulas, or a formula prefixed by. Just as the familiar first-order existential and universal quantifiers are duals to each other (in the sense that xα x α), we have the dual operator ( diamond ) for our box which is defined as φ := φ. We also make use of the classical abbreviations for conjunction, implication, bi-implication and the constant true ( top ): φ ψ := ( φ ψ), φ ψ = φ ψ, φ ψ := (φ ψ) (ψ φ) and :=. Definition 1.3. A modal similarity type is a pair T = (O, τ) where O is a non-empty set, and τ is a function τ : O N. The elements of O are called modal operators; we use ( triangle ), 0, 1,..., to denote elements of O. The function τ assigns to each operator O a finite arity, indicating the number of arguments can be applied to. The similarity type of the basic modal language is called T 0, where O 0 = and τ( ) = 1. We will now define frames, models and the satisfaction relation for the basic modal language. 7

8 Definition 1.4. A frame for the basic modal language is a pair F = (W, R) such that (i) W is a non-empty set. (ii) R is a binary relation on W, so R W W. That is, a frame for the basic modal language is simply a relational structure bearing a single binary relation. A model for the basic modal language is a pair M = (F, V ), where F is a frame for the basic modal language, and V is a function called a valuation. It assigns to each proposition letter p in Φ a subset V (p) of W. Formally, V is a map: Φ P(W ), where P(W ) denotes the powerset of W. Informally, we think of V (p) as the set of points in our model where p is true. Given a model M = (F, V ), we say that M is based on the frame F, or that F is the frame underlying M. In the remaining part of this section we will learn how to interpret the basic modal language in models by defining the following satisfaction relation = M. Definition 1.5. Suppose w is a point in a model M = (F, V ). Then we inductively define the notion of a formula being satisfied (or true) in M at a state w as follows: w = M p iff w V (p), where p Φ, w = M never, w = M φ iff not w = M φ, w = M φ ψ iff w = M φ or w = M ψ, w = M φ iff for all v W such that Rvw we have v = M φ. It follows from this definition that w = M φ if and only if there exists some v W such that Rvw and v = M φ. Notice how and are actually local versions of the usual - for all and - there exists quantifiers. The following example should help the reader to become familiar with this semantics. Example 1.6. et M = (F, V ) be such that F = (W, R), W = {w, v, u}, R = {(w, v), (v, w), (w, u), (v, u)} and V is such that V (p) = {w, v, u} and V (q) = {v}. 8

9 In this model, we have the following: w = p w = p since w points to both u and v, and v = p and u = p. w = p but w = p. Indeed, Rwv and v = q, but Rwu as well and u = q. w = (p q). Indeed, for every state that w points to (u and v), either p is true or q is true. These are just examples of formulas true in this model Correspondence theory In this section we will present some correspondence theory results that we will use later on in the completeness proof of the subset space logic. We refer to [2, chapter 3] for the proofs of these results. First, let us give the needed axioms together with their traditional names: (T) p p (4) p p (B) p p (5) p p Proposition 1.7. For any frame F = (W, R), the axiom (T) corresponds to reflexivity of the relation R: F = T iff F = xrxx. Proposition 1.8. For any frame F = (W, R), the axiom (4) corresponds to transitivity of the relation R: F = 4 iff F = x y z(rxy Ryz Rxz). Proposition 1.9. For any frame F = (W, R), the axiom (B) corresponds to symmetry of the relation R: F = B iff F = x y(rxy Ryx). Proposition For any frame F = (W, R), the axiom (5) corresponds to euclideanness of the relation R: F = 5 iff F = x y z((rxy Ryz) Ryz). 9

10 1.3. Canonical model In this section we will define the modal canonical model in the basic language and give the steps of the standard completeness proof. We will not prove any of the results, and again refer to [2, section 4.2] for the strongly interested reader. First we need to define modal logics and strong completeness. Definition (Modal ogics) A modal logic Λ is a set of modal formulas that contains all propositional tautologies and is closed under modus ponens (that is, if φ Λ and φ ψ Λ then ψ Λ) and uniform substitution (that is, if φ belongs to Λ then so do all of its substitution instances). If φ Λ we say that φ is a theorem of Λ and write Λ φ; if not, we write Λ φ. Definition A logic Λ is strongly complete with respect to some class of structures S if and only if every Λ-consistent set of formulas is satisfiable on some G S. Definition A modal logic is normal if it contains Kripke s axiom and the axiom for the duality of and : (K) (Dual) (p q) ( p q), p p, and is closed under generalization (that is, if Λ φ then Λ φ). Canonical models are introduced in order to prove completeness. Given a normal logic Λ, its strong completeness is proven with respect to some class of structures by showing that every Λ-consistent set of formulas can be satisfied in some suitable model. The suitable models are built out of maximal consistent sets of formulas, and in particular, canonical models are built. Definition A set of formulas Γ is maximal Λ-consistent if Γ is Λ-consistent, and any set of formulas properly containing Γ is Λ-consistent. If Γ is maximal Λ-consistent we say it is a Λ-msc. Proposition (Properties of maximal consistent sets) If Λ is a logic and Γ is a Λ-msc then: (i) Γ is closed under modus ponens; (ii) Λ Γ; (iii) for all formulas φ: φ Γ or φ Γ; (iv) for all formulas φ, ψ: φ ψ Γ iff φ Γ or ψ Γ. The following lemma is to ensure that there are enough maximal Λ-consistent sets. This is important because they are the building blocks of this construction. emma (indenbaum s emma) If U is a Λ-consistent set of formulas then there is a Λ-msc U + such that U U +. 10

11 Now we can define the canonical model. Definition The canonical model M Λ for a normal modal logic Λ in the basic language is the triple (W Λ, R Λ, V Λ ) where: (i) W Λ is the set of all Λ-mscs; (ii) R Λ is the binary relation on W Λ defined by R Λ wu if for all formulas ψ, ψ u implies ψ w. R Λ is called the canonical relation; (iii) V Λ is the valuation defined by V Λ (p) = {w W Λ : p w}. V Λ is called the canonical (ornatural) valuation. The pair F Λ = (W Λ, R Λ ) is called the canonical frame for Λ. The following lemma ensures that there exist enough coherently related mcss. emma (Existence emma) For any normal modal logic Λ and any state w W Λ, if φ w then there is a state v W Λ such that R Λ wv and φ v. emma (Truth emma) For any normal modal logic Λ and any formula φ, w = M Λ φ if and only if φ w. Theorem (Canonical Model Theorem) Any normal modal logic is strongly complete with respect to its canonical model. 11

12 2. ogic of subset spaces & topologic In this chapter we will stuy the logic of subset spaces and the logical system topologic. We will begin with the syntax and semantics, then define the logic of subset spaces and prove completeness of the base axioms for subset spaces. We will finish this chapter by having a look at topologic Syntax Definition 2.1. A subset space is a pair X = (x, O) where X is a set and O is a set of subsets of X. We assume that X O for convenience, though this is not really necessary. The space X is closed under intersection, respectively union, if the following holds: if S, T O, then S T O, respectively S T O. Definition 2.2. et A be an arbitrary countable set of atomic formulas. Define as the smallest set containing each A A that is closed under the following rules: if φ, ψ, then also φ ψ and φ, if φ, then Kφ and φ. Definition 2.3. et X be a subset space. Thinking of X as a Kripke frame, we can give the semantics of as an interpretation of the atomic formulas. By this we mean a map α : A P(X). The pair (X, α) is a model, denoted by M Semantics Definition 2.4. For p X and p u O, we define the satisfaction relation = M on (X O) by recursion on φ: p, u = M A iff p α(a) p, u = M φ ψ iff p, u = M φ and p, u = M ψ p, u = M φ iff p, u = M φ p, u = M Kφ if for all q u, q, u = M φ p, u = M φ if for all v O such that p v u, p, v = M φ 12

13 As usual, the dual of K is, with φ K φ, and the dual of is, with φ φ. The satisfaction relation for and is defined dually as follows: p, u = M φ if there exists some q u such that q, u = M φ p, u = M φ if there exists v O such that p v u, p, v = M φ. This semantics can be represented visually, as is shown in the following picture. Figure 2.1.: This is a visual representation: p, u = Kφ iff for every q u, q, u = φ, and p, u = φ iff for every v u s.t. p v, p, v = φ. When M is clear from the context, we write p, u = φ. et T. We write T = φ if for all models M, all p X and all u O, the following holds: if p, u = ψ for each ψ T, then p, u = φ. Formulas of special interest Given a model M and a formula φ, φ is local in M if for all p, u, v we have p, u = φ iff p, v = π. In other words, the truth of φ depends only on the point. φ is local if it is local for all points in M. φ is persistent in M if φ φ is valid in M. φ is persistent if φ φ is persistent in every M. Note that ever formula of the form K φ is persistent. Also, if v u, then every persistent formula satisfied by p, u is also satisfied by p, v. 13

14 These formulas are of special interest because their truth is dependent only on the points and not of the opens around them. Their interpretation in this language is similar to usual modal interpretation. It also implies that if φ A, it can be interpreted as a set of points Axioms, inference rules & soundness The following are the base axioms. Propositional tautologies (A A) ( A A), for A A; K φ Kφ, this is the Cross axiom; Kφ (φ KKφ), this axiom combines reflexivity and transitivity for K; φ Kφ, Brouwer s axiom for symmetry. This is used in [8, 6, 5]; φ Kφ, the axiom for euclideanness. This is used in [1, 3]; For K to be S5-like, there needs an axiom for symmetry. This axiom is found in two versions throughout the literature, as the Brouwer s axiom or the axiom for euclideanness. However, they are equivalent in the presence of the reflexivity and transitivity axioms. Indeed: suppose we have φ Kφ as an axiom. The reflexivity axiom is equivalent to φ φ, so with the Eucliddeanness axiom φ Kφ we have φ Kφ. Suppose now that we have Brouwer s axiom. Use uniform substitution with φ = φ, so we get φ Kφ. Now K-generalization on the transitivity axiom gives K(φ φ) and by the K-axiom and modus ponens it follows that Kφ Kφ. Thus φ Kφ. φ (φ φ), this axiom combines reflexivity and transitivity for ; (φ ψ) ( φ ψ), this is Kripke s axiom for the modality; K(φ ψ) (Kφ Kψ), this is the Kripke s axiom for the K modality. We add the following rules of inference: modus ponens, K-necessitation and -necessitation. Definition 2.5. The logic of subset spaces is the logic described by the above axioms and rules of inference. We will now check the soundness of these axioms for subset spaces. 14

15 (A A) ( A A), for A A This axiom was added because the semantics for atomic formulas ignores the observation whether p, u = A depends only on p. It is an axiom that we need to include for atomic preservation. Moss and Parikh put it in the following words: Our framework is based on the intuition that the interpretations of atomic formulas are absolute in this sense. Obviously this axiom is sound. K φ Kφ et s check the soundness of the Cross axiom. To do so, fix a subset space X, let p X and u O and suppose p, u = K φ. Then for all p u, we have p, u = φ. So for all u u such that p u u we have p, u = φ. To show: p, u = Kφ. So let u be arbitrary such that u u and p u u. Now we want: p, u = Kφ. So let p u be arbitrary, we want: p, u = φ. So let us use the premise. et p := p and u := u. It follows that p, u = φ, thus we are done. Note that the converse is not valid. (a) The premise (b) the conclusion Figure 2.2.: Visual representation of the Cross axiom Kφ (φ KKφ) Fix a subset space X, let p X and u O and suppose p, u = Kφ. Then for all q u we have q, u = φ. So obviously p, u = φ. We need to show that p, u = KKφ, i.e., for all q u, q, u = Kφ, i.e. for all q u and for all r u : r, u = φ, which is obviously true. So the axiom for reflexivity and transitivity of K is sound. 15

16 φ Kφ We will show the soundness of Brouwer s axiom. So fix a subset space X, let p X and u O and suppose p, u = φ. We want to show that p, u = Kφ, i.e., that for all q u there exists a r u such that r, u = φ, which is true since p, u = φ. So the Brouwer s axiom is sound. φ (φ φ) Fix a subset space X, let p X and u O and suppose p, u = φ. Then for all v u we have p, v = φ. So clearly p, u = φ as well, which shows the soundness of the reflexivity of. We still need to show that p, u = φ, i.e., for all v u we have p, v = φ. So let v u, and let w v. Then also w u, so p, w = φ. So p, v = φ. So the axiom for reflexivity and transitivity of is sound. (φ ψ) ( φ ψ) Fix a subset space X, let p X and u O. Suppose p, u = (φ ψ) and p, u = φ. We want to show that p, u = ψ. So let v u, then p, v = φ ψ, and p, v = φ. By modus ponens we have p, v = ψ, thus p, u = ψ. So the (K)-axiom for is sound. K(φ ψ) (Kφ Kψ) Fix a subset space X, let p X and u O. Suppose p, u = K(φ ψ) and p, u = Kφ. We want to show that p, u = Kψ. So let q u, then q, u = φ ψ, and q, u = φ. By modus ponens we have q, u = ψ, thus p, u = Kψ. So the (K)-axiom for K is sound. Note how these axioms resemble the axioms given in section 1.2. The axiom with the -modality that combines reflexivity and transitivity is actually a combination of (4) and (T) (which is equivalent to p p). The axiom with the K-modality that combines reflexivity and transitivity is actually just the same as for, but with a different modality, namely the K-modality, which is also a local version of the -quantifier. We have already called the axiom for the symmetry for Brouwer s axiom, and with good reason: it is just the (B) of section 1.2. The axiom for euclideanness is, similarly, the (5) of this section. Now that we have shown soundness for all of the base axioms, the following lemma holds. emma 2.6. (Soundness) If T and T φ, then T = φ. 16

17 2.4. Completeness for subset spaces In this section we will prove completeness of the base axioms for subset spaces Properties of theories The proof uses maximal consistent subsets of, they are called m-theories. Fix a language, and let T H be the set of m-theories in. Use U, V,... to denote m-theories. In order to prove that the proof system is complete, we need to show that for every m-theory T, there exists a subset space model X = (X, O, α), a point p X, and a subset u O such that p, u = X T. First we need to define relations on m-theories: Definition 2.7. Define the relations and on m-theories by: U V iff if φ V, then φ U, and U V iff if φ V, then φ V. Because of the maximal consistency of m-theories there are equivalent definitions, such as: U V if when Kφ U, then φ V, and U V if when φ U, then φ V. Further, define U V if for some W, U some W, U W V. Proposition 2.8. Concerning the relations is an equivalence relation. is a reflexive and transitive relation. W and V, and define U : 3. If φ T, then there is some U such that φ U and T U. 4. If φ T, then there is some U such that φ U and T U. V if for Proof. The following proofs are all simple consequences of the S4-ness of and the S5-ness of. Note that 3. and 4. are equivalents of the Existence emma in standard modal completeness proofs as in section To show : is an equivalence relation. Reflexive. To show : if φ U, then φ U. Suppose φ U. By the reflexivity axiom (equivalent to φ φ) and the maximal consistency of U it follows that φ U. So is a reflexive relation. Symmetric. Suppose U V. To show : V U, i.e. if φ U then φ V. Suppose φ U and φ / V. Since V is a maximal consistent set, V, so K φ V. Then from U V it follows that K φ U. So by 17

18 the symmetry axiom φ Kφ (equivalent to Kφ φ) we have φ U. This is in contradiction with the consistency of U, so φ V. So is symmetric. Transitive. Suppose U V W. To show : U W, i.e. if φ W then φ U. Suppose φ W, then φ V and φ U. Thus by the transitivity axiom (equivalent to φ φ) φ U. So U W, which means is transitive. 2. To show : is reflexive and transitive. Reflexive. To show : if φ U then φ U. So suppose φ U. By the reflexivity axiom (equivalent to φ φ) and the maximal consistency of U it follows that φ U. So is reflexive. Transitive. Suppose U V W. To show : U W, i.e. if φ W then φ U. So suppose φ W. Then φ V and φ U. By the transitivity axiom (equivalent to φ φ) it follows that φ U. Thus is transitive. 3. Suppose φ T, and let A = {φ} {ψ : Kψ T }. We want to show that A is consistent, so suppose towards a contradiction that it is not. Then there exist ψ 1,..., ψ n A such that Kψ 1,..., Kψ n T and ψ 1 ψ n φ. The maximal theory T is closed under conjunction, so Kψ 1 Kψ n T. We are working in a normal modal logic since we have Kripke s axiom for K, the dual of K and K-necessitation [2, p. 33], so (Kψ 1... Kψ n ) K(ψ 1 ψ n ). Therefore K(ψ 1 ψ n ) T. Now apply necessitation to ψ 1 ψ n φ to obtain K(ψ 1 ψ n φ) T. Followed by the use of Kripke s axiom and modus ponens we get K(ψ 1 ψ n ) K φ T, which by modus ponens again gives K φ T. This is in contradiction to T being consistent since φ T. So A is consistent, and by indenbaum s lemma [2, p. 197] there exists a maximally consistent U A. It is easy to see that φ U and T U. 4. Suppose φ T, and let A = {φ} {ψ : ψ T }. We want to show that A is consistent, so suppose towards a contradiction that it is not. Then there exist ψ 1,..., ψ n A such that ψ 1,..., ψ n T and ψ 1 ψ n φ. The maximal theory T is closed under conjunction, so ψ 1 ψ n T. We are working in a normal modal logic since we have Kripke s axiom for, the dual of and -necessitation [2, p. 33], so ( ψ 1... ψ n ) (ψ 1 ψ n ). Therefore (ψ 1 ψ n ) T. Now apply necessitation to ψ 1 ψ n φ to obtain (ψ 1 ψ n φ) T. Followed by the use of Kripke s axiom and modus ponens we get (ψ 1 ψ n ) φ T, which by modus ponens again gives φ T. This is in contradiction to T being consistent since φ T. So A is consistent, and by indenbaum s lemma [2, p. 197] there exists a maximally consistent U A. It is easy to see that φ U and T U. 18

19 Proposition 2.9. If U V, then U V. In other words, if U and V are m- theories, and W is such that U W V, then there exists some T such that U T V. Proof. This proof makes use of the Cross Axiom. Suppose we have U, V and W such that U W V. et S := {φ : φ V } {ψ : Kψ U}. Suppose towards a contradiction that S is not consistent. Then there exists a finite subset S 0 of S which is inconsistent. Write S 0 = {φ 1, φ 2,..., φ n } {ψ 1, ψ 2,..., ψ m }, where for each i and j, φ i belongs to V and Kψ j belongs to U. et φ := φ 1 φ n and similarly ψ := ψ 1 ψ m. Since V is closed under conjunction, φ V. And because Kψ is equivalent to a conjunction K(ψ 1... ψ m ) from U, we have Kψ U. Finally φ φ i for all i, and ψ ψ j for all j. Since S 0 is inconsistent, φ ψ. By the (K)-axiom and modus ponens, it follows that φ ψ, so this sentence belongs to U. Also, φ V, so φ W and φ U. By the Cross axiom, φ U. By modus ponens on φ ψ, it follows that ψ U. But this is in contradiction with U being consistent since Kψ U. So S is consistent. By indenbaum s lemma [2, p. 197] there exists T S maximally consistent. By construction U T V, which proves the proposition. Dabrowski, Moss and Parikh put the meaning of this proposition in such a nice wording that we would like to cite them here: This proposition is the embodiment of the Cross Axiom scheme in the realm of m-theories. [3, p. 85] The next proposition is a generalization and will be used in the proof of completeness. Proposition Suppose that T 1 Then there are U 1 U 2... T 2... T n, and suppose T n U n such that 4U n = U, and for all i, T i U. U i. T 1 T 2... U 1 U 2... T n U n Proof. Given T 1, T 2,..., T n and U as above, let U n = U. Since T n 1 get U n 1 from proposition 2.9 such that T n 1 U n 1 in this way in order to get U n 2,..., U 1. T n U n, we U n. We continue backwards This concludes this subsection, as we now know sufficiently about m-theories in order to be able to prove completeness. 19

20 The proof In this section we will give the proof of completeness. It contains the strategy of construction and the Truth emma, from which completeness will follow. We will follow the proof of Dabrowski, Moss and Parikh [3] and give additional explanations along the way. Theorem (Completeness for subset spaces) The base axioms are complete for interpretation in subset spaces. That is, if T = φ, then T φ. Strategy ike standard modal completeness proofs, the first idea in proving completeness is to consider the canonical model of the logic. This is the set T H of m-theories with the relations and as defined in subsection It would be best if we could define a family O of subsets of T H in order to obtain a subset space. Then we would hope to show that every theory T is the theory of some pair p, u from that model, indeed we might hope that P is T itself. Unfortunately, this idea does not seem to work. The problem is that we do not know any way to define a subset space structure on T H which leads to completeness. For this reason, we do not approach completeness via the canonical model. Our strategy is to build a space X of abstract points. The opens will also be given in an abstract way, via a poset P and an order-reversing map i : P P (X), where P (X) is the set of non-empty subsets of X. The points are abstract since they are not theories. But with each x and each p so that x i(p) there will be a target m-theory t(x, p). The goal of the construction is to arrange that in the overall model, th(x, i(p)) = t(x, p). Even though we are not using the canonical model, the completeness proof reminds us of the standard modal completeness proofs that do use the canonical modal of a logic. We shall have a Truth emma and existence lemmas (proposition 2.8) as described in section 1.3. By keeping this in mind when reading the proof, the reader can have a better intuitive notion of what is happening. We only need to prove that every m-theory T has a model in order to prove completeness. et T be an m-theory. We build (1) A set X containing a designated element x 0. (2) A poset P, with least element. (3) A function i : P P (X) such that p q iff i(q) i(p), and i( ) = X. So i is an order-reversing homomorphism from P,, to P (X),, X. (4) A partial function t : X P T H such that t(x, p) is defined iff x i(p). The following properties need to be ensured for all p P, x i(p) and φ: 20

21 (a1) If y i(p), then t(x, p) t(y, p). (a2) If φ t(x, p), then for some y i(p), φ t(y, p). (b1) If q p, then t(x, p) t(x, q). (b2) If φ t(x, p), then for some q p, φ t(x, q). (c) t(x 0, ) = T, where T is the m-theory that we fixed and want to make a model for. Suppose we have X, P, i and t, and that they satisfy these properties. consider the subset space model Then we X = x, {i(p) : p P}, α where the valuation is as follows α(a) = {x : A t(x, )}. Truth emma The Truth emma is key in proving completeness for subset spaces. emma (The Truth emma). Assume conditions (1)-(4) for X,P,i and t. Then for all x X and all p P such that x i(p), th X (x, i(p)) = t(x, p). Proof. The atomic case holds by definition of X. Indeed, suppose A th X (x, i(p)), i.e. x, i(p) = A. This means that x α(a), and from the definition of the valuation α, this is equivalent to A t(x, ). So what we want to show is the following: A t(x, ) A t(x, p). So suppose A t(x, ). By the axiom for atomic preservation A t(x, ), since t(x, ) T H. Since is the least element of the poset P, we have p. By property (b1), this implies t(x, ) t(x, p). Now we use the equivalent definition of to conclude that A t(x, p). Suppose now that A t(x, p). As before, we have t(x, ) t(x, p), so A t(x, ). By the axiom for atomic preservation A A, we have A t(x, ). The induction steps for Boolean connectives are consequences of the fact that the sets in T H are m-theories. So let us do the induction step for. Suppose that φ th X (x, i(p)), i.e. that x, i(p) = φ. Then there is some y i(p) such that y, i(p) = φ, i.e. φ th X (y, i(p)). By the induction hypothesis φ t(y, p). By condition (a1), t(x, p) t(y, p). Therefore φ t(x, p). Now suppose φ t(x, p). Then by property (a2) there is some y i(p) such that φ t(y, p). By the induction hypothesis we have φ th X (y, i(p)), i.e. y, i(p) = φ. Thus x, i(p) = φ, i.e. φ th X (x, i(p)). 21

22 Next we do the induction step for. First suppose that φ th X (x, i(p)), i.e. x, i(p) = φ. Then there is some i(q) i(p) such that x i(q) and x, i(q) = φ. By the induction hypothesis we have φ t(x, q). By property (3), p q, so by property (b1) it follows that t(x, p) t(x, q). So φ t(x, p). Now suppose that φ t(x, p). Then by (b2) there is some q p, φ t(x, q). From the induction hypothesis it follows that φ th X (x, q), i.e. that x, i(q) = φ. But i(q) i(p) by property (3), so x, i(p) = φ, i.e. φ th X (x, p). This proves the Truth emma. Now we have almost proven completeness. All we need to do is note that our m-theory T has a model: by the Truth emma and property (4c) we have T = t(x 0, ) = th X (x 0, i( )) = th X (x 0, X), thus the base axioms are complete for subset space models Construction In this section we will build X, P, i and t by recursion. That is, we build approximations X n, P n, i n and t n satisfying certain local and global properties. Fix two objects x 0 and. The following local properties will be satisfied by the construction. (1) X n is a finite set containing x 0. (2) P n is a finite poset with as minimum, and with the property that for each p P n, the lower set of p, {q P n : q p}, is linearly ordered. (3) The map i n : P n P (X n ), where P (X) is the collection of subsets of X n with at least two elements, has the property that p q iff i n (q) i n (p). Also, i n ( ) = X n. (4) The function t n : X n P n T H is partial with the property that t n (x, p) is defined iff x i n (p). Furthermore, we assume the following properties for all x X n and p P n : (a) If x, y i n (p), then t n (x, p) t n (y, p). (b) If x i n (q) and q p, then t n (x, p) (c) t 0 (x 0, ) = T. t n (x, q). These conditions are required for specific reasons. In (2), the requirement that the lower sets of points is linear is essential to our construction. By maintaining this property throughout the construction, we can use proposition 2.9 to add points to the model. The condition in (3) that each i n (p) has at least two elements is not really necessary, but it leads to a simplification of the overall construction. 22

23 There will also be global properties satisfied by the construction: (G1) X n X n+1. (G2) P n+1 is an end extension of P n. This means that P n is a suborder of P n+1 and if p P n+1, q P n and p q, then p P n. (G3) For all p P n+1, i n+1 (p) X n = i n (p). (G4) The restriction of t n+1 to X n P n is t n. Finally, our construction has to satisfy some overall requirements: (R4a) If φ t n (x, p), then for some m > n, there is some y i m (p) such that φ t m (y, p). (R4b) If φ t n (x, p), then for some m > n, there is some q p in P m such that φ t m (x, q). Suppose we build X n, P n, i n and t n in such a way they satisfy the (), (G) and (R) requirements. Definition et X = n N X n, and let P be the limit of the posets P n. et i be defined by i(p) = n>m i n+1(p), where m is the least number such that p P m. Finally, define t(x, p) = t n (x, p) where n is any number such that t n (x, p) is defined. The construction has arranged that t n (x, p) = t n+1 (x, p) whenever the latter is defined. Proposition Suppose we build X n, P n, i n and t n in accordance with the (), (G) and (R) requirements. Then X, P, i and t as defined above satisfy conditions (1)-(4) above. Proof. It is clear that (1) holds as we have constructed X in this way. In the same way we have constructed P to have as its least element so (2) also holds. To check (3), first note that x i(p) if and only for some n, x i n (p). Now suppose p q and let x i(p). Then there is an n such that x i n (p). By (3) i n (p) i n (q). So x i n (q), i.e. x i(q). Thus i(p) i(q). On the other hand, if p q, then let n be such that both p, q X n. By (G2), p q in X n, so by (3), i n (p) i n (q). This means there is some x such that x i n (p) \ i n (q). By (G3), i n (q) = X n i(q), so since x i n (p) X n we see that x / i(q). Hence i(p) i(q). Furthermore, it is easy to see that i( ) = X. This completes the verification of (3). The verifications of (4) are consequences of the overall (R) requirements and conditions (G4) and (4a). 23

24 Details of the construction In this section we will go deeper into the construction and give the details. At the beginning we fix a map v : ω {, } ω ω with the property that if v(n) = (x, m, k) then m < n, and for all (x, m, k) there is some n > m such that v(n) = (x, m, k). Note that we have used a suggestive notation with {, }, it could have been {1, 2} just as well. We define by recursion on n a tuple X n, P n, i n, t n, α n, β n. The last two, α n and β n, are functions whose purpose is to ensure all of the (R) requirements are met in an orderly fashion in countably many steps: α n : ω {(x, p, φ) X n P n : φ t n (x, p)} β n : ω {(x, p, φ) X n P n : φ t n (x, p)} et X 0 be a two-element set {x 0, x 1 }, let P 0 be the trivial poset { }, let i 0 ( ) = X and t 0 (x 0, ) = T, where T is the m-theory that we started with. All of these () requirements are satisfied for n = 0. We also fix functions α 0 and β 0, but this is just to make sure that they are not arbitrary. Now suppose we are at stage n+1 of the construction. There are two cases, depending on the value of v(n + 1). Case 1: v(n + 1) = (, m, k) for some m and k. Consider α m (k), and let it be equal to the triple (x, p, φ). This means that φ t m (x, p). et y be a new point not in X n. et X n+1 := X n {y}. et P n+1 := P n. et { i n (q) {y} if q p, i n+1 (q) = i n (q) otherwise. Note that conditions (1), (2), (G1),(G2) and (G3) are trivial. Indeed, X n+1 is still finite (1) and X n X n+1 (G1), and P n+1 = P n (2,G2). For (G3), let q P n+1. Then if q p, i n+1 (q) = i n (q) {y} so i n+1 (q) X n = i n (q) since y X n+1 \ X n. Otherwise i n+1 (q) = i n (q) so obviously i n+1 (q) X n = i n (q). Also, i n+1 ( ) = X n+1. Next we will check condition (3), i.e. that q r if and only if i n+1 (r) i n+1 (q). For this we look at three cases. First, if neither q p nor r p, then i n+1 (q) = i n (q) and i n+1 (r) = i n (r). So we are done, since (3) holds for n. Second, if both q and r are below p, then i n+1 (q) = i n (q) {y} and i n+1 (r) = i n (r) {y}. So i n+1 (r) i n+1 (q) if and only if i n (r) i n (q), which, again, is equivalent to q r. For the last case, suppose that q p but r p. Then also r q. And since y / i n+1 (r), we have i n+1 (q) i n+1 (r). Now, since i n+1 (q) = i n (q) {y}, we again have that i n+1 (r) i n+1 (q) if and only if i n (r) i n (q). Now we have completed the verification of (3). Next we wish to define t n+1. For this, we stipulate that (G4) holds. Remember that t n+1 is a map from X n+1 P n+1 to T H, so we only need to define t n+1 (y, q) for q p, 24

25 since y is the only point in X n+1 that is not in X n, and when q p we have i n (q) i n (p), so no changes are made. We use condition (2) to see that {q : q p} is a finite, linearly ordered set. So write this set as q 1 q 2 q N = p. Then i(p) i(q 2 ) i(q 1 ) by (3), so for all i, x i(q i ). Thus we can write t(x, q 1 ) t(x, q 2 )... t(x, q N ). Note that we have used the notation t(x, q i ) here instead of t n+1 (x, q i ). This is because of condition (G4) combined with the definition of t: the restriction of t n+1 to X n P n is t n, and t(z, r) = t n (z, r), where n is any number such that t n (z, r) is defined. Now let U be such that t(x, q N ) U and φ U. This exists by 3. of proposition 2.8. To be able to apply this proposition we need to realize that φ t(x, q N ). We began with the triple α m (k) which told us that φ t m (x, p). We know that p = q N so φ t m (x, q N ). And of course by the definition of t again t(x, q N ) = t m (x, q n ). So φ t(x, q N ) and we can use the proposition. To give a visual representation: t(x, q 1 ) t(x, q 2 )... t(x, q N ) U By proposition 3.12 there exist m-theories U 1, U 2,..., U N such that U = U N, t(x, q i ) U i for all i, and U 1 U 2... U N. et t(y, q i ) := U i for 1 i N. So we get: t(x, q 1 ) t(x, q 2 )... t(x, q N ) t(y, q 1 ) t(y, q 2 )... t(y, q N ) and this definition of t n+1 ensures (4b). Next we need to check whether (4a) holds for this definition of t n+1. So suppose a, b i n+1 (q). We might as well assume that a = y, and hence that q p. If b = y as well, then we have t n+1 (a, q) t n+1 (b, q) by the reflexivity of. So assume now that b y, i.e. that b i ( q). From q p it follows that i n+1 (q) i n+1 (p); indeed we have shown above that (3) holds for n+1. Also, x i m (p) and m < n+1 by the property of v, so by (G3) it follows that x i n+1 (p), thus x i n+1 (q). Now t n+1 (x, q) is defined. Since both x and b are not equal to y, we know that t n+1 (x, q) = t n (x, q) and t n+1 (b, q) = t n (b, q). Also, (4a) holds for n, so t n+1 (x, q) t n+1 (b, q). By construction of t n+1, and by (G4), t n+1 (x, q) t n+1 (y, q). So by symmetry and transitivity, t n+1 (y, q) t n+1 (b, q). This completes the verification of (4a). Case 2: v(n + 1) = (, m, k) for some m and k. Now consider β m (k), and let it be equal to the triple (x, p, φ). This means that φ t m (x, p). 25

26 et y be a new point not in X n, and let q be a new point not in P n. et X n+1 = X n {y}. et P n+1 = P n {q} and extend the partial order so that for al r P n, r < q in P n+1 if and only if r q. Then the new point q is not below any element of P n, and its lower set is a chain which means it is linearly ordered. We now have (1) and (G1) since X n+1 is still finite and X n X n+1. And we have (2) and (G2) since P n+1 is still finite and P n is a suborder of P n+1 such that if p P n+1, q P n and p q, then p P n. et i n+1 (q) = {x, y}, and for r X n, let { i n (r) {y} if r p, i n+1 (r) = i n (r) otherwise. Now we wish to check property (3). First note that p q, so we wish to show that i n+1 (q) i n+1 (p). But from p p it follows that i n+1 (p) = i n (p) {y}. Since (3) holds for n, i n (p) contains at least two points, so it contains a point x x. Also y x, thus i n+1 (q) = {x, y} is a proper subset of i n+1 (p). Second, if r q in X n+1 then either r = q or r < q. In the first case we are done; in the second case we have r p i n+1 (q) i n+1 (r) for the same reason as above, so we are done as well. Now if r q, then r p so i n+1 (r) = i n (r) hence y / i n+1 (r). So in this case, i n+1 (q) i n+1 (r). This verifies most of (3), and the remainder of the verification is as in Case 1 above. So all we need to do now is define t n+1. As in subcase 1b we stipulate that (G4) holds for n. For all r p in P n, let t n+1 (y, r) = t n+1 (x, r) and for all z?? t n+1 (z, r) = t n+1 (r, x). In addition, let t n+1 (y, q) = t n+1 (x, q) = U, where U is any m-theory so that t n (x, p) U and φ U, which exists by proposition The fact that both (4a) and (4b) hold for n implies easily that they hold for n + 1 as well. This concludes the definition of X n+1, P n+1, i n+1 and t n+1 for both case 1 and case 2. Now fix enumerations α n+1 and β n+1 as above to complete the construction in both cases. The construction is now complete. At each step, the () and (G) requirements have been satisfied. It remains to check the (R) requirements on the overall construction. So let us check (R4a). First suppose that φ t n (x, p). et k be such that α n (k) = (x, p, φ). et N be such that v(n) = (, n, k). Then at stage N we ensure that there is some y X N such that φ t n (y, p). This verifies condition (R4a). This completes the completeness proof of the base axioms for subset spaces. 26

27 2.5. The system topologic In this section we will define the topologic system, which is an extension of the logic of subset spaces. It brings us closer to topology since it ensures closure of O under intersection and union. Definition A directed space is one where for every p, u, v with p u and p v there is a w O such that p w and w (u v). An intersection space is one where we can take w = u v. A union space is a space that is closed under union: for any u, v O: u v O. Definition The system whose axioms are the base axioms together with the Weak-Directedness axiom and the Union axiom will be called topologic. The idea is that topologic will be strong enough to support elementary topological reasoning. [1, p. 313] Definition The two following axioms, the Weak-Directedness axiom and the Union axiom, ensure closure under union and intersection. (WD) (Un) φ φ φ ψ (φ ψ K(φ ψ)) However, (WD) does not lead to a complete axiomatisation of the valid sentences on intersection spaces. This is discussed here [2, p. 316]. Proposition The Weak-Directedness Axiom is sound for directed spaces. Proof. Fix a directed space X = (X, O) and let x X and u O. Suppose that x, u = φ. Then there is a u u such that x, u = φ. Thus for all u u, x, u = φ. Now let v u such that x v. et w O be such that x w u v. Since w u, x, w = φ. Also w v, hence x, v = φ. We chose v to be arbitrary, thus x, u = φ. (a) The premise (b) the conclusion Figure 2.3.: Visual representation of the (WD) axiom 27

28 Proposition The Union Axioms are sound on union spaces. (Un) φ ψ (φ ψ K(φ ψ)) Proof. Fix a union space X = (X, O), let x X and u O. Suppose x, u = φ φ. Then: - There exists v such that x v u and x, v = φ, - there exists x u such that x, u = φ, and there exists v u such that x v and x, v = ψ. To show: x, u = (φ ψ K(φ ψ)). Figure 2.4.: Visual representation of the (Un) axiom - et w := v v, then w u. - Now v w and x, v = φ so x, w = φ, - v w and x v w, so from x, v = ψ follows x, w = ψ. - et y w be arbitrary. Then either y v or y v. If y v, then since x v and x, v = φ we have y, v = φ, thus y, v = (φ ψ). If y v, then since x v and x, v = ψ we have y, v = ψ, thus y, v = (φ ψ). So y, w = (φ ψ). Since y was arbitrary it follows that x, w = K(φ ψ). So x, w = φ ψ K(φ ψ). Thus x, u = (φ ψ K(φ ψ)). 28

29 The following completeness result was proven by Georgatos [6]. Theorem (Georgatos 1993 [6]) The topologic axioms are complete for topological spaces, indeed for complete lattice spaces. There are different versions of the (Un) axiom to be found in the literature: the version above is to be found in [1, p. 312] and in [3, p. 79], whereas the following version is used by Moss and Parikh in [8, p. 105]: (Un) (φ Kχ) (ψ Kχ) (φ ψ Kχ) and Georgatos uses the following for his completeness proof in [6, p. 7]: (Un) (Kφ ψ) (Kφ χ) (ψ ψ χ) All these versions look very similar, and there is a reason for this not to matter. Georgatos notes that any formula, sound in the class of subset spaces closed under finite union and intersection, which implies the formula (Kφ ψ) (Kφ χ) (Kφ ψ χ) where φ φ, ψ ψ and χ χ are theorems, can replace the Union axiom in the completeness proof of topologic. An equivalent form of the Weak-Directedness axiom appeared here [8, p. 103]: φ φ (φ ψ). With this we conclude the chapter on the logic of subset spaces and topologic. 29

30 3. The local difference modality K This chapter is dedicated to the local difference modality K. Previous work has been done on combining topology with a difference modality. Andrey Kudinov has studied propositional modal logic with two modal operators and [ ] [7], and has shown that some important topological properties are expressible in this language. He notes that the [ ] modality and its interpretation in Kripke frames have been studied more deeply by Maarten de Rijke [4]. We are interested in doing some topological reasoning, and found the need for a difference modality. However, in our system, the truth of a formula depends on a point and an open around that point, which is why the local difference modality K will the result of localizing the standard difference modality by restricting to the current open. This is how our work differs from previous work. Before introducing the new modality K, let us do some elementary topological reasoning in the language of the previous chapter Elementary topological reasoning et X = (X, O) be a subset frame such that O is a topology on X. Now we can express certain topological notions such as open, closed and dense sets. The above definitions are actually motivated by topology, as the following observations will show. Claim. The set α(a) is open if and only if the formula A KA is valid in the model. Proof. Suppose p, u = A KA. We want to show that A is an open set. So let p A (this is equivalent to p, u = A. Then by modus ponens p, u = KA, i.e., there exists a v u such that p v u and p, v = KA. So there exists v u such that for all q v : q A (i.e., q, v = A). But this means that p A implies that there exists an open v such that p V A, i.e., p Int(A. So A Int(A), and of course Int(A) A, so A = Int(A), thus A is an open set. Suppose α(a) is open. et p X and u O be arbitrary such that p u O. Suppose p, u = A, i.e. that p α(a). Now consider α(a) u, this is an open set since O is a topology on X, and obviously α(a) u u. Also, p α(a) u, and p, α(a) u = KA since clearly α(a) u α(a). Thus p, u = KA. 30

31 Claim. The set α(a) is closed if and only if the formula A A is valid in the model. Proof. We use the duality of and in this proof. A is closed X\A is open A K A is valid K A A is valid A A is valid Claim. The set α(a) is dense iff the formula A is valid, and it is nowhere dense if the formula K A is valid. Proof. The set α(a) is dense if and only if for every point x X, every open neighborhood around x has a nonempty intersection with α(a). This is exactly what A being valid means. Claim. A point x belongs to the boundary of α(a) if p, X = (A A). Proof. A point x belongs to the boundary of α(a) if and only if every open neighborhood around x has a nonempty intersection with both α(a) and X \ α(a). This is exactly what p, X = (A A) means. Next we considered expressing the topological derivative: Definition 3.1. The derivative of A X is defined as the set of limit points of A. It is denoted by d(a). A point x is called a limit point of A X is for each open neighborhood U of x, the set A (U \ {x}) is nonempty. However, before we can begin doing so, we have to realize that we have no ability to express difference in our language, which is necessary for expressing things like the derivative. Consider therefore the following language The local difference modality K We will now introduce a new modality that will enable us to express difference between two points Syntax Definition 3.2. Define a subset space as in definition 2.1. et A be an arbitrary countable set of atomic formulas. Define as the smallest set containing each A A that is closed under the following rules: 31