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Modal Logic with Variable Modalities & its Applications to Querying Knowledge Bases Evgeny Zolin The University of Manchester

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2/17 Talk Outline Part 1. Logic with variable modalities –Standard modal logic –Variable modalities: Syntax & Semantics Expressivity & Complexity Part 2. Querying KBs using ML –Answering unary queries –Answering boolean queries –50% + 25% + 10%

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3/17 Standard Modal Logic (Multi-)modal language: –propositional variables: p 0, p 1, … –boolean connectives: ?, ! –modal operators (“modalities”): ¤ 1, …, ¤ m Modal formulas: Other connectives are definable:

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4/17 Kripke Semantics Frame: F = h W, R 1, …, R m i, where R i µ W £ W Model: M = h F, i, where a valuation ( p i ) µ W A formula is true at a point e of a model M : M, e ² Validity of a formula at a point e of a frame F : F, e ² iff M, e ² for any model M based on F F ² iff F, e ² for all points e in the frame F

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5/17 Expressive power Typical questions: – What property of frames does a modal formula express? – Which properties of frames are modally expressible? etc. Typical answers: p ◊ p ! xRx (reflexivity) ◊ p ◊◊ p ! xRy yRz xRz (transitivity) ¤ ( ¤ p p ) ¤ p ! transitivity no infinite ascending chains Only relational first- or second-order properties…

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6/17 Introducing Variable Modalities The language is extended in two ways: Modal formulas: The dual variable modalities are defined as: propositional variables: p 0, p 1, … variable modalities: ¡ 0, ¡ 1, … propositional constants: A 1,…, A n constant modalities: ¤ 1,…, ¤ m

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7/17 Semantics for Variable Modalities Frame: F = h W ; V 1,…, V n ; R 1,…, R m i, V i µ W, R i µ W £ W Model: M = h F, ; S 0, S 1,… i, ( p i ) µ W ; S i µ W £ W A formula is true at a point e of a model M : M, e ² Validity of a formula at a point e of a frame F : F, e ² iff M, e ² for any model M based on F In other words: is true at e for any interpretation of propositional variables p i and variable modalities ¡ i

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8/17 What can we express now? Ex.1: Formula ¤ p ! ¡ p. Frame for it: F = h W, R i. Thus, “ R is a universal relation” is expressible! Ex.2: Formula p ! ¡ p. Frame for it: F = h W i. Ex.3: Question: complexity of reasoning for the new language?

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9/17 Complexity and further examples Theorem. Satisfiability is PSPACE- complete. Just because the minimal logic K’ coincides with K. Ex.4: “Any element from A is reflexive” Ex.5: Ex.6: “All elements in A are visible from the point e ”

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10/17 Part 2. Querying KBs using ML Task 1: Find all individuals a such that KB ² a : C, i.e. answer the query q ( x ) Ã x : C over a given KB. Solution: KB ² a : C, KB [ { a : : C } is unsatisfiable Task 2: Find all individuals a such that KB ² aRa, i.e. answer the query q ( x ) Ã xRx over a given KB. Solution a: KB ² aRa, KB ² a : 9 R.{ a } Recall that q ( x ) (reflexivity) is expressed by p ! ◊ p Solution b: KB ² aRa, KB ² a :( : P t 9 R. P ) ( P fresh)

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11/17 Answering unary queries Task 3: Answer the query q ( x ) over a KB: q ( x ) Ã 9 y ( xRy xSy y : A ) This q ( x ) is expressed by a modal formula: ¤ R p ! § S ( p Æ A ) (where p is a variable, A a constant) Solution: KB ² q ( a ), KB ² a : :8 R. P t 9 S.( P u A ) Idea: Given q ( x ), find a corresponding modal formula , and replace each p i with P i (fresh concept names), ¤ i with 8 R i and ¡ i with 8 S i (fresh role names). The resulting concept C will answer your query! x R y S A

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12/17 50%+25%+10%, for unary queries Definition. q ( x ) locally corresponds to : if for any frame F and its point e, Definition. A query q ( x ) is answered by a concept C : q ( x ) ¼ C, if for any KB and a, KB ² q ( a ), KB ² a : C Theorem (50%) Theorem (25%) If then for any F and e, Theorem (10%) If (and no ¡ in ), then for finitely branching frames:

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13/17 Answering boolean queries Task 1. How to check whether KB ² Reflexive( R ) ? Solution 1: check KB [ { : aRa } for unsatisfiability ( a fresh), where : aRa is a shortcut for a : :9 R.{ a } Solution 2: KB ² a : : P t 9 R. P ( a, P are fresh) Task 2. How to check whether KB ² Transitive( R ) ? Solution: KB ² a : :9 R. P t 9 R. 9 R. P ( a, P are fresh) Task 3. How to check whether KB ² R v S ? Solution: KB ² a : :9 R. P t 9 S. P ( a, P are fresh) And so on: R 1 ± R 2 v R 3 ± R 4 ± R 5 ; Commute( R, S ); … Recall that “global” reflexivity is expressed by p ! ◊ p Recall that transitivity is expressed by ◊ p ! ◊◊ p Recall that role inclusion is expressed by ◊ R p ! ◊ S p

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14/17 50%+25%+10%, for boolean queries Definition. q globally corresponds to : if for any frame F, we have: Definition. A concept C answers a boolean query q : q ¼ C, if for any KB, KB ² q, KB ² a : C ( a – fresh) Theorem (50%) Theorem (25%) If then for any F, Theorem (10%) If then for any finite frame F,

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15/17 Mary Likes All Cats Task: KB ² “Mary likes all cats” Mary (individual), Likes (role), Cat (concept) Solution 1: KB ² Cat v 9 Likes —.{Mary} Need to introduce inverse roles and nominals… Solution 2: KB ² Mary: 8: Likes. : Cat Need to introduce role complement (ExpTime) Recall: Solution 3: KB ² Mary: :8 Likes. P t 8 S.( : Cat t P )

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16/17 Modal validity vs. entailment from a KB Validity of a modal formula ≈ closed world assumption Example: F = h W, R i, where W = { a, b, c, d }, R = { h a, b i, h a, c i, h c, d i }. F, b ² : ◊ > ( b has no R -successors) F, c ² ◊ p ! □ p ( R is functional at the point c ) Entailment from a KB ≈ open world assumption KB= h T, A i, TBox T is empty, Abox A = { aRb, aRc, cRd } a c b d

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17/17 Conclusions and outlook New modal language, more expressive, but the same complexity Its expressive power can be used for querying KBs Questions left open: Whether the remaining 15% holds? –I–In particular, any negative results? “Genuinely” cyclic queries? Automatic correspondence: given q ( x ), how to build ? –E–Extension to Sahlqvist & Kracht theorem, etc. Thank you!

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