Modal Logicwith Variable Modalities

&its Applications to

Querying Knowledge Bases

Evgeny ZolinThe University of Manchester

zolin@cs.man.ac.uk

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%

3/17

Standard Modal Logic• (Multi-)modal language:

– propositional variables: p0 , p1 , …

– boolean connectives: ?, !

– modal operators (“modalities”): ¤1 , … , ¤m

• Modal formulas:

• Other connectives are definable:

4/17

Kripke Semantics

• Frame: F = hW ,R1, … ,Rmi, where Ri µ W£W

• Model: M = hF , i, where a valuation (pi)µW

• 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

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…

6/17

Introducing Variable Modalities• The language is extended in two ways:

• Modal formulas:

• The dual variable modalities are defined as:

propositional variables: p0 , p1 , …

variable modalities: ¡0, ¡1, …

propositional constants: A1 ,…,An constant modalities: ¤1 ,…, ¤m

7/17

Semantics for Variable Modalities• Frame: F =hW ; V1 ,…,Vn; R1,…,Rm i, Vi µ W, Ri µ W£W

• Model: M =hF ,; S0,S1 ,…i, (pi)µ W ; SiµW£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 pi and variable modalities ¡i

8/17

What can we express now?

Ex.1: Formula ¤p ! ¡p. Frame for it: F = hW,Ri.

Thus, “R is a universal relation” is expressible!

Ex.2: Formula p ! ¡p. Frame for it: F = hW i.

Ex.3:

Question: complexity of reasoning for the new language?

9/17

Complexity and further examplesTheorem. 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”

10/17

Part 2. Querying KBs using MLTask 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:9R.{a}

Recall that q(x) (reflexivity) is expressed by p ! ◊p

Solution b: KB ² aRa , KB ² a:(:P t 9R.P ) (P fresh)

11/17

Answering unary queriesTask 3: Answer the query q(x) over a KB:

q(x) Ã 9y ( 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: :8R.P t 9S.(P u A)

Idea: Given q(x), find a corresponding modal formula , and replace each pi with Pi (fresh concept names),

¤i with 8Ri and ¡i with 8Si (fresh role names). The resulting concept C will answer your query!

xR

yS A

12/17

50%+25%+10%, for unary queriesDefinition. 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:

13/17

Answering boolean queriesTask 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: :9R.{a}

Solution 2: KB ² a: :P t 9R.P (a,P are fresh)

Task 2. How to check whether KB ² Transitive(R) ?

Solution: KB ² a: :9R.P t 9R.9R.P (a,P are fresh)

Task 3. How to check whether KB ² R v S ?

Solution: KB ² a: :9R.P t 9S.P (a,P are fresh)

And so on: R1±R2 v R3±R4±R5; 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

14/17

50%+25%+10%, for boolean queriesDefinition. 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,

15/17

Mary Likes All CatsTask: 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: :8Likes.P t 8S.(:Cat t P )

16/17

Modal validity vs. entailment from a KB

• Validity of a modal formula ≈ closed world assumption

Example: F = hW,Ri, where W = {a,b,c,d },

R = {ha,b i, ha,c i, hc,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= hT, A i, TBox T is empty, Abox A = { aRb, aRc, cRd }

a

c

b

d

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?

– In particular, any negative results? “Genuinely” cyclic queries?

• Automatic correspondence: given q(x), how to build ?– Extension to Sahlqvist & Kracht theorem, etc.

Thank you!