foundations of description logics

Foundations of Description Logics

Arina Britz

[email protected]

ISAO, Cape Town 2018

[email protected]

Origins

Logic in the East – Buddhist and Islamic traditions

Logic in the West – Aristotle (300s BC)

Logicism – Boole, Frege (late 1800s), Russell (early 1900s)

Symbolic logic – Godel and Tarski (1930s)

Computation and IT – Von Neumann and Turing (1940s)

Classical logic – Idealist modelling and reasoning

Non-classical logics

Logic engineering

Among all the liberal arts, the first is logic – John of Salisbury

Syllogisms

Logical argument forms

Statements made about predicates / subjects / categories:I All A are BI No A are BI Some A are BI Not all A are B

Let’s write ¬ whenever we mean ‘not’:I All A are BI All A are ¬BI Some A are BI Some A are ¬B

Let’s write v for ‘all ... are’, and 6v instead of ‘not all ... are’:I A v BI A v ¬BI A 6v ¬BI A 6v B

From traditional to modern logic

Predicates, argument forms, limited negation, concept inclusion,assertions 3

Fully symbolic language with compositional semantics 7

Boolean connectives:I and: ∧I or: ∨I not: ¬I only if: →I ... but, unless, if and only if, ...

Objects:I Named objects (constant symbols): a, b, c , . . .I Placeholders for unnamed objects (variables): x , y , z , . . .

Predicates: A,B, . . . (unary, binary, ..., n-ary)

Quantifiers:I all: ∀I some: ∃

Translating sentences to predicate logic

Simple statements:I All A are B: ∀x(A(x)→ B(x))I No A are B: ¬∃x(A(x) ∧ B(x))I Some A are B: ∃x(A(x) ∧ B(x))I Not all A are B: ¬∀x(A(x)→ B(x))

Sentences with multiple quantifiers and binary predicates:I Every boy loves a girl:

∀x(B(x)→ ∃y(G (y) ∧ L(x , y)))

I No girl who loves a boy is not loved by some boy:

¬∃x(G (x) ∧ ∃y(B(y) ∧ L(x , y)) ∧ ¬∃z(B(z) ∧ L(z , x)))

I There is a cycle of n alternating boys and girls holding hands.

Building complex formulas

Definition (The language of predicate logic1)

An atomic formula is an n-ary predicate symbol A followed by narguments, which can be constants or variables

An atomic formula is a formula

If φ and ψ are formulas, then so are: ¬φ, φ ∧ ψ, φ ∨ ψ, and φ→ ψ

If φ is a formula and x is a variable, then ∀x(φ) and ∃x(φ) areformulas

A sentence is a formula in which there are no free variables

Die Grenzen meiner Sprache bedeuten die Grenzen meiner Welt –Wittgenstein

1without function symbols

Exercise

Which of the following are predicate formulas?

¬A(c)

∀x(P(x) ∨ R(x , y , z))

R(x , y) ∧ R(x , y , z)

∃y∀x(P(x) ∨ Q(x , y))

∃P(P(x))

∃x(R(x ,A(c))→ A(x))

A mind all logic is like a knife all blade; it makes the hand bleed that usesit – Rabindranath Tagore

Exercise


¬A(c) 3

∀x(P(x) ∨ R(x , y , z))

R(x , y) ∧ R(x , y , z)

∃y∀x(P(x) ∨ Q(x , y))

∃P(P(x))

∃x(R(x ,A(c))→ A(x))


Exercise


¬A(c) 3

∀x(P(x) ∨ R(x , y , z)) 3

R(x , y) ∧ R(x , y , z)

∃y∀x(P(x) ∨ Q(x , y))

∃P(P(x))

∃x(R(x ,A(c))→ A(x))


Exercise


¬A(c) 3

∀x(P(x) ∨ R(x , y , z)) 3

R(x , y) ∧ R(x , y , z) 7

∃y∀x(P(x) ∨ Q(x , y))

∃P(P(x))

∃x(R(x ,A(c))→ A(x))


Exercise


¬A(c) 3

∀x(P(x) ∨ R(x , y , z)) 3

R(x , y) ∧ R(x , y , z) 7

∃y∀x(P(x) ∨ Q(x , y)) 3

∃P(P(x))

∃x(R(x ,A(c))→ A(x))


Exercise


¬A(c) 3

∀x(P(x) ∨ R(x , y , z)) 3

R(x , y) ∧ R(x , y , z) 7

∃y∀x(P(x) ∨ Q(x , y)) 3

∃P(P(x)) 7

∃x(R(x ,A(c))→ A(x))


Exercise


¬A(c) 3

∀x(P(x) ∨ R(x , y , z)) 3

R(x , y) ∧ R(x , y , z) 7

∃y∀x(P(x) ∨ Q(x , y)) 3

∃P(P(x)) 7

∃x(R(x ,A(c))→ A(x)) 7


Exercise

Translate to predicate logic:

Every small dog travelling with its owner is happy.

Tintin owns a small, happy dog.

Milou is a small dog who travels with Tintin.

Does it follow that Milou is happy?

∀x∀y((Small(x) ∧ Dog(x) ∧ TravelsWith(x , y) ∧ Owns(y , x))→Happy(x))

∃x(Small(x) ∧ Dog(x) ∧ Happy(x) ∧ Owns(tintin, x))

Dog(milou) ∧ Small(milou) ∧ TravelsWith(milou, tintin)

Semantic intuition

Dog

Happy

Small

x1

emo

tintin

nemo

jane

laika

milou

Owns

Owns

Owns

TravelsWith

Semantic intuition

Dog

Happy

Small

x1

emo

tintin

nemo

jane

milou

laika

Owns

Owns

Owns

TravelsWith

Semantics

A sentence is either true or false in any given interpretation

The truth of a complex sentence in an interpretation is determinedonly by the truth of its components

Semantic entailment

Formalisation of valid argument forms

φ1, . . . , φk |= ψ:I From premises φ1, . . . , φk the conclusion ψ followsI In each interpretation where all the premises φ1, . . . , φk are true, so is

the conclusion ψI Every model of φ1, . . . , φk is a model of ψ

The sentence ‘snow is white’ is true if, and only if, snow is white – AlfredTarski

Validity Problem

Check if if φ is valid, written |= φ

Look for a counterexample

Predicate logic is semi-decidable – there is no guaranteed method totest validity, if the answer is ‘No’ the method may not terminate

Predicate logic is axiomatisable

Predicate logic has many useful decidable fragments:I Monadic predicate logic – adding a single binary predicate makes it

undecidableI Two-variable fragmentI Guarded fragment – closed under Boolean composition and guarded

quantification

Example

∀y(G (x , y)→ C (y))

∃y(G (x , y) ∧ C (y))

∀x(G (x)→ C (x))

Validity Problem

Check if if φ is valid, written |= φ

Look for a counterexample

Predicate logic is semi-decidable – there is no guaranteed method totest validity, if the answer is ‘No’ the method may not terminate

Predicate logic is axiomatisablePredicate logic has many useful decidable fragments:

I Monadic predicate logic – adding a single binary predicate makes itundecidable

I Two-variable fragmentI Guarded fragment – closed under Boolean composition and guarded

quantification

Example

∀y(G (x , y)→ C (y))

∃y(G (x , y) ∧ C (y))

∀x(G (x)→ C (x))

Guarded Fragment

Definition (The guarded fragment GF of predicate logic)

> and ⊥ are in GF

If φ is an atomic formula, then φ is in GF

GF is closed under Boolean composition ∧, ∨, ¬ and →If φ is in GF and A(x) is an atomic formula for which every freevariable of φ is among the arguments of A, then ∀y(A(x)→ φ) and∃y(A(x) ∧ φ) are in GF for any sequence y of variables

Example

∀x∀y(R(x , y)→ ¬R(y , x)) (asymmetry) 3

∀x∀y(MWC (x , y)→ M(x , y) ∧ ∃z(C (z , x) ∧ C (z , y))) 7

Exercise

Translate the following statements to predicate logic:

Anyone’s aunt is a sister of their parent.Jane’s aunts are her parents’ sisters.

Which of these are in the 2-variable fragment of predicate logic?

Which are in the guarded fragment?

∀x∀y(A(x , y)→ ∃z(S(x , z) ∧ P(z , y)))

∀x(A(x , jane)→ ∃z(S(x , z) ∧ P(z , jane)))

Next up – Choosing a fragment of predicate logic to work with

Description Logics

Family of knowledge representation languages for authoring ontologies

Designed to represent knowledge about things, categories of things,and relationships between things and categories

Foundation for the Web Ontology Language (OWL), built on W3CResource Description Framework (RDF) standard for objects

Good tradeoff between expressiveness and complexity of reasoning

Amenable to implementation

From predicate to description logic

Description Logics are syntactic variants of (usually) decidable fragmentsof predicate logic:

Only unary and binary predicates

All variables are hidden from the syntax

No free variables

Restricted use of quantifiers, mostly within GF

Only universal sentences as terminological axioms

Concept building blocks

Atomic building blocks:I Individual names I (objects), e.g. tintin, tibetI Concept names C (classes), e.g. Dog, CountryI Role names R (relations), e.g. owns, travelsTo

Boolean constructors:I conjunction: u (class intersection)I disjunction: t (class union)I negation: ¬ (class complement)

Role restrictions:I existential restriction: ∃ (some)I universal restriction: ∀ (only)

and more: cardinality constraints, inverse roles, role composition, ...

Building complex concepts

Definition (Concept language of ALC)

Every concept name A ∈ C is a concept

> and ⊥ are concepts

If C and D are concepts, then so are ¬C , C u D, C t D

If r ∈ R and C is a concept, then ∀r .C and ∃r .C are concepts

Which of the following are concepts?

> u r .>

∃r .>

C t ¬¬∃D

∃r .∀s.C u D

∀r .(C u ¬D)








> u r .> 7

∃r .>

C t ¬¬∃D

∃r .∀s.C u D

∀r .(C u ¬D)








> u r .> 7

∃r .> 3

C t ¬¬∃D

∃r .∀s.C u D

∀r .(C u ¬D)








> u r .> 7

∃r .> 3

C t ¬¬∃D 7

∃r .∀s.C u D

∀r .(C u ¬D)








> u r .> 7

∃r .> 3

C t ¬¬∃D 7

∃r .∀s.C u D 3

∀r .(C u ¬D)








> u r .> 7

∃r .> 3

C t ¬¬∃D 7

∃r .∀s.C u D 3

∀r .(C u ¬D) 3

Concept language semantics

Definition (Interpretation)

An ALC interpretation I consists of a nonempty domain ∆, and aninterpretation function ·I such that:

for each individual name a ∈ I, aI ∈ ∆

for each concept name A ∈ C, AI ⊆ ∆

for each role name r ∈ R, rI ⊆ ∆×∆

∆ICanineI

MaleI

DogI

x

emoI

tintinI

nemoI

janeI

laikaI

milouI

ownsI

ownsI

ownsI

ownsI


Extend interpretations to complex concept expressions:

>I = ∆

⊥I = ∅(¬C )I = ∆\CI

(C u D)I = CI ∩ DI

(C t D)I = CI ∪ DI

(∃r .C )I = {x ∈ ∆ | x is related by rI to some element in CI}= {x ∈ ∆ | ∃y(rI(x , y) ∧ CI(y))}

(∀r .C )I = {x ∈ ∆ | x is related by rI only to elements in CI}= {x ∈ ∆ | ∀y(rI(x , y)→ CI(y))}


Arbitrary concept

A class in the domain

CI ⊆ ∆I

∆I

CI

Concept negation

The complement of a class

(¬C )I = ∆I \ CI

∆I

CI


Concept conjunction

The intersection of two classes

(C u D)I = CI ∩ DI

∆I

CI DI

Concept disjunction

The union of two classes

(C t D)I = CI ∪ DI

∆I

CI DI


Existential restriction

(∃r .C )I = {x ∈ ∆ | ∃y(rI(x , y) ∧ CI(y))}

∆I

(∃r .C )I CI

rI

Universal restriction

(∀r .C )I = {x ∈ ∆ | ∀y(rI(x , y)→ CI(y))}

∆I

(∀r .C )I CI

rI

Example

∆ICanineI

MaleI

DogI

x

emoI

tintinI

nemoI

janeI

laikaI

milouI

ownsI

ownsI

ownsI

ownsI

(Canine uMale)I = {. . .}(∃owns.(Canine uMale))I = {. . .}(¬Canine u ∀owns.¬(Male u Canine))I = {. . .}

Mapping to predicate logic

Same notion of domain of interpretation

Individual names mapped to constant symbols

Concept names mapped to unary atomic formulas

Role names mapped to binary atomic formulas

Complex concepts mapped to predicate formulas with a single freevariable

Some unanswered questions – how to

relate or compare complex concepts?

bind the hidden free variable in complex concepts?

assert knowledge about individuals?

determine the veracity of statements?

Making statements

Terminological axioms (TBox axioms):I All owners of a male dog are maleI In predicate logic:

∀x(∃y(Owns(x , y) ∧Male(y) ∧ Dog(y))→ Male(x))

I This is a sentence in predicate logic, so no free variables !

Assertions (ABox assertions):I assert categories to which individuals or pairs of individuals belongI Nemo is a dogI Emo owns NemoI In predicate logic:

Dog(nemo) ∧ Owns(emo, nemo)

Axioms

C v D

Concept inclusion / subsumption:I C is subsumed by DI C is more specific than DI D generalises CI All C are DI Every C is a D

Formalises the syllogism “All ... are ...”

Predicate logic translation:I Binds the free variable in C and in DI ∀x(C (x)→ D(x))

Example

∃owns.(Dog uMale) v Male

Assertions

a : C ; (a, b) : r

Concept and role assertions:I a is an instance of CI a and b are related by r

Predicate logic translation:I No free variablesI C (a)I r(a, b)

Example

nemo : Dognemo : ¬Male(emo, nemo) : owns

Truth in an interpretation

Recall – An interpretation I is a tuple 〈∆, ·I〉 with domain ∆ of objectsand interpretation function ·I such that:

for each individual name a ∈ I, aI ∈ ∆

for each concept name A ∈ C, AI ⊆ ∆

for each role name r ∈ R, rI ⊆ ∆×∆

Definition (Satisfaction)

I C v D if CI ⊆ DI (read: I satisfies C v D)

I a : C if aI ∈ CI

I (a, b) : r if (aI , bI) ∈ rI

Truth in an interpretation

I C v D if CI ⊆ DI

∆I

CI

DI

I a : C if aI ∈ CI

∆I

aI

CI

I (a, b) : r if (aI , bI) ∈ rI

∆I

aI bI

rI

Example

∆ICanineI

MaleI

DogI

x

emoI

tintinI

nemoI

janeI

laikaI

milouI

ownsI

ownsI

ownsI

ownsI

I |= laika : Canine ?

I |= > v Canine t ∃owns.Canine ?

I |= ∃owns.(Male u Canine) v Male ?

Example

∆ICanineI

MaleI

DogI

x

emoI

tintinI

nemoI

janeI

laikaI

milouI

ownsI

ownsI

ownsI

ownsI

I |= laika : Canine ? 3

I |= > v Canine t ∃owns.Canine ?


Example

∆ICanineI

MaleI

DogI

x

emoI

tintinI

nemoI

janeI

laikaI

milouI

ownsI

ownsI

ownsI

ownsI


I |= > v Canine t ∃owns.Canine ? 3


Example

∆ICanineI

MaleI

DogI

x

emoI

tintinI

nemoI

janeI

laikaI

milouI

ownsI

ownsI

ownsI

ownsI


I |= > v Canine t ∃owns.Canine ? 3

I |= ∃owns.(Male u Canine) v Male ? 3

Satisfiability

A knowledge base is a tuple K := 〈T ,A〉, for some TBox T andABox ACaptures both structural knowledge about the domain and explicitknowledge about individuals

Definition (Satisfaction extended)

I T if I C v D for each C v D ∈ TI A if I a : C for each a : C ∈ A and I (a, b) : r for each(a, b) : r ∈ A

If I T ∪ A, we say I is a model of K := 〈T ,A〉K is consistent if it has a model

Reasoning

Each TBox axiom C v D is either true or false in IEach assertion a : C and (a, b) : r is either true or false in IEach I is a complete description of what is true and false in I

But ...

There are infinitely many different interpretations

The domain of interpretation ∆I can itself be infinite

See first, think later, then test. But always see first. Otherwise you willonly see what you were expecting – Douglas Adams

Reasoning

Open World AssumptionI “lack of knowledge to the contrary does not imply falsity”I reasoning across all interpretations of a knowledge base

Definition (Entailment)

An axiom or assertion α follows from a knowledge base K, written K |= α,if every model of K is also a model of α.

Consistency – does K have a model?

Entailment – does the axiom α follow from K?

Instance checking – does the assertion α follow from K?

Entailment

Example

What follows from K = 〈T ,A〉2?

T =

EmployedStudent ≡ Student u Employee,

Student u ¬Employee v ¬∃pays.Tax,

EmployedStudent u ¬Parent v ∃pays.Tax,

EmployedStudent u Parent v ¬∃pays.Tax,

∃worksFor.Company v Employee

A =

ibm : Company,

mary : Parent,

john : Student,

(john, ibm) : worksFor

T |= Student u ∃worksFor.Company v EmployedStudent

T 6|= Employee v ∃worksFor.Company

K |= john : EmployedStudent

K 6|= mary : ¬∃pays.Tax

2Example adapted from ESLLI 2018 tutorial by Ivan Varzinczak

Query Answering

For which individuals does the query answer α follow from K?

q(x) := EmployedStudent(x) ∧ ¬Parent(x)

Conjunctive queries

q(x , y) := EmployedStudent(x) ∧ worksFor(x , y) ∧ Company(y)

q(x) := ∃y(EmployedStudent(x) ∧ worksFor(x , y) ∧ Company(y))

First-order queries

q(x , y) := EmployedStudent(x) ∧ worksFor(x , y) ∧ Company(y)

foundations of description logics

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