non standard logics & modal logics

Models and FormalismsS. GarlattiMR2A Informatique2012

Progress

1 Introduction

2 Modal Logics

3 Model Theory

4 Axiomatic Theory

5 Main Modal Systems

6 Axioms and Class of Models

7 A Knowledge and Belief Logic

2 / 41 SG Models and Formalisms

Non-classical Logics

Classical logics: propositional logic, first-order logic

Non-classical logics can be classified in two main categories:Extended logics

New logical constants are addedThe set of well-formed formulas (wff) is a proper superset of theset of well-formed formulas in classical logic.The set of theorems generated is a proper superset of the set oftheorems generated by classical logic,Added theorems are only the result of the new wff.

Classical logics: propositional logic, first-order logic

Non-classical logics can be classified in two main categories:

Deviant logicsThe usual logical constants are kept, but with a different meaningOnly a subset of the theorems from the classical logic hold

A non-exclusive classification, a logic could be in both.

Extended LogicsModal LogicsTense LogicsCombination of Tense and ModalityIntensional LogicDynamic LogicDeontic Logic (obligatory, permitted)Conditional logic...

Deviant LogicsIntuitionistic logicsMultivalued logicsFuzzy logicsParaconsistent logicsProbabilistic logic...

Progress

1 Introduction

2 Modal Logics

3 Model Theory

4 Axiomatic Theory

Modal Logics

Traditional alethic modalities or modalities of truthPropositional logic

New logical constantsNecessity: �Possibility: ♦

AlphabetA set of propositional symbols S = {p1, p2, ..., pn}A set of connectors: ¬,→,↔,∨,∧A set of modalities: �,♦

The Language

Well-Formed FormulaeLet pi be a propositional symbol, (pi ∈ S), pi is an AtomicformulaIf G and H are wff(G → H), (G ↔ H), (G ∨ H), (G ∧ H), ¬G ,�G and ♦Gare wff .

Let P and Q be wff(�P → ♦P)(�♦P → ♦♦Q)((�♦P ∨ ♦�♦Q) ∧�P)

Progress

1 Introduction

2 Modal Logics

3 Model Theory

4 Axiomatic Theory

Model Theory

In Propositional and Predicate LogicsThe Interpretation function I is truth-functional: the truth orfalsity of a well-formed formula is determined by the truth orfalsity of its components.

In Modal Logics, is-it the same?

P ¬P �PV F ?F V F

In Modal logics, the Interpretation function I is nottruth-functional

Model Theory

A standard Model is a structureM = (W ,R,P) whereW is a set of possible worlds

R is a binary relation on W (R ⊆ W ×W)R is an accessibility relationship between possible worlds.We can write: α R β

P is a mapping from natural numbers to subsets of W(Pn ⊆ W , for each natural number n)P is an assignement of truth value to atomic formulae at possibleworlds.It is a function on the set {0, 1, 2, ..., } of natural numbers suchthat for each such number n, Pn is a subset of W.It represents an assignment of sets of possible worlds to atomicformulae (for pi ∈ S,Pi ⊆ W,∀α such that α ∈ Pi , pi is true),

Model Theory

A standard Model is a structureM = (W ,R,P) whereW is a set of possible worldsR is a binary relation on W (R ⊆ W ×W)

R is an accessibility relationship between possible worlds.We can write: α R β

Model Theory

A standard Model is a structureM = (W ,R,P) whereW is a set of possible worldsR is a binary relation on W (R ⊆ W ×W)R is an accessibility relationship between possible worlds.We can write: α R β

Model Theory

P is a mapping from natural numbers to subsets of W(Pn ⊆ W , for each natural number n)

P is an assignement of truth value to atomic formulae at possibleworlds.It is a function on the set {0, 1, 2, ..., } of natural numbers suchthat for each such number n, Pn is a subset of W.It represents an assignment of sets of possible worlds to atomicformulae (for pi ∈ S,Pi ⊆ W,∀α such that α ∈ Pi , pi is true),

Model Theory

Possible Worlds, Truth, Validity

Truth in a possible world, α ∈ W inM = (W ,R,P)

For propositional symbols pi , |=Mα pi iff α ∈ Pi for

n = {1, 2, 3, ..., n}|=M

α A means that A is true for the world α in the standardmodelM|=M

α ¬A iff not |=Mα A

|=Mα (A ∨ B) iff either |=M

α A or |=Mα B), or both

|=Mα (A ∧ B) iff both |=M

α A and |=Mα B)

n = {1, 2, 3, ..., n}

|=Mα A means that A is true for the world α in the standard

modelM|=M

α A and |=Mα B)

n = {1, 2, 3, ..., n}|=M

α A means that A is true for the world α in the standardmodelM

|=Mα ¬A iff not |=M

α A and |=Mα B)

n = {1, 2, 3, ..., n}|=M

α A and |=Mα B)

n = {1, 2, 3, ..., n}|=M

α A and |=Mα B)

n = {1, 2, 3, ..., n}|=M

α A and |=Mα B)

Truth in a possible world, α ∈ W inM = (W ,R,P)|=M

α �A iff for every β inM such that αRβ, |=Mβ A

|=Mα ♦A iff for some β inM such that αRβ, |=M

Truth in a Model|=M A iff for every world α inM, |=M

Validity in a class of Models|=C A iff for every modelM in C , |=M A

Progress

1 Introduction

2 Modal Logics

3 Model Theory

4 Axiomatic Theory

Axiomatic Theory

Modal Logic KTAxioms

Necessity axiom, called T(�P → P)

Axiom K(�(P → Q)→ (�P → �Q))

Inference RulesNecessity Rule

P ` �PModus Ponens

P, (P → Q) ` Q

Axiomatic Theory

Modal Logic KTAxiomsNecessity axiom, called T

(�P → P)

Axiom K(�(P → Q)→ (�P → �Q))

P, (P → Q) ` Q

Axiomatic Theory

(�P → P)Axiom K

(�(P → Q)→ (�P → �Q))

P, (P → Q) ` Q

Axiomatic Theory

(�P → P)Axiom K

(�(P → Q)→ (�P → �Q))

P ` �P

Modus PonensP, (P → Q) ` Q

Axiomatic Theory

(�P → P)Axiom K

(�(P → Q)→ (�P → �Q))

P, (P → Q) ` Q

KT System: Axiomatic Theory

Properties of the Modal Logic KT

Consistency: KT is ConsistentSoundness: KT is SoundCompleteness: KT is Complete

Consequences: Axioms K and T are Valid Formulae.

Progress

1 Introduction

2 Modal Logics

3 Model Theory

4 Axiomatic Theory

Main Modal Systems

The main modal systems are composed from a small set of axioms:

Axiom T : (�P → P)Axiom D: (�P → ♦P)Axiom B: (P → �♦P)Axiom 4: (�P → ��P)Axiom 5: (♦P → �♦P)

From these axioms, it possible to build the following systems: KT,KD, KB, K4, K5, K45, KB4, KD4, KTB, KT4, KT5, etc.

Main Modal Systems

The main modal systems are composed from a small set of axioms:

Axiom T : (�P → P)Axiom D: (�P → ♦P)Axiom B: (P → �♦P)Axiom 4: (�P → ��P)Axiom 5: (♦P → �♦P)

From these axioms, it possible to build the following systems: KT,KD, KB, K4, K5, K45, KB4, KD4, KTB, KT4, KT5, etc.

Relations between Axiomatic Modal Systems

KD K4 K5 KB

KTKDB KD4KD5 K45

KTB KT4 KD45 KB4

Inclusion of K Axiomatic Systems

For instance, we can demonstrate that KD is included in KT,We have to deduce (�P → ♦P) from (�P → P)

(�P → P)

(¬P → ¬�P)(¬P → ♦¬P)

T contraposition

(P → ♦P)(�P → ♦P)

Transitivity

Progress

1 Introduction

2 Modal Logics

3 Model Theory

4 Axiomatic Theory

Axioms and Class of Models

Theorem: None of the Axioms D, T, B, 4 and 5 is Valid in theclass of all standards modelsProof: it is enough for each axiom to exhibit a counter-model thatfalsifies it. For the axiom D: (�P → ♦P):

1 LetM = (W ,R,P) be a standard model in which W = {α},R = ∅, Pn = ∅ for n > 0

2 M contains one world to which no world is related and at whichevery atomic formula is false.

3 W is a set, R is a binary relation(the empty relation) and P is amapping from natural numbers to (empty) subsets of W

4 Every β inM such that αRβ is such that |=Mβ P and there is

not some β inM such that αRβ and |=Mβ P

Theorem: None of the Axioms D, T, B, 4 and 5 is Valid in theclass of all standards modelsProof: it is enough for each axiom to exhibit a counter-model thatfalsifies it. For the axiom D: (�P → ♦P):1 LetM = (W ,R,P) be a standard model in which W = {α},R = ∅, Pn = ∅ for n > 0

The necessity Axiom (�P → P) in the following Standard Model:

w1¬p1

Under What Conditions, The necessity Axiom (�P → P) inthe KT system will be valid:

Hypothesis |=Mα �P, we have to demonstrate |=M

1 By definition |=Mα �P iff for every β inM such that αRβ, |=M

2 Consequently, |=Mα P iff R is reflexive, that is to say αRα.

In the KT system, the necessity Axiom is valid under the followingcondition:|=C (�P → P), C is a class of models in which R isReflexive

Definition: LetM = (W ,R,P) be a standard model, therelation R, with α, β, γ ∈ W is:

Serial iff for every α inM there is a β inM such that α R β

Reflexive iff for every α inM, α R α

Symmetric iff for every α and β inM, if α R β then β R α

Transitive iff for every α, β and γ inM, if α R β and β R γthen α R γ

Euclidian iff for every α, β and γ inM, if α R β and α R γthen β R γ

Definition: LetM = (W ,R,P) be a standard model, therelation R, with α, β, γ ∈ W is:Serial iff for every α inM there is a β inM such that α R β

Theorem: the following axioms are valid respectively in theindicated class of standard models:

D: SerialT : ReflexiveB: Symmetric4: Transitive5: Euclidian

Systems Serial Reflexive Symmetric Transitive EuclidianKD xKT xKB xK4 xK5 xKDB x xKD4 x xKD5 x xKD45 x xKB4 x xKB4 x xKTB x xKT4 x xKT5 x xKT5 x x xKT5 x x xKT5 x x x

Progress

1 Introduction

2 Modal Logics

3 Model Theory

4 Axiomatic Theory

Knowledge and Belief Logic

Two modal operators, B for Belief and K for Knowledge,Knowledge Properties for an agent λ:Axioms1 Axiom 1 (Distribution Axiom, K): (KλP ∧ Kλ(P → Q))→ KλQ

- Same as (Kλ(P → Q))→ (KλP → KλQ))2 Axiom 2 (Knowledge Axiom or T): (KλP → P)

3 Axiom 3 (Positive Introspection Axiom, 4): (KλP → KλKλP)

4 Axiom 4 (Negative Introspection Axiom): (¬KλP → Kλ¬KλP)

- (¬�P → �¬�P) ` (♦¬P → �♦¬P)- (♦¬P → �♦¬P) ` (♦Q → �♦Q) (axiom 5)

Two modal operators, B for Belief and K for Knowledge,Knowledge Properties for an agent λ:

Inference RulesNecessitation: P ` KλP

Omniscience Logic: if P ` Q and KλP then ` KλQ (axiom 1 +necessitation rule)

The axiomatic theory correspond with the KT45 modal system inwhich the accessibility relation is symmetric, reflexive andtransitive.

Three-Wise-Men Problem

A king wishes to know whether his three advisors are as wise asthey claim to be.

Three chairs are lined up, all facing the same direction, with onebehind the other. The wise men are instructed to sit down.The wise man in the back (wise man 3) can see the backs of theother two men.The man in the middle (wise man 2) can only see the one wiseman in front of him (wise man 1).The wise man in front (wise man 1) can see neither wise man 3nor wise man 2.

A king wishes to know whether his three advisors are as wise asthey claim to be.Three chairs are lined up, all facing the same direction, with onebehind the other. The wise men are instructed to sit down.

The wise man in the back (wise man 3) can see the backs of theother two men.The man in the middle (wise man 2) can only see the one wiseman in front of him (wise man 1).The wise man in front (wise man 1) can see neither wise man 3nor wise man 2.

A king wishes to know whether his three advisors are as wise asthey claim to be.Three chairs are lined up, all facing the same direction, with onebehind the other. The wise men are instructed to sit down.The wise man in the back (wise man 3) can see the backs of theother two men.

The man in the middle (wise man 2) can only see the one wiseman in front of him (wise man 1).The wise man in front (wise man 1) can see neither wise man 3nor wise man 2.

A king wishes to know whether his three advisors are as wise asthey claim to be.Three chairs are lined up, all facing the same direction, with onebehind the other. The wise men are instructed to sit down.The wise man in the back (wise man 3) can see the backs of theother two men.The man in the middle (wise man 2) can only see the one wiseman in front of him (wise man 1).

The wise man in front (wise man 1) can see neither wise man 3nor wise man 2.

A king wishes to know whether his three advisors are as wise asthey claim to be.Three chairs are lined up, all facing the same direction, with onebehind the other. The wise men are instructed to sit down.The wise man in the back (wise man 3) can see the backs of theother two men.The man in the middle (wise man 2) can only see the one wiseman in front of him (wise man 1).The wise man in front (wise man 1) can see neither wise man 3nor wise man 2.

The king informs the wise men that he has three cards, all ofwhich are either black or white, at least one of which iswhite. He places one card, face up, behind each of the threewise men.

Each wise man must determine the color of his own card andannounce what it is as soon as he knows. The first to correctlyannounce the color of his own card will be aptly rewarded. Allknow that this will happen. The room is silent; then, afterseveral minutes, wise man 1 says: My card is white!.

A king wishes to know whether his three advisors are as wise asthey claim to be.The king informs the wise men that he has three cards, all ofwhich are either black or white, at least one of which iswhite. He places one card, face up, behind each of the threewise men.

We assume in this puzzle that the wise men do not lie, that theyall have the same reasoning capabilities, and that they can all thinkat the same speed. We then can postulate that the followingreasoning took place.

Each wise man knows there is at least one white card. If thecards of wise man 2 and wise man 1 were black, then wise man3 would have been able to announce immediately that his cardwas white.They all realize this (they are all truly wise). Since wise man 3kept silent, either wise man 2’s card is white, or wise man 1’s is.At this point wise man 2 would be able to determine, if wise man1’s were black, that his card was white. They all realize this.Since wise man 2 also remains silent, wise man 1 knows his cardmust be white.

Each wise man knows there is at least one white card. If thecards of wise man 2 and wise man 1 were black, then wise man3 would have been able to announce immediately that his cardwas white.

They all realize this (they are all truly wise). Since wise man 3kept silent, either wise man 2’s card is white, or wise man 1’s is.At this point wise man 2 would be able to determine, if wise man1’s were black, that his card was white. They all realize this.Since wise man 2 also remains silent, wise man 1 knows his cardmust be white.

Each wise man knows there is at least one white card. If thecards of wise man 2 and wise man 1 were black, then wise man3 would have been able to announce immediately that his cardwas white.They all realize this (they are all truly wise). Since wise man 3kept silent, either wise man 2’s card is white, or wise man 1’s is.

At this point wise man 2 would be able to determine, if wise man1’s were black, that his card was white. They all realize this.Since wise man 2 also remains silent, wise man 1 knows his cardmust be white.

Each wise man knows there is at least one white card. If thecards of wise man 2 and wise man 1 were black, then wise man3 would have been able to announce immediately that his cardwas white.They all realize this (they are all truly wise). Since wise man 3kept silent, either wise man 2’s card is white, or wise man 1’s is.At this point wise man 2 would be able to determine, if wise man1’s were black, that his card was white. They all realize this.

Since wise man 2 also remains silent, wise man 1 knows his cardmust be white.

Each wise man knows there is at least one white card. If thecards of wise man 2 and wise man 1 were black, then wise man3 would have been able to announce immediately that his cardwas white.They all realize this (they are all truly wise). Since wise man 3kept silent, either wise man 2’s card is white, or wise man 1’s is.At this point wise man 2 would be able to determine, if wise man1’s were black, that his card was white. They all realize this.Since wise man 2 also remains silent, wise man 1 knows his cardmust be white.

If we reduce the problem with two advisors A and B and have thefollowing axioms:

1 A and B know that each one can see the card of the other one

If A does not have a white card, B knows that A does not have awhite card- (¬Blanc(A)→ KB¬Blanc(A))

A knows that if A does not have a white card, B knows that Adoes not have a white card- KA(¬Blanc(A)→ KB¬Blanc(A))

A knows that if A does not have a white card, B knows that Adoes not have a white card

- KA(¬Blanc(A)→ KB¬Blanc(A))

2 A and B know that at least one of them has a white card andeach one knows that the other knows it.

A knows that B knows that if A does not have a white card thenB has a white card- KAKB(¬Blanc(A)→ Blanc(B))

B declare that he does not say if he has a white card, then Aknows that B does not knows the color of his card.- KA¬KBBlanc(B)

A knows that B knows that if A does not have a white card thenB has a white card

- KAKB(¬Blanc(A)→ Blanc(B))

B declare that he does not say if he has a white card, then Aknows that B does not knows the color of his card.

- KA¬KBBlanc(B)

If we can reduce the problem with two advisors and have thefollowing axioms:

KA(¬Blanc(A)→ KB¬Blanc(A)) [1]KAKB(¬Blanc(A)→ Blanc(B)) [2]KA¬KBBlanc(B) [3]

[1] (KA(¬Blanc(A) → KB¬Blanc(A)) (KλP → P) [Axiom2][4] (¬Blanc(A) → KB¬Blanc(A))

[2] KAKB(¬Blanc(A) → Blanc(B)) (KλP → P) [Axiom2][5] (KB(¬Blanc(A) → Blanc(B))

(Kλ(P → Q) → (KλP → KλQ)) [Axiom1][6] (KB¬Blanc(A) → KBBlanc(B))

If we can reduce the problem with two advisors and have thefollowing axioms:KA(¬Blanc(A)→ KB¬Blanc(A)) [1]

KAKB(¬Blanc(A)→ Blanc(B)) [2]KA¬KBBlanc(B) [3]

If we can reduce the problem with two advisors and have thefollowing axioms:KA(¬Blanc(A)→ KB¬Blanc(A)) [1]KAKB(¬Blanc(A)→ Blanc(B)) [2]KA¬KBBlanc(B) [3]

Omniscience

Contraposition

Transitivity[4] (¬Blanc(A) → KB¬Blanc(A))

(KB¬Blanc(A) → KBBlanc(B)) [6]

[7] (¬Blanc(A) → KBBlanc(B))

[8] (¬KBBlanc(B) → Blanc(A))[1] [2] [4] [5]

[9] (KA(¬KBBlanc(B) → Blanc(A)))

(Kλ(P → Q) → (KλP → KλQ)) [Axiom1]

(KA¬KBBlanc(A) → KABlanc(A))

KA¬KBBlanc(B) [3][11] KABlanc(A)

Two modal operators, B for Belief and K for Knowledge, BeliefProperties for an agent λ:Axioms

An agent cannot belief a contradiction: ¬Bλ(False)

Positive Introspection Axiom (axiom 4): (BλP → BλBλP)an agent believes what he believes

(BλBλP → BλP) (converse of the previous axiom)

(BλBαP → BλP)an agent can belief what another agent believes

Modal Logic Theorem Prover

Some Modal Logic Theorem ProversMOLTAP, a Modal Logic Tableau Prover :http://twan.home.fmf.nl/moltap/index.htmlAiML.NET Advances in Modal Logic :http://www.cs.man.ac.uk/ schmidt/tools/MOLLE : http://molle.sourceforge.net/LoTREC: possible worlds finally made accessible :http://www.irit.fr/Lotrec/A Theorem Prover for Intuitionistic Modal Logic S5 (IS5):http://pl.postech.ac.kr/IS5/MleanTAP: A Modal Theorem Prover (in Prolog) :http://www.leancop.de/mleantap/

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