Introduction sur les logiques non standards. étude des logiques modales

1. Models and Formalisms
S. Garlatti
MR2A Informatique
2012

2. 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 / 40 SG Models and Formalisms

3. 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 added
The set of well-formed formulas (wff) is a proper superset of the
set of well-formed formulas in classical logic.
The set of theorems generated is a proper superset of the set of
theorems generated by classical logic,
Added theorems are only the result of the new wff.
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4. Non-classical Logics
Classical logics: propositional logic, first-order logic
Non-classical logics can be classified in two main categories:
Deviant logics
The usual logical constants are kept, but with a different meaning
Only a subset of the theorems from the classical logic hold
A non-exclusive classification, a logic could be in both.
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5. Non-classical Logics
Extended Logics
Modal Logics
Tense Logics
Combination of Tense and Modality
Intensional Logic
Dynamic Logic
Deontic Logic (obligatory, permitted)
Conditional logic
...
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6. Non-classical Logics
Deviant Logics
Intuitionistic logics
Multivalued logics
Fuzzy logics
Paraconsistent logics
Probabilistic logic
...
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7. 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
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8. Modal Logics
Traditional alethic modalities or modalities of truth
Propositional logic
New logical constants
Necessity:
Possibility: ♦
Alphabet
A set of propositional symbols S = {p1, p2, ..., pn}
A set of connectors: ¬, →, ↔, ∨, ∧
A set of modalities: , ♦
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9. The Language
Well-Formed Formulae
Let pi be a propositional symbol, (pi ∈ S), pi is an Atomic
formula
If G and H are wff
(G → H), (G ↔ H), (G ∨ H), (G ∧ H), ¬G, G and ♦G
are wff .
Let P and Q be wff
( P → ♦P)
( ♦P → ♦♦Q)
(( ♦P ∨ ♦ ♦Q) ∧ P)
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10. 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
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11. Model Theory
In Propositional and Predicate Logics
The Interpretation function I is truth-functional: the truth or
falsity of a well-formed formula is determined by the truth or
falsity of its components.
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12. Model Theory
In Propositional and Predicate Logics
The Interpretation function I is truth-functional: the truth or
falsity of a well-formed formula is determined by the truth or
falsity of its components.
In Modal Logics, is-it the same?
P ¬P P
V F ?
F V F
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13. Model Theory
In Propositional and Predicate Logics
The Interpretation function I is truth-functional: the truth or
falsity of a well-formed formula is determined by the truth or
falsity of its components.
In Modal Logics, is-it the same?
P ¬P P
V F ?
F V F
In Modal logics, the Interpretation function I is not
truth-functional
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14. Model Theory
A standard Model is a structure M = (W, R, P) where
W is a set of possible worlds
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15. Model Theory
A standard Model is a structure M = (W, R, P) where
W is a set of possible worlds
R is a binary relation on W (R ⊆ W × W)
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16. Model Theory
A standard Model is a structure M = (W, R, P) where
W 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
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17. Model Theory
A standard Model is a structure M = (W, R, P) where
W 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)
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18. Model Theory
A standard Model is a structure M = (W, R, P) where
W 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 possible
worlds.
It is a function on the set {0, 1, 2, ..., } of natural numbers such
that for each such number n, Pn is a subset of W.
It represents an assignment of sets of possible worlds to atomic
formulae (for pi ∈ S, Pi ⊆ W, ∀α such that α ∈ Pi , pi is true),
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19. Possible Worlds, Truth, Validity
Truth in a possible world, α ∈ W in M = (W, R, P)
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20. Possible Worlds, Truth, Validity
Truth in a possible world, α ∈ W in M = (W, R, P)
For propositional symbols pi , |=M
α pi iff α ∈ Pi for
n = {1, 2, 3, ..., n}
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21. Possible Worlds, Truth, Validity
Truth in a possible world, α ∈ W in M = (W, R, P)
For propositional symbols pi , |=M
α pi iff α ∈ Pi for
n = {1, 2, 3, ..., n}
|=M
α A means that A is true for the world α in the standard
model M
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22. Possible Worlds, Truth, Validity
Truth in a possible world, α ∈ W in M = (W, R, P)
For propositional symbols pi , |=M
α pi iff α ∈ Pi for
n = {1, 2, 3, ..., n}
|=M
α A means that A is true for the world α in the standard
model M
|=M
α ¬A iff not |=M
α A
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23. Possible Worlds, Truth, Validity
Truth in a possible world, α ∈ W in M = (W, R, P)
For propositional symbols pi , |=M
α pi iff α ∈ Pi for
n = {1, 2, 3, ..., n}
|=M
α A means that A is true for the world α in the standard
model M
|=M
α ¬A iff not |=M
α A
|=M
α (A ∨ B) iff either |=M
α A or |=M
α B), or both
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24. Possible Worlds, Truth, Validity
Truth in a possible world, α ∈ W in M = (W, R, P)
For propositional symbols pi , |=M
α pi iff α ∈ Pi for
n = {1, 2, 3, ..., n}
|=M
α A means that A is true for the world α in the standard
model M
|=M
α ¬A iff not |=M
α A
|=M
α (A ∨ B) iff either |=M
α A or |=M
α B), or both
|=M
α (A ∧ B) iff both |=M
α A and |=M
α B)
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25. Possible Worlds, Truth, Validity
Truth in a possible world, α ∈ W in M = (W, R, P)
|=M
α A iff for every β in M such that αRβ, |=M
β A
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26. Possible Worlds, Truth, Validity
Truth in a possible world, α ∈ W in M = (W, R, P)
|=M
α A iff for every β in M such that αRβ, |=M
β A
|=M
α ♦A iff for some β in M such that αRβ, |=M
β A
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27. Possible Worlds, Truth, Validity
Truth in a possible world, α ∈ W in M = (W, R, P)
|=M
α A iff for every β in M such that αRβ, |=M
β A
|=M
α ♦A iff for some β in M such that αRβ, |=M
β A
Truth in a Model
|=M
A iff for every world α in M, |=M
α A
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28. Possible Worlds, Truth, Validity
Truth in a possible world, α ∈ W in M = (W, R, P)
|=M
α A iff for every β in M such that αRβ, |=M
β A
|=M
α ♦A iff for some β in M such that αRβ, |=M
β A
Truth in a Model
|=M
A iff for every world α in M, |=M
α A
Validity in a class of Models
|=C A iff for every model M in C, |=M
A
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29. 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
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30. Axiomatic Theory
Modal Logic KT
Axioms
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31. Axiomatic Theory
Modal Logic KT
Axioms
Necessity axiom, called T
( P → P)
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32. Axiomatic Theory
Modal Logic KT
Axioms
Necessity axiom, called T
( P → P)
Axiom K
( (P → Q) → ( P → Q))
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33. Axiomatic Theory
Modal Logic KT
Axioms
Necessity axiom, called T
( P → P)
Axiom K
( (P → Q) → ( P → Q))
Inference Rules
Necessity Rule
P P
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34. Axiomatic Theory
Modal Logic KT
Axioms
Necessity axiom, called T
( P → P)
Axiom K
( (P → Q) → ( P → Q))
Inference Rules
Necessity Rule
P P
Modus Ponens
P, (P → Q) Q
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35. KT System: Axiomatic Theory
Properties of the Modal Logic KT
Consistency: KT is Consistent
Soundness: KT is Sound
Completeness: KT is Complete
Consequences: Axioms K and T are Valid Formulae.
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36. 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
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37. 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)
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38. 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.
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39. Relations between Axiomatic Modal Systems
K
KD K4 K5 KB
KTKDB KD4KD5 K45
KTB KT4 KD45 KB4
KT5
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40. 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
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41. 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
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42. Axioms and Class of Models
Theorem: None of the Axioms D, T, B, 4 and 5 is Valid in the
class of all standards models
Proof: it is enough for each axiom to exhibit a counter-model that
falsifies it. For the axiom D: ( P → ♦P):
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43. Axioms and Class of Models
Theorem: None of the Axioms D, T, B, 4 and 5 is Valid in the
class of all standards models
Proof: it is enough for each axiom to exhibit a counter-model that
falsifies it. For the axiom D: ( P → ♦P):
1 Let M = (W, R, P) be a standard model in which W = {α},
R = ∅, Pn = ∅ for n 0
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44. Axioms and Class of Models
Theorem: None of the Axioms D, T, B, 4 and 5 is Valid in the
class of all standards models
Proof: it is enough for each axiom to exhibit a counter-model that
falsifies it. For the axiom D: ( P → ♦P):
1 Let M = (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 which
every atomic formula is false.
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45. Axioms and Class of Models
Theorem: None of the Axioms D, T, B, 4 and 5 is Valid in the
class of all standards models
Proof: it is enough for each axiom to exhibit a counter-model that
falsifies it. For the axiom D: ( P → ♦P):
1 Let M = (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 which
every atomic formula is false.
3 W is a set, R is a binary relation(the empty relation) and P is a
mapping from natural numbers to (empty) subsets of W
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46. Axioms and Class of Models
Theorem: None of the Axioms D, T, B, 4 and 5 is Valid in the
class of all standards models
Proof: it is enough for each axiom to exhibit a counter-model that
falsifies it. For the axiom D: ( P → ♦P):
1 Let M = (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 which
every atomic formula is false.
3 W is a set, R is a binary relation(the empty relation) and P is a
mapping from natural numbers to (empty) subsets of W
4 Every β in M such that αRβ is such that |=M
β P and there is
not some β in M such that αRβ and |=M
β P
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47. Axioms and Class of Models
The necessity Axiom ( P → P) in the following Standard Model:
w1
¬p1
w2
p1
w3
p1
w4
p1R
R
R
R
R
R
RR
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48. Axioms and Class of Models
Under What Conditions, The necessity Axiom ( P → P) in
the KT system will be valid:
Hypothesis |=M
α P, we have to demonstrate |=M
α P
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49. Axioms and Class of Models
Under What Conditions, The necessity Axiom ( P → P) in
the KT system will be valid:
Hypothesis |=M
α P, we have to demonstrate |=M
α P
1 By definition |=M
α P iff for every β in M such that αRβ, |=M
β P
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50. Axioms and Class of Models
Under What Conditions, The necessity Axiom ( P → P) in
the KT system will be valid:
Hypothesis |=M
α P, we have to demonstrate |=M
α P
1 By definition |=M
α P iff for every β in M such that αRβ, |=M
β P
2 Consequently, |=M
α P iff R is reflexive, that is to say αRα.
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51. Axioms and Class of Models
Under What Conditions, The necessity Axiom ( P → P) in
the KT system will be valid:
Hypothesis |=M
α P, we have to demonstrate |=M
α P
1 By definition |=M
α P iff for every β in M such that αRβ, |=M
β P
2 Consequently, |=M
α P iff R is reflexive, that is to say αRα.
In the KT system, the necessity Axiom is valid under the following
condition:
|=C ( P → P), C is a class of models in which R is
Reflexive
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52. Axioms and Class of Models
Definition: Let M = (W, R, P) be a standard model, the
relation R, with α, β, γ ∈ W is:
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53. Axioms and Class of Models
Definition: Let M = (W, R, P) be a standard model, the
relation R, with α, β, γ ∈ W is:
Serial iff for every α in M there is a β in M such that α R β
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54. Axioms and Class of Models
Definition: Let M = (W, R, P) be a standard model, the
relation R, with α, β, γ ∈ W is:
Serial iff for every α in M there is a β in M such that α R β
Reflexive iff for every α in M, α R α
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55. Axioms and Class of Models
Definition: Let M = (W, R, P) be a standard model, the
relation R, with α, β, γ ∈ W is:
Serial iff for every α in M there is a β in M such that α R β
Reflexive iff for every α in M, α R α
Symmetric iff for every α and β in M, if α R β then β R α
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56. Axioms and Class of Models
Definition: Let M = (W, R, P) be a standard model, the
relation R, with α, β, γ ∈ W is:
Serial iff for every α in M there is a β in M such that α R β
Reflexive iff for every α in M, α R α
Symmetric iff for every α and β in M, if α R β then β R α
Transitive iff for every α, β and γ in M, if α R β and β R γ
then α R γ
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57. Axioms and Class of Models
Definition: Let M = (W, R, P) be a standard model, the
relation R, with α, β, γ ∈ W is:
Serial iff for every α in M there is a β in M such that α R β
Reflexive iff for every α in M, α R α
Symmetric iff for every α and β in M, if α R β then β R α
Transitive iff for every α, β and γ in M, if α R β and β R γ
then α R γ
Euclidian iff for every α, β and γ in M, if α R β and α R γ
then β R γ
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58. Axioms and Class of Models
Theorem: the following axioms are valid respectively in the
indicated class of standard models:
D: Serial
T: Reflexive
B: Symmetric
4: Transitive
5: Euclidian
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59. Axioms and Class of Models
Systems Serial Reflexive Symmetric Transitive Euclidian
KD x
KT x
KB x
K4 x
K5 x
KDB x x
KD4 x x
KD5 x x
KD45 x x
KB4 x x
KB4 x x
KTB x x
KT4 x x
KT5 x x
KT5 x x x
KT5 x x x
KT5 x x x
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60. 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
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61. Knowledge and Belief Logic
Two modal operators, B for Belief and K for Knowledge,
Knowledge Properties for an agent λ:
Axioms
1 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)
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62. Knowledge and Belief Logic
Two modal operators, B for Belief and K for Knowledge,
Knowledge Properties for an agent λ:
Inference Rules
Necessitation: P KλP
Omniscience Logic: if P Q and KλP then KλQ (axiom 1 +
necessitation rule)
The axiomatic theory correspond with the KT45 modal system in
which the accessibility relation is symmetric, reflexive and
transitive.
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63. Three-Wise-Men Problem
A king wishes to know whether his three advisors are as wise as
they claim to be.
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64. Three-Wise-Men Problem
A king wishes to know whether his three advisors are as wise as
they claim to be.
Three chairs are lined up, all facing the same direction, with one
behind the other. The wise men are instructed to sit down.
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65. Three-Wise-Men Problem
A king wishes to know whether his three advisors are as wise as
they claim to be.
Three chairs are lined up, all facing the same direction, with one
behind the other. The wise men are instructed to sit down.
The wise man in the back (wise man 3) can see the backs of the
other two men.
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66. Three-Wise-Men Problem
A king wishes to know whether his three advisors are as wise as
they claim to be.
Three chairs are lined up, all facing the same direction, with one
behind the other. The wise men are instructed to sit down.
The wise man in the back (wise man 3) can see the backs of the
other two men.
The man in the middle (wise man 2) can only see the one wise
man in front of him (wise man 1).
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67. Three-Wise-Men Problem
A king wishes to know whether his three advisors are as wise as
they claim to be.
Three chairs are lined up, all facing the same direction, with one
behind the other. The wise men are instructed to sit down.
The wise man in the back (wise man 3) can see the backs of the
other two men.
The man in the middle (wise man 2) can only see the one wise
man in front of him (wise man 1).
The wise man in front (wise man 1) can see neither wise man 3
nor wise man 2.
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68. Three-Wise-Men Problem
A king wishes to know whether his three advisors are as wise as
they claim to be.
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69. Three-Wise-Men Problem
A king wishes to know whether his three advisors are as wise as
they claim to be.
The king informs the wise men that he has three cards, all of
which are either black or white, at least one of which is
white. He places one card, face up, behind each of the three
wise men.
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70. Three-Wise-Men Problem
A king wishes to know whether his three advisors are as wise as
they claim to be.
The king informs the wise men that he has three cards, all of
which are either black or white, at least one of which is
white. He places one card, face up, behind each of the three
wise men.
Each wise man must determine the color of his own card and
announce what it is as soon as he knows. The first to correctly
announce the color of his own card will be aptly rewarded. All
know that this will happen. The room is silent; then, after
several minutes, wise man 1 says: My card is white!.
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71. Three-Wise-Men Problem
A king wishes to know whether his three advisors are as wise as
they claim to be.
We assume in this puzzle that the wise men do not lie, that they
all have the same reasoning capabilities, and that they can all think
at the same speed. We then can postulate that the following
reasoning took place.
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72. Three-Wise-Men Problem
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73. Three-Wise-Men Problem
Each wise man knows there is at least one white card. If the
cards of wise man 2 and wise man 1 were black, then wise man
3 would have been able to announce immediately that his card
was white.
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74. Three-Wise-Men Problem
Each wise man knows there is at least one white card. If the
cards of wise man 2 and wise man 1 were black, then wise man
3 would have been able to announce immediately that his card
was white.
They all realize this (they are all truly wise). Since wise man 3
kept silent, either wise man 2’s card is white, or wise man 1’s is.
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75. Three-Wise-Men Problem
Each wise man knows there is at least one white card. If the
cards of wise man 2 and wise man 1 were black, then wise man
3 would have been able to announce immediately that his card
was white.
They all realize this (they are all truly wise). Since wise man 3
kept 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 man
1’s were black, that his card was white. They all realize this.
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76. Three-Wise-Men Problem
Each wise man knows there is at least one white card. If the
cards of wise man 2 and wise man 1 were black, then wise man
3 would have been able to announce immediately that his card
was white.
They all realize this (they are all truly wise). Since wise man 3
kept 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 man
1’s were black, that his card was white. They all realize this.
Since wise man 2 also remains silent, wise man 1 knows his card
must be white.
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77. Three-Wise-Men Problem
If we reduce the problem with two advisors A and B and have the
following axioms:
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78. Three-Wise-Men Problem
If we reduce the problem with two advisors A and B and have the
following axioms:
1 A and B know that each one can see the card of the other one
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79. Three-Wise-Men Problem
If we reduce the problem with two advisors A and B and have the
following 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 a
white card
- (¬Blanc(A) → KB¬Blanc(A))
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80. Three-Wise-Men Problem
If we reduce the problem with two advisors A and B and have the
following 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 a
white card
- (¬Blanc(A) → KB¬Blanc(A))
A knows that if A does not have a white card, B knows that A
does not have a white card
36 / 40 SG Models and Formalisms

81. Three-Wise-Men Problem
If we reduce the problem with two advisors A and B and have the
following 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 a
white card
- (¬Blanc(A) → KB¬Blanc(A))
A knows that if A does not have a white card, B knows that A
does not have a white card
- KA(¬Blanc(A) → KB¬Blanc(A))
36 / 40 SG Models and Formalisms

82. Three-Wise-Men Problem
If we reduce the problem with two advisors A and B and have the
following axioms:
2 A and B know that at least one of them has a white card and
each one knows that the other knows it.
37 / 40 SG Models and Formalisms

83. Three-Wise-Men Problem
If we reduce the problem with two advisors A and B and have the
following axioms:
2 A and B know that at least one of them has a white card and
each one knows that the other knows it.
A knows that B knows that if A does not have a white card then
B has a white card
37 / 40 SG Models and Formalisms

84. Three-Wise-Men Problem
If we reduce the problem with two advisors A and B and have the
following axioms:
2 A and B know that at least one of them has a white card and
each one knows that the other knows it.
A knows that B knows that if A does not have a white card then
B has a white card
- KAKB(¬Blanc(A) → Blanc(B))
37 / 40 SG Models and Formalisms

85. Three-Wise-Men Problem
If we reduce the problem with two advisors A and B and have the
following axioms:
2 A and B know that at least one of them has a white card and
each one knows that the other knows it.
A knows that B knows that if A does not have a white card then
B has a white card
- KAKB(¬Blanc(A) → Blanc(B))
B declare that he does not say if he has a white card, then A
knows that B does not knows the color of his card.
37 / 40 SG Models and Formalisms

86. Three-Wise-Men Problem
If we reduce the problem with two advisors A and B and have the
following axioms:
2 A and B know that at least one of them has a white card and
each one knows that the other knows it.
A knows that B knows that if A does not have a white card then
B has a white card
- KAKB(¬Blanc(A) → Blanc(B))
B declare that he does not say if he has a white card, then A
knows that B does not knows the color of his card.
- KA¬KBBlanc(B)
37 / 40 SG Models and Formalisms

87. Three-Wise-Men Problem
If we can reduce the problem with two advisors and have the
following axioms:
38 / 40 SG Models and Formalisms

88. Three-Wise-Men Problem
If we can reduce the problem with two advisors and have the
following axioms:
KA(¬Blanc(A) → KB¬Blanc(A)) [1]
38 / 40 SG Models and Formalisms

89. Three-Wise-Men Problem
If we can reduce the problem with two advisors and have the
following axioms:
KA(¬Blanc(A) → KB¬Blanc(A)) [1]
KAKB(¬Blanc(A) → Blanc(B)) [2]
KA¬KBBlanc(B) [3]
38 / 40 SG Models and Formalisms

90. Three-Wise-Men Problem
If we can reduce the problem with two advisors and have the
following axioms:
KA(¬Blanc(A) → KB¬Blanc(A)) [1]
KAKB(¬Blanc(A) → Blanc(B)) [2]
KA¬KBBlanc(B) [3]
38 / 40 SG Models and Formalisms

91. Three-Wise-Men Problem
If we can reduce the problem with two advisors and have the
following 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))
38 / 40 SG Models and Formalisms

92. Three-Wise-Men Problem
If we can reduce the problem with two advisors and have the
following 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))
38 / 40 SG Models and Formalisms

93. Three-Wise-Men Problem
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)
39 / 40 SG Models and Formalisms

94. Knowledge and Belief Logic
Two modal operators, B for Belief and K for Knowledge, Belief
Properties for an agent λ:
Axioms
40 / 40 SG Models and Formalisms

95. Knowledge and Belief Logic
Two modal operators, B for Belief and K for Knowledge, Belief
Properties for an agent λ:
Axioms
An agent cannot belief a contradiction: ¬Bλ(False)
40 / 40 SG Models and Formalisms

96. Knowledge and Belief Logic
Two modal operators, B for Belief and K for Knowledge, Belief
Properties 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
40 / 40 SG Models and Formalisms

97. Knowledge and Belief Logic
Two modal operators, B for Belief and K for Knowledge, Belief
Properties 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
40 / 40 SG Models and Formalisms