epistemic logics - profs.info.uaic.ro · outline epistemic logics kt45 logic and relations with...

Outline Epistemic Logics

Epistemic Logics

Cosmin Varlan

March 20, 2015

Epistemic Logics Cosmin Varlan


1 Epistemic LogicsKT45 Logic and relations with other Epistemic LogicsSyntax and SemanticsReasoning in groups of agentsMuddy Children Example - 3 children



KT45 Logic and relations with other Epistemic Logics

K modal logic

• Is the weakest Modal Logic, it basically has only K scheme• �(ϕ→ ψ) ∧�ϕ→ �ψ (K )• �(ϕ→ ψ)→ (�ϕ→ �ψ) ((K ) in equivalent form)

• However, a necessitation rule is usually present: that all thefacts that are true in the system are known by all the agents.

• ϕ→ �ϕ




Possible axioms

• T: �ϕ→ ϕ - R is reflexive (so, if R is reflexive,M |= �ϕ→ ϕ)

• B: ϕ→ �♦ϕ - R is symmetric

• D: �ϕ→ ♦ϕ - R is serial

• 4: �ϕ→ ��ϕ - R is transitive

• 5: ♦ϕ→ �♦ϕ - R is Euclidean

• �ϕ↔ ♦ϕ - R is functional

• �(ϕ ∧�ϕ→ ψ) ∨�(ψ ∧�ψ → ϕ) - R is linear




KT45 (S5) - Modal Logic of Knowledge

• K (which means omniscience): the knowledge Q has areclosed under semantic implication.

• Knowledge generalization rule: |= ϕ→ �ϕ

• T: - Q knows only true facts.

• 4. Positive introspection - If agent Q knows something, heknows that he knows it.

• 5. Negative introspection - If agent Q does not knowssomething, he knows that he does not knows it.




KT45 (S5) - Modal Logic of Knowledge

• This is idealisations of Knowledge ! Human knowledge hasnone of the properties above !!! Even computer agents canlack some of them.

• |= �ϕ→ ϕ (T is the difference between knowledge modallogic and beliefs modal logic).

• A relation that is reflexive, transitive and Euclidean, is anequivalence relation.




KT4 - Modal Logic of Knowledge

• K (which means omniscience): the knowledge agent Q has areclosed under semantic implication.

• Knowledge generalization rule: |= ϕ→ �ϕ

• T: - Q knows only true facts.

• 4. Positive introspection - If agent Q knows something, heknows that he knows it.

• Useful in computer science.




Some interesting theorems

Theorem

Any sequence of modal operators and negations in KT45 isequivalent to one of the following: −, �, ♦, ¬, ¬� and ¬♦, where− means the lack of any negation of modality.

Theorem

Any sequence of modal operators and negations in KT4 isequivalent to one of the following: −, �, ♦, �♦, ♦�, �♦�, ♦�♦,¬, ¬�, ¬♦, ¬�♦, ¬♦�, ¬�♦� and ¬♦�♦.




Relations between different modal logics

K

K4 K5 KD KT ≡ KDT

K45 KD5 KD4 KT4 ≡ KDT4

KD45 KT5 ≡ KT45 ≡ KDT5 ≡ KDT45



Syntax and Semantics

Multiagent systems

• KT45 can be applied to only one agent. That is usually notthe case, in a multiagent system there are more then oneagent.

• Reasoning in a multiagent system implies not only theknowledge an agent can extract directly from the systemwhere he resides but also reasoning about what the otheragents know.

• Ex. Muddy children puzzle, Prisoner dilemma etc.




Modal Logic KT45n - an epistemic logic for multiagentsystems

• We consider now a set of agents: A = {1, 2, 3, ..., n}• Instead �, Ki will be used for expressing what i knows

(i ∈ {1..n}).

• If we have At = {p, q, r , ...}, Kip means that the agent iknows that p is true.

• Ex: K1p ∧ K1¬K2K1p




Syntax of Epistemic Logic for multiagent systems

Even though we started the Epistemic logic by looking at it’ssemantics, it’s time to correctly define the Syntax of it:

Definition

Let A = {1, 2, ...n} be a set of n agents. The formulas ofepistemic logic are defined by the following BNF:

ϕ ::= ⊥|>|p|¬ϕ|ϕ ∧ ϕ|ϕ ∨ ϕ|ϕ→ ϕ|ϕ↔ ϕ|Kiϕ(i ∈ A)




Kripke structures for Epistemic Logic in multiagent systems

Definition

A Kripke structure for the epistemic logic is a t-upleK = 〈W , π,K1,K2, ...Kn〉 where:

• W is a nonempty set of possible worlds;

• π is the labeling function that assigns a truth value of everyprimitive proposition in each state(π : W × P → {true, false});

• Ki is a set of binary relations on W , ∀i ∈ 1..n (a set of pairs,each pairs being formed from two elements of W ).




Properties of accessibility relation

• A pair (w1,w2) ∈ Ki means that the agent i cannotdistinguish using his own sensors which of the world (w1 orw2) is the reality.

• Each relation between worlds is an equivalence relation:• reflexive (the agent cannot see differences in the same world)• symmetric (if the agent cannot distinguish between the worlds

w1 and w2 then he also will not be able to distinguish betweenworlds w2 and w1)

• transitive (if the agent cannot distinguish between worlds w1

and w2 and cannot distinguish the world w2 from w3, he willnot be able to differentiate world w1 from w3)

The result above is natural since the epistemic Logic formultiagent systems is KT45 applied to each agent.



Reasoning in groups of agents

Operators for groups of agents - EGϕ

• Some new connectors are found in Epistemic Logic formultiagent systems. Considering G ⊆ A then we write:

• EGp to symbolize that all agents in group G know that p istrue in the current system configuration.

• EGϕ is read Everybody in the group G knows ϕ

• If G = {1, 2, 3, ..., n} then• (K, x) |= EGϕ⇔ (K, x) |= K1ϕ ∧ K2ϕ ∧ ...Knϕ




Operators for groups of agents - CGϕ

• If G ⊂ A then:

• CGϕ means that all agents in G know ϕ and each of themknows that all of them know it (this can be achieved only overa secure communication channels).

• CGϕ is read Is common knowledge among agents in G that ϕ• (K, x) |= CGϕ⇔ (K, x) |= E k

Gϕ,∀k ∈ N (E 0Gϕ ≡ ϕ,

E 1Gϕ ≡ EGϕ, E 2

Gϕ ≡ EGEGϕ etc.)




Operators for groups of agents - DGϕ

• DGϕ means that all agents in G can infer ϕ if they all puttheir knowledge in common.

• DGϕ is read Is distributed knowledge among agents in G thatϕ

• (K, x) |= DGϕ⇔ (K, x ′) |= ϕ,∀x ′ ∈ {x ′|(x , x ′) ∈ ∩i∈GKi}• Basically, the distributed knowledge means filtering the worlds

and remaining with a world that all agents agree upon.




DGϕ - Example

• K = 〈W , π,K1,K2〉 where:• W = {w1,w2,w3};• K1 = {(w1,w3)}, K2 = {(w1,w2)}; (also (k1, k1) ∈ K1. Why

?)• π(w1) = {p, q, r}, π(w2) = {p, q}, π(w3) = {p, r}.

Which of the following is true: (K,w1) |= K1p, (K,w1) |= K1q,(K,w1) |= K1r , (K,w1) |= K2p, (K,w1) |= K2q, (K,w1) |= K2r ,(K,w1) |= D{1,2}p, (K,w1) |= D{1,2}q, (K,w1) |= D{1,2}r ?

What can you say about E{1,2} or C{1,2} for the atomicpropositions for w1?




KT45n

Theorem

For each formula ϕ, ψ it is true that:

• (Kiϕ ∧ Ki (ϕ→ ψ))→ ψ (K)

• ϕ→ Kiϕ (gen rule)

• Kiϕ→ ϕ (T)

• Kiϕ→ KiKiϕ (4)

• ¬Kiϕ→ Ki¬Kiϕ (5)

Proof:Like in basic modal logic and based on the properties of theepistemic logic (how it was engineered).




KT45n

Theorem

For two formulas ϕ , ψ, ∀x ∈W , and each subset G of agents, itis true that:

• |= EGϕ↔ ∧i∈GKiϕ;

• |= CGϕ↔ EG (ϕ ∧ CGϕ) (Fixed Point Axiom);

• If |= ϕ→ EG (ψ ∧ ϕ) then |= ϕ→ CGψ (Induction Rule).



Muddy Children Example - 3 children

Muddy Children - 3 agents

• P = {p1, p2, p3} (Atomic propositions)

• K = 〈W , π,K1,K2,K3〉;• W = {(0, 0, 0), (0, 0, 1), ....(1, 1, 1)}; where (0, 0, 1) means

that only 3rd children has mud on his forehead. (|W | = 8)

• π((0, 0, 0)) = ∅, π((0, 0, 1)) = {p3},. . .π((1, 1, 1)) = {p1, p2, p3}.

K1 = {((0, 0, 0), (1, 0, 0)), ((0, 1, 0), (1, 1, 0)), ((0, 0, 1), (1, 0, 1)),((0, 1, 1), (1, 1, 1))}

K2 = {((0, 0, 0), (0, 1, 0)), ((0, 0, 1), (0, 1, 1)), ((1, 0, 0), (1, 1, 0)),((1, 0, 1), (1, 1, 1))}

K3 = {((0, 0, 0), (0, 0, 1)), ((1, 0, 0), (1, 0, 1)), ((0, 1, 0), (0, 1, 1)),((1, 1, 0), (1, 1, 1))}





(0, 0, 0)

(0, 0, 1) (0, 1, 0) (1, 0, 0)

(0, 1, 1) (1, 0, 1) (1, 1, 0)

(1, 1, 1)

32

1

21

3 1

32

12

3

Figure: Graphical representation of the Kripke structure for MuddyChildren puzzle





The Kripke structure does not have arrows anymore (why?) thereare no nodes with arcs to themselves (why) ? Each edge will belabeled with the name of the agent considering the two statesequivalent.





Common Knowledge (C{1,2,3}) - facts that are true in each state:

• |= C{1,2,3}(pi → Kjpi ), ∀i , j with i 6= j ;

• |= C{1,2,3}(¬pi → Kj¬pi ), ∀i , j with i 6= j ;





ϕ = p1 ∨ p2 ∨ p3 then (K , (1, 0, 1)) |= E{1,2,3}ϕ ?

ϕ = p1 ∨ p2 ∨ p3 then (K , (1, 0, 1)) |= E{1,2,3}E{1,2,3}ϕ ?

ϕ = p1 ∨ p2 ∨ p3 then (K , (1, 0, 1)) |= C{1,2,3}ϕ ?





After first round (father announces them that one is dirty):

That means that, in the current state of the system (let’s call itwi1), (K,wi1) |= ϕ (ϕ = p1 ∨ p2 ∨ p3). and allows to each agentsto deduce that (0, 0, 0) is not the current system’s state(C{1,2,3}ϕ).

In the Kripke structure, node representing the state (0, 0, 0) andalso edges to this state can be ignored / removed.




Muddy Children - 3 agentsAfter first round (father announces them that one is dirty):

That means that, in the current state of the system (let’s call itwi1), (K,wi1) |= ϕ (ϕ = p1 ∨ p2 ∨ p3). and allows to each agentsto deduce that (0, 0, 0) is not the current system’s state(C{1,2,3}ϕ).

In the Kripke structure, node representing the state (0, 0, 0) andalso edges to this state can be ignored / removed.

K1 = {((0, 1, 0), (1, 1, 0)), ((0, 0, 1), (1, 0, 1)), ((0, 1, 1), (1, 1, 1))}

K2 = {((0, 0, 1), (0, 1, 1)), ((1, 0, 0), (1, 1, 0)), ((1, 0, 1), (1, 1, 1))}

K3 = {((1, 0, 0), (1, 0, 1)), ((0, 1, 0), (0, 1, 1)), ((1, 1, 0), (1, 1, 1))}





(0, 0, 1) (0, 1, 0) (1, 0, 0)

(0, 1, 1) (1, 0, 1) (1, 1, 0)

(1, 1, 1)

21

3 1

32

12

3

Figure: Kripke structure after 1st step of muddy children puzzle





If only one child would have been muddy (suppose is the 3rd one),then the real system’s state is (0, 0, 1) and because child 3 nolonger thinks the system state is (0, 0, 0) or (0, 0, 1) because(0, 0, 0) was eliminated, he can successfully announce that he isdirty.

Supposing that there were two muddy children, none of thechildren would be able to correctly reason that he is dirty. Afterthe second round, however they will understand that there aremore then one muddy children (because otherwise the muddy onewould have been capable to be sure that he is muddy).

So, if none of them deduce that he is muddy, in the second round,the worlds labeled with one atomic proposition are removed.





K1 = {((0, 1, 1), (1, 1, 1))}K2 = {((1, 0, 1), (1, 1, 1))}K3 = {((1, 1, 0), (1, 1, 1))}

(0, 1, 1) (1, 0, 1) (1, 1, 0)

(1, 1, 1)

12

3

Figure: Kripke structure after 2nd step of muddy children puzzleEpistemic Logics Cosmin Varlan




If the two that ar dirty still aren’t able to deduce that they are thedirty one, then, maybe, each of them expect the other two to comeforward and since they don’t do that, they all are simultaneouslyable (in the erd round) to tell that all of them are muddy.


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