epistemic assumptions in unboundedly many...

IntroductionKnowledge of ’n’ Agent

Logicians in the barUnboundedly many agents

Epistemic assumptions in unboundedly manyagents

Anantha Padmanabha MS

Institute of Mathematical Sciences, Chennai

Anantha Padmanabha MS Epistemic assumptions in unboundedly many agents



Motivation

How does rumour spread in a social network ?

How does a player make a move, based on the knowledge ofprevious actions of her opponent?

In this talk

To study questions like these, we need to model knowledge.

We shall see how to model knowledge and what are itsproperties (axioms).

The players/participants are generally referred to as “agents”.




Modelling Knowledge

Let Ag = {1, 2, · · · n} be the set of agents.

We have Ki operator for each i , where Kiα is intended tomean that i th agent knows α.

Examples

i knows that “A is taller than B”.− Ki (A is taller than B).

Suppose there is a match between India and Australia andagent i did not watch the match but i knows that agent jwatched the match. Now i knows that “j knows whetherIndia won or not” AND i does not know whether “India wonor not”.(∨ := or ; ∧ := and)- Ki (Kjw ∨ Kj¬w) ∧ ¬(Kiw ∨ Ki¬w)where w means “India won”.




Modelling Knowledge



Examples


Suppose there is a match between India and Australia andagent i did not watch the match but i knows that agent jwatched the match.

Now i knows that “j knows whetherIndia won or not” AND i does not know whether “India wonor not”.(∨ := or ; ∧ := and)- Ki (Kjw ∨ Kj¬w) ∧ ¬(Kiw ∨ Ki¬w)where w means “India won”.




Modelling Knowledge



Examples


Suppose there is a match between India and Australia andagent i did not watch the match but i knows that agent jwatched the match. Now i knows that “j knows whetherIndia won or not” AND i does not know whether “India wonor not”.

(∨ := or ; ∧ := and)- Ki (Kjw ∨ Kj¬w) ∧ ¬(Kiw ∨ Ki¬w)where w means “India won”.




Modelling Knowledge



Examples


Suppose there is a match between India and Australia andagent i did not watch the match but i knows that agent jwatched the match. Now i knows that “j knows whetherIndia won or not” AND i does not know whether “India wonor not”.(∨ := or ; ∧ := and)- Ki (Kjw ∨ Kj¬w) ∧ ¬(Kiw ∨ Ki¬w)where w means “India won”.




Properties of Knowledge

Now let us see what properties should the Ki operatorsatisfy.

(⇒:= implies)

Knowledge is closed under conjunction.(Kiα ∧ Kiβ)⇒ Ki (α ∧ β).

Agent cannot have false knowledge.− Ki (2 + 3 = 6) is absurd. If agent i knows α then α has tobe true.Formally: Kiα⇒ α. (T axiom)

Agent can use her knowledge for deductions.− Suppose, Ki (Australia lost ⇒ India won) andKi (Australia lost) then agent i should be able to deduce thatKi (India won).

Formally:Ki (α⇒ β) ∧ Ki (α)

Ki (β).(K axiom)





Now let us see what properties should the Ki operatorsatisfy.(⇒:= implies)





Ki (β).(K axiom)







Agent cannot have false knowledge.

− Ki (2 + 3 = 6) is absurd. If agent i knows α then α has tobe true.Formally: Kiα⇒ α. (T axiom)



Ki (β).(K axiom)







Agent cannot have false knowledge.− Ki (2 + 3 = 6) is absurd. If agent i knows α then α has tobe true.

Formally: Kiα⇒ α. (T axiom)



Ki (β).(K axiom)










Ki (β).(K axiom)








Agent can use her knowledge for deductions.

− Suppose, Ki (Australia lost ⇒ India won) andKi (Australia lost) then agent i should be able to deduce thatKi (India won).


Ki (β).(K axiom)








Agent can use her knowledge for deductions.− Suppose, Ki (Australia lost ⇒ India won) andKi (Australia lost)

then agent i should be able to deduce thatKi (India won).


Ki (β).(K axiom)










Ki (β).(K axiom)




Properties of Knowledge (for an Ideal agent)

Agent is logically omniscient.

− Ki (India won ∨ ¬India won) with out actually knowing theresult of the match. In general, agent i knows all tautologies(Formulas that are always true).

Formally:α is a tautology

Ki (α)(necessitation)

Agent can positively introspect.− if “i knows that India won” then “i knows that (i knowsIndia won)”.Formally: Kiα⇒ Ki (Kiα). (Axiom 4)

Agent can negatively introspect.− if “i does not know that India won” then “i knows that (idoes not know that India won)”.Formally: ¬Ki (α)⇒ Ki (¬Ki (α)). (Axiom 5)





Agent is logically omniscient.− Ki (India won ∨ ¬India won) with out actually knowing theresult of the match.

In general, agent i knows all tautologies(Formulas that are always true).









Agent is logically omniscient.− Ki (India won ∨ ¬India won) with out actually knowing theresult of the match. In general, agent i knows all tautologies(Formulas that are always true).












Agent can positively introspect.

− if “i knows that India won” then “i knows that (i knowsIndia won)”.Formally: Kiα⇒ Ki (Kiα). (Axiom 4)










Agent can negatively introspect.

− if “i does not know that India won” then “i knows that (idoes not know that India won)”.Formally: ¬Ki (α)⇒ Ki (¬Ki (α)). (Axiom 5)




Logicians in a bar

We shall model this setting

How did the last logician figure out the answer ?

We shall modelthis situation in our logic and find it out.




Logicians in a bar

We shall model this setting

How did the last logician figure out the answer ? We shall modelthis situation in our logic and find it out.




A puzzle

Logicians in a bar

Let us consider 2 logicians to make the notations simple. The storystill remains the same. How did the 2nd(last) agent know theanswer ?




Formalizing the puzzle

Let {1, 2} be the agents.

Let a = Logician 1 wants beer ; b = Logician 2 wants beer .

The waiter asks “Does everyone wants beer” which translatesto “Is (a ∧ b) true”?

Now for any agent i ∈ {1, 2}i th agent answers “yes” iff Ki (a ∧ b).i th agent answers “no” iff Ki (¬(a ∧ b)).i th agent answers “I dont know” iff¬Ki (a ∧ b) and ¬Ki (¬(a ∧ b)).




In this setting, initially the first logician says “I dont know”.

After hearing this, agent 2 reasons as follows:1 Suppose ¬a. It means “1 does not want beer” and hence 1

knows that 1 does not want beer K1¬a.2 Also, (¬a⇒ ¬(a ∧ b)) is a tautology. Hence (by necessitation)

K1(¬a⇒ ¬(a ∧ b)).3 Now (by K axiom on (1) and (2)) we have K1(¬(a ∧ b)).

(Ki (α⇒ β) ∧ Ki (α)

Ki (β).)

4 Now, K1(¬(a ∧ b)) iff agent 1 would have answered “No”.This is a contradiction to the announcement made by Agent 1.

Hence 2 deduces that “¬a is impossible. Therefore it has tobe the case that a” and hence K2a.

Now, Logician 2 can answer “No” if she does not want beer andshe can answer “yes” if she wants beer since she has figured outthat Logician 1 wants beer. Since she answered yes, in this case“she also wants beer”.




In this setting, initially the first logician says “I dont know”.After hearing this, agent 2 reasons as follows:

1 Suppose ¬a. It means “1 does not want beer”

and hence 1knows that 1 does not want beer K1¬a.

2 Also, (¬a⇒ ¬(a ∧ b)) is a tautology. Hence (by necessitation)K1(¬a⇒ ¬(a ∧ b)).

3 Now (by K axiom on (1) and (2)) we have K1(¬(a ∧ b)).

(Ki (α⇒ β) ∧ Ki (α)

Ki (β).)








1 Suppose ¬a. It means “1 does not want beer” and hence 1knows that 1 does not want beer K1¬a.



(Ki (α⇒ β) ∧ Ki (α)

Ki (β).)











(Ki (α⇒ β) ∧ Ki (α)

Ki (β).)


Hence 2 deduces that “¬a is impossible. Therefore it has tobe the case that a”

and hence K2a.









(Ki (α⇒ β) ∧ Ki (α)

Ki (β).)











(Ki (α⇒ β) ∧ Ki (α)

Ki (β).)



Now, Logician 2 can answer “No” if she does not want beer

andshe can answer “yes” if she wants beer since she has figured outthat Logician 1 wants beer. Since she answered yes, in this case“she also wants beer”.








(Ki (α⇒ β) ∧ Ki (α)

Ki (β).)



Now, Logician 2 can answer “No” if she does not want beer andshe can answer “yes” if she wants beer since she has figured outthat Logician 1 wants beer.

Since she answered yes, in this case“she also wants beer”.








(Ki (α⇒ β) ∧ Ki (α)

Ki (β).)







Complete Axiomatization

Complete Axiomatization

Note we used only K axiom and necessitation in the last proof.

In fact for any model in this logic, all deductions aboutknowledge of agents can be arrived using only KT45 axioms+ necessitation + MP (also called S5 axiom system). (andwe can prove that it is indeed the case).

Hence S5 forms a complete axiom system for n ideal agentssetting.




Unboundedly many agents

For the above logic, we need to know the number of agentsparticipating in the system beforehand.

This might not be the case always. We might not be able tobound the number of agents participating apriori.

Client - Server models.Social networks.Large games (voting).

We would like to have logic that can handle arbitrary manynumber of agents (finite unbounded or infinite).

One approach is to use quantification over agents so that wecan express statements like ∃xKx(α).

My work includes exploring various logics suitable to modelknowledge of unboundedly many agents.




Any questions ? ;)


epistemic assumptions in unboundedly many...

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