DynamicEpistemicLogic—anovervie · DynamicEpistemicLogic—anoverview Hans van Ditmarsch Logic, University of Sevilla, Spain, [email protected] personal.us.es/hvd/ the logic of knowledge

Dynamic Epistemic Logic — an overview

Hans van DitmarschLogic, University of Sevilla, Spain, [email protected]

personal.us.es/hvd/

◮ the logic of knowledge and belief

◮ public announcement logic

◮ non-public actions

◮ factual change

◮ Moore-sentences

◮ knowability

◮ epistemic protocol synthesis

personal.us.es/hvd/

Sevilla

Three agents: Anne, Bill, Cath draw cards 0, 1, and 2

012

The state 012 represents the actual deal of cards.


012 021a

The state 012 represents the actual deal of cards.Anne is uncertain between card deals 012 and 021.


012 021

210

a

b

The state 012 represents the actual deal of cards.Anne is uncertain between card deals 012 and 021.Bill is uncertain between card deals 012 and 210.


102

012 021

210

a

b

c

The state 012 represents the actual deal of cards.Anne is uncertain between card deals 012 and 021.Bill is uncertain between card deals 012 and 210.Cath is uncertain between card deals 012 and 102.


102

012 021

210

a

b

c

The state 012 represents the actual deal of cards.Anne is uncertain between card deals 012 and 021.Bill is uncertain between card deals 012 and 210.Cath is uncertain between card deals 012 and 102.But...


102

012 021

210

a

b

c

The state 012 represents the actual deal of cards.Anne is uncertain between card deals 012 and 021.Bill is uncertain between card deals 012 and 210.Cath is uncertain between card deals 012 and 102.But... Anne holds 0, but Anne considers is possible that Billconsiders it possible that Cath holds 0, namely in deal 210...


201

102

012 021

210

120

a

a

a

b b

bc

c c

◮ Anne knows that Bill knows that Cath knows her own card:KaKb(Kc0c ∨ Kc1c ∨ Kc2c )

◮ Anne has card 0, but she considers it possible that Billconsiders it possible that Cath knows that Anne does not havecard 0: 0a ∧ KaKbKc¬0a

Multi-agent Epistemic Logic – Syntax & Semantics

Language ϕ ::= p | ¬ϕ | (ϕ ∧ ϕ) | Kaϕ

Structures

A Kripke model is a structure M = 〈S ,R ,V 〉, where

◮ domain S is a nonempty set of states;

◮ R yields an accessibility relation Ra ⊆ S × S for every a ∈ A;

◮ V is a valuation (function) V : P → P(S).

If all Ra are equivalence relations ∼a, M is an epistemic model.A pointed epistemic model is an epistemic state (M, s).

Semantics

M, s |= p iff s ∈ V (p)M, s |= (ϕ ∧ ψ) iff M, s |= ϕ and M, s |= ψ

M, s |= ¬ϕ iff not (M, s |= ϕ)M, s |= Kaϕ iff for all t such that s ∼a t it holds that M, t |= ϕ

Example

201

102

012 021

210

120

a

a

a

b b

bc

c c

Hexa, 012 |= KaKbKc¬0a⇐012 ∼a 021 and Hexa, 021 |= KbKc¬0a⇐021 ∼b 120 and Hexa, 120 |= Kc¬0a⇐∼c (120) = {120, 210}, Hexa, 120 |= ¬0a and Hexa, 210 |= ¬0a⇐Hexa, 120 6|= 0a and Hexa, 210 6|= 0a⇐120, 210 6∈ V (0a) = {012, 021}

Axiomatization

all instantiations of propositional tautologiesKa(ϕ→ ψ) → (Kaϕ→ Kaψ)Kaϕ→ ϕ

Kaϕ→ KaKaϕ

¬Kaϕ→ Ka¬Kaϕ

From ϕ and ϕ→ ψ, infer ψFrom ϕ, infer Kaϕ

Intermezzo — Common knowledge

◮ language: ϕ ::= p | ¬ϕ | (ϕ ∧ ϕ) | Kaϕ | CBϕ

◮ accessibility: ∼B := (⋃

a∈B ∼a)∗

◮ semantics:M, s |= CBϕ iff for all t : s ∼B t implies M, t |= ϕ

Common knowledge has the properties of individual knowledge,and the axiomatization can be extended, e.g., with induction:

CB(ϕ→∧

a∈B

Kaϕ) → (ϕ→ CBϕ)

Recent technical innovation: conditional common knowledge CψBϕ

‘along all the B-paths satisfying ψ it holds that ϕ.’

We have that C⊤B ϕ iff CBϕ.

Public announcements: Example

201

102

012 021

210

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a

a

a

b b

bc

c c

◮ After Anne says that she does not have card 1, Cath knowsthat Bill has card 1.

◮ After Anne says that she does not have card 1, Cath knowsAnne’s card.

◮ Bill still doesn’t know Anne’s card after that.

Public announcements: Example

201

102

012 021

210

120

a

a

a

b b

bc

c c

◮ After Anne says that she does not have card 1, Cath knowsthat Bill has card 1.[¬1a]Kc1b

◮ After Anne says that she does not have card 1, Cath knowsAnne’s card.[¬1a](Kc0a ∨ Kc1a ∨ Kc2a)

◮ Bill still doesn’t know Anne’s card after that:[¬1a]¬(Kb0a ∨ Kb1a ∨ Kb2a)

Public Announcement Logic: language

ϕ ::= p | ¬ϕ | (ϕ ∧ ϕ) | Kaϕ | CBϕ | [ϕ]ϕ

Write 〈ϕ〉ψ for ¬[ϕ]¬ψ

For [ϕ]ψ read “after the announcement of ϕ, ψ (is true).”

For 〈ϕ〉ψ read “ϕ is true and after the announcement of ϕ, ψ.”

Public Announcement Logic: semantics

The effect of the public announcement of ϕ is the restriction of theepistemic state to all states where ϕ holds. So, ‘announce ϕ’ canbe seen as an epistemic state transformer, with a correspondingdynamic modal operator [ϕ].

‘ϕ is the announcement’means‘ϕ is publicly and truthfully announced’.

M, s |= [ϕ]ψ iff (M, s |= ϕ implies M|ϕ, s |= ψ)

M|ϕ := 〈S ′,∼′,V ′〉:

S ′ := [[ϕ]]M := {s ∈ S | M, s |= ϕ}∼′

a := ∼a ∩ ([[ϕ]]M × [[ϕ]]M)V ′(p) := V (p) ∩ [[ϕ]]M

Example announcement in Hexa

201

102

012 021

210

120

a

a

a

b b

bc

c c

⇒ 201

012 021

210

a

a

bc

Hexa, 012 |= 〈¬1a〉Kc0a⇐Hexa, 012 |= ¬1a and Hexa|¬1a, 012 |= Kc0a⇐Hexa, 012 |= ¬1a & (Hexa|¬1a, 012 |= 0a & ∼c (012) = {012})⇐012 6= V (1a) and 012 ∈ V ′(0a)

A dynamic epistemic logic classic

Muddy Children

A group of children has been playing outside and are called backinto the house by their father. The children gather round him. Asone may imagine, some of them have become dirty from the playand in particular: they may have mud on their forehead. Childrencan only see whether other children are muddy, and not if there isany mud on their own forehead. All this is commonly known, andthe children are, obviously, perfect logicians. Father now says: “Atleast one of you has mud on his or her forehead.” And then: “Willthose who know whether they are muddy step forward.” If nobodysteps forward, father keeps repeating the request. What happens?

Muddy Children

000 100

010 110

001 101

011 111

a

a

c c

b b

b b

ac c

a

Given: The children can see each other

Muddy Children

100

010 110

001 101

011 111

a

c

b

b b

ac c

a

After: At least one of you has mud on his or her forehead.ma ∨mb ∨mc

Muddy Children

110

101

011 111

b

c

a

After: Will those who know whether they are muddy step forward?¬( (Kama ∨ Ka¬ma) ∨ (Kbmb ∨ Kb¬mb) ∨ (Kcmc ∨ Kc¬mc) )

Muddy Children

110

After: Will those who know whether they are muddy step forward?(Kama ∨ Ka¬ma) ∨ (Kbmb ∨ Kb¬mb) ∨ (Kcmc ∨ Kc¬mc)

On the origin of Muddy Children

On the origin of Muddy Children

German translation of Rabelais’ Gargantua et Pantagruel:Gottlob Regis, Meister Franz Rabelais der Arzeney DoctorenGargantua und Pantagruel, usw., Barth, Leipzig, 1832.

Ungelacht pfetz ich dich. Gesellschaftsspiel. Jeder zwickt seinenrechten Nachbar an Kinn oder Nase; wenn er lacht, giebt er einPfand. Zwei von der Gesellschaft sind namlich im Complot undhaben einen verkohlten Korkstopsel, woran sie sich die Finger, undmithin denen, die sie zupfen, die Gesichter schwarzen. Diesewerden nun um so lacherlicher, weil jeder glaubt, man lache uberden anderen.

I pinch you without laughing. Parlour game. Everybody pinches hisright neighbour into chin or nose; if one laughs, one must give apledge. Two in the round have secretly blackened their fingers on acharred piece of cork, and hence will blacken the faces of theirneighbours. These neighbours make a fool of themselves, sincethey both think that everybody is laughing about the other one.

Axiomatization of public announcement logic[ϕ]p ↔ (ϕ→ p)[ϕ]¬ψ ↔ (ϕ→ ¬[ϕ]ψ)[ϕ](ψ ∧ χ) ↔ ([ϕ]ψ ∧ [ϕ]χ)[ϕ]Kaψ ↔ (ϕ→ Ka[ϕ]ψ)[ϕ][ψ]χ ↔ [ϕ ∧ [ϕ]ψ]χFrom ϕ, infer [ψ]ϕFrom χ→ [ϕ]ψ and χ ∧ ϕ→ EBχ, infer χ→ [ϕ]CBψ

Expressivity (Plaza, Gerbrandy): Every formula in the language ofpublic announcement logic without common knowledge isequivalent to a formula in the language of epistemic logic.

Axiomatization of public announcement logic[ϕ]p ↔ (ϕ→ p)[ϕ]¬ψ ↔ (ϕ→ ¬[ϕ]ψ)[ϕ](ψ ∧ χ) ↔ ([ϕ]ψ ∧ [ϕ]χ)[ϕ]Kaψ ↔ (ϕ→ Ka[ϕ]ψ)[ϕ][ψ]χ ↔ [ϕ ∧ [ϕ]ψ]χFrom ϕ, infer [ψ]ϕFrom χ→ [ϕ]ψ and χ ∧ ϕ→ EBχ, infer χ→ [ϕ]CBψ

Expressivity (Plaza, Gerbrandy): Every formula in the language ofpublic announcement logic without common knowledge isequivalent to a formula in the language of epistemic logic.

Announcement and relativized common knowledge

[ϕ]CχBψ ↔ C

ϕ∧[ϕ]χB [ϕ]ψ

Sequence of announcements

[ϕ][ψ]χ ↔ [ϕ ∧ [ϕ]ψ]χ

Anne does not have card 1, and Cath now knows Anne’s card.Sequence of two announcements:

¬1a ; (Kc0a ∨ Kc1a ∨ Kc2a)

Single announcement:

¬1a ∧ [¬1a](Kc0a ∨ Kc1a ∨ Kc2a)

201

102

012 021

210

120

a

a

a

b b

bc

c c

⇒ 201

012 021

210

a

a

bc

⇒

012

210

b

Intermezzo — More complex dynamics (= non-public)

(Anne holds 0, Bill holds 1, and Cath holds 2.) Anneshows (only) Bill her card. (She shows card 0.) Cathcannot see the face of the shown card, but notices that acard is being shown.

201

102

012 021

210

120

a

a

a

b b

bc

c c

?

Intermezzo — More complex dynamics (= non-public)

(Anne holds 0, Bill holds 1, and Cath holds 2.) Anneshows (only) Bill her card. (She shows card 0.) Cathcannot see the face of the shown card, but notices that acard is being shown.

201

102

012 021

210

120

a

a

a

b b

bc

c c

201

102

012 021

210

120

a

a

a

c

c c

Intermezzo — Anne shows card 0 to Bill

201

102

012 021

210

120

a

a

a

b b

bc

c c

×0 1

2

c

cc

201, 2

102, 1

012, 0 021, 0

210, 2

120, 1

a

a

a

c

c c

Muddy Children — No, not again!!

000 100

010 110

001 101

011 111

a

a

c c

b b

b b

ac c

a

100

010 110

001 101

011 111

a

c

b

b b

ac c

a

110

101

011 111

b

ca

110

◮ At least one of you has mud on his or her forehead.

◮ Will those who know whether they are muddy step forward?

◮ Will those who know whether they are muddy step forward?

Muddy Children — with cleaning!!

Suppose that after telling the children that at least one of them ismuddy... father empties a bucket of water over Anne (splash!).

Muddy Children — with cleaning

000 100

010 110

001 101

011 111

a

a

c c

b b

b b

ac c

a

100

010 110

001 101

011 111

a

c

b

b b

ac c

a

???

◮ At least one of you has mud on his or her forehead.

◮ Father empties a bucket of water over Anne (splash!). ?

◮ Will those who know whether they are muddy step forward? ?

◮ Will those who know whether they are muddy step forward? ?

000 100

010 110

001 101

011 111

a

a

c c

b b

b b

ac c

a

100

010 110

001 101

011 111

a

c

b

b b

ac c

a

000

010 010

001 001

011 011

a

c

b

b b

ac c

a

000

010

001

011 011

c

b

b

ca

000

010

001

011

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b

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c

ma ∨mb ∨mc ma := ⊥

¬(Kbmb ∨ Kb¬mb) ∧ ¬(Kcmc ∨ Kc¬mc)

¬(Kbmb ∨ Kb¬mb) ∧ ¬(Kcmc ∨ Kc¬mc)

◮ Last step: Cath learns that Anne knows that she was muddy.

◮ Assignment reduction principle: [p := ϕ]p ↔ ϕ

◮ Applications to the computational frame problem.

Unsuccessful updates and Moore sentences


You do not know that this is my first visit to Brazil.



So what?



So what?




So what?


Liar!



So what?


Liar!

◮ You do not know that this is my first visit to Brazil: p ∧ ¬Kp.

◮ After announcing this you know that p: [p ∧ ¬Kp]Kp is valid.

◮ And therefore also [p ∧ ¬Kp](¬p ∨ Kp).

◮ And therefore also [p ∧ ¬Kp]¬(p ∧ ¬Kp).

◮ I cannot truthfully announce p ∧ ¬Kp twice.


Postulate of success:ϕ→ 〈ϕ〉Kϕ

Announcement of a fact always makes it public:

|= [p]Kp

Announcements of non-facts do not have to make them public:

6|= [ϕ]Kϕ

It can be even worse, as we have seen:

|= [p ∧ ¬Kp]¬(p ∧ ¬Kp)|= [p ∧ ¬Kp]¬K (p ∧ ¬Kp)

0 1 1p ∧ ¬Kp

Fitch paradox of knowability

Fitch’s paradox is that:there is an unknown truth is incons. with all truths are knowable.∃p(p ∧ ¬Kp) is inconsistent with ∀q(q → 3Kq).

Substitute p∧¬Kp for q and you get: (p∧¬Kp) → 3K (p∧¬Kp).For any ‘reasonable’ interpretation of 3, this can’t be.

Fitch paradox of knowability

Fitch’s paradox is that:there is an unknown truth is incons. with all truths are knowable.∃p(p ∧ ¬Kp) is inconsistent with ∀q(q → 3Kq).

Substitute p∧¬Kp for q and you get: (p∧¬Kp) → 3K (p∧¬Kp).For any ‘reasonable’ interpretation of 3, this can’t be.

But what is reasonable? Consider the following:

Interpretϕ→ 3KϕasIf ϕ is true, then there is an announcement after which ϕ is known.

This is false when ϕ is p ∧ ¬Kp.

Arbitrary announcement logic – 3(Kp ∨ K¬p) is valid

3ϕ is true in a model, iffthere is an epistemic ψ such that 〈ψ〉ϕ is true, iffthere is a ... model restriction such that ϕ is true in the restriction.

1——0 ⇒ 1p

3(Kp ∨ K¬p), 〈p〉(Kp ∨ K¬p) p,Kp

1——0 ⇒ 0¬p

3(Kp ∨ K¬p), 〈¬p〉(Kp ∨ K¬p) ¬p,K¬p

Moore-sentence: p ∧ ¬Kp

p ∧ ¬Kp ⇒ Kp,¬p ∨ Kp,¬(p ∧ ¬Kp)

Further issues in dynamic epistemic logic

◮ AGM-type belief revision(from belief in ¬p to belief in p).

◮ Logics with quantification over information change(as in ϕ→ 3Kϕ)

◮ Dynamic epistemic logic and temporal epistemic logic(replace [ϕ]ψ by Xψ)

◮ Public announcement protocols for three agents(sender, receiver, eavesdropper)

◮ Protocol synthesis and knowledge progression.

Feria de Sevilla

DynamicEpistemicLogic—anovervie · DynamicEpistemicLogic—anoverview Hans van Ditmarsch Logic, University of Sevilla, Spain, [email protected] personal.us.es/hvd/ the logic of knowledge

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