Distributed knowledge & beliefs Lennart v. Luijk Tijs Zwinkels Jeroen Kuijpers Jelle Prins

Overview Recapitulation Proving a system Implicit knowledge Message logic (ML) Belief Discussion

Recapitulation Commonly used symbols ├ (single flubber) used for axiom systems(K) ╞ (double flubber) used for world models( K ) Seriality Euclidicity: s1 s2 s3 s1 s2 s3

Proving a system You prove a System S by proving: S├ φ  M╞ φ Soundness Completeness

Soundness Definition Let S be an axiom system for epistemic formulas, and let M be a class of Kripke models. Then S is called sound with respect to M, if S├ φ => M ╞ φ “Everything that can be proven with the axiom system is actually true.”

Completeness Definition Let S be an axiom system for epistemic formulas, and let M be a class of Kripke models. Then S is called complete with respect to M, if M ╞ φ => S├ φ “Everything that actually is true can be proven with the axiom system”

(M,s)╞ I φ (M,t)╞ φ for all t such that (s,t) є R 1 ∩ … ∩ R m If φ is true in every world which can be reached by all agent from the current world w1, then φ is implicit knowledge in w1 ( (M,w1)╞ I φ ) (A11) K i φ  Iφ (i=1,…,m) Implicit knowledge (§2.3)

Distributed Knowledge: R4:

Implicit knowledge (§2.3) W1 W2 W3 R1,R2 R1 R2 p,~q ~p, ~q You are here ╞ K 1 p ╞ K 2 (p  q) ╞ Iq (and also Ip) p,q Distributed Knowledge: (M,s)╞ Iφ  (M,t)╞ φ for all t such that (s,t) є R1∩ … ∩ Rm

In normal human language: Iφ : ‘A clever man knows φ’ such as a detective If one agent knows a b and another knows a then together they know b. Compare this to C : ‘Any fool knows φ’ Compare this to Ki : ‘Person i knows φ’ Implicit knowledge (§2.3)

Examples Distributed knowledge: Universe Background radiation: Arno Penzias and Robert Wilson have noise in their satellite dish. Thinks this is because of ‘white dielectric material’ (bird droppings) This radiation has been predicted years earlier by George Gamow, but didn’t have the instruments to measure the radiation.

Implicit knowledge (§2.3) Some axioms and systems with I Axioms: (A11) K i φ  Iφ(i=1,…,m) (R4) Systems: KI (m) = K (m) + (A11) + K I TI (m) = T (m) + (A11) + T I S4I (m) = S4 (m) + (A11) + S4 I S5I (m) = S5 (m) + (A11) + S5 I

Implicit knowledge (§2.3) Proof: Soundness (A11): Kip  Ip: Suppose (M,s) |= p. If t is such that (s,t) є (R1 ^.. ^ RN), then ofcourse Rist, so (M,t) |= ip. Mention Completeness

Message Logic ML axiom is added to S5I (m) (ML) “The axiom(ML) says that, if it is implicit knowledge that a state is impossible, then the stronger formula is true that some agent knows that the state is impossible.” Counter example: W1 W2 W3 R1,R2 R1 R2 q q ~q ╞ I~q but also ~(K~q)

Belief (§2.4) (M,s)╞ B i φ (M,t)╞ φ for all t with (s,t) є R i © Gummbah The escaped Knock-knock canary brought false hope to many a lonely citizen Come in! Knock Knock!

Belief (§2.4) (D) ¬B i ( ┴ ) (axiom: a knowledge base is not inconsistent) Same as : ¬ B i (φ ^ ¬φ) Same as S5 but no (A3), instead we have (D) KD45 (m) = (R1)+(R2)+(A1)+(A2)+(D)+(A4)+(A5) stRiRi φBiφBiφ

Belief (§2.4) Proof soundness of KD45 We know that the canonical model M c (KD45 (m) ) posesses accessibility relations R i c that are serial, transitive and euclidian. We may combine this with the observation that serial, transitive and euclidian Kripke models are models for (D), (A4) and (A5), respectively. For (A4) and (A5) we know this already. Therefore, we only have to consider the soundness of the Axiom (D).

Belief (§2.4) Proof soundness of KD45 Suppose KD45 (m) ╞ ¬B i ( ┴ ). Then there would be an KD45 (m) -model M with a state s such that (M,s)╞ B i ┴. This would mean that all R i -successors of s would verify ┴, which is only possible if s does not have any R i successor. However, by seriality, we know that s does have them, so our assumption, i.e. that KD45 (m) ╞ ¬B i ( ┴ ), must be false. Hence we have KD45 (m) ╞ ¬B i ( ┴ ). Completeness possible to prove, not of interest here.

Belief (§2.4) W1 W2 W3 R1 R2 p ~p~p p R1 ╞ B 1 p ╞ B 2 ¬p

Discussion

Logical Omniscience (§2.5) L01-L010 given, give criticism on L05-L09

Knowledge & Belief (§2.13) “logics gone bad” Combining knowledge & beliefs (axiom system KL) Both sound systems Both systems have axioms that are good, but not watertight Combination of the two shows the flaws in the axioms Result: Wrong example: T4: K i p ↔ B i K i p Is this a valid theorem in KL?

Proof: K i φ  B i K i φ 1) KL (i) ├ K i φ  K i K i φ (A4) 2) KL (i) ├ K i φ  B i φ (KL14) 3) KL (i) ├ (K i φ  B i φ)  (K i K i φ  B i K i φ) (A1) 4) KL (i) ├ K i K i φ  B i K i φ (MP 2,3) 5) KL (i) ├ (K i φ  K i K i φ)  (K i φ  B i K i φ) (HS short) 6) KL (i) ├ K i φ  B i K i φ (MP 4,5)

Short proof: B i K i φ  K i φ 1) KL (i) ├ B i K i φ  ¬B i ¬K i φ (D “¬B i ( ┴ )” in its form ¬(B i φ ^ B i ¬φ) and prop. logic) 2) KL (i) ├ ¬B i ¬K i φ  ¬K i ¬K i φ (KL14 “K i φ  B i φ” and prop. logic: contraposition) 3) KL (i) ├ ¬K i ¬K i φ  K i φ (A5/KL3 “¬K i φ  K i ¬K i φ” and prop. logic: contraposition) 4) KL (i) ├ B i K i φ  K i φ (from 1,2,3 by prop. logic: hypothetical syllogism)

Problems with K&B Example: Homeopathic dilution Two persons live in axiom system KL (Hippie Tijs and scientist Lennart) Both take the same homeopathic medicine to releave them from extreme fatigue due to too much work at their university Tijs believes he knows it works (B t K t w) Lennart believes he knows it doesn’t work (B L K L ¬w) One dies and one survives. Who will survive?

Problems with K&B Another Example: Two persons live in axiom system KD45 (m) (Hippie Tijs and scientist Lennart) Both take the same homeopathic medicine to releave them from extreme fatigue due to too much work at their university Tijs believes he knows it works (B t K t w) Lennart believes he knows it doesn’t work (B L K L ¬w) What happens now?

FIN

Full Proof of: B i K i p  K i p

To prove: B i K i p  K i p 1) KL (i) ├ ¬B i ( ┴ ) (D) 2) KL (i) ├ ¬B i ( ┴ )  ¬(B i φ ^ B i ¬φ) (A1) 3) KL (i) ├ ¬(B i φ ^ B i ¬φ) (mp 1,2) 4) KL (i) ├ ¬(B i φ ^ B i ¬φ)  ¬(B i K i φ ^ B i ¬K i φ) (A1) 5) KL (i) ├ ¬(B i K i φ ^ B i ¬K i φ ) (MP 3,4) 6) KL (i) ├ ¬(B i K i φ ^ B i ¬K i φ)  ¬B i K i φ V ¬B i ¬K i φ (A1) 7) KL (i) ├ ¬B i K i φ V ¬B i ¬K i φ (MP 5,6) 8) KL (i) ├ (¬B i K i φ V ¬B i ¬K i φ)  (B i K i φ  ¬B i ¬K i φ) (A1) 9) KL (i) ├ B i K i φ  ¬B i ¬K i φ (MP 7,8) 10) KL (i) ├ (K i φ  B i φ) (KL 14) 11) KL (i) ├ (K i φ  B i φ)  (K i ¬ K i φ  B i ¬ K i φ) (A1)

To prove: B i K i p  K i p 9) KL (i) ├ B i K i φ  ¬B i ¬K i φ 10) KL (i) ├ (K i φ  B i φ) 11) KL (i) ├ (K i φ  B i φ)  (K i ¬K i φ  B i ¬K i φ) 12) KL (i) ├ (K i ¬K i φ  B i ¬K i φ) (MP 10,11) 13) KL (i) ├ ¬B i ¬K i φ  ¬K i ¬K i φ (Contraposition of 12) 14) KL (i) ├ ¬K i φ  K i ¬K i φ (A5/ KL3) 15) KL (i) ├ ¬K i ¬K i φ  K i φ (Contraposition of 14) | 16) KL (i) ├ B i K i φ (Assumption) | 17) KL (i) ├ ¬B i ¬K i φ (MP 16, 9) | 18) KL (i) ├ ¬K i ¬K i φ (MP 17, 13) | 19) KL (i) ├ K i φ (MP 18, 15)

To prove: B i K i p  K i p | 16) KL (i) ├ B i K i φ (Assumption) | 17) KL (i) ├ ¬B i ¬K i φ (MP 16, 9) | 18) KL (i) ├ ¬K i ¬K i φ (MP 17, 13) | 19) KL (i) ├ K i φ (MP 18, 15) 20) KL (i) ├ B i K i φ  K i φ (  intro 16-19)