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Where do informational partitions come from? Tho saugh I... alle the mervelous signals Of the goddys celestials Chaucer, The House of Fame (c1384).......

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Presentation on theme: "Where do informational partitions come from? Tho saugh I... alle the mervelous signals Of the goddys celestials Chaucer, The House of Fame (c1384)......."— Presentation transcript:

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2 Where do informational partitions come from?

3 Tho saugh I... alle the mervelous signals Of the goddys celestials Chaucer, The House of Fame (c1384) States Signal = player’s information = player’s knowledge

4 E From partition to knowledge K(E) Here the player knows the element of the partition that contains the states....and also E. The event that she knows E. Partitional knowledge ω (ω)(ω) Ω – a state space  – a partition of Ω  (ω) – the element of  that contains state ω. ω  K(E) when  (ω)  E. Ω – a state space  – a partition of Ω  (ω) – the element of  that contains state ω. ω  K(E) when  (ω)  E.

5 From knowledge to partition Syntax Bob knows that G.W. Bush is the president of the US and he does not know that Barbara is G.W.’s wife. The set of all sentences form a Boolean algebra (w.r.t “and”, “or”, “not”) with an operator (know) Language: An algebra of subsets, A with an operator K: A  A. Language: An algebra of subsets, A with an operator K: A  A. A family of sets closed under intersection and complement Sentences Deduction rules vs. Semantics

6 From knowledge to partition The axioms of S5 knowledge: 0. The axioms of Boolean algebra 1. K(  ) =  2. K(E)  K(F) = K(E  F) 3. K(E)  E 4. ¬ K(E) = K(¬ K(E)) 0. The axioms of Boolean algebra 1. K(  ) =  2. K(E)  K(F) = K(E  F) 3. K(E)  E 4. ¬ K(E) = K(¬ K(E))

7 British Panel Gives Rumsfeld 'Foot In Mouth' Award Defense Secretary's Comment On 'Known Unknowns' Is Most Baffling LONDON -- He may not know it -- or he may know that he knows it -- but Secretary of Defense Donald Rumsfeld has won this year's "Foot in Mouth" award for the by a public figure. Britain's Plain English Campaign cited Rumsfeld's comment on Iraq, when he said the following during a Pentagon briefing: "Reports that say that something hasn't happened are always interesting to me, because as we know,, there are things we know we know. "We also know. That is to say we know there are some things we do not know. But there are also most baffling statement there are known knowns there are known unknowns unknown unknowns: the ones we don't know we don't know.”

8 From knowledge to partition The axioms of S5 knowledge: 0. The axioms of Boolean algebra 1. K(  ) =  2. K(E)  K(F) = K(E  F) 3. K(E)  E 4. ¬ K(E) = K(¬ K(E)) 0. The axioms of Boolean algebra 1. K(  ) =  2. K(E)  K(F) = K(E  F) 3. K(E)  E 4. ¬ K(E) = K(¬ K(E)) Is S5-knowledge partitional? Partitional knowledge satisfies S5.

9 Fairy tale: S5 implies partition Fairy tale: S5 implies partition Hard facts: Words alone do not bring about partitions Hard facts: Words alone do not bring about partitions. S5 

10 A 0 – the algebra generated by all arithmetic sequences. An arithmetic sequence: {a + zd | z  Z} for d  0. Ω = the set of integers, Z. Example 1 e.g z K: A 0  A 0 is the identity. K is generated by the partition into singletons.

11 A – the algebra generated by A 0 and the set P. K(E) = F P Example 1 Each E  A can be uniquely decomposed:  E = X  F  Y P¬ P  A0A0 1. K(  ) =  2. K(E)  K(E’) = K(E  E’) 3. K(E)  E 4. ¬ K(E) = K(¬ K(E)) K is the identity on A 0 K (P) =  2 K satisfies S5. The candidate for a partition: singletons. If K is partitional then at state 3 the player knows P (because it contains {3}). Therefore K is not partitional.

12 From partition to knowledge s common  1 -  2 -  c - coarser than  1 and  2 finest than any such partition K c - narower than K 1 and K 2 For each E, K c (E)  K 1 (E), K 2 (E) broader than any such knowledge the common knowledge partition. Agreeing to disagree, Aumann (1976)

13 From partition to knowledge s common  1 -  2 -  c - coarser than  1 and  2 finest than any such partition K c - narower than K 1 and K 2 For each E, K c (E)  K 1 (E), K 2 (E) broader than any such knowledge the common knowledge partition. Agreeing to disagree, Aumann (1976) Is it possible to prove syntactically the existence of K c for S5- knowledge K 1 and K 2 ? NO!

14 Is there a set S of axioms of knowledge such that K satisfies SK is partitional For S = S5: For S = {K(E) = E}: NO!


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