Probabilistic Justification Logic

Transcription

1 Ioannis Kokkinis Logic and Theory Group Institute of Computer Science University of Bern joint work with Petar Maksimović, Zoran Ognjanović and Thomas Studer Logic and Applications Dubrovnik September 23, 2015

2 Overview 1 Motivation The Justification Logic J 2 3 Ioannis Kokkinis 2 / 26

3 Motivation The Justification Logic J Overview 1 Motivation The Justification Logic J 2 3 Ioannis Kokkinis 3 / 26

4 Motivation The Justification Logic J Overview 1 Motivation The Justification Logic J 2 3 Ioannis Kokkinis 4 / 26

5 Motivation The Justification Logic J From Classical Logic to Justification Logic c.p.l. + α (Propositional) Modal Logic Ioannis Kokkinis 5 / 26

6 Motivation The Justification Logic J From Classical Logic to Justification Logic c.p.l. + α (Propositional) Modal Logic Modal Logic α α Ioannis Kokkinis 5 / 26

7 Motivation The Justification Logic J From Classical Logic to Justification Logic c.p.l. + α (Propositional) Modal Logic Modal Logic α Justification Logic α since t α t : α Ioannis Kokkinis 5 / 26

8 Motivation The Justification Logic J Justification Logic Ioannis Kokkinis 6 / 26

9 Motivation The Justification Logic J Justification Logic t : α Ioannis Kokkinis 6 / 26

10 Motivation The Justification Logic J Justification Logic t : α t : α Ioannis Kokkinis 6 / 26

11 Motivation The Justification Logic J Justification Logic t : α t : α Ioannis Kokkinis 6 / 26

12 Motivation The Justification Logic J Justification Logic t : α t : α t : α Ioannis Kokkinis 6 / 26

13 Motivation The Justification Logic J Justification Logic t : α t : α t : α t : α Ioannis Kokkinis 6 / 26

14 Motivation The Justification Logic J t : α P 1 (t : α) 10 t : α P 8 (t : α) 10 Ioannis Kokkinis 6 / 26

15 Motivation The Justification Logic J Overview 1 Motivation The Justification Logic J 2 3 Ioannis Kokkinis 7 / 26

16 Motivation The Justification Logic J Justification Terms and Formulas Definition (Justification Terms) t ::= c x (t t) (t + t)!t Definition (Justification Formulas-Language L J ) α ::= p α α α t : α where t is a term and p is an atomic proposition. Ioannis Kokkinis 8 / 26

17 Motivation The Justification Logic J Axiomatization for J Axioms of J: (P) finitely many axiom schemes for c.p.l. in the language L J (J) u : (α β) (v : α u v : β) (+) (u : α v : α) u + v : α Ioannis Kokkinis 9 / 26

18 Motivation The Justification Logic J Axiomatization for J Axioms of J: (P) finitely many axiom schemes for c.p.l. in the language L J (J) u : (α β) (v : α u v : β) (+) (u : α v : α) u + v : α CS { (c, α) } c is a constant and α is an instance of some axiom of J Ioannis Kokkinis 9 / 26

19 Motivation The Justification Logic J Axiomatization for J Axioms of J: (P) finitely many axiom schemes for c.p.l. in the language L J (J) u : (α β) (v : α u v : β) (+) (u : α v : α) u + v : α CS { (c, α) c is a constant and α is an instance of some axiom of J } For some CS the system J CS is: axioms of J + (AN!)!! } {{!! } c : :!c : c : α, where (c, α) CS and n N n times (MP) if T α and T α β then T β Ioannis Kokkinis 9 / 26

20 Motivation The Justification Logic J Semantics for J For a given CS, we define the function (basic CS-evaluation): : Prop {T, F} : Tm P(L J ) [T = true and F = false] [P stands for powerset] such that: ( α β u and α v ) = β (u v) 1 2 u v (u + v) 3 for (c, α) CS: α c c : α (!c)!c : c : α (!!c). Ioannis Kokkinis 10 / 26

21 Motivation The Justification Logic J Truth under a Basic CS-Evaluation Let be a basic CS-evaluation. We have: p p = T t : α α t α ( α ) α β α and β Ioannis Kokkinis 11 / 26

22 Overview 1 Motivation The Justification Logic J 2 3 Ioannis Kokkinis 12 / 26

23 Overview 1 Motivation The Justification Logic J 2 3 Ioannis Kokkinis 13 / 26

24 Probabilistic Formulas Definition (Probabilistic Formulas-Language L P ) for s Q [0, 1] and α L J. A ::= P s α A A A Ioannis Kokkinis 14 / 26

25 Probabilistic Formulas Definition (Probabilistic Formulas-Language L P ) for s Q [0, 1] and α L J. A ::= P s α A A A P <s α P s α P s α P 1 s α P >s α P s α P =s α P s α P s α Ioannis Kokkinis 14 / 26

26 Semantics = P s α? Ioannis Kokkinis 15 / 26

27 Semantics Probability Space: M = W, H, µ = P s α? Ioannis Kokkinis 15 / 26

28 Semantics Probability Space: M = W, H, µ M = P s α [α] α α 5 α Ioannis Kokkinis 15 / 26

29 Semantics Probability Space: M = W, H, µ M = P s α µ([α]) s α α 5 α Ioannis Kokkinis 15 / 26

30 Semantics Probability Space: M = W, H, µ M = P s α µ([α]) s 8 M = A M = A M = A B ( M = A and M = B ) α 10 α 5 α Ioannis Kokkinis 15 / 26

31 Axiomatization Axioms of PJ: (P) finitely many axiom schemes for c.p.l. in the language L P (PI) P 0 α (WE) P r α P <sα, where s > r (LE) P <sα P s α (DIS) P r α P s β P 1 (α β) P min(1,r+s) (α β) (UN) P r α P <sβ P <r+s(α β), where r + s 1 For some CS the system PJ CS is: Axioms of PJ + (MP) if T A and T A B then T B (CE) if JCS α then PJCS P 1 α (ST) if T A P s 1 α for every integer k 1 and s > 0 k s then T A P s α Ioannis Kokkinis 16 / 26

32 Soundness and Strong Completeness Theorem Any PJ CS is sound and strongly complete with respect to PJ CS -models, i.e.: T PJCS A T PJCS A Ioannis Kokkinis 17 / 26

33 Complexity It has been proved that, under some restrictions on the CS, the J CS -satisfiability problem belongs to Σ p 2. We can prove that, under the same restrictions on the CS, the PJ CS -satisfiability problem belongs to the same complexity class. Thus, probability operators do not increase the complexity of the justification logic J. Ioannis Kokkinis 18 / 26

34 Iterations of the Probability Operator Can we have formulas like P s P r α or t : P r α? Ioannis Kokkinis 19 / 26

35 Iterations of the Probability Operator Can we have formulas like P s P r α or t : P r α? The logic PPJ is defined over the following language: α ::= p α α α t : α P s α Ioannis Kokkinis 19 / 26

36 Iterations of the Probability Operator Can we have formulas like P s P r α or t : P r α? The logic PPJ is defined over the following language: α ::= p α α α t : α P s α J + PJ PPJ soundness: completeness: decidability : complexity: open Ioannis Kokkinis 19 / 26

37 Overview 1 Motivation The Justification Logic J 2 3 Ioannis Kokkinis 20 / 26

38 The Lottery Paradox Fair Lottery: with 1000 tickets exactly 1 winning ticket Ioannis Kokkinis 21 / 26

39 The Lottery Paradox Fair Lottery: with 1000 tickets exactly 1 winning ticket w i = ticket i wins Bel(α) = it is rational to believe α Deg bel (α) = degree of belief of α Ioannis Kokkinis 21 / 26

40 The Lottery Paradox Fair Lottery: with 1000 tickets exactly 1 winning ticket w i = ticket i wins Bel(α) = it is rational to believe α Deg bel (α) = degree of belief of α Postulates of Rational Belief: 1 Deg bel (α) > 0.99 Bel(α) [The Lockean Thesis] ( ) 2 Bel(α) and Bel(β) = Bel(α β) Ioannis Kokkinis 21 / 26

41 The Lottery Paradox Fair Lottery: with 1000 tickets exactly 1 winning ticket w i = ticket i wins Bel(α) = it is rational to believe α Deg bel (α) = degree of belief of α Postulates of Rational Belief: 1 Deg bel (α) > 0.99 Bel(α) [The Lockean Thesis] ( ) 2 Bel(α) and Bel(β) = Bel(α β) Paradox: ( 1 i 1000)[Deg bel ( w i ) = ] Ioannis Kokkinis 21 / 26

42 The Lottery Paradox Fair Lottery: with 1000 tickets exactly 1 winning ticket w i = ticket i wins Bel(α) = it is rational to believe α Deg bel (α) = degree of belief of α Postulates of Rational Belief: 1 Deg bel (α) > 0.99 Bel(α) [The Lockean Thesis] ( ) 2 Bel(α) and Bel(β) = Bel(α β) Paradox: ( 1 i 1000)[Deg bel ( w i ) = ] = ( 1 i 1000)[Bel( w i )] Ioannis Kokkinis 21 / 26

43 The Lottery Paradox Fair Lottery: with 1000 tickets exactly 1 winning ticket w i = ticket i wins Bel(α) = it is rational to believe α Deg bel (α) = degree of belief of α Postulates of Rational Belief: 1 Deg bel (α) > 0.99 Bel(α) [The Lockean Thesis] ( ) 2 Bel(α) and Bel(β) = Bel(α β) Paradox: ( 1 i 1000)[Deg bel ( w i ) = ] = ( 1 i 1000)[Bel( w i )] = Bel( w 1... w 1000 ) Ioannis Kokkinis 21 / 26

44 The Lottery Paradox Fair Lottery: with 1000 tickets exactly 1 winning ticket w i = ticket i wins Bel(α) = it is rational to believe α Deg bel (α) = degree of belief of α Postulates of Rational Belief: 1 Deg bel (α) > 0.99 Bel(α) [The Lockean Thesis] ( ) 2 Bel(α) and Bel(β) = Bel(α β) Paradox: ( 1 i 1000)[Deg bel ( w i ) = ] = ( 1 i 1000)[Bel( w i )] = Bel( w 1... w 1000 ) Deg bel (w 1... w 1000 ) = 1 Ioannis Kokkinis 21 / 26

45 The Lottery Paradox Fair Lottery: with 1000 tickets exactly 1 winning ticket w i = ticket i wins Bel(α) = it is rational to believe α Deg bel (α) = degree of belief of α Postulates of Rational Belief: 1 Deg bel (α) > 0.99 Bel(α) [The Lockean Thesis] ( ) 2 Bel(α) and Bel(β) = Bel(α β) Paradox: ( 1 i 1000)[Deg bel ( w i ) = ] = ( 1 i 1000)[Bel( w i )] = Bel( w 1... w 1000 ) Deg bel (w 1... w 1000 ) = 1 = Bel(w 1... w 1000 ) Ioannis Kokkinis 21 / 26

46 in PPJ Formalization of the Postulates of Rational Belief in PPJ: 1 For every term t we add to our assumptions: t : P >0.99 (α) pb(t) : α [The Lockean Thesis] 2 For some CS we have: PPJCS s 1 : α s 2 : β con(s 1, s 2 ) : α β Ioannis Kokkinis 22 / 26

47 in PPJ Formalization of the Postulates of Rational Belief in PPJ: 1 For every term t we add to our assumptions: t : P >0.99 (α) pb(t) : α [The Lockean Thesis] 2 For some CS we have: PPJCS Let 1 i There is a term t i : and by the Lockean Thesis: s 1 : α s 2 : β con(s 1, s 2 ) : α β t i : (P = w i) pb(t i ) : w i Ioannis Kokkinis 22 / 26

48 in PPJ Thus there is a term t: t : ( w 1... w 1000 ) Ioannis Kokkinis 23 / 26

49 in PPJ Thus there is a term t: t : ( w 1... w 1000 ) But there also exists a term u such that: u : (P =1 (w 1... w 1000 )) thus, by the Lockean Thesis: pb(u) : (w 1... w 1000 ) Ioannis Kokkinis 23 / 26

50 in PPJ Thus there is a term t: t : ( w 1... w 1000 ) But there also exists a term u such that: thus, by the Lockean Thesis: Avoiding the paradox in PPJ: Restrict the CS such that: PPJCS u : (P =1 (w 1... w 1000 )) pb(u) : (w 1... w 1000 ) s 1 : α s 2 : β con(s 1, s 2 ) : α β holds only if con(s 1, s 2 ) does not contain two different subterms of the form pb(t). [Formalization of an idea by Leitgeb (2014)] Ioannis Kokkinis 23 / 26

51 Overview 1 Motivation The Justification Logic J 2 3 Ioannis Kokkinis 24 / 26

52 Further Work: Statistical Evidence Obtain justifications from formulas? α h(α) Ioannis Kokkinis 25 / 26

53 Further Work: Statistical Evidence Obtain justifications from formulas? α h(α) CP(β α) s interprets to something like: P s (h(α) : β) Ioannis Kokkinis 25 / 26

54 Summary Probabilistic justification logic can be used to model the idea PJ = P s + J different kinds of evidence for α lead to different degrees of belief in α PJ is sound and strongly complete complexity of PJ is no worse than the complexity of J sound, complete and decidable PPJ Complexity of PPJ? Statistical evidence? Ioannis Kokkinis 26 / 26

55 Summary Probabilistic justification logic can be used to model the idea PJ = P s + J different kinds of evidence for α lead to different degrees of belief in α PJ is sound and strongly complete complexity of PJ is no worse than the complexity of J sound, complete and decidable PPJ Complexity of PPJ? Statistical evidence? Thank you for your attention! Ioannis Kokkinis 26 / 26