1. 1 Derivation Schemes for Topological Logics

2. 2 Derived Logics What Are They? Why Do We Need Them? How Can We Use Them? Colleague: Michael Westmoreland

3. 3 History 1936 Von Neumann and Birkhoff a lattice of propositions based on the closed subspaces of Hilbert space now known as “quantum logic” based on measurement non-Boolean (fails to meet distributive properties) No satisfying way to do implication

4. 4 A Topological Logic A proposition is an equivalence class of sets S  S’ iff int(S) = int(S’)  [S] = [ (int S) c ] [S]  [S’] = [ (int S)  (int S’) ] [S]  [S’] = [ (int S)  (int S’) ] Most Boolean properties hold Law of noncontradiction: [S]   [S] = [int S]  [ (int S) c ]= [S  (int S c ) ] choosing canonical representation= [  ] But not all:   [S] =? [S] No!   [S] =  [ (int S) c ] So [S]  [int( (int S c ) c ] =  [ int S c ] =   [S]

5. 5 Logic Properties No tertium non datur: [S]   [S]  [U] where U is the universal set. What about truth assignment? A measurement (open set) m verifies a proposition P iff m  P i  P i  P. Example: the real line with the standard topology. P = [ (-3, 5) ]. m = (0, 4) verifies P since (0,4)  (-3, 5), [-3, 5), [-3,5], (-3,5] We speak of “verification” rather than truth. Rationale: Let S be a classical system and P a proposition about S with P 0 as the canonical representative of [P]. Then P 0 = int P j  P j  [P]. A measurement m that contains points of P 0 but does not lie entirely in P 0 would not verify P.

6. 6 More Properties P = (-3, 5). m = (0,6) does not verify P. Should we conclude P is false? The state of S could lie in P 0 and still be consistent with the result of the measurement m. In fact, there is a more precise measurement, say m’ that lies entirely in P 0 and the result of m. Hence, we cannot conclude that P is false. New concept for assigning truth values: associated with a given measurement (set), three possibilities: verifiability set, falsifiability set, indeterminate. Twin Open Set Phase Space Logic (TOSPS) A measurement m verifies P if m  P 0 where P 0 is the canonical rep of P. A measurement falsifies a proposition if m  Cl(P 0 ) c.

7. 7 Twin Open Set Phase Space Logic Definition: P is a proposition in TOSPS logic if P = ( [V 0 ], [F 0 ] ), where V 0 and F 0 are disjoint open sets. Definition: Let P = ( [V 0 ], [F 0 ] ) be a proposition in TOSPS logic and m be a measurement. P will be assigned the truth value true if m  V 0 ; false if m  F 0 ; indeterminate otherwise. Logical Operators P = ( [P V ], [P F ] ) Q = ( [Q V ], [Q F ] ) P  Q = ([int P V  int Q V ], [int P F  int Q F ] ) P  Q = ( [int P V  int Q V ], [int P F  int Q F ] )  P = ( [int P F ], [int P V ] )

8. 8 Properties  P =  ([P F ], [Q F ]) = ( [P V ], [P F ] ) = P P   P = ( [P V ], [P F ] )  ( [P F ], [P V ] ) = ( [int P V  int P F ], [int P F  int P V ] )= ([  ], [U]) P   P = ([U], [  ]) DeMorgan’s laws Ditributivity All Boolean properties, but tertium non datur.

9. 9 Note:  fails to be truth functional P = [(-1,2), (5,9)] Q = [(1,3), (8,11)] P  Q = [(-1,3), (8,9)] The measurement m = (0, 2.5) assigns I to P, I to Q, and T to P  Q, since m  P V  Q V m = (0,4) assigns I to P  Q and I to P and I to Q, since m ⊈ P V  Q V

10. 10 Twin Open Set Logic Based on Exact Measurement (Discrete) PQ PQPQ TTT TIT TFT ITT III IFI FTT FII FFF

11. 11 Truth Tables for TOPSL PQ PQPQ TTT TIT TFT ITT IIT or I IFI FTT FII FFF

12. 12 PQ P  Q TTT TII TFF ITI IIF or I IFF FTF FIF FFF

13. 13 P PP TF II FT

14. 14 Example Example to illustrate lack of truth functionality for disjunction P = [(-2, 2), (5,9) ] Q = [(1, 3), (8, 11) ] P  Q = [(-2, 3), (8,9) ] Suppose m = (0, 2.5) m assigns “I” to each of P and Q, “T” to P  Q since m  P V  Q V Now suppose m = ( 0, 4) m assigns “I” to P  Q as well as to P and to Q since m  (P V  Q V )

15. 15 PQ P  Q =  P  Q TTT TII TFF ITT IIT or I IFI FTT FIT FFT

16. 16 Applications to Billiard Ball Model of Computation

17. 17

18. 18

19. 19

20. 20 OR-Gate

21. 21 AND-Gate

22. 22 NOT-Gate

23. 23 Derivation Gate Input the value of P and the value of P → Q

24. 24 Derivation For a Boolean lattice, define P ≤ Q when P  Q is valid where ≤ is the lattice ordering Modus Ponens

25. 25 Three Questions to Consider 1. What is a proper ordering for the propositions in twin open set logic? 2. What is a proper implication operator in twin open set logic? 3. What derivation method can be implemented given the answers to 1 and 2?

26. 26 Characterization Theorem Let (A, , ,  ) be a DeMorgan algebra. If we define an ordering ≼ on the algebra by P ≼ Q  def P  Q = Q, then P  (  P  Q) ≼ Q iff (A, , ,  ) is a boolean algebra. Reminder: TOPSL is a DeMorgan algebra.

27. 27 Proof Need: (A, , ,  ) satisfies the law of non contradiction. In any DeMorgan algebra satisfying our hypothesis, 0 ≼ P   P. Substituting Q = 0 in the modus ponens scheme, P  (  P  0) ≼0

28. 28 Using distributivity, P  (  P  0) ≼ 0 (P   P)  (P  0) ≼ 0 Since P  0 = 0 and Q  0 = Q for any Q, P   P ≼ 0 By antisymmetry of ≼, P   P = 0 and so (A, , ,  ) is boolean.

29. 29 Implications of the theorem: Any DeMorgan algebra in which 1. Entailment is given by  (  ), 2. The implication operator is given by  P  Q, and 3. Modus ponens is satisfied must be a Boolean algebra.

30. 30 Non Standard Derivation TOSPL is a DeMorgan algebra, but not a boolean algebra. At least one of the three properties above must fail.

31. 31 Modus Ponens Fails Ordering for TOSL (suggested by  or  ) P  Q  P V  Q V and Q F  P F motivated by either P  Q = P  P V  Q V and Q F  P F P  Q = Q  P V  Q V and Q F  P F Q is more readily verified and less easily falsified than P.

32. 32 Implication P→Q  def  P  Q So P→Q =  P  Q = [(P F  Q V ),(P V  Q F )] Previous Theorem tells us that modus ponens fails. Why does it?

33. 33 Theorem: With the ordering given by , it is not the case that P  ( P→Q )  Q Proof: P  ( P→Q ) = P  (  P  Q ) = ( P   P)  (P  Q) = [(P V  P F ), (P V  P F )]  [(P V  Q V ), (P F  Q F )] = [ , (P V  P F )]  [(P V  Q V ), (P F  Q F )] = [(P V  Q V ),, ((P V  P F )  (P F  Q F ))] = [(P V  Q V ), ((P V  Q F )  P F )]

34. 34 For P  ( P→Q ) ≼ Q, Q F  (P V  Q F )  P F But whenever P V  P F  X (the whole space), the containment fails. In any nondiscrete topology we have disjoint open sets P V and P F such that P V  P F  X and the claim is established.

35. 35 Need: a proposition that contains [(P V  Q V ), ((P V  Q F )  P F )] One possibility [Q V,  ] Given P and P→Q [Q V,  ]

36. 36 Good Point: It works. Not so good: So does any proposition of the form [Q V, Y] where Y is any open subset of int(Q V C ) Cannot falsify

37. 37 Modus Tollens P→Q =  P  Q =  (  Q)  P   P =  Q →  P Consider  Q  ( P→Q ) = [( P F  Q F), ((P V  Q F )  Q V )] Analog to modus ponens: [P F,  ]

38. 38 Another Possibility For Modus Ponens: Given P and P→Q, Conclude [ Q V, P V  Q F ] = def Q P For Modus Tollens: Given  Q  ( P→Q ), Conclude [ P F, P V  Q F ] = def P  Q Now P  ( P→Q ) ≼ Q P

39. 39 Moreover, P V  Q V  Q V and P V  Q F  (P V  Q F )  P F thereby respecting entailment

40. 40 Non Standard Entailment P  Q  def P v  Q v  Not antisymmetric, but is reflexive and transitive (a quasi ordering relation) Theorem:  satisfies: P  ( P→Q )  Q

41. 41 What about falsifiability? P  Q  def Q F  P F Does not give a valid modus ponens!

42. 42 Both Verifiability and Falsifiability Quasi ordering: Reminder: P S = P V  P F P ≤ Q  def

43. 43 theorem The quasi ordering defined gives P  ( P→Q ) ≤ Q

44. 44 Non Standard Implication Instead of P→Q =  P  Q P ↪ Q = def [P V  Q V, Q F \ ]

45. 45 Motivation sup(X | P  X ≤ Q) well defined for any orthonormal lattice. Propositions in TOSL make a lattice, but not orthonormal sup(X | P  X ≲ Q) where X = [X V, X F ] and X V = sup(Y | P V  Y  Q V ) and X F = inf(Y | Q F  P F  Y)

46. 46 To get existence need: P F  Q V  X V This blocks inf(Y | Q F  P F  Y) Leading to X F = (Q F \ )

47. 47 Theorem: P  (P ↪ Q) ≴ Q

48. 48 Why ? 1. We get the usual implication operator when considering the discrete twin logic. 2. Natural interpretation of implication when measurement P verifies P and P ↪ Q, whatever form ↪ may take.

49. 49 Discoveries In any derivation scheme that is given by the lattice theoretic entailment, an implication P  Q that is equivalent to  P  Q must be Boolean. Define P  Q =  P  Q = [(P F  Q V ), (P V  Q F )] m will assign a value of true to P  Q iff m assigns a value of true to either  P or Q. i.e., m  (P F  Q V ). Alternately, m will assign a value of false to P  Q iff m  (P V  Q F ). m assigns indeterminate to P  Q iff m  (P V  Q F ) and m  (P F  Q V )

50. 50 Derivation in Collision Models Replace modus ponens by P and P → Q yield [Q V, P V  Q F ] Replace modus tollens by  Q and P → Q yield [Q V, P V  Q F ]