Computational Logic and Applications KRAKÓW On density of truth of infinite logic. Zofia Kostrzycka University of Technology, Opole, Poland

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1 Computational Logic and Applications KRAKÓW 2008 On density of truth of infinite logic Zofia Kostrzycka University of Technology, Opole, Poland

2 By locally infinite logic, we mean a logic, which in some language with a finite number of variables, has infinitely many classes of non-equivalent formulas. Examples: Int k, Cl k, Int, k Cl, k locally finite Cl,, k Int,, k locally finite locally infinite

3 L- given logic Definition 1. ϕ L ψ if both ϕ ψ T aut L and ψ ϕ T aut L. Definition 2. L/ = {[α], α F orm} Definition 3. The order of classes [α] is defined as [α] [β] iff α β T aut L.

4 System Cl,, 1 [p] [p p] [ ] [ p] where p := p.

5 System Int,, 1 α 0 = α 1 = p α 2 = p α 2n+1 = α 2n α 2n 1 α 2n+2 = α 2n α 2n 1 for n 1

6 Rieger - Nishimura lattice A ω A 13 A 9 A 5 A 1 A 12 A 11 A 10 A 8 A 7 A 6 A 4 A 3 A 2 A 0 [α k ] = A k

7 Intuitionistic logic - motivation To exclude non-constructive proofs.

8 1. A proof of α β consists of a proof of α and a proof of β. 2. A proof of α β is given by presenting either a proof of α or a proof of β. 3. A proof of α β is a construction which, giving a proof of α, returns a proof of β. 4. has no proof and a proof of α is a construction which, given a proof of α, would return a proof of.

9 States of knowledge Our knowledge is developing discretely, passing from one state to another. Let x 1 Rx 2. If x 1 : α = 1, then x 2 : α = 1. But it is possible: x 1 : α = 0 and x 2 : α = 1.

10 Kripke frames and models for Int Definition 4. An intuitionistic Kripke frame is a pair F = W, R consisting of non-empty set W and a partial order R on W. That means that R is reflexive, transitive and antisymmetric. Elements of W are called the points, and xry is read y is accessible from x. A valuation in F is a function V : (x, p j ) {0, 1}. If V (x, p j ) = 1 and xry, then V (y, p j ) = 1.

11 V (x, α β) = 1 iff V (x, α) = 1 and V (x, β) = 1. V (x, α β) = 1 iff V (x, α) = 1 or V (x, β) = 1. V (x, α β) = 1 iff for all y such that xry V (y, α) = 1 implies V (y, β) = 1. V (x, ) = 0.

12 Example 1 p p T aut Int where p := p x 2 p = 1 x 1 p = 0

13 x 2 p = 1, p = 0 x 1 p = 0

14 x 2 p = 1, p = 0 x 1 p = 0, p = 0

15 x 2 p = 1, p = 0 x 1 p = 0, p = 0, p (p ) = 0

16 Example 2 p p T aut Int where p := p x 2 p = 1 x 1 p = 0

17 x 2 p = 1, p = 0 x 1 p = 0

18 x 2 p = 1, p = 0 x 1 p = 0, p = 0

19 x 2 p = 1, p = 0 x 1 p = 0, p = 0 (p ) = 1

20 x 2 p = 1, p = 0 x 1 p = 0, p = 0 ((p ) ) p = 0

21 Example 3 (p q) (q p) T aut Int x 2 p = 1, q = 0 x 3 x 1 p = 0, q = 1 x 1 Rx 2 and x 1 Rx 3

22 x 2 p = 1, q = 0 p q = 0 x 3 x 1 p = 0, q = 1 q p = 0

23 x 2 p = 1, q = 0 p q = 0 x 3 x 1 p q = 0, q p = 0 p = 0, q = 1 q p = 0

24 x 2 p = 1, q = 0 p q = 0 x 3 x 1 p q = 0, q p = 0 (p q) (q p) = 0 p = 0, q = 1 q p = 0

25 Rieger - Nishimura lattice A ω A 13 A 9 A 5 A 1 A 12 A 11 A 10 A 8 A 7 A 6 A 4 A 3 A 2 A 0 [α k ] = A k

26 We associate the density µ(a) with a subset A of formulas as: µ(a) = lim n #{t A : t = n} #{t F orm : t = n} (1) if the appropriate limit exists. Asymptotic density is finitely additive: for A B =. µ(a B) = µ(a) + µ(b).

27 Asymptotic density is not countably additive: µ A i i=0 i=0 µ (A i ) But: µ A i i=0 i=0 µ (A i )

28 The case of the Rieger - Nishimura lattice Because: i=0 A i A ω = F orm then 1 = µ i=0 A i A ω i=0 Hence if the densities exist, then i=0 µ ( A i) + µ(a ω ). µ ( A i) 1 and µ(a ω ) 1. If µ(a ω ) exists?

29 Problem 5. Does the density of truth of Int,, 1 (or Int,, 1 ) as a limit, exist? Problem 6. Does the density of truth of Int,, k (or Int,, k ) as a limit, exist?

30 Generating functions The Drmota-Lalley-Woods theorem Theorem 7. Consider a nonlinear polynomial system, defined by a set of equations {y = Φ j (z, y 1,..., y m )}, 1 j m which is a-proper, a-positive, a-irreducible and a-aperiodic. Then 1. All component solutions y i have the same radius of convergence ρ <.

31 2. There exist functions h j analytic at the origin such that y j = h j ( 1 z/ρ), (z ρ ). (2) 3. All other dominant singularities are of the form ρω with ω being a root of unity. 4. If the system is a-aperiodic then all y j have ρ as unique dominant singularity. In that case, the coefficients admit a complete asymptotic expansion of the form: [z n ]y j (z) ρ n d k n 1 k/2. (3) k 1

32 Application of the Drmota-Lalley-Woods theorem Suppose we have two functions f T and f F enumerating the tautologies of some logic and all formulas. Suppose they have the same dominant singularity ρ and there are the suitable constants α 1, α 2, β 1, β 2 such that: f T (z) = α 1 β 1 1 z/ρ + O(1 z/ρ), (4) f F (z) = α 2 β 2 1 z/ρ + O(1 z/ρ). (5) Then the density of truth (probability that a random formula is a tautology) is given by: µ(t ) = lim n [z n ]f T (z) [z n ]f F (z) = β 1 β 2. (6)

33 The main generating function Language: p,,,. Lemma 8. The generating function f for the numbers F n is the following: f(z) = z. 4

34 Finite quotient sub-lattices obtained from the Rieger - Nishimura lattice R Definition 9. Let (B, ) be a pseudo-boolean algebra (PBA). A nonempty set D B is a filter if for any a, b D it holds: 1) a b D, 2) if a D and a c, then c D. [A 2n 1 ) = {α F orm : α 2n 1 α A ω } -generated filter Sequence of finite quotient algebras AL 4 := R/ [A 3 ), AL 6 := R/ [A 5 ), AL 8 := R/ [A 7 ),... AL 2n := R/ [A 2n 1 ),...

35 A 13 A ω A 9 A 5 A 1 A 12 A 11 A 10 A 8 A 7 A 6 A 4 A 3 A 2 A 0

36 Algebra AL 4 := R/ [A3 ) [A 3 ) A 1 A 4 A 2 A 0

37 A 13 A ω A 9 A 5 A 1 A 12 A 11 A 10 A 8 A 7 A 6 A 4 A 3 A 2 A 0

38 Algebra AL 6 := R/ [A5 ) [A 5 ) A 1 A 2 A 4 A 3 A 6 A 0

39 Algebra AL 8 := R/ [A7 ) A 1 A 5 A 8 A 4 A 3 A 0 [A 7 ) A 2 A 6

40 Decreasing sequence of filters: [A 3 ) [A 5 )... [A 2n 1 )... A ω Theorem 10. The density µ(a k ) exists for any k N. Corollary 11. The densities µ([a 2n 1 )) exist for any n N.

41 Proof. Density of classes from AL 4. The operations {, } in the algebra are given by the following truth-tables: A 0 A 1 A 4 A 2 [A 3 ) A 0 [A 3 ) [A 3 ) [A 3 ) [A 3 ) A 1 A 4 A 2 [A 3 ) A 2 [A 3 ) A 2 A 1 A 4 A 1 A 4 [A 3 ) [A 3 ) [A 3 ) A 0 A 1 A 4 A 2 [A 3 ) A 0 A 1 A 4 A 2 [A 3 ) A 0 A 0 A 1 A 4 A 2 [A 3 ) A 1 A 4 A 1 A 4 A 1 A 4 [A 3 ) [A 3 ) A 2 A 2 [A 3 ) A 2 [A 3 ) [A 3 ) [A 3 ) [A 3 ) [A 3 ) [A 3 )

42 f 0 (z) = f [3) (z)f 0 (z) + [f 0 (z)] 2 + z (f 1 + f 4 )(z) = f [3) (z)(f 1 + f 4 )(z)+ f 2 (z)[f 0 (z) + (f 1 + f 4 )(z)] + 2f 0 (z)(f 1 + f 4 )(z) +[(f 1 + f 4 )(z)] 2 + z f 2 (z) = f [3) (z)f 2 (z) + (f 1 + f 4 )(z)[f 0 (z)+ f 2 (z)] + 2f 0 (z)f 2 (z) + [f 2 (z)] 2 f [3) (z) = f(z) [f 0 (z) + (f 1 + f 4 )(z) + f 2 (z)] The system is a-proper, a-positive, a-irreducible and a- aperiodic. All the functions have the same as the function f unique dominant singularity z 0 = 1/16 and the densities of the classes A 0, A 1 A 4, A 2, [A 3 ) exist. For the function f [3) there is a strictly positive formula built from the other functions. We use the another one for simplicity

43 Analogous situation holds for each algebra AL 2n for any n N. In any case we obtain a system of 2n equations, which is a-proper, a-positive, a-irreducible and a-aperiodic. So, the densities again exist.

44 Calculation of the basic functions From system of four equations we calculate: f 0 = 1 (1 + 3f 0 4 f (1 + 3f0 f)2 8z f 1 + f 4 = 2f 0 f 0 f 2 = f0 f 0 f [3) = f f 0 (f 1 + f 4 ) f 2. where f = z 4 and f 0 = z 1 f )

45 Lemma 12. Expansions of functions f, f 0, f 1 + f 4, f 2 and f [3) in a neighborhood of z 0 = 1/16 are as follows: f(z) = z +... f 0 (z) = a 0 + a z +... (f 1 + f 4 )(z) = b 0 + b z +... f 2 (z) = c 0 + c z +... f [3) (z) = d 0 + d z +... a , a , b , b c , c , d , d

46 Lemma 13. The densities of the classes of formulas from the algebra AL 4 exist and are the following: µ(a 0 ) µ(a 1 A 4 ) µ(a 2 ) µ([a 3 ))

47 Observation 14. The algebra AL 4 is a Lindenbaum algebra of the classical logic with one variable. Hence µ(cl,, p ) = µ([a 3 )) 0.736

48 Densities of classes from AL 6 Lemma 15. The densities of the classes from the algebra AL 6 exist and are the following:

49 An upper estimation of density of Int,, p We consider the algebra AL 2k = R/ [A2n 1 ).

50 A 2k 2 A 2k 5 A 2k 6 [A 2k 1 ) A 2k 3 A 2k A 2k 7 A 2k 4 A 2k 8 A 5 A 1 A 4 A 3 A 2 A 0

51 [A 2k 1 ) n = A 0 i F n i + A 2 i ( F n i A 0 n i A1 n i A4 n i ) +... ( A 2k 3 i + A 2k i A 1 i ( F n i A 0 n i A2 n i ) ) ( A 2k 3 n i + A 2k n i + [A2k 1 ) n i ) + Ai 2k 2 ( A 2k 2 n i + [A 2k 1 ) n i ) + A 2k 2 i ( A 2k 4 + A 2k 3 + A 2k n i n i [A 2k 1 ) i [A 2k 1 ) n i + [A 2k 1 ) i ( F n i [A 2k 1 ) n i ) n i )

52 Of course, the sum written above as... is finite in the case of finite algebra AL 2k. After simplification, the above formula may by transformed into the following one: f [2k 1) = [f 0 f + f 1 (f f 0 f 2 ) + f 2 (f f 0 f 1 f 4 ) + +f 3 (f f 0 f 1 f 2 f 4 ) f 2k 2 (f f 0 f 1... f 2k 3 f 2k ) + (f 2k 3 + f (f f 0 f 1... f 2k 4 f 2k 2 ) + +2f 2k 2 (f 2k 4 + f 2k 3 + f 2k )]/(1 2f)

53 µ([a 3 )) µ([a 5 )) µ([a 7 )) µ([a 9 )) µ([a 3 )) µ([a 5 ))... µ([a 2n 1 ))... µ(a ω ) Observation 16. If µ(a ω ) exists, then µ(a ω ) <

54 A lower estimation of µ(a ω ).

55 A 13 A ω A 9 A 5 A 1 A 12 A 11 A 10 A 8 A 7 A 6 A 4 A 3 A 2 A 0

56 A ω n = A 0 i F n i + A 1 i ( F n i ( A 0 n i + A2 n i )) + A 2 i ( F n i ( A 0 n i + A1 n i + A4 n i )) + A 3 i ( F n i ( A 0 n i + A1 n i + A2 n i + A4 n i )) + A 4 i ( F n i ( A 0 n i + A1 n i + A2 n i + A3 n i + A6 n i )) A 5 i ( F n i ( A 0 n i + A1 n i A4 n i + A6 n i )) + A ω n 1 i Aω n i + 2 A ω i ( F n i A ω n i )

57 Observation 17. Let (c n ), (d n ) and (e n ) be three sequences of natural numbers, such that c n d n for all n N. Suppose two new sequences are defined recursively as follows: x n = c n + Then x n y n for any n N. e i x n i, y n = d n + e i y n i

58 A ω n = A 0 i F n i + A 1 i ( F n i ( A 0 n i + A2 n i ))+ A 2 i ( F n i ( A 0 n i + A1 n i + A4 n i ))+ A 3 i ( F n i ( A 0 n i + A1 n i + A2 n i + A4 n i ))+ A 4 i ( F n i ( A 0 n i + A1 n i + A2 n i + A3 n i + A6 n i )) A 5 i ( F n i ( A 0 n i + A1 n i A4 n i + A6 n i ))+ A ω n 1 i Aω n i + 2 A ω i ( F n i A ω n i )

59 Smaller numbers B 5 n :

60 B 5 n = A 0 i F n i + A 1 i ( F n i ( A 0 n i + A2 n i )) + A 2 i ( F n i ( A 0 n i + A1 n i + A4 n i )) + A 3 i ( F n i ( A 0 n i + A1 n i + A2 n i + A4 n i )) + A 4 i ( F n i ( A 0 n i + A1 n i + A2 n i + A3 n i + A6 n i )) A 5 i ( F n i ( A 0 n i A4 n i + A6 n i )) + Bi 5 n 1 B5 n i + 2 B 5 i ( F n i B 5 n i )

61 g 5 = (f 0 f + f 1 (f f 0 f 2 ) + f 2 (f f 0 f 1 f 4 ))+ (7 f 3 (f f 0 f 1 f 2 f 4 ) + f 4 (f f 0 f 1 f 2 f 3 f 6 )+ + f 5 (f f 0 f 1 f 2 f 3 f 4 f 6 )) /(1 2f). Lemma 18. The density of the class B 5 exists and is the following: µ(b 5 ) (8)

62 Theorem 19. If the density of the class A ω exists, then it is estimated as follows: µ(a ω )

63 The existence of µ(a ω ). Bn 2k n 1 = A 0 i F n i + A 1 i ( F n i A 0 n i A2 n i ) + A 2 i ( F n i A 0 n i A1 n i A4 n i ) +... A 2k i ( F n i A 0 n i A1 n i... A2k 1 n i A 2k+2 n i ) Bi 2k Bn i 2k n Bi 2k ( F n i Bn i 2k ).

64 Observation: B 4 B 6... A ω... [A 5 ) [A 3 ). Sequence of compartments [µ(b 2k ), µ([a 2k 1 ))], for k 2. Problem: to show that their lengths tend to 0. Lemma 20. ( µ([a 2k 1 )) µ(b 2k ) ) = 0 lim k

65 Proof. We consider the numbers [A 2k 1 ) n B 2k n. [A 2k 1 ) n Bn 2k = A 2k 2 i A 2k i ( A 2k 4 + A 2k 3 + A 2k n i ) + n i n i ( A 2k 3 n i + A 2k 1 n i + A 2k+2 n i ) + F i ( [A 2k 1 ) n i B 2k n i ) A 2k 1 i [A 2k 1 ) n i The numbers n 1 A2k 1 i [A 2k 1 ) n i are non-negative, so on the base of Observation we may consider larger

66 numbers C k n : C k n = 2 + A 2k 2 i A 2k i ( A 2k 4 + A 2k 3 + A 2k n i ) + n i n i ( A 2k 3 n i + A 2k 1 n i + A 2k+2 n i ) + 2 F i Cn k The numbers C k n characterize the set C k consisting of formulas being disjunctions between formulas from A 2k 2 and A 2k 4 A 2k 3 A 2k, and implications from A 2k to A 2k 3 A 2k 1 A 2k+2, and disjunctions between formulas from C k and formulas from F orm. From the above we obtain formulas defining the generat-

67 ing functions f C k for the numbers C k n : f C k = [f 2k (f 2k 3 + f 2k 1 ) + 2f 2k 2 (f 2k 4 + f 2k 3 + f 2k )]. 1 2f The function f C k is defined by functions with dominant singularity at z 0 = 1/16 (see proof of Theorem 10). So, it has the same dominant singularity. The density of the class C k can be computed as follows: µ(c k ) = f C k( 16 1 ) f ( 16 1 ). We show that the values of f C k( 16 1 ) tend to 0 when k tends to infinity. For simplicity, we introduce a new symbol

68 h k := f 2k (f 2k 3 + f 2k 1 ) + 2f 2k 2 (f 2k 4 + f 2k 3 + f 2k ). Then, from (10), we have: f C k ( 1 16 ) = h k ( 16 1 ) (1 2f( 1 16 )) h k( 16 1 ) ( 2f ( 16 1 )) (1 2f( ))2 The values of f( 16 1 ) and f ( 16 1 ) exist and are constant. To prove that ( ) ( ) 1 1 lim k h k = 0 and lim h k = 0. (10) 16 k 16 we prove that lim k f k ( 1 16 ) ( ) 1 = 0 and lim f k k 16 = 0. (11) It is straightforward to observe that (11) yields (10). From Theorem 10 it follows that µ(a k ) exists for each k N.

69 The series k=0 µ ( A k) is convergent and hence lim k µ ( A k) = 0. So, from the transfer lemma (we know that the functions f k have the same dominant singularity) we obtain: ( ) lim f k 1 k 16 = 0. (12) Similarly, let us consider the series k=0 f k ( 116 ). This series is bounded by f ( 1 16 ) = 1 2 and the values f k ( 1 16 ) are

70 non-negative, so it also must be convergent. Hence: lim k f k ( 1 16 ) = 0. (13) By Theorem 19 and Lemma 20 we get: Theorem 21. The density of the class A ω exists and is about 70%. It could be justified as follows: for each i N, f i (1/16) 0 because f i (z) = n=0 a inz n and the series is convergent at z 0 = 1 and then 16 the sum n=0 a in( 1 16 )n is also non-negative.

71 Problem 1 Investigate the logics Int,, k and Cl,, k and give an answer if they are asymptotically identical.