Recursion Theory. Joost J. Joosten

Transcription

1 Recursion Theory Joost J. Joosten Institute for Logic Language and Computation University of Amsterdam Plantage Muidergracht TV Amsterdam Room P 3.26, jjoosten@phil.uu.nl jjoosten Recursion Theory p.1/15

2 The Turing Universe Consider again degrees D of Turing equivalent sets Recursion Theory p.2/15

3 The Turing Universe Consider again degrees D of Turing equivalent sets is well defined on these degrees by deg(a) deg(b) A T B Recursion Theory p.2/15

4 The Turing Universe Consider again degrees D of Turing equivalent sets is well defined on these degrees by deg(a) deg(b) A T B Thus, defines a partial ordering on D Recursion Theory p.2/15

5 The Turing Universe Consider again degrees D of Turing equivalent sets is well defined on these degrees by deg(a) deg(b) A T B Thus, defines a partial ordering on D Hamkins: what does D look like? Recursion Theory p.2/15

6 The Turing Universe Consider again degrees D of Turing equivalent sets is well defined on these degrees by deg(a) deg(b) A T B Thus, defines a partial ordering on D Hamkins: what does D look like? First simple question: how many degrees are there? Recursion Theory p.2/15

7 The Turing Universe Consider again degrees D of Turing equivalent sets is well defined on these degrees by deg(a) deg(b) A T B Thus, defines a partial ordering on D Hamkins: what does D look like? First simple question: how many degrees are there? Answer: there are uncountably many degrees Recursion Theory p.2/15

8 Countability revisited Cantor: 2 N is not countable Recursion Theory p.3/15

9 Countability revisited Cantor: 2 N is not countable Proof: consider S := {x x / X x } Recursion Theory p.3/15

10 Countability revisited Cantor: 2 N is not countable Proof: consider S := {x x / X x } If S = X e for some e, we would have a problem Recursion Theory p.3/15

11 Countability revisited Cantor: 2 N is not countable Proof: consider S := {x x / X x } If S = X e for some e, we would have a problem Again, draw the diagram to see why this is called a diagonal argument! Recursion Theory p.3/15

12 Countability revisited Cantor: 2 N is not countable Proof: consider S := {x x / X x } If S = X e for some e, we would have a problem Again, draw the diagram to see why this is called a diagonal argument! Each degree contains precisely ℵ 0 many sets (deg(a) {Φ A i i N}) Recursion Theory p.3/15

13 Countability revisited Cantor: 2 N is not countable Proof: consider S := {x x / X x } If S = X e for some e, we would have a problem Again, draw the diagram to see why this is called a diagonal argument! Each degree contains precisely ℵ 0 many sets (deg(a) {Φ A i i N}) Each set is contained in precisely one degree Recursion Theory p.3/15

14 Countability revisited Cantor: 2 N is not countable Proof: consider S := {x x / X x } If S = X e for some e, we would have a problem Again, draw the diagram to see why this is called a diagonal argument! Each degree contains precisely ℵ 0 many sets (deg(a) {Φ A i i N}) Each set is contained in precisely one degree However, a countable union of countable sets is again countable via coding Recursion Theory p.3/15

15 Countability revisited Likewise, we have {b b a} ℵ 0 Recursion Theory p.4/15

16 Countability revisited Likewise, we have {b b a} ℵ 0 Is this an indication that D is necessarily broad? Recursion Theory p.4/15

17 Countability revisited Likewise, we have {b b a} ℵ 0 Is this an indication that D is necessarily broad? No: look at the set of all countable ordinals Recursion Theory p.4/15

18 Countability revisited Likewise, we have {b b a} ℵ 0 Is this an indication that D is necessarily broad? No: look at the set of all countable ordinals This is a linearly ordered uncountable set where each element only has ℵ 0 many predecessors Recursion Theory p.4/15

19 Countability revisited Likewise, we have {b b a} ℵ 0 Is this an indication that D is necessarily broad? No: look at the set of all countable ordinals This is a linearly ordered uncountable set where each element only has ℵ 0 many predecessors However, {b b a} ℵ 0 does imply that there is no maximal element Recursion Theory p.4/15

20 Countability revisited Likewise, we have {b b a} ℵ 0 Is this an indication that D is necessarily broad? No: look at the set of all countable ordinals This is a linearly ordered uncountable set where each element only has ℵ 0 many predecessors However, {b b a} ℵ 0 does imply that there is no maximal element Aside: also in D there is a minimal element 0 Recursion Theory p.4/15

21 Relativizing continued Structure gets inherited through relativization Recursion Theory p.5/15

22 Relativizing continued Structure gets inherited through relativization We can consider A-c.e. sets: Recursion Theory p.5/15

23 Relativizing continued Structure gets inherited through relativization We can consider A-c.e. sets: the sets that can be computably enumerated using queries to A Recursion Theory p.5/15

24 Relativizing continued Structure gets inherited through relativization We can consider A-c.e. sets: the sets that can be computably enumerated using queries to A Lemma: B is A-c.e., iff B = W A e for some e Recursion Theory p.5/15

25 Relativizing continued Structure gets inherited through relativization We can consider A-c.e. sets: the sets that can be computably enumerated using queries to A Lemma: B is A-c.e., iff B = W A e for some e We can now again consider the A-computable approximations Recursion Theory p.5/15

26 Relativizing continued Structure gets inherited through relativization We can consider A-c.e. sets: the sets that can be computably enumerated using queries to A Lemma: B is A-c.e., iff B = W A e for some e We can now again consider the A-computable approximations Φ A e,s and W e,s Recursion Theory p.5/15

27 Relativizing continued Structure gets inherited through relativization We can consider A-c.e. sets: the sets that can be computably enumerated using queries to A Lemma: B is A-c.e., iff B = W A e for some e We can now again consider the A-computable approximations Φ A e,s and W e,s Likewise, we define the notions Σ A 1 and ΠA 1 Recursion Theory p.5/15

28 Relativazed complementation X is A-computable X is A-c.e. Recursion Theory p.6/15

29 Relativazed complementation X is A-computable X is A-c.e. X is A-computable both X and X are A-c.e. Recursion Theory p.6/15

30 Relativazed complementation X is A-computable X is A-c.e. X is A-computable both X and X are A-c.e. X A 1 Recursion Theory p.6/15

31 Relativazed complementation X is A-computable X is A-c.e. X is A-computable both X and X are A-c.e. X A 1 As A-c.e. coincides with Σ A 1 Recursion Theory p.6/15

32 Relativized Normal Form Summarizing the above, we can state the relativized NFT: Recursion Theory p.7/15

33 Relativized Normal Form Summarizing the above, we can state the relativized NFT: The following three statements are equivalent Recursion Theory p.7/15

34 Relativized Normal Form Summarizing the above, we can state the relativized NFT: The following three statements are equivalent X is A-computably enumerable Recursion Theory p.7/15

35 Relativized Normal Form Summarizing the above, we can state the relativized NFT: The following three statements are equivalent X is A-computably enumerable X is W A e for some e Recursion Theory p.7/15

36 Relativized Normal Form Summarizing the above, we can state the relativized NFT: The following three statements are equivalent X is A-computably enumerable X is W A e for some e X is Σ A 1 Recursion Theory p.7/15

37 Halting problem relativized We can define A = { x,y x W A y } (= K A 0 ) Recursion Theory p.8/15

38 Halting problem relativized We can define A = { x,y x W A y } (= K A 0 ) A is A-c.e., but not A-computable Recursion Theory p.8/15

39 Halting problem relativized We can define A = { x,y x W A y } (= K A 0 ) A is A-c.e., but not A-computable K A 0 is called the jump of A and is also denoted by A Recursion Theory p.8/15

40 Halting problem relativized We can define A = { x,y x Wy A } (= K0 A) A is A-c.e., but not A-computable K A 0 is called the jump of A and is also denoted by A We can iterate jumps Recursion Theory p.8/15

41 The Jump Theorem A is A-c.e. Recursion Theory p.9/15

42 The Jump Theorem A is A-c.e. B is A-c.e. iff B m A Recursion Theory p.9/15

43 The Jump Theorem A is A-c.e. B is A-c.e. iff B m A A T A Recursion Theory p.9/15

44 The Jump Theorem A is A-c.e. B is A-c.e. iff B m A A T A So we have proved once more that there is no maximal element in D Recursion Theory p.9/15

45 Jumps on degrees Can we lift the Jump operation to degrees Recursion Theory p.10/15

46 Jumps on degrees Can we lift the Jump operation to degrees If a is a degree, is a well defined? Recursion Theory p.10/15

47 Jumps on degrees Can we lift the Jump operation to degrees If a is a degree, is a well defined? Direct approach would use a false assumption: Recursion Theory p.10/15

48 Jumps on degrees Can we lift the Jump operation to degrees If a is a degree, is a well defined? Direct approach would use a false assumption: A = We A = We B so A is B-c.e., whence m B Recursion Theory p.10/15

49 Jumps on degrees Can we lift the Jump operation to degrees If a is a degree, is a well defined? Direct approach would use a false assumption: A = We A = We B so A is B-c.e., whence m B It is not true that: A T B implies We A = We B! Recursion Theory p.10/15

50 Jumps on degrees Can we lift the Jump operation to degrees If a is a degree, is a well defined? Direct approach would use a false assumption: A = W A e = W B e so A is B-c.e., whence m B It is not true that: A T B implies W A e = W B e! However: A T B and X is A-c.e., implies that X is B-c.e. Recursion Theory p.10/15

51 Jumps on degrees Can we lift the Jump operation to degrees If a is a degree, is a well defined? Direct approach would use a false assumption: A = W A e = W B e so A is B-c.e., whence m B It is not true that: A T B implies W A e = W B e! However: A T B and X is A-c.e., implies that X is B-c.e. This yields the required: A T B A T B Recursion Theory p.10/15

52 Jumps on degrees Can we lift the Jump operation to degrees If a is a degree, is a well defined? Direct approach would use a false assumption: A = W A e = W B e so A is B-c.e., whence m B It is not true that: A T B implies W A e = W B e! However: A T B and X is A-c.e., implies that X is B-c.e. This yields the required: A T B A T B In particular: 0 < 0 < 0 < 0 < 0... Recursion Theory p.10/15

53 Higher up in the hierarchy We define the Σ 0 n and the Π 0 n, and often omit the superscript 0. Recursion Theory p.11/15

54 Higher up in the hierarchy We define the Σ 0 n and the Π 0 n, and often omit the superscript 0. Example: Tot is a Π 0 2 -set Recursion Theory p.11/15

55 Higher up in the hierarchy We define the Σ 0 n and the Π 0 n, and often omit the superscript 0. Example: Tot is a Π 0 2 -set We shall see that there is a tight connection between the (n) and the Σ n definable sets. Recursion Theory p.11/15

56 Higher up in the hierarchy We define the Σ 0 n and the Π 0 n, and often omit the superscript 0. Example: Tot is a Π 0 2 -set We shall see that there is a tight connection between the (n) and the Σ n definable sets. This is one of Post s famous theorems: Recursion Theory p.11/15

57 Higher up in the hierarchy We define the Σ 0 n and the Π 0 n, and often omit the superscript 0. Example: Tot is a Π 0 2 -set We shall see that there is a tight connection between the (n) and the Σ n definable sets. This is one of Post s famous theorems: (n+1) is Σ n+1 -complete (a generalization of m-completeness) Recursion Theory p.11/15

58 Higher up in the hierarchy We define the Σ 0 n and the Π 0 n, and often omit the superscript 0. Example: Tot is a Π 0 2 -set We shall see that there is a tight connection between the (n) and the Σ n definable sets. This is one of Post s famous theorems: (n+1) is Σ n+1 -complete (a generalization of m-completeness) A set A is Σ n complete if it is Σ n, and for any other Σ n set B we have that B m A Recursion Theory p.11/15

59 Higher up in the hierarchy We define the Σ 0 n and the Π 0 n, and often omit the superscript 0. Example: Tot is a Π 0 2 -set We shall see that there is a tight connection between the (n) and the Σ n definable sets. This is one of Post s famous theorems: (n+1) is Σ n+1 -complete (a generalization of m-completeness) A set A is Σ n complete if it is Σ n, and for any other Σ n set B we have that B m A To prove Post s Theorem we need the following lemma Recursion Theory p.11/15

60 (n) -Relativizing Principle The (n) -Relativizing Principle: Recursion Theory p.12/15

61 (n) -Relativizing Principle The (n) -Relativizing Principle: A is Σ 0 n+1 A is c.e. in (n). Recursion Theory p.12/15

62 (n) -Relativizing Principle The (n) -Relativizing Principle: A is Σ 0 n+1 A is c.e. in (n). Proof: by induction on n Recursion Theory p.12/15

63 (n) -Relativizing Principle The (n) -Relativizing Principle: A is Σ 0 n+1 A is c.e. in (n). Proof: by induction on n n = 0 is already established Recursion Theory p.12/15

64 (n) -Relativizing Principle The (n) -Relativizing Principle: A is Σ 0 n+1 A is c.e. in (n). Proof: by induction on n n = 0 is already established Assume A Σ n+2 Recursion Theory p.12/15

65 (n) -Relativizing Principle The (n) -Relativizing Principle: A is Σ 0 n+1 A is c.e. in (n). Proof: by induction on n n = 0 is already established Assume A Σ n+2 that is, for some Π n+1 relation R we have x A y R(x,y) Recursion Theory p.12/15

66 (n) -Relativizing Principle The (n) -Relativizing Principle: A is Σ 0 n+1 A is c.e. in (n). Proof: by induction on n n = 0 is already established Assume A Σ n+2 that is, for some Π n+1 relation R we have x A y R(x,y) That is A Σ R 1, where R is Σ n+1 Recursion Theory p.12/15

67 (n) -Relativizing Principle The (n) -Relativizing Principle: A is Σ 0 n+1 A is c.e. in (n). Proof: by induction on n n = 0 is already established Assume A Σ n+2 that is, for some Π n+1 relation R we have x A y R(x,y) That is A Σ R 1, where R is Σ n+1 IH: R is c.e. in n Recursion Theory p.12/15

68 (n) -Relativizing Principle The (n) -Relativizing Principle: A is Σ 0 n+1 A is c.e. in (n). Proof: by induction on n n = 0 is already established Assume A Σ n+2 that is, for some Π n+1 relation R we have x A y R(x,y) That is A Σ R 1, where R is Σ n+1 IH: R is c.e. in n By Jump-Theorem: R m n+1 Recursion Theory p.12/15

69 (n) -Relativizing Principle The (n) -Relativizing Principle: A is Σ 0 n+1 A is c.e. in (n). Proof: by induction on n n = 0 is already established Assume A Σ n+2 that is, for some Π n+1 relation R we have x A y R(x,y) That is A Σ R 1, where R is Σ n+1 IH: R is c.e. in n By Jump-Theorem: R m n+1 So, A Σ n+1 1 and by NFT c.e. in n+1 Recursion Theory p.12/15

70 (n) -Relativizing Principle A is Σ 0 n+1 A is c.e. in (n). Recursion Theory p.13/15

71 (n) -Relativizing Principle A is Σ 0 n+1 A is c.e. in (n). Suppose A is c.e. in (n+1), i.e., A = W (n+1) i. Recursion Theory p.13/15

72 (n) -Relativizing Principle A is Σ 0 n+1 A is c.e. in (n). Suppose A is c.e. in (n+1), i.e., A = W (n+1) i. As (n+1) is c.e. in (n), by the IH: (n+1) Σ n+1. Recursion Theory p.13/15

73 (n) -Relativizing Principle A is Σ 0 n+1 A is c.e. in (n). Suppose A is c.e. in (n+1), i.e., A = W (n+1) i. As (n+1) is c.e. in (n), by the IH: (n+1) Σ n+1. Now, x A iff s and some oracle queries to (n+1) and its complement such that: x W (n+1) i,a Recursion Theory p.13/15

74 (n) -Relativizing Principle A is Σ 0 n+1 A is c.e. in (n). Suppose A is c.e. in (n+1), i.e., A = W (n+1) i. As (n+1) is c.e. in (n), by the IH: (n+1) Σ n+1. Now, x A iff s and some oracle queries to (n+1) and its complement such that: x W (n+1) i,a Bringing this into prenex normal form gives us A Σ n+2. Recursion Theory p.13/15

75 Post s Theorem A n+1 A,A T (0) Recursion Theory p.14/15

76 Post s Theorem A n+1 A,A T (0) Proof: By the relativized Complementation Lemma and using that A is Σ 0 n+1 A is c.e. in (n). Recursion Theory p.14/15

77 Post s Theorem (n+1) is Σ n+1 -complete Recursion Theory p.15/15

78 Post s Theorem (n+1) is Σ n+1 -complete Proof: If A Σ n+1 then, by previous lemma: A is c.e. in (n) Recursion Theory p.15/15

79 Post s Theorem (n+1) is Σ n+1 -complete Proof: If A Σ n+1 then, by previous lemma: A is c.e. in (n) By Jump Theorem: A m ( (n) ) (= (n+1) ) Recursion Theory p.15/15

80 Post s Theorem (n+1) is Σ n+1 -complete Proof: If A Σ n+1 then, by previous lemma: A is c.e. in (n) By Jump Theorem: A m ( (n) ) (= (n+1) ) Each quantifier adds new complexity! Recursion Theory p.15/15

81 Post s Theorem (n+1) is Σ n+1 -complete Proof: If A Σ n+1 then, by previous lemma: A is c.e. in (n) By Jump Theorem: A m ( (n) ) (= (n+1) ) Each quantifier adds new complexity! Informational content grows Recursion Theory p.15/15

82 Post s Theorem (n+1) is Σ n+1 -complete Proof: If A Σ n+1 then, by previous lemma: A is c.e. in (n) By Jump Theorem: A m ( (n) ) (= (n+1) ) Each quantifier adds new complexity! Informational content grows To go beyond ω we need hyperarithmetic sets and second order logic Recursion Theory p.15/15