Proof Theory of Induction

Transcription

1 1/ 70 Proof Theory of Induction Stefan Hetzl Institute of Discrete Mathematics and Geometry Vienna University of Technology Summer School for Proof Theory in First-Order Logic Funchal, Madeira August 2017

2 2/ 70 Outline Gentzen s consistency proof The omega rule Cyclic proofs

3 3/ 70 Outline Gentzen s consistency proof Background Peano arithmetic Reduction of cut and induction Ordinals The consistency proof The omega rule Cyclic proofs

4 4/ 70 Background Hilbert s programme (1920ies) Formalisation of mathematics Proof of consistency by finitary methods

5 4/ 70 Background Hilbert s programme (1920ies) Formalisation of mathematics Proof of consistency by finitary methods Gödel s (2nd) incompleteness theorem (1931) Theorem. For a (consistent, axiomatisable, and sufficiently strong) first-order theory T, T Con T.

6 4/ 70 Background Hilbert s programme (1920ies) Formalisation of mathematics Proof of consistency by finitary methods Gödel s (2nd) incompleteness theorem (1931) Theorem. For a (consistent, axiomatisable, and sufficiently strong) first-order theory T, T Con T. Gentzen s approach (1936-): split consistency proof for T into: 1. A cut-elimination procedure (in weak theory) 2. A termination assumption (transcends theory)

7 5/ 70 Cut-elimination and consistency Why is cut-elimination relevant for consistency? Definition. A theory T is inconsistent if there is some formula ϕ s.t. T ϕ and T ϕ. Suppose there are T -proofs π 1 of ϕ and π 2 of ϕ, then π = (π 2 ) ϕ (π 1 ) ϕ ϕ l cut is a proof of. By cut-elimination, there is a cut-free proof π of. Contradiction.

8 6/ 70 Outline Gentzen s consistency proof Background Peano arithmetic Reduction of cut and induction Ordinals The consistency proof The omega rule Cyclic proofs

9 7/ 70 Peano arithmetic The language L = {0, s, +,, =} Basic arithmetic BA consists of the axioms: x y (s(x) = s(y) x = y) x 0 s(x) x x + 0 = x x y x + s(y) = s(x + y) x x 0 = 0 x y x s(y) = x y + x Peano arithmetic PA consists of the axioms of BA together with, for every formula ϕ(x, z), the induction axiom z ( (ϕ(0, z) y(ϕ(y, z) ϕ(s(y), z)) x ϕ(x, z) ).

10 8/ 70 A sequent calculus for PA A sequent calculus for FOL with equality in L plus inital sequents and the induction rule s(t) = s(u) t = u 0 s(t) t + 0 = t t + s(u) = s(t + u) t 0 = 0 t s(u) = t u + t ϕ(α), Γ, ϕ(s(α)) ϕ(0), Γ, ϕ(t) where α does not appear in ϕ(0), Γ, ϕ(t).

11 9/ 70 Simple proofs Definition. A PA-proof is called simple if it consists of only of initial sequents, atomic cuts, and structural inference.

12 9/ 70 Simple proofs Definition. A PA-proof is called simple if it consists of only of initial sequents, atomic cuts, and structural inference. Simple proof lemma. There is no simple proof of. Proof. Let π be a simple proof of. W.l.o.g. π is variable-free. Every formula in π is of the form s = t with s, t variable-free. Evaluate formulas and sequents to or in N. Every inital sequent evaluates to. Every rule preserves. evaluates to.

13 10/ 70 Outline Gentzen s consistency proof Background Peano arithmetic Reduction of cut and induction Ordinals The consistency proof The omega rule Cyclic proofs

14 11/ 70 Interaction between induction and cut: example. Γ ϕ(0). Γ, ϕ(α) ϕ(s(α)) ind Γ, ϕ(0) ϕ(t) cut Γ ϕ(t)

15 11/ 70 Interaction between induction and cut: example. Γ ϕ(0). Γ, ϕ(α) ϕ(s(α)) ind Γ, ϕ(0) ϕ(t) cut Γ ϕ(t) = eliminating cuts means eliminating inductions too

16 12/ 70 Numerals and evaluation Definition. For n N define the L-term n = s n (0). A term of the form s n (0) is called numeral. Evaluation lemma. Let t is a variable free L-term. Then there is an n N s.t. BA t = n, and for any formula ϕ(x) there is an induction-free proof of ϕ(n) ϕ(t)

17 13/ 70 Cut-elimination and induction (π(α)) ϕ(α), Γ, ϕ(s(α)) ϕ(0), Γ, ϕ(t) ind

18 13/ 70 Cut-elimination and induction (π(α)) ϕ(α), Γ, ϕ(s(α)) ϕ(0), Γ, ϕ(t) ind If t is variable-free, there is n N s.t. BA t = n

19 13/ 70 Cut-elimination and induction (π(α)) ϕ(α), Γ, ϕ(s(α)) ϕ(0), Γ, ϕ(t) ind If t is variable-free, there is n N s.t. BA t = n (π(0)) ϕ(0), Γ, ϕ(1) ϕ(0), Γ, ϕ(2) (π(1)) ϕ(1), Γ, ϕ(2) cut. Eval. Lem. ϕ(0), Γ, ϕ(n) ϕ(n) ϕ(t) ϕ(0), Γ, ϕ(t) cut

20 13/ 70 Cut-elimination and induction (π(α)) ϕ(α), Γ, ϕ(s(α)) ϕ(0), Γ, ϕ(t) ind If t is variable-free, there is n N s.t. BA t = n (π(0)) ϕ(0), Γ, ϕ(1) ϕ(0), Γ, ϕ(2) (π(1)) ϕ(1), Γ, ϕ(2) cut. Eval. Lem. ϕ(0), Γ, ϕ(n) ϕ(n) ϕ(t) ϕ(0), Γ, ϕ(t) Under which conditions does this work? cut

21 14/ 70 The end-piece Definition. A logical inference ι in a PA-proof π is called explicit if it is ancestor of the end-sequent, and implicit if it is ancestor of a cut formula.

22 14/ 70 The end-piece Definition. A logical inference ι in a PA-proof π is called explicit if it is ancestor of the end-sequent, and implicit if it is ancestor of a cut formula. Definition. The end-piece of a PA-proof π: all sequents which are not above an implicit logical inference.

23 14/ 70 The end-piece Definition. A logical inference ι in a PA-proof π is called explicit if it is ancestor of the end-sequent, and implicit if it is ancestor of a cut formula. Definition. The end-piece of a PA-proof π: all sequents which are not above an implicit logical inference. Example.. ψ(0, 0) y ψ(0, y) r. ψ(α, β) y ψ(s(α), y) y ψ(α, y) y ψ(s(α), y) l ind y ψ(0, y) y ψ(t, y) y ψ(t, y). ψ(t, α) y ϕ(t, y) cut y ψ(t, y) y ϕ(t, y) l cut y ϕ(t, y) x y ϕ(x, y) r

24 14/ 70 The end-piece Definition. A logical inference ι in a PA-proof π is called explicit if it is ancestor of the end-sequent, and implicit if it is ancestor of a cut formula. Definition. The end-piece of a PA-proof π: all sequents which are not above an implicit logical inference. Example.. ψ(0, 0) y ψ(0, y) r. ψ(α, β) y ψ(s(α), y) y ψ(α, y) y ψ(s(α), y) l ind y ψ(0, y) y ψ(t, y) y ψ(t, y). ψ(t, α) y ϕ(t, y) cut y ψ(t, y) y ϕ(t, y) l cut y ϕ(t, y) x y ϕ(x, y) r

25 15/ 70 Σ 1 -sequents Definition. A Σ 1 -sequent is a sequent of the form x 1 ϕ 1,..., x k ϕ k x k+1 ϕ k+1,..., x n ϕ n s.t. ϕ 1,..., ϕ n quantifier-free, ϕ i contains only variables from x i.

26 15/ 70 Σ 1 -sequents Definition. A Σ 1 -sequent is a sequent of the form x 1 ϕ 1,..., x k ϕ k x k+1 ϕ k+1,..., x n ϕ n s.t. ϕ 1,..., ϕ n quantifier-free, ϕ i contains only variables from x i. Lemma. Let π be a PA-proof of a Σ 1 -sequent Γ and let ϕ(α), Γ, ϕ(s(α)) ϕ(0), Γ, ϕ(t) ind be a lowermost induction in the end-piece of π. Then t is variable-free.

27 15/ 70 Σ 1 -sequents Definition. A Σ 1 -sequent is a sequent of the form x 1 ϕ 1,..., x k ϕ k x k+1 ϕ k+1,..., x n ϕ n s.t. ϕ 1,..., ϕ n quantifier-free, ϕ i contains only variables from x i. Lemma. Let π be a PA-proof of a Σ 1 -sequent Γ and let ϕ(α), Γ, ϕ(s(α)) ϕ(0), Γ, ϕ(t) ind be a lowermost induction in the end-piece of π. Then t is variable-free. Proof. W.l.o.g. all variables in π are eigenvariables. End-piece does not contain r, l. End-piece contains only eigenvariables of inductions. The term of a lowermost induction is variable-free.

28 16/ 70 Termination Let π be a proof of a Σ 1 -sequent Γ. If end-piece of π contains ind: = reduce lowermost ind If end-piece of π contains non-atomic cut: = reduce suitable non-atomic cut Otherwise: π contains only atomic cuts (Then Γ = = implies that π is simple)

29 16/ 70 Termination Let π be a proof of a Σ 1 -sequent Γ. If end-piece of π contains ind: = reduce lowermost ind If end-piece of π contains non-atomic cut: = reduce suitable non-atomic cut Otherwise: π contains only atomic cuts (Then Γ = = implies that π is simple) Have: π 1 π 2 π 3 Do we ever enter the Otherwise -case?

30 16/ 70 Termination Let π be a proof of a Σ 1 -sequent Γ. If end-piece of π contains ind: = reduce lowermost ind If end-piece of π contains non-atomic cut: = reduce suitable non-atomic cut Otherwise: π contains only atomic cuts (Then Γ = = implies that π is simple) Have: π 1 π 2 π 3 Do we ever enter the Otherwise -case? Want: well-founded (X, <) and mapping o s.t. o(π 1 ) > o(π 2 ) > o(π 3 ) >

31 17/ 70 Outline Gentzen s consistency proof Background Peano arithmetic Reduction of cut and induction Ordinals The consistency proof The omega rule Cyclic proofs

32 18/ 70 Ordinals, informally The order of the natural numbers 0 1 2

33 18/ 70 Ordinals, informally Add limit element ω, i.e., n N : n < ω ω

34 18/ 70 Ordinals, informally Add another successor element after that ω ω+1

35 18/ 70 Ordinals, informally and so on ω ω+1 ω+2

36 18/ 70 Ordinals, informally Add a new limit element again ω ω+1 ω+2 ω 2

37 18/ 70 Ordinals, informally The ordinals ω ω ω+1 ω+2 ω 2

38 18/ 70 Ordinals, informally The ordinals ω ω ω+1 ω+2 ω 2 Repeat the above 0 ω ω 2 ω 3

39 18/ 70 Ordinals, informally The ordinals ω ω ω+1 ω+2 ω 2 And add a new limit element again 0 ω ω 2 ω 3 ω ω=ω 2

40 18/ 70 Ordinals, informally The ordinals ω ω ω+1 ω+2 ω 2 The ordinals ω 2 0 ω ω 2 ω 3 ω ω=ω 2

41 18/ 70 Ordinals, informally The ordinals ω ω ω+1 ω+2 ω 2 The ordinals ω 2 0 ω ω 2 ω 3 ω ω=ω 2 Repeat the repetition 0 ω ω 2 ω 3

42 18/ 70 Ordinals, informally The ordinals ω ω ω+1 ω+2 ω 2 The ordinals ω 2 0 ω ω 2 ω 3 ω ω=ω 2 And add a new limit element again 0 ω ω 2 ω 3 ω ω

43 18/ 70 Ordinals, informally The ordinals ω ω ω+1 ω+2 ω 2 The ordinals ω 2 0 ω ω 2 ω 3 ω ω=ω 2 The ordinals ω ω 0 ω ω 2 ω 3 ω ω

44 18/ 70 Ordinals, informally The ordinals ω ω ω+1 ω+2 ω 2 The ordinals ω 2 0 ω ω 2 ω 3 ω ω=ω 2 The ordinals ω ω 0 ω ω 2 ω 3 ω ω Iterate one more time 0 ω ω ω ω ω ω

45 18/ 70 Ordinals, informally The ordinals ω ω ω+1 ω+2 ω 2 The ordinals ω 2 0 ω ω 2 ω 3 ω ω=ω 2 The ordinals ω ω 0 ω ω 2 ω 3 ω ω And add a new limit element 0 ω ω ω ω ω ω ε0

46 18/ 70 Ordinals, informally The ordinals ω ω ω+1 ω+2 ω 2 The ordinals ω 2 0 ω ω 2 ω 3 ω ω=ω 2 The ordinals ω ω 0 ω ω 2 ω 3 ω ω The ordinals ε 0 0 ω ω ω ω ω ω ε0

47 19/ 70 The ordinals < ε 0 Ordinal arithmetic addition +, multiplication, exponentiation less-than <

48 19/ 70 The ordinals < ε 0 Ordinal arithmetic addition +, multiplication, exponentiation less-than < Cantor normal form: if 0 < α < ε 0, then α can be written as α = ω α ω αn where α 1 α n, 0 < α i < ε 0, α i in normal form. This normal form is unique.

49 19/ 70 The ordinals < ε 0 Ordinal arithmetic addition +, multiplication, exponentiation less-than < Cantor normal form: if 0 < α < ε 0, then α can be written as α = ω α ω αn where α 1 α n, 0 < α i < ε 0, α i in normal form. This normal form is unique. The natural sum: for α = ω α ω αn, β = ω β ω βm let α # β = ω λ1 ω λm+n λ 1 λ m+n, {λ 1,..., λ m+n } = {α 1,..., α n, β 1,..., β m }.

50 The ordinals < ε 0 Ordinal arithmetic addition +, multiplication, exponentiation less-than < Cantor normal form: if 0 < α < ε 0, then α can be written as α = ω α ω αn where α 1 α n, 0 < α i < ε 0, α i in normal form. This normal form is unique. The natural sum: for α = ω α ω αn, β = ω β ω βm let α # β = ω λ1 ω λm+n λ 1 λ m+n, {λ 1,..., λ m+n } = {α 1,..., α n, β 1,..., β m }. Formalisation Term signature O = {0/0, ω/1, +/2} Modulo equality Operations +,, exp, # and relation < 19/ 70

51 20/ 70 Outline Gentzen s consistency proof Background Peano arithmetic Reduction of cut and induction Ordinals The consistency proof The omega rule Cyclic proofs

52 21/ 70 The height of a sequent Definition. Logical complexity of a formula, cut, induction.

53 21/ 70 The height of a sequent Definition. Logical complexity of a formula, cut, induction. Definition. Height of a sequent S in a proof π, h(s, π), is maximum of log. complexities of cut or induction below S in π.

54 21/ 70 The height of a sequent Definition. Logical complexity of a formula, cut, induction. Definition. Height of a sequent S in a proof π, h(s, π), is maximum of log. complexities of cut or induction below S in π. Observation. If S 1 and S 2 are premises of a binary inference, then h(s 1, π) = h(s 2, π).

55 21/ 70 The height of a sequent Definition. Logical complexity of a formula, cut, induction. Definition. Height of a sequent S in a proof π, h(s, π), is maximum of log. complexities of cut or induction below S in π. Observation. If S 1 and S 2 are premises of a binary inference, then h(s 1, π) = h(s 2, π). Notation. h(s) for h(s, π) if π is clear

56 22/ 70 The ordinal assignment Definition. Let S be a sequent in a proof π. Define o(s): Initial sequent S: o(s) = 1 Structural inference S S : o(s) = o(s ) Unary logical inference S Binary logical inference S 1 S 2 S S 1 S 2 S S : o(s) = o(s ) + 1 : o(s) = o(s 1 ) # o(s 2 ) cut : o(s) = ω h(si ) h(s)(o(s 1 ) # o(s 2 )) S S ind : o(s) = ω h(s ) h(s)+1(α 1 + 1) where o(s ) = ω α1 + + ω αn with α 1 α n.

57 22/ 70 The ordinal assignment Definition. Let S be a sequent in a proof π. Define o(s): Initial sequent S: o(s) = 1 Structural inference S S : o(s) = o(s ) Unary logical inference S Binary logical inference S 1 S 2 S S 1 S 2 S S : o(s) = o(s ) + 1 : o(s) = o(s 1 ) # o(s 2 ) cut : o(s) = ω h(si ) h(s)(o(s 1 ) # o(s 2 )) S S ind : o(s) = ω h(s ) h(s)+1(α 1 + 1) where o(s ) = ω α1 + + ω αn with α 1 α n. Definition. Let π be a proof of S, then o(π) = o(s, π).

58 23/ 70 The consistency proof Reduction Lemma. Let Γ be a Σ 1 -sequent, let π be a proof of Γ that contains a non-atomic cut or an induction. Then there is a proof π of Γ with o(π ) < o(π).

59 23/ 70 The consistency proof Reduction Lemma. Let Γ be a Σ 1 -sequent, let π be a proof of Γ that contains a non-atomic cut or an induction. Then there is a proof π of Γ with o(π ) < o(π). Theorem. PA is consistent. Proof. Suppose PA is inconsistent, then there is a proof π of. is a Σ 1 -sequent o(π) < ε0 Induction on o(π) (reduction lemma): obtain π of s.t. π does not contain induction nor non-atomic cut π is a simple proof of Contradiction.

60 23/ 70 The consistency proof Reduction Lemma. Let Γ be a Σ 1 -sequent, let π be a proof of Γ that contains a non-atomic cut or an induction. Then there is a proof π of Γ with o(π ) < o(π). Theorem. PA is consistent. Proof. Suppose PA is inconsistent, then there is a proof π of. is a Σ 1 -sequent o(π) < ε0 Induction on o(π) (reduction lemma): obtain π of s.t. π does not contain induction nor non-atomic cut π is a simple proof of Contradiction. Remark. Formalisation in PRA: PRA + TI(ϕ(x), <ε0 ) Con PA for quantifier-free ϕ(x) In particular: PRA proves Reduction Lemma, Simple Proof Lemma

61 24/ 70 More general end-sequents (1/2) Let Exp(x, y) be a representation of n 2 n, in particular: PA Exp(0, 1) PA Exp(x, y) Exp(s(x), y 2) Then PA x y Exp(x, y):. Exp(0, 1) y Exp(0, y) r. Exp(β, γ) Exp(s(β), γ 2) Exp(β, γ) y Exp(s(β), y) r y Exp(β, y) y Exp(s(β), y) l ind y Exp(0, y) y Exp(α, y) cut y Exp(α, y) x y Exp(x, y) r

62 24/ 70 More general end-sequents (1/2) Let Exp(x, y) be a representation of n 2 n, in particular: PA Exp(0, 1) PA Exp(x, y) Exp(s(x), y 2) Then PA x y Exp(x, y):. Exp(0, 1) y Exp(0, y) r. Exp(β, γ) Exp(s(β), γ 2) Exp(β, γ) y Exp(s(β), y) r y Exp(β, y) y Exp(s(β), y) l ind y Exp(0, y) y Exp(α, y) cut y Exp(α, y) x y Exp(x, y) r But cut-elimination procedure does not work (Eigenvariable α in end-piece not introduced by induction)

63 25/ 70 More general end-sequents (2/2) Proposition. Every PA-proof of x y Exp(x, y) contains an induction inference. Proof. Let π be a PA-proof of x y Exp(x, y). Then w.l.o.g. π ends with r. Suppose there is induction-free proof of y Exp(α, y), then by cut-elimination / Herbrand s theorem, there are t 1,..., t n L {α} s.t. Contradiction. BA Exp(α, t 1 ) Exp(α, t n ), i.e., BA α ( Exp(α, t 1 ) Exp(α, t n ) ).

64 26/ 70 Summary of 1st part Gentzen s consistency proof of PA Sequent calculus with induction rule Cut- and induction-elimination Except termination formalisable in PRA Ordinals (up to ε 0 in Cantor normal form) Proof of Σ 1 -sequent: lowermost induction has variable-free term. Impossible for Π 2 -sequents

65 27/ 70 Outline Gentzen s consistency proof The omega rule Infinitary propositional logic Infinitary proofs The consistency proof Cyclic proofs

66 28/ 70 Infinitary propositional logic Observation: N = x ϕ(x) iff N = n N ϕ(n) N = x ϕ(x) iff N = n N ϕ(n)

67 28/ 70 Infinitary propositional logic Observation: N = x ϕ(x) iff N = n N ϕ(n) N = x ϕ(x) iff N = n N ϕ(n) The ω-rule Γ, ϕ(0) Γ, ϕ(1) Γ, ϕ(2) Γ, n N ϕ(n)

68 28/ 70 Infinitary propositional logic Observation: N = x ϕ(x) iff N = n N ϕ(n) N = x ϕ(x) iff N = n N ϕ(n) The ω-rule Γ, ϕ(0) Γ, ϕ(1) Γ, ϕ(2) Γ, n N ϕ(n) Definition. The formulas of infinitary propositional logic: Variable-free atoms (in L = {0, s, +,, =}) Negated variable-free atoms (in L) If, for i I, ϕi is a formula, then i I ϕ i is a formula If, for i I, ϕ i is a formula, then i I ϕ i is a formula Here: I countable

69 29/ 70 Translation of first-order L-sentences Definition. Translation of first-order L-sentences: A = A for an atom A [ x ϕ(x)] = ϕ (n) [ x ϕ(x)] = ϕ (n) n N n N [ϕ ψ] = ϕ ψ [ϕ ψ] = ϕ ψ [ϕ ψ] = ϕ ψ [ ϕ] = ϕ

70 29/ 70 Translation of first-order L-sentences Definition. Translation of first-order L-sentences: A = A for an atom A [ x ϕ(x)] = ϕ (n) [ x ϕ(x)] = ϕ (n) n N n N [ϕ ψ] = ϕ ψ [ϕ ψ] = ϕ ψ [ϕ ψ] = ϕ ψ [ ϕ] = ϕ Where dualisation is defined as: A = A A = A for an atom A for a formula ϕ ϕ i ϕ i = ϕ i = i I i I i I i I ϕ i

71 30/ 70 Example: induction axiom The translation of the induction axiom [I x ϕ] = [ ϕ(0) y (ϕ(y) ϕ(s(y))) x ϕ(x) ]

72 30/ 70 Example: induction axiom The translation of the induction axiom [I x ϕ] = [ ϕ(0) y (ϕ(y) ϕ(s(y))) x ϕ(x) ] = [ ϕ(0) y (ϕ(y) ϕ(s(y))) ] ϕ (n) n N

73 30/ 70 Example: induction axiom The translation of the induction axiom [I x ϕ] = [ ϕ(0) y (ϕ(y) ϕ(s(y))) x ϕ(x) ] = [ ϕ(0) y (ϕ(y) ϕ(s(y))) ] ϕ (n) n N = ϕ (0) ϕ (n) ϕ (n + 1) ϕ (n) n N n N

74 30/ 70 Example: induction axiom The translation of the induction axiom [I x ϕ] = [ ϕ(0) y (ϕ(y) ϕ(s(y))) x ϕ(x) ] = [ ϕ(0) y (ϕ(y) ϕ(s(y))) ] ϕ (n) n N = ϕ (0) ϕ (n) ϕ (n + 1) ϕ (n) n N n N = ϕ (0) ϕ (n) ϕ (n + 1) ϕ (n) n N n N

75 31/ 70 Outline Gentzen s consistency proof The omega rule Infinitary propositional logic Infinitary proofs The consistency proof Cyclic proofs

76 32/ 70 Sequent calculus LK Definition. The set of axioms is the smallest set which contains A, A for any atom A, S if S is an axiom of BA (as a sequent). and is closed under atomic cut, i.e., If Γ, A and, A are axioms, then so is a subset of Γ,.

77 32/ 70 Sequent calculus LK Definition. The set of axioms is the smallest set which contains A, A for any atom A, S if S is an axiom of BA (as a sequent). and is closed under atomic cut, i.e., If Γ, A and, A are axioms, then so is a subset of Γ,. Definition. LK works on sequents of inf. prop. logic. Γ, if is an axiom Γ, ϕ j for a j I Γ, ϕ j for all j I Γ, i I ϕ i Γ, ϕ Γ, ϕ Γ cut Γ, i I ϕ i Proofs of infinite width but finite height

78 33/ 70 Translation of BA-proofs Def. Translation of BA-proof of Γ to LK -proof of Γ, Axioms r (π 1 ) (π 2 ) Γ, ϕ Π Λ, ψ Γ, Π, Λ, ϕ ψ r (π 1 ) Γ,, ϕ (π 2 ) Π, Λ, ψ Γ, Π,, Λ, ϕ ψ r (π(α)) Γ, ϕ(α) Γ, x ϕ(x) r (π(n) ) Γ,, ϕ (n) for all n N Γ,, n N ϕ (n) Similarily for other rules

79 33/ 70 Translation of BA-proofs Def. Translation of BA-proof of Γ to LK -proof of Γ, Axioms r (π 1 ) (π 2 ) Γ, ϕ Π Λ, ψ Γ, Π, Λ, ϕ ψ r (π 1 ) Γ,, ϕ (π 2 ) Π, Λ, ψ Γ, Π,, Λ, ϕ ψ r (π(α)) Γ, ϕ(α) Γ, x ϕ(x) r (π(n) ) Γ,, ϕ (n) for all n N Γ,, n N ϕ (n) Similarily for other rules Formalisation: primitive recursive function

80 34/ 70 LK proves induction Informally: Assume ϕ(0) and ϕ(n) ϕ(n + 1), then ϕ(0) ϕ(0) ϕ(0) ϕ(1) ϕ(1) ϕ(0) ϕ(0) ϕ(1) ϕ(1) n N ϕ(n) ϕ(1) ϕ(2) ϕ(2)

81 34/ 70 LK proves induction Informally: Assume ϕ(0) and ϕ(n) ϕ(n + 1), then ϕ(0) ϕ(0) ϕ(0) ϕ(1) ϕ(1) ϕ(0) ϕ(0) ϕ(1) ϕ(1) n N ϕ(n) ϕ(1) ϕ(2) ϕ(2) Hence LK [I x ϕ(x)]

82 34/ 70 LK proves induction Informally: Assume ϕ(0) and ϕ(n) ϕ(n + 1), then ϕ(0) ϕ(0) ϕ(0) ϕ(1) ϕ(1) ϕ(0) ϕ(0) ϕ(1) ϕ(1) n N ϕ(n) ϕ(1) ϕ(2) ϕ(2) Hence LK [I x ϕ(x)] One application of the ω-rule Formalisation: prim. rec. function mapping Ix ϕ(x) to LK -proof

83 35/ 70 Completeness of LK Obs. For a first-order L-sentence σ: if N = σ, then LK σ.

84 35/ 70 Completeness of LK Obs. For a first-order L-sentence σ: if N = σ, then LK σ. Proof. Induction on the logical complexity of σ: If σ is atom, then BA σ with atomic cuts, so {σ} is axiom.

85 35/ 70 Completeness of LK Obs. For a first-order L-sentence σ: if N = σ, then LK σ. Proof. Induction on the logical complexity of σ: If σ is atom, then BA σ with atomic cuts, so {σ} is axiom. If σ = x ϕ(x), then N = ϕ(n) for all n N, so by IH. ϕ (0) ϕ (1) ϕ (2) n N ϕ (n)

86 35/ 70 Completeness of LK Obs. For a first-order L-sentence σ: if N = σ, then LK σ. Proof. Induction on the logical complexity of σ: If σ is atom, then BA σ with atomic cuts, so {σ} is axiom. If σ = x ϕ(x), then N = ϕ(n) for all n N, so ϕ (0) ϕ (1) ϕ (2) n N ϕ (n) by IH. If σ = x ϕ(x), then there is a k N s.t. N = ϕ(k), so ϕ (k) n N ϕ (n) by IH....

87 36/ 70 Outline Gentzen s consistency proof The omega rule Infinitary propositional logic Infinitary proofs The consistency proof Cyclic proofs

88 37/ 70 The height of a proof Proof height in a finite system, example: π = (π 1 ) (π 2 ) Γ, A Π Λ, B Γ, Π, Λ, A B r Then h(π) = 1 + max{h(π 1 ), h(π 2 )} = sup{h(π i ) + 1 i {1, 2}}.

89 37/ 70 The height of a proof Proof height in a finite system, example: π = (π 1 ) (π 2 ) Γ, A Π Λ, B Γ, Π, Λ, A B r Then h(π) = 1 + max{h(π 1 ), h(π 2 )} = sup{h(π i ) + 1 i {1, 2}}. Definition. For LK -proof π with direct subproofs π i, i I define h(π) = sup{h(π i ) + 1 i I }

90 38/ 70 The cut rank of a proof Definition. For a formula ϕ with direct subformulas ϕ i, i I define the depth of ϕ as d(ϕ) = sup{d(ϕ i ) + 1 i I }. Observation. The depth of σ is the depth of σ.

91 38/ 70 The cut rank of a proof Definition. For a formula ϕ with direct subformulas ϕ i, i I define the depth of ϕ as d(ϕ) = sup{d(ϕ i ) + 1 i I }. Observation. The depth of σ is the depth of σ. Definition. For LK -proof π with direct subproofs π i, i I and last inference ι define the cut rank of π: If ι is cut on some ϕ, then ρ(π) = max{d(ϕ), ρ(π1 ), ρ(π 2 )}. Otherwise, ρ(π) = sup{ρ(πi ) i I }.

92 38/ 70 The cut rank of a proof Definition. For a formula ϕ with direct subformulas ϕ i, i I define the depth of ϕ as d(ϕ) = sup{d(ϕ i ) + 1 i I }. Observation. The depth of σ is the depth of σ. Definition. For LK -proof π with direct subproofs π i, i I and last inference ι define the cut rank of π: If ι is cut on some ϕ, then ρ(π) = max{d(ϕ), ρ(π1 ), ρ(π 2 )}. Otherwise, ρ(π) = sup{ρ(πi ) i I }. Notation. LK ρ α S if there is an LK -proof of S with cut rank ρ and height α.

93 39/ 70 The proof translation revisited Lemma. If PA σ, then there is an r < ω s.t. LK r ω 2 σ. Proof. If PA σ, then there is a BA-proof π of I 1,..., I n σ.

94 39/ 70 The proof translation revisited Lemma. If PA σ, then there is an r < ω s.t. LK r ω 2 σ. Proof. If PA σ, then there is a BA-proof π of I 1,..., I n σ. ψ = (π n ) I n (π 1 ) (π ) I1 I1,..., I n, σ I2,..., I n, σ cut σ h(π ) < ω, h(π i ) = ω for 1 i n I n., σ cut

95 39/ 70 The proof translation revisited Lemma. If PA σ, then there is an r < ω s.t. LK r ω 2 σ. Proof. If PA σ, then there is a BA-proof π of I 1,..., I n σ. ψ = (π n ) I n (π 1 ) (π ) I1 I1,..., I n, σ I2,..., I n, σ cut σ h(π ) < ω, h(π i ) = ω for 1 i n h(ψ) = w + n < w 2. I n., σ cut

96 40/ 70 Cut-elimination in LK Theorem. If r < ω and LK r α S, then LK 0 2 α r S.

97 40/ 70 Cut-elimination in LK Theorem. If r < ω and LK r α S, then LK 0 2 α S. r Proof Sketch. Standard cut-elimination argument.

98 40/ 70 Cut-elimination in LK Theorem. If r < ω and LK r α S, then LK 0 2 α S. r Proof Sketch. Standard cut-elimination argument. Corollary. If PA σ, then LK 0 α σ for some α < ε 0.

99 40/ 70 Cut-elimination in LK Theorem. If r < ω and LK r α S, then LK 0 2 α S. r Proof Sketch. Standard cut-elimination argument. Corollary. If PA σ, then LK 0 α σ for some α < ε 0. Proof. If PA σ, then there is r < ω s.t. LK r ω 2 σ. By the cut-elimination theorem, LK 0 2 σ and 2 ω 2 ω 2 r < ω r+2 < ε 0. r

100 40/ 70 Cut-elimination in LK Theorem. If r < ω and LK r α S, then LK 0 2 α S. r Proof Sketch. Standard cut-elimination argument. Corollary. If PA σ, then LK 0 α σ for some α < ε 0. Proof. If PA σ, then there is r < ω s.t. LK r ω 2 σ. By the cut-elimination theorem, LK 0 2 σ and 2 ω 2 ω 2 r < ω r+2 < ε 0. r Corollary. PA is consistent.

101 40/ 70 Cut-elimination in LK Theorem. If r < ω and LK r α S, then LK 0 2 α S. r Proof Sketch. Standard cut-elimination argument. Corollary. If PA σ, then LK 0 α σ for some α < ε 0. Proof. If PA σ, then there is r < ω s.t. LK r ω 2 σ. By the cut-elimination theorem, LK 0 2 σ and 2 ω 2 ω 2 r < ω r+2 < ε 0. r Corollary. PA is consistent. Proof. Suppose PA, then LK 0 α for some α < ε 0, but there is not cut-free proof of in LK.

102 41/ 70 Summary of 2nd part The ω-rule The calculus LK Proofs of infinite width, no infinite branches Depth measured by ordinals Cut-elimination Consistency proof

103 42/ 70 Outline Gentzen s consistency proof The omega rule Cyclic proofs Infinite proofs Cyclic proofs Proof by induction Cyclic proofs vs. proofs by induction

104 43/ 70 Infinite descent Fermat: descente infinie There is no infinite descending chain a 1 > a 2 > with a i N Example. 2 is irrational. Proof. Suppose there are p, q 1 s.t. 2 = p q, i.e., p2 = 2q 2. Then: 2 p 2,

105 43/ 70 Infinite descent Fermat: descente infinie There is no infinite descending chain a 1 > a 2 > with a i N Example. 2 is irrational. Proof. Suppose there are p, q 1 s.t. 2 = p q, i.e., p2 = 2q 2. Then: 2 p 2, 4 p 2,

106 43/ 70 Infinite descent Fermat: descente infinie There is no infinite descending chain a 1 > a 2 > with a i N Example. 2 is irrational. Proof. Suppose there are p, q 1 s.t. 2 = p q, i.e., p2 = 2q 2. Then: 2 p 2, 4 p 2, 4 2q 2,

107 43/ 70 Infinite descent Fermat: descente infinie There is no infinite descending chain a 1 > a 2 > with a i N Example. 2 is irrational. Proof. Suppose there are p, q 1 s.t. 2 = p q, i.e., p2 = 2q 2. Then: 2 p 2, 4 p 2, 4 2q 2, 2 q 2,

108 43/ 70 Infinite descent Fermat: descente infinie There is no infinite descending chain a 1 > a 2 > with a i N Example. 2 is irrational. Proof. Suppose there are p, q 1 s.t. 2 = p q, i.e., p2 = 2q 2. Then: 2 p 2, 4 p 2, 4 2q 2, 2 q 2, 4 q 2

109 43/ 70 Infinite descent Fermat: descente infinie There is no infinite descending chain a 1 > a 2 > with a i N Example. 2 is irrational. Proof. Suppose there are p, q 1 s.t. 2 = p q, i.e., p2 = 2q 2. Then: 2 p 2, 4 p 2, 4 2q 2, 2 q 2, 4 q 2 So 2 p, 2 q, let p = p 2, q = q 2.

110 43/ 70 Infinite descent Fermat: descente infinie There is no infinite descending chain a 1 > a 2 > with a i N Example. 2 is irrational. Proof. Suppose there are p, q 1 s.t. 2 = p q, i.e., p2 = 2q 2. Then: 2 p 2, 4 p 2, 4 2q 2, 2 q 2, 4 q 2 So 2 p, 2 q, let p = p 2, q = q 2. Then p 2 = p2 4 = 2 q2 4 = 2q 2.

111 43/ 70 Infinite descent Fermat: descente infinie There is no infinite descending chain a 1 > a 2 > with a i N Example. 2 is irrational. Proof. Suppose there are p, q 1 s.t. 2 = p q, i.e., p2 = 2q 2. Then: 2 p 2, 4 p 2, 4 2q 2, 2 q 2, 4 q 2 So 2 p, 2 q, let p = p 2, q = q 2. Then p 2 = p2 4 = 2 q2 4 = 2q 2. So there is an infinitely desceding sequence p > p >... Contradiction.

112 43/ 70 Infinite descent Fermat: descente infinie There is no infinite descending chain a 1 > a 2 > with a i N Example. 2 is irrational. Proof. Suppose there are p, q 1 s.t. 2 = p q, i.e., p2 = 2q 2. Then: 2 p 2, 4 p 2, 4 2q 2, 2 q 2, 4 q 2 So 2 p, 2 q, let p = p 2, q = q 2. Then p 2 = p2 4 = 2 q2 4 = 2q 2. So there is an infinitely desceding sequence p > p >... Contradiction. 3rd part: formalisation of cyclic proofs [Brotherston, Simpson 11]

113 Inductive definitions First-order signature Σ Inductive predicate symbols Σ I Σ For each P Σ I a finite set of productions of the form Q 1 (u 1 ) Q k (u k ) P(t) with Q 1,..., Q k Σ and t, u 1,..., u k term vectors. Example. N(0) N(x) N(s(x)) L(nil) N(x) L(z) L(cons(x, z)) E(0) E(x) O(s(x)) O(x) E(s(x)) 44/ 70

114 45/ 70 Case split rules For inductive predicate P, case split rule is: where each production cases P(u), Γ case P Q 1 (u 1 [x]) Q k (u k [x]) P(t[x]) of P gives rise to a case (premise) Γ, u = t[x], Q 1 (u 1 [x]),..., Q k (u k [x]) where x is fresh in each case.

115 46/ 70 Case split rules: examples t = 0, Γ t = s(x), N(x), Γ N(t), Γ case N

116 46/ 70 Case split rules: examples t = 0, Γ t = s(x), N(x), Γ N(t), Γ t = 0, Γ t = s(x), O(x), Γ E(t), Γ case N case E

117 46/ 70 Case split rules: examples t = 0, Γ t = s(x), N(x), Γ N(t), Γ t = 0, Γ t = s(x), O(x), Γ E(t), Γ case N case E t = s(x), E(x), Γ O(t), Γ case O

118 46/ 70 Case split rules: examples t = 0, Γ t = s(x), N(x), Γ N(t), Γ t = 0, Γ t = s(x), O(x), Γ E(t), Γ case N case E t = s(x), E(x), Γ O(t), Γ case O t = nil, Γ t = cons(x, z), N(x), L(z), Γ L(t), Γ case L

119 47/ 70 Right-introduction rules Definition. Let P be inductive predicate and a production for P. Then Q 1 (u 1 [x]) Q k (u k [x]) P(t[x]) Γ, Q 1 (u 1 [v]) Γ, Q k (u k [v]) Γ, P(t[v]) P r is a right-introduction rule for P, where v is a vector of terms.

120 48/ 70 Right-introduction rules: examples Γ, N(0) N r Γ, N(t) Γ, N(s(t)) N r

121 48/ 70 Right-introduction rules: examples Γ, N(0) N r Γ, N(t) Γ, N(s(t)) N r Γ, E(0) E r Γ, O(t) Γ, E(s(t)) E r Γ, E(t) Γ, O(s(t)) O r

122 48/ 70 Right-introduction rules: examples Γ, N(0) N r Γ, N(t) Γ, N(s(t)) N r Γ, E(0) E r Γ, O(t) Γ, E(s(t)) E r Γ, E(t) Γ, O(s(t)) O r Γ, L(nil) L r Γ, N(t) Γ, L(u) Γ, L(cons(t, u)) L r

123 49/ 70 LKID ω pre-proofs Definition. Set D of inductive definitions, rules of D-LKID ω are: usual LK for FOL with equality the substitution rules Γ Γσ σ subst The case split rules for ind. predicates of D The right-introduction rules for ind. predicates of D Notation. LKID ω instead of D-LKID ω.

124 49/ 70 LKID ω pre-proofs Definition. Set D of inductive definitions, rules of D-LKID ω are: usual LK for FOL with equality the substitution rules Γ Γσ σ subst The case split rules for ind. predicates of D The right-introduction rules for ind. predicates of D Notation. LKID ω instead of D-LKID ω. Definition. An LKID ω pre-proof is a (possibly infinite) tree built from these rules.

125 49/ 70 LKID ω pre-proofs Definition. Set D of inductive definitions, rules of D-LKID ω are: usual LK for FOL with equality the substitution rules Γ Γσ σ subst The case split rules for ind. predicates of D The right-introduction rules for ind. predicates of D Notation. LKID ω instead of D-LKID ω. Definition. An LKID ω pre-proof is a (possibly infinite) tree built from these rules. Remark. LKID ω pre-proofs are not sound.

126 50/ 70 LKID ω proofs Definition. Path (Γ i i ) 1 i<α of sequents (for some α ω)

127 50/ 70 LKID ω proofs Definition. Path (Γ i i ) 1 i<α of sequents (for some α ω) Definition. (τ i ) 1 i<α is a trace in (Γ i i ) 1 i<α if every τi is a Pt in Γ i for an inductive predicate P, and τ i is successor of τ i+1.

128 50/ 70 LKID ω proofs Definition. Path (Γ i i ) 1 i<α of sequents (for some α ω) Definition. (τ i ) 1 i<α is a trace in (Γ i i ) 1 i<α if every τi is a Pt in Γ i for an inductive predicate P, and τ i is successor of τ i+1. Definition. i is a progress point in (τ i ) 1 i<α if τ i+1 obtained from τ i by case split.

129 50/ 70 LKID ω proofs Definition. Path (Γ i i ) 1 i<α of sequents (for some α ω) Definition. (τ i ) 1 i<α is a trace in (Γ i i ) 1 i<α if every τi is a Pt in Γ i for an inductive predicate P, and τ i is successor of τ i+1. Definition. i is a progress point in (τ i ) 1 i<α if τ i+1 obtained from τ i by case split. Definition. An LKID ω proof is an LKID ω pre-proof that satisfies the global trace condition: every infinite path contains a trace with infinitely many progress points.

130 51/ 70 LKID ω : example E(0) E r x 0 = 0 E(0), O(x 0 ) w x 0 = 0 E(x 0 ), O(x 0 ) =. N(x 1 ) E(x 1 ), O(x 1 ) case N N(x 1 ) O(x 1 ), O(s(x 1 )) O r N(x 1 ) E(s(x 1 )), O(s(x 1 )) E r x 0 = s(x 1 ), N(x 1 ) E(s(x 1 )), O(s(x 1 )) w l = x 0 = s(x 1 ), N(x 1 ) E(x 0 ), O(x 0 ) case N N(x 0 ) E(x 0 ), O(x 0 )

131 52/ 70 Properties of LKID ω Definition. Standard model... inductive predicates are interpreted as least fixed points (of semantics of their productions)

132 52/ 70 Properties of LKID ω Definition. Standard model... inductive predicates are interpreted as least fixed points (of semantics of their productions) Theorem. LKID ω is sound w.r.t. standard models.

133 52/ 70 Properties of LKID ω Definition. Standard model... inductive predicates are interpreted as least fixed points (of semantics of their productions) Theorem. LKID ω is sound w.r.t. standard models. Theorem. Cut-free LKID ω is complete w.r.t. standard models. Proof Sketch. Construct proof search tree as LKID ω proof If S not valid, search tree of S has infinite branch without progress

134 53/ 70 Outline Gentzen s consistency proof The omega rule Cyclic proofs Infinite proofs Cyclic proofs Proof by induction Cyclic proofs vs. proofs by induction

135 54/ 70 CLKID ω Consider derivation trees built from LKID ω -rules. (derivation tree: branch may end with non-axiom) Definition. A bud in D is a leaf which is not an axiom.

136 54/ 70 CLKID ω Consider derivation trees built from LKID ω -rules. (derivation tree: branch may end with non-axiom) Definition. A bud in D is a leaf which is not an axiom. Definition. For a bud S in D an internal node S in D is called companion of S if S = S.

137 54/ 70 CLKID ω Consider derivation trees built from LKID ω -rules. (derivation tree: branch may end with non-axiom) Definition. A bud in D is a leaf which is not an axiom. Definition. For a bud S in D an internal node S in D is called companion of S if S = S. Definition. A CLKID ω pre-proof is a pair (D, γ) where D is a finite derivation tree, and γ assigns a companion to each bud.

138 54/ 70 CLKID ω Consider derivation trees built from LKID ω -rules. (derivation tree: branch may end with non-axiom) Definition. A bud in D is a leaf which is not an axiom. Definition. For a bud S in D an internal node S in D is called companion of S if S = S. Definition. A CLKID ω pre-proof is a pair (D, γ) where D is a finite derivation tree, and γ assigns a companion to each bud. CLKID ω pre-proof unfolds to LKID ω -proof by identifying each bud with its companion. Definition. A CLKID ω proof is a CLKID ω pre-proof whose unfolding satisfies the global trace condition.

139 55/ 70 CLKID ω : Example O(x) N(x) O(x ) N(x ) subst N(0) x = 0 N(0) w x = s(x ), O(x ) N(x ) w l l x = 0 N(x) = x = s(x ), O(x ) N(s(x )) Nr = x = s(x ), O(x ) N(x) case E E(x) N(x) E(x) O(x) N(x) E(x) N(x) E(x ) N(x ) subst x = s(x ), E(x ) N(x ) w l x = s(x ), E(x ) N(s(x )) Nr = x = s(x ), E(x ) N(x) case O O(x) N(x) l

140 56/ 70 Decidability Theorem. The following problem is decidable: given CLKID ω pre-proof (D, γ), is (D, γ) a CLKID ω proof?

141 56/ 70 Decidability Theorem. The following problem is decidable: given CLKID ω pre-proof (D, γ), is (D, γ) a CLKID ω proof? Definition. Büchi automaton is NFA accepting an infinite word w Σ ω if w has a path visiting an accepting state infinitely often.

142 56/ 70 Decidability Theorem. The following problem is decidable: given CLKID ω pre-proof (D, γ), is (D, γ) a CLKID ω proof? Definition. Büchi automaton is NFA accepting an infinite word w Σ ω if w has a path visiting an accepting state infinitely often. Proof Sketch. P = (D, γ) induces Büchi automata Ball s.t. L(B all ) is set of all infinite paths, and Bacc s.t. L(B acc ) is set of infinite paths satisfying prog. cond. L(B all ) L(B acc ) is decidable.

143 57/ 70 Outline Gentzen s consistency proof The omega rule Cyclic proofs Infinite proofs Cyclic proofs Proof by induction Cyclic proofs vs. proofs by induction

144 58/ 70 Induction rules Definition. Inductive predicate P, induction rule for P: minor premises Γ, ϕ P (u) Γ, P(u) ind P where minor premises from productions of inductive predicates mutually dependent with P one induction formula ϕq for each such predicate Q

145 59/ 70 Induction rules: examples Γ, ϕ N (0) ϕ N (x), Γ, ϕ N (s(x)) ϕ N (t), Γ N(t), Γ ind N

146 59/ 70 Induction rules: examples Γ, ϕ N (0) ϕ N (x), Γ, ϕ N (s(x)) ϕ N (t), Γ N(t), Γ ind N Γ, ϕ E (0) ϕ E (x), Γ, ϕ O (s(x)) ϕ O (x), Γ, ϕ E (s(x)) ϕ E (t), Γ E(t), Γ ind E

147 59/ 70 Induction rules: examples Γ, ϕ N (0) ϕ N (x), Γ, ϕ N (s(x)) ϕ N (t), Γ N(t), Γ ind N Γ, ϕ E (0) ϕ E (x), Γ, ϕ O (s(x)) ϕ O (x), Γ, ϕ E (s(x)) ϕ E (t), Γ E(t), Γ ind E Γ, ϕ L (nil) N(x), ϕ L (z), Γ, ϕ L (cons(x, z)) ϕ L (t), Γ L(t), Γ ind L

148 60/ 70 Properties of LKID Henkin model... least fixed points formed in subset of power set

149 60/ 70 Properties of LKID Henkin model... least fixed points formed in subset of power set Theorem. LKID is sound w.r.t. Henkin models.

150 60/ 70 Properties of LKID Henkin model... least fixed points formed in subset of power set Theorem. LKID is sound w.r.t. Henkin models. Theorem. LKID without cut is complete w.r.t. Henkin models.

151 60/ 70 Properties of LKID Henkin model... least fixed points formed in subset of power set Theorem. LKID is sound w.r.t. Henkin models. Theorem. LKID without cut is complete w.r.t. Henkin models. Lemma. PA can be interpreted in LKID plus BA-axioms.

152 60/ 70 Properties of LKID Henkin model... least fixed points formed in subset of power set Theorem. LKID is sound w.r.t. Henkin models. Theorem. LKID without cut is complete w.r.t. Henkin models. Lemma. PA can be interpreted in LKID plus BA-axioms. Corollary. PA is consistent.

153 61/ 70 Outline Gentzen s consistency proof The omega rule Cyclic proofs Infinite proofs Cyclic proofs Proof by induction Cyclic proofs vs. proofs by induction

154 62/ 70 Translation from LKID to CLKID ω (1/2) Theorem. If LKID Γ then CLKID ω Γ.

155 Translation from LKID to CLKID ω (1/2) Theorem. If LKID Γ then CLKID ω Γ. Proof. Translate inductions into cycles. For example: (π b ) Γ, ϕ(0) (π s ) ϕ(x), Γ, ϕ(s(x)) N(t), Γ (π c ) ϕ(t), Γ ind N 62/ 70

156 Translation from LKID to CLKID ω (1/2) Theorem. If LKID Γ then CLKID ω Γ. Proof. Translate inductions into cycles. For example: (π b ) Γ, ϕ(0) (π s ) ϕ(x), Γ, ϕ(s(x)) N(t), Γ (π c ) ϕ(t), Γ ind N Let A = {ϕ(0), x (ϕ(x) ϕ(s(x)))}, translate to N(z), A ϕ(z) N(t), A ϕ(t) subst N(t) A ϕ(t) (π b, π s ) Γ A, (π c ) ϕ(t), Γ Γ, A ϕ(t) cut N(t), Γ l 62/ 70

157 63/ 70 Translation from LKID to CLKID ω (2/2) Proof (cont.) reminder: A = {ϕ(0), x (ϕ(x) ϕ(s(x)))} where z = 0, A ϕ(z) A, N(z) ϕ(z) A, N(y) ϕ(y) subst ϕ(s(y)) ϕ(s(y)) A, N(y), ϕ(y) ϕ(s(y)) ϕ(s(y)) c l, l A, N(y) ϕ(s(y)) z = s(y), A, N(y) ϕ(z) =, w l case N A, N(z) ϕ(z) l What about the other direction?

158 64/ 70 Brotherston/Simpson conjecture Conjecture [Brotherston/Simpson 11]. If CLKID ω Γ then LKID Γ.

159 64/ 70 Brotherston/Simpson conjecture Conjecture [Brotherston/Simpson 11]. If CLKID ω Γ then LKID Γ. Theorem [Simpson 17]. True for PA.

160 64/ 70 Brotherston/Simpson conjecture Conjecture [Brotherston/Simpson 11]. If CLKID ω Γ then LKID Γ. Theorem [Simpson 17]. True for PA. Theorem [Berardi/Tatsuta 17]. False in general.

161 64/ 70 Brotherston/Simpson conjecture Conjecture [Brotherston/Simpson 11]. If CLKID ω Γ then LKID Γ. Theorem [Simpson 17]. True for PA. Theorem [Berardi/Tatsuta 17]. False in general. Theorem [Berardi/Tatsuta 17]. True for calculi containing PA.

162 65/ 70 Cyclic arithmetic is equivalent to Peano arithmetic Let L = {0, s, +,, <} Definition. Cyclic arithmetic (CA) is set of first-order L-sentences σ s.t. σ has CLKID ω proof from basic arithmetic axioms where N is the only inductive predicate.

163 65/ 70 Cyclic arithmetic is equivalent to Peano arithmetic Let L = {0, s, +,, <} Definition. Cyclic arithmetic (CA) is set of first-order L-sentences σ s.t. σ has CLKID ω proof from basic arithmetic axioms where N is the only inductive predicate. Theorem [Simpson 17]. CA = PA.

164 65/ 70 Cyclic arithmetic is equivalent to Peano arithmetic Let L = {0, s, +,, <} Definition. Cyclic arithmetic (CA) is set of first-order L-sentences σ s.t. σ has CLKID ω proof from basic arithmetic axioms where N is the only inductive predicate. Theorem [Simpson 17]. CA = PA. Proof Sketch. PA CA

165 65/ 70 Cyclic arithmetic is equivalent to Peano arithmetic Let L = {0, s, +,, <} Definition. Cyclic arithmetic (CA) is set of first-order L-sentences σ s.t. σ has CLKID ω proof from basic arithmetic axioms where N is the only inductive predicate. Theorem [Simpson 17]. CA = PA. Proof Sketch. PA CA CA PA: Formalisation of unfolding CLKID ω LKID ω in ACA 0 (incl. theory of Büchi automata) Formalisation of soundness of LKID ω in ACA 0 Truth reflection principle for Σn -sentences Conservativity of ACA0 over PA

166 66/ 70 Extended equivalence Definition. Given set of inductive definitions D: D-LKID + PA: add basic arithmetic axioms and ind. pred. N D-CLKID ω + PA: add basic arithmetic axioms and ind. pred. N

167 66/ 70 Extended equivalence Definition. Given set of inductive definitions D: D-LKID + PA: add basic arithmetic axioms and ind. pred. N D-CLKID ω + PA: add basic arithmetic axioms and ind. pred. N Theorem [Berardi/Tatsuta 17]. D-LKID + PA = D-CLKID ω + PA

168 66/ 70 Extended equivalence Definition. Given set of inductive definitions D: D-LKID + PA: add basic arithmetic axioms and ind. pred. N D-CLKID ω + PA: add basic arithmetic axioms and ind. pred. N Theorem [Berardi/Tatsuta 17]. D-LKID + PA = D-CLKID ω + PA Proof Sketch. D-LKID + PA D-CLKID ω + PA

169 66/ 70 Extended equivalence Definition. Given set of inductive definitions D: D-LKID + PA: add basic arithmetic axioms and ind. pred. N D-CLKID ω + PA: add basic arithmetic axioms and ind. pred. N Theorem [Berardi/Tatsuta 17]. D-LKID + PA = D-CLKID ω + PA Proof Sketch. D-LKID + PA D-CLKID ω + PA D-CLKID ω + PA D-LKID + PA: 1. Cut D-CLKID ω + PA proof π into cycle-free parts 2. Prove induction principle on order < π in PA 3. Combine 1. and 2. to D-LKID + PA proof.

170 67/ 70 CLKID ω is not equivalent to LKID (1/2) 2-Hydra statement provable in CLKID ω but not in LKID Hydra: mythical monster: cut off one head, grows two new heads

171 67/ 70 CLKID ω is not equivalent to LKID (1/2) 2-Hydra statement provable in CLKID ω but not in LKID Hydra: mythical monster: cut off one head, grows two new heads 2-Hydra: Let a, b N, then (a+1, b+2) (a, b), (0, b+2) (b+1, b), (a+2, 0) (a+1, a) Terminate if none of these rules apply.

172 67/ 70 CLKID ω is not equivalent to LKID (1/2) 2-Hydra statement provable in CLKID ω but not in LKID Hydra: mythical monster: cut off one head, grows two new heads 2-Hydra: Let a, b N, then (a+1, b+2) (a, b), (0, b+2) (b+1, b), (a+2, 0) (a+1, a) Terminate if none of these rules apply. Formalisation: let L = {0/0, s/1, N/1, p/2}, N defined inductively (H) H 1 H 2 H 3 H 4 x, y N p(x, y) (H 1 ) p(0, 0) p(s(0), 0) x N p(x, s(0)) (H 2 ) x, y N ( p(x, y) p(s(x), s(s(y))) ) (H 3 ) y N ( p(s(y), y) p(0, s(s(y))) ) (H 4 ) x N ( p(s(x), x) p(s(s(x)), 0) )

173 68/ 70 CLKID ω is not equivalent to LKID (2/2) Lemma. CLKID ω H Proof. Short and straightforward cyclic proof.

174 68/ 70 CLKID ω is not equivalent to LKID (2/2) Lemma. CLKID ω H Proof. Short and straightforward cyclic proof. Theorem [Berardi/Tatsuta 17]. LKID H Proof Sketch. Counter-Henkin-structure: Domain N Z Suitable infinite sequence of pairs in Z.

175 69/ 70 Summary of 3rd part Inductive definitions Infinitely deep proofs LKID ω Sound and complete w.r.t. standard models Cyclic subsystem CLKID ω of LKID ω Finite proofs Sound and complete w.r.t. Henkin models Proofs by induction LKID LKID CLKID ω CLKID ω LKID if PA is included

176 70/ 70 Summary Three proof-theoretic approaches to induction: Induction rules The ω-rule Cyclic proofs What this talk did not contain: The incompleteness theorems Program extraction (and consistency proofs based on that) Bounded arithmetic (and connections to computational complexity) Inductive theorem proving...