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(1)

Florian Pelupessy

Department of Mathematics, Ghent University, Belgium

FACULTY OF SCIENCES

(2)

Overview

1 Introduction

What are phase transitions for unprovability?

2 Small examples

Ackermann function, Growing trees

3 Orders

Monomial ideals

4 Colours

unordered regressive Ramsey numbers, adjacent Ramsey

5 Concluding remarks

Florian Pelupessy Phase transitions for unprovability 2 / 49

(3)

Pressure

Temperature Liquid

Solid

Gas q

Figure: Phase transitions in physics

(4)

Phase transition:

at a threshold point a small change has a big consequence.

Florian Pelupessy Phase transitions for unprovability 4 / 49

(5)

Phase transition:

at a threshold point a small change has a big consequence. Phase transition in logic:

1 Take an interesting theorem with a parameter.

(6)

Phase transition:

at a threshold point a small change has a big consequence. Phase transition in logic:

1 Take an interesting theorem with a parameter.

2 Find values of that parameter for which the theorem is provable and values for which it is unprovable.

Florian Pelupessy Phase transitions for unprovability 4 / 49

(7)

Phase transition:

at a threshold point a small change has a big consequence. Phase transition in logic:

1 Take an interesting theorem with a parameter.

2 Find values of that parameter for which the theorem is provable and values for which it is unprovable.

3 Determine a threshold for that parameter where the theorem changes from unprovable to provable.

(8)

Peano Arithmetic

Take language L = {+, ×, <, 0, 1}. The first order theory of Peano Arithmetic consists of some ordinary rules for arithmetic:

Distributivity, associativity and commutativity of + and × with neutral elements 0 and 1 respectively.

<is a discrete linear order, with minimal element 0, 1 is the successor of 0.

x < y → x + z < y + z. And the axiom scheme for induction:

Florian Pelupessy Phase transitions for unprovability 5 / 49

(9)

Peano Arithmetic

Take language L = {+, ×, <, 0, 1}. The first order theory of Peano Arithmetic consists of some ordinary rules for arithmetic:

Distributivity, associativity and commutativity of + and × with neutral elements 0 and 1 respectively.

<is a discrete linear order, with minimal element 0, 1 is the successor of 0.

x < y → x + z < y + z.

And the axiom scheme for induction: for every L-formula ϕ : [ϕ(0, ¯y) ∧ ∀x(ϕ(x, ¯y) → ϕ(x + 1, ¯y))] → ∀xϕ(x, ¯y).

(10)

fragments

Σn-formulas are formulas of the form:

∃x1. . . xi1∀y1. . . yi2∃ . . . ϕ

where there are n quantifiers and ϕ contains only bounded quantifiers.nis PA with the induction scheme restricted to Σn-formulas.

Florian Pelupessy Phase transitions for unprovability 6 / 49

(11)

Unprovability

G¨odel (1931): There exist L-theorems which are unprovable in PA. Since then interesting PA-unprovable theorems have been found, among them: Paris-Harrington strong Ramsey theorem, Paris and Kirby Hydra battles, Goodstein sequences, miniaturised Kruskal theorem and the Kanamori-McAloon principle (regressive Ramsey).

(12)

ordinals below ε

0

Intuitive description of the ordinals below ε0: 0, 1, 2, 3, . . .

Florian Pelupessy Phase transitions for unprovability 8 / 49

(13)

ordinals below ε

0

Intuitive description of the ordinals below ε0: 0, 1, 2, 3, . . . , ω

(14)

ordinals below ε

0

Intuitive description of the ordinals below ε0: 0, 1, 2, 3, . . . , ω, ω + 1, ω + 2, . . .

Florian Pelupessy Phase transitions for unprovability 8 / 49

(15)

ordinals below ε

0

Intuitive description of the ordinals below ε0:

0, 1, 2, 3, . . . , ω, ω + 1, ω + 2, . . . , ω + ω = ω · 2,

(16)

ordinals below ε

0

Intuitive description of the ordinals below ε0:

0, 1, 2, 3, . . . , ω, ω + 1, ω + 2, . . . , ω + ω = ω · 2,

ω· 3, ω · 4, . . . , ω · ω = ω2,

Florian Pelupessy Phase transitions for unprovability 8 / 49

(17)

ordinals below ε

0

Intuitive description of the ordinals below ε0:

0, 1, 2, 3, . . . , ω, ω + 1, ω + 2, . . . , ω + ω = ω · 2,

ω· 3, ω · 4, . . . , ω · ω = ω2, ω3, ω4, . . .

(18)

ordinals below ε

0

Intuitive description of the ordinals below ε0:

0, 1, 2, 3, . . . , ω, ω + 1, ω + 2, . . . , ω + ω = ω · 2,

ω· 3, ω · 4, . . . , ω · ω = ω2, ω3, ω4, . . .

ωω= ω1,

Florian Pelupessy Phase transitions for unprovability 8 / 49

(19)

ordinals below ε

0

Intuitive description of the ordinals below ε0:

0, 1, 2, 3, . . . , ω, ω + 1, ω + 2, . . . , ω + ω = ω · 2,

ω· 3, ω · 4, . . . , ω · ω = ω2, ω3, ω4, . . .

ωω= ω1, ωω

ω = ω2,

(20)

ordinals below ε

0

Intuitive description of the ordinals below ε0:

0, 1, 2, 3, . . . , ω, ω + 1, ω + 2, . . . , ω + ω = ω · 2,

ω· 3, ω · 4, . . . , ω · ω = ω2, ω3, ω4, . . .

ωω= ω1, ωω

ω = ω2, ωω

ωω

= ω3, . . . ,

Florian Pelupessy Phase transitions for unprovability 8 / 49

(21)

ordinals below ε

0

Intuitive description of the ordinals below ε0:

0, 1, 2, 3, . . . , ω, ω + 1, ω + 2, . . . , ω + ω = ω · 2,

ω· 3, ω · 4, . . . , ω · ω = ω2, ω3, ω4, . . .

ωω= ω1, ωω

ω = ω2, ωω

ωω

= ω3, . . . , ωω= ε0

(22)

ordinals below ε

0

Important properties of these ordinals: They are well ordered.

Any ordinal α can be written in the Cantor Normal Form: α= ωα1· m1+ · · · + ωαn· mn, where α > α1>· · · > αn and the mi>0.

Florian Pelupessy Phase transitions for unprovability 9 / 49

(23)

ordinals below ε

0

Important properties of these ordinals: They are well ordered.

Any ordinal α can be written in the Cantor Normal Form: α= ωα1· m1+ · · · + ωαn· mn, where α > α1>· · · > αn and the mi>0.

They are used as a way to measure the logical strength of a theory.

(24)

Fast growing hierarchy

Define:

ωα+1[x] = ωα· x ωγ[x] = ωγ[x] (α + ωβ)[x] = α+ (ωβ[x])

Florian Pelupessy Phase transitions for unprovability 10 / 49

(25)

Fast growing hierarchy

Define:

ωα+1[x] = ωα· x ωγ[x] = ωγ[x] (α + ωβ)[x] = α+ (ωβ[x]) and

F0(i) = i+ 1 Fα+1(i) = Fαi(i)

Fγ(i) = Fγ[i](i) Where Fαi denotes the i-times iteration of Fα.

(26)

Important fact

If a computable function is provably total in IΣn then it can be bounded by some function Fαwith α < ωn−1.

If such a function is provably total in PA then it can be bounded by some function Fαwith α < ε0.

Florian Pelupessy Phase transitions for unprovability 11 / 49

(27)

Ackermann

A0(i) = i+ 1 An+1(i) = Ain(i)

A(i) = Ai(i)

Where Aindenotes the i-times iteration of An.

(28)

Ackermann

A0(i) = i+ 1 An+1(i) = Ain(i)

A(i) = Ai(i)

Where Aindenotes the i-times iteration of An.

Ais total, but not provably so in IΣ1(notice that A = Fω).

Florian Pelupessy Phase transitions for unprovability 12 / 49

(29)

Ackermann with parameter

Af,0(i) = i+ 1 Af,n+1(i) = Aff(i),n(i)

Af(i) = Af,i(i)

(30)

Ackermann with parameter

Af,0(i) = i+ 1 Af,n+1(i) = Aff(i),n(i)

Af(i) = Af,i(i)

Theorem (Kojman, Lee, Omri, Weiermann) Take fc(i) =ci :

Afc is not provably total in IΣ1, but Alog is provably total in IΣ1

Florian Pelupessy Phase transitions for unprovability 13 / 49

(31)

Growing trees

Examine the following process for l:

(32)

Growing trees

Examine the following process for l:

l+1 new leaves

z }| {

Step 1

Florian Pelupessy Phase transitions for unprovability 15 / 49

(33)

Growing trees

Examine the following process for l:

l+2 new leaves

z }| {

(34)

Growing trees

Examine the following process for l:

l+3 new leaves

z }| {

Step 3

Florian Pelupessy Phase transitions for unprovability 17 / 49

(35)

Growing trees

Examine the following process for l:

l+4 new leaves

z }| {

(36)

Growing trees

Examine the following process for l:

l+5 new leaves

z }| {

Step 5

Florian Pelupessy Phase transitions for unprovability 19 / 49

(37)

Growing trees

Examine the following process for l:

l+6 new leaves

z }| {

(38)

Growing trees

Examine the following process for l:

l+7 new leaves

z }| {

Step 7

Florian Pelupessy Phase transitions for unprovability 21 / 49

(39)

Growing trees

Examine the following process for l:

l+8 new leaves

z }| {

(40)

Growing trees

Examine the following process for l:

Step i: select a leaf

Florian Pelupessy Phase transitions for unprovability 23 / 49

(41)

Growing trees

Examine the following process for l:

l+i new leaves

z }| {

(42)

Growing trees

Lemma (MKL)

For every h and l there exists a K such that every growing tree reaches height h within K steps.

Intuition: Every growing tree reaches any height.

Florian Pelupessy Phase transitions for unprovability 25 / 49

(43)

Growing trees

Lemma (MKL)

For every h and l there exists a K such that every growing tree reaches height h within K steps.

Intuition: Every growing tree reaches any height. Theorem

MKL is unprovable in IΣ1.

(44)

Growing trees with parameter

Examine the following process for l and f :

Step i: select a leaf

Florian Pelupessy Phase transitions for unprovability 26 / 49

(45)

Growing trees with parameter

Examine the following process for l and f :

f(l+i) new leaves

z }| {

(46)

Growing trees with parameter

Lemma (MKLf)

For every f , h and l there exists a K such that every growing tree reaches height h within K steps.

Florian Pelupessy Phase transitions for unprovability 28 / 49

(47)

Growing trees with parameter

Lemma (MKLf)

For every f , h and l there exists a K such that every growing tree reaches height h within K steps.

Theorem

1 MKLid is unprovable in IΣ1,

2 MKLc is provable in IΣ1for constant function c.

(48)

Growing trees with parameter

Theorem Take fc(i) =ci :

1 MKLf

c is unprovable in IΣ1, but

2 MKL

log is provable in IΣ1.

Florian Pelupessy Phase transitions for unprovability 29 / 49

(49)

Monomial ideals

Given field K and polynomial ring K [Xd, . . . , X0, Y]. A monomial ideal is an ideal that is generated by monomials.

The degree of a monomial is the total degree.

The degree of a set of generators is the maximum of the degrees of its elements.

deg(I ) is the minimum of the degrees of the sets that generate I . Intuition: the degree of an ideal is the minimum degree necessary to be able to generate it.

(50)

Monomial ideals

Theorem (Maclagan 2001)

For every l , d there exists M such that for every sequence I1, . . . , IM of monomial ideals in K[Xd, . . . , X0, Y] with deg(In) ≤ l + n there are i < j with Ii⊇ Ij.

Intuition: Any sufficiently long linearly bounded sequence contains an ideal that is a subset of an earlier one.

Florian Pelupessy Phase transitions for unprovability 31 / 49

(51)

Maclagan with parameter

Theorem (MMf)

For every f , l , d there exists M such that for every sequence I1, . . . , IM of monomial ideals in K[Xd, . . . , X0, Y] with deg(In) ≤ l + f (n) there are i < j with Ii⊇ Ij.

(52)

Maclagan with parameter

Theorem (MMf)

For every f , l , d there exists M such that for every sequence I1, . . . , IM of monomial ideals in K[Xd, . . . , X0, Y] with deg(In) ≤ l + f (n) there are i < j with Ii⊇ Ij.

Take fc(i) =pc log(i): Theorem (Pelupessy) MMf is

1 unprovable in IΣ2for f = fc, but

2 provable for f = log log.

Florian Pelupessy Phase transitions for unprovability 32 / 49

(53)

Proof sketch:

First show that MMidis unprovable in the following manner: Associate with each ordinal < ωωmonomials in variables Xi.

(54)

Proof sketch:

First show that MMidis unprovable in the following manner: Associate with each ordinal < ωωmonomials in variables Xi.

Define Mα(l) as the maximal length of a bad sequence of ideals with generators from monomials that are associated with ordinals ≤ α.

Florian Pelupessy Phase transitions for unprovability 33 / 49

(55)

Proof sketch:

First show that MMidis unprovable in the following manner: Associate with each ordinal < ωωmonomials in variables Xi.

Define Mα(l) as the maximal length of a bad sequence of ideals with generators from monomials that are associated with ordinals ≤ α. Notice, using induction, that Mα(h(l)) ≥ Fα(l) for some primitive recursive function h.

(56)

Proof sketch:

Then use this result to show that MMclogis unprovable: Take bad sequence I1, . . . IM from previously,

Florian Pelupessy Phase transitions for unprovability 34 / 49

(57)

Proof sketch:

Then use this result to show that MMclogis unprovable: Take bad sequence I1, . . . IM from previously,

Define new sequence:

Ifc(1), . . . , Ifc(M).

(58)

Proof sketch:

Then use this result to show that MMclogis unprovable: Take bad sequence I1, . . . IM from previously,

Define new sequence:

Ifc(1), . . . , Ifc(M).

Use extra variables to ensure that the sequence becomes a bad sequence. (2c + 3 extra variables suffice).



Florian Pelupessy Phase transitions for unprovability 34 / 49

(59)

Kanamori McAloon

Identify R with the set of its predecessors: R = {0, . . . , R − 1}. [R]d is the set of d-element subsets of R.

(60)

Kanamori McAloon

Identify R with the set of its predecessors: R = {0, . . . , R − 1}. [R]d is the set of d-element subsets of R.

For a colouring C : [R]2→ N a set H ⊆ R is homogeneous if C is constant on [H]2,

Florian Pelupessy Phase transitions for unprovability 35 / 49

(61)

Kanamori McAloon

Identify R with the set of its predecessors: R = {0, . . . , R − 1}. [R]d is the set of d-element subsets of R.

For a colouring C : [R]2→ N a set H ⊆ R is homogeneous if C is constant on [H]2,

min-homogeneous if C (x, y ) = C (x, z) for all x, y , z ∈ H with x < y , x < z,

(62)

Kanamori McAloon

Identify R with the set of its predecessors: R = {0, . . . , R − 1}. [R]d is the set of d-element subsets of R.

For a colouring C : [R]2→ N a set H ⊆ R is homogeneous if C is constant on [H]2,

min-homogeneous if C (x, y ) = C (x, z) for all x, y , z ∈ H with x < y , x < z,

min-homogeneous for a linear order ≺ on H if C (x, y) = C (x, z) for all x, y , z ∈ H with x ≺ y, x ≺ z.

Florian Pelupessy Phase transitions for unprovability 35 / 49

(63)

Kanamori McAloon

Theorem (Ramsey)

For all l , c there exists R such that for any colouring C: [R]2→ c there exists H of size l that is homogeneous for C .

(64)

Kanamori McAloon

Theorem (Ramsey)

For all l , c there exists R such that for any colouring C: [R]2→ c there exists H of size l that is homogeneous for C .

Theorem (KM2)

For all l there exists R such that for any colouring C: [R]2→ N with C(x, y ) ≤ min(x, y) there exists H of size l that is min-homogeneous for C .

Florian Pelupessy Phase transitions for unprovability 36 / 49

(65)

unordered Kanamori McAloon

Theorem (uKM2)

For all l there exists R such that for any colouring C: [R]2→ N with C(x, y ) ≤ min(x, y) there exists H of size l with linear order ≺ that is min-homogeneous for C .

(66)

unordered Kanamori McAloon

Theorem (uKM2)

For all l there exists R such that for any colouring C: [R]2→ N with C(x, y ) ≤ min(x, y) there exists H of size l with linear order ≺ that is min-homogeneous for C .

Theorem

uKM2is unprovable in IΣ1.

Florian Pelupessy Phase transitions for unprovability 37 / 49

(67)

unordered Kanamori McAloon with parameter

Theorem (uKM2f)

For all f , l there exists R such that for any colouring C : [R]2→ N with C(x, y ) ≤ f (min(x, y)) there exists H of size l with linear order ≺ that is min-homogeneous for C .

(68)

unordered Kanamori McAloon with parameter

Theorem (uKM2f)

For all f , l there exists R such that for any colouring C : [R]2→ N with C(x, y ) ≤ f (min(x, y)) there exists H of size l with linear order ≺ that is min-homogeneous for C .

Theorem (Pelupessy, Weiermann) Take f(i) = log i and fd(i) = di :

1 uKM2

fd is unprovable in IΣ1, but

2 uKM2

f is provable in IΣ1.

Florian Pelupessy Phase transitions for unprovability 38 / 49

(69)

Proof sketch

Show:

uKMd (Rc(m + 4)) ≥ Ac+1d (m)

(70)

Proof sketch

Show:

uKMd (Rc(m + 4)) ≥ Ac+1d (m)

↓ ↓ ↓

fast ⇐ slow fast

Florian Pelupessy Phase transitions for unprovability 39 / 49

(71)

Proof sketch

uKMd (Rc(m + 4)) ≥ Ac+1d (m)

Get two nice colourings C , D involving Ac+1d .

Take the H of size m + 4 that is min-homogeneous for C and homogeneous for D

Use the nice properties to see that one of the four <-largest elements of H is larger than Ac+1d (m).



(72)

unordered Kanamori McAloon with parameter

Theorem (uKMf)

For all f , d , l there exists R such that for any colouring C : [R]d→ N with C(x, y ) ≤ f (min(x, y)) there exists H of size l with linear order ≺ that is min-homogeneous for C .

Florian Pelupessy Phase transitions for unprovability 41 / 49

(73)

unordered Kanamori McAloon with parameter

Theorem (uKMf)

For all f , d , l there exists R such that for any colouring C : [R]d→ N with C(x, y ) ≤ f (min(x, y)) there exists H of size l with linear order ≺ that is min-homogeneous for C .

Define log to be the inverse of the tower function. Theorem (Pelupessy, Bovykin)

1 uKM

logk is unprovable in PA, but

2 uKM

log is provable in PA.

(74)

Proof sketch

Construct a model of PA + ¬uKMlogk.



Florian Pelupessy Phase transitions for unprovability 42 / 49

(75)

We call a colouring C : [R]d → Nr limitedif max C (x) ≤ max x. Theorem (AR)

For every d , r there exists R such that for every limited colouring C: [R]d→ Nr there exist x1<· · · < xd+1 with

C(x1, . . . , xd) ≤ C (x2, . . . , xd+1).

(76)

We call a colouring C : [R]d → Nr limitedif max C (x) ≤ max x. Theorem (AR)

For every d , r there exists R such that for every limited colouring C: [R]d→ Nr there exist x1<· · · < xd+1 with

C(x1, . . . , xd) ≤ C (x2, . . . , xd+1).

Theorem (Friedman)

Adjacent Ramsey is unprovable in PA.

Florian Pelupessy Phase transitions for unprovability 43 / 49

(77)

We call a colouring C : [R]d → Nr f -limitedif max C (x) ≤ f (max x). Theorem (ARf)

For every d , r there exists R such that for every f -limited colouring C: [R]d→ Nr there exist x1<· · · < xd+1 with

C(x1, . . . , xd) ≤ C (x2, . . . , xd+1).

(78)

We call a colouring C : [R]d → Nr f -limitedif max C (x) ≤ f (max x). Theorem (ARf)

For every d , r there exists R such that for every f -limited colouring C: [R]d→ Nr there exist x1<· · · < xd+1 with

C(x1, . . . , xd) ≤ C (x2, . . . , xd+1).

Theorem (Pelupessy)

1 AR

logk is unprovable in PA, but

2 ARlog is provable in PA.

Florian Pelupessy Phase transitions for unprovability 44 / 49

(79)

Proof sketch

Take a limited colouring C : [R]d→ Nr (d > k). Define a new colouring D(x) = C (logkx1, . . . ,logkxd). This colouring will be logk-limited, but may contain undesired identical elements. This can be solved by adding an extra coordinate, using estimates for ARc (roughly a tower of exponentials of height k).



(80)

Some of these results have been sharpened: Theorem

1 MKLf is unprovable in IΣ1 if f(i) = A−1(i)i , but

2 MKLf is provable in IΣ1if f(i) = A−c1(i)i .

Florian Pelupessy Phase transitions for unprovability 46 / 49

(81)

Some of these results have been sharpened: Theorem

1 MKLf is unprovable in IΣ1 if f(i) = A−1(i)i , but

2 MKLf is provable in IΣ1if f(i) = A−c1(i)i . Theorem

Take fα(i) = F −1α (i)plog(i), then MMfα is

1 unprovable in IΣ2for α= ωω, but

2 provable for α < ωω.

(82)

Some of these results have been sharpened: Theorem

1 MKLf is unprovable in IΣ1 if f(i) = A−1(i)i , but

2 MKLf is provable in IΣ1if f(i) = A−c1(i)i . Theorem

Take fα(i) = F −1α (i)plog(i), then MMfα is

1 unprovable in IΣ2for α= ωω, but

2 provable for α < ωω.

Theorem

1 uKM2

f is unprovable in IΣ1for f(i) = A−1(i)i , but

2 uKM2

f is provable in IΣ1for f(i) = A−c1(i)i .

Florian Pelupessy Phase transitions for unprovability 46 / 49

(83)

Take fα(i) = logFα−1(i)(i). Conjecture

1 uKMf

ε0 is unprovable in PA, but

2 uKMf

α is provable in PA if for α < ε0.

(84)

Take fα(i) = logFα−1(i)(i). Conjecture

1 uKMf

ε0 is unprovable in PA, but

2 uKMf

α is provable in PA if for α < ε0.

Motivation: this would correspond with existing results for Kanamori McAloon.

Florian Pelupessy Phase transitions for unprovability 47 / 49

(85)

The proof of PA-unprovability of adjacent Ramsey can be adapted into showing that for fixed k > 1 this theorem is unprovable in IΣk−1. What are the corresponding phase transitions?

(86)



Thank you for listening.

Florian Pelupessy Phase transitions for unprovability 49 / 49

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