G odel & Recursivity - HEC Lausanne › logique › enseignement › goedel_and... · 2 2 412 40 replace all 4’s by 5’s, then subtract 1: 2 52 2 512 50 1 60 G4 p3q 60 :::etc Amazingly,

Godel & Recursivity

JACQUES DUPARC

Batiment InternefCH - 1015 Lausanne

[email protected]

[email protected]

2

Contents

Introduction 7

I Recursivity 9

1 Towards Turing Machines 11

1.1 Deterministic Finite Automata . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.2 Nondeterministic Finite Automata . . . . . . . . . . . . . . . . . . . . . . . . . . 12

1.3 Regular Expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

1.4 Non-Regular Languages . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

1.5 Pushdown Automata . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

1.6 Context-Free Grammar . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

2 Turing Machines 27

2.1 Deterministic Turing Machines . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

2.2 Non-Deterministic Turing Machines . . . . . . . . . . . . . . . . . . . . . . . . . 32

2.3 The Concept of Algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

2.4 Universal Turing Machine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

2.5 The Halting Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

2.6 Turing Machine with Oracle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

3 Recursive Functions 51

3.1 Primitive Recursive Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

3.2 Variable Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

3.3 Bounded Minimisation and Bounded Quantification . . . . . . . . . . . . . . . . 58

3.4 Coding Sequences of Integers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

3.5 Partial Recursive Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

II Arithmetic 75

4 Representing Functions 77

4.1 Robinson Arithmetic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

4.2 Representable Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

4 CONTENTS

5 Godel’s First Incompleteness Theorem 111

5.1 Godel Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

5.2 Coding the Proofs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123

5.3 Undecidability of Robinson Arithmetic . . . . . . . . . . . . . . . . . . . . . . . . 139

6 Godel’s Second Incompleteness Theorem 145

6.1 Peano Arithmetic and IΣ01 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145

6.2 The Arithmetical Hierarchy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153

6.3 A first glance at Godel’s second incompleteness theorem . . . . . . . . . . . . . . 157

6.4 The core of the proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161

Introduction

Introduction 7

The basic requirements for this course are contained in the ”Mathematical Logic” course. Among

other things, you should have a clear understanding of each of the following: first order language,

signature, terms, formulas, theory, proof theory, models, completeness theorem, compactness

theorem, Lowenheim-Skolem theorem.

It makes no sense to take this course without this solid background on first order logic.

The title of the course is ”Godel and Recursivity” but it should rather be ”Recursivity and

Godel” since that is the way we are going to go through these topics

(1) Recursivity

(2) Godel’s incompleteness theorems (there are two of them)

Recursivity is at the heart of computer science, it represents the mathematical side of what

computing is like. It is related to arithmetics and to proof theory.

Godel’s incompleteness theorems are concerned with number theory (arithmetics) which itself

lies at the core of mathematics. They contradict the commonly shared idea that everything that

is true can be proved. It ruins the plan, for every mathematical statement ϕ to either prove it

or disprove it (by proving ϕ).

Godel’s first incompleteness theorem says that there exists a formula ϕ from number theory such

that neither ϕ nor ϕ is provable. More precisely, it says that in Peano Arithmetics (which is

a first order axiomatization of arithmetics) there exists a formula ϕ that cannot be proved nor

disproved, and if we were to add this formula to Peano Arithmetics, one would find a second

one that would not be provable nor disprovable inside their first extension of Peano Arithmetics.

And if this new formula would be added again, we could find a third one and so on and so

forth. To put it differently, if we want to extend Peano Arithmetics to a larger theory which is

complete in the sense that it proves or disproves any given formula, there would not have any

understanding of this theory, we would not get hold of it, for it would not be recursive, meaning

that we would not have any efficient way of figuring out whether a given closed formula is part of

the theory or not (not provable from the theory, but simply part of the theory!). This is precisely

where the notion of recursivity plays a crucial role. One does not have the right comprehension

of Godel’s incompleteness theorems without a proper understanding of what recursivity is like.

The formula that we will construct (the one that is not provable nor disprovable in Peano Arith-

metics) is rather odd. There is no chance that one might tumble over such a formula during the

usual mathematical practice.

However, since Godel’s incompleteness theorem was proved, there have been several examples

of real arithmetic mathematical formulas that are not provable nor disprovable in Peano Arith-

metics, although the are formulated in the language of arithmetics.

A good example of such a formula is the one related to Goodstein sequences (1944). A Goodstein

sequence is of the form Gmp0q, G

mp1q, G

mp2q, . . ., etc, where m is a positive integer. It is defined the

following way (we take m “ 4 as an example, the general case being obtained by replacing

G4p0q “ 4 by Gm

p0q “ m and gathering the other values Gmp1q, G

mp2q, etc the same way):

8 Godel & Recursivity

˝ G4p0q “ 4 write 4 in hereditary base 2: 4 “ 22

replace all 2’s by 3’s, then subtract 1: 33 “ 26

˝ G4p1q “ 26 write 26 in hereditary base 3: 26 “ 2 ¨ 32 ` 2 ¨ 312 ¨ 30

replace all 3’s by 4’s, then subtract 1: 2 ¨ 42 ` 2 ¨ 412 ¨ 40 ´ 1 “ 41

˝ G4p2q “ 41 write 41 in hereditary base 4: 41 “ 2 ¨ 42 ` 2 ¨ 412 ¨ 40

replace all 4’s by 5’s, then subtract 1: 2 ¨ 52 ` 2 ¨ 512 ¨ 50 ´ 1 “ 60

˝ G4p3q “ 60 . . . etc

Amazingly, G4pnq increases until n reaches the value 3 ¨ 2402653209 where it reaches the maximum

of 3 ¨ 2402653210´ 1, it stays there for the next 3 ¨ 2402653209 steps then starts its final descent and

eventually reaches 0.

Amazingly, for every integer m, the Goodstein sequence pGmpnqqnPN is ultimately constant with

value 0, i.e.

limnÑ8

Gmpnq “ 0.

However this statement which is easily formalizable in the language of arithmetics is not provable

in Peano Arithmetics (Kirby and Paris 1982). It requires a stronger theory to be proved (for

instance record order arithmetics).

Godel’s second incompleteness theorem than says that mathematics cannot prove its own con-

sistency (unless it is inconsistent in which case it can prove its own consistency for it can prove

everything). More precisely, in any recursive extension J of Peano Arithmetics, the formula

”ConspJq” (which is a formula from number theory that asserts that there is no proof of K from

J) is not provable unless J is inconsistent i.e.

IfJ &c thenJ &c ConspJq

We will present three different approaches:

˝ Computer Science ÝÑ Turing Machine (one of the abstract model of computer)

˝ Arithmetic ÝÑ Recursive functions which are particular functions Nk Ñ N

Part I

Recursivity

Chapter 1

Towards Turing Machines

The whole chapter is highly inspired by Michael Sipser’s book: “Introduction to the Theory of

Computation” [43]. It is a dashing introduction to the notions of Finite Automata, PushDown

Automata, Turing Machines.

We also recommend “Introduction to automata theory, languages, and computation” by John E.

Hopcroft, Rajeev Motwani et Jeffrey D. Ullman [29]; “Computational complexity” by Christos

H. Papadimitriou [36] and “A mathematical introduction to logic” by Herbert B. Enderton [16].

1.1 Deterministic Finite Automata

We will see that any finite automaton can be regarded as a rudimentary Turing machine: a

Turing machine that never writes anything and only goes one direction.

Definition 1 A deterministic finite automaton (DFA) is a 5-tuple pQ,Σ, δ, q0, F q, where

(1) Q is a finite set called the states,

(2) Σ is a finite set called the alphabet,

(3) δ : Qˆ Σ ÝÑ Q is the transition function,

(4) q0 P Q is the initial state, and

(5) F Ď Q is the set of accepting states.1

We denote by Σăω (or equivalently by Σ˚) the set of finite words on Σ and by ε the empty

sequence.

1Accept states sometimes are called final states.


Definition 2 A DFA A “ pQ,Σ, δ, q0, F q on an alphabet Σ accepts the word w P Σăω if and

only if

˝ either w “ ε (the empty sequence) and q0 P F

˝ or w “ xa0, . . . , any with each ai P Σ, and there is a sequence of states r0, . . . , rn`1 such

that:

‚ r0 “ q0

‚ @i ă n, δpri, aiq “ ri`1

‚ rn`1 P F .

Notation 3 Given any DFA A,

LpAq “ tw P Σăω : w is accepted by Au .

LpAq denotes the language accepted by A.

Definition 4 Any language recognised by some deterministic finite automata (DFA) is called

regular.

1.2 Nondeterministic Finite Automata

Given any alphabet Σ, we both assume that ε R Σ holds and write Σε for ΣY tεu.

Definition 5 A nondeterministic finite automaton (NFA) is a 5-tuple pQ,Σ, δ, q0, F q, where

(1) Q is a finite set of states,

(2) Σ is a finite alphabet,

(3) δ : Qˆ Σε ÝÑ PpQq is the transition function,


(5) F Ď Q is the set of accepting states.

Definition 6 Let N “ pQ,Σ, δ, q0, F q be an NFA and w P Σăω. We say that N accepts w if

and only if

˝ either w “ ε the empty sequence and q0 P F

˝ or w can be written as w “ xa0, . . . , any with each ai P Σε, and there is and a sequence of

states r0, . . . , rn`1 such that:

‚ r0 “ q0

‚ @i ă n, ri`1 P δpri, aiq,

Recursivity 13

‚ rn`1 P F .

Proposition 7 Every NFA has an equivalent DFA. i.e. given any NFA N there exists some

DFA D such that

LpN q “ LpDq.

Proof of Proposition 7: Given any NFA N “ xQ,Σ, δ, q0, F y, we build some DFA D “

xQ1,Σ, δ1, q10, F1y that recognises the same language.

(1) Q1 “ PpQq

(2) For S Ď Q and a P Σ we set

δ1pS, aq “ tq1 P Q | Dq P S qε˚aε˚ÝÝÝÝÑ q1u

where qε˚aε˚ÝÝÝÝÑ q1 stands for the existence of a path in the graph of N that goes through

exactly one edge labelled with ”a”, the others being labelled with ”ε”.

(3) q10 “ tq0u

(4) F 1 “ tS Ď Q | S X F ‰ Hu.

% 7

Definition 8 Let A and B be languages. We define the regular operations union, concatenation,

and star as follows.

˝ Union: AYB “ tx | x P A or x P Bu.

˝ Concatenation: A ˝B “ txy | x P A and y P Bu.

˝ Star: A˚ “ tx1x2 . . . xk | k ě 0 and each xi P Au.

Theorem 9 Regular languages are closed under union, concatenation and star.

Proof of Theorem 9: Let N 1 “ pQ1,Σ,∆1, q1, F1q, N 2 “ pQ2,Σ,∆2, q2, F2q be two NFAs

recognising respectively A1 and A2.

N1 N2


Union We need an NFA N such that N recognises a string if and only if N 1 or N 2 recognises

it. By working nondeterministically, the automaton N is allowed to split into two copies:

we construct N in such a way that N 1 and N 2 work in parallel at the same time. We

assume Q1 XQ2 “ H and q0 R Q1 YQ2. Define N “ pQ,Σ,∆, q0, F q where

(1) Q “ tq0u YQ1 YQ2.

(2) ∆ Ă Qˆ Σε ˆQ is defined by: pp, s, rq P ∆ if and only if one of the following is true

(a) p “ q0

s “ ε

r P tq1, q2u

(b) p, r P Q1

pp, s, rq P ∆1

(c) p, r P Q2

pp, s, rq P ∆2.

(3) F “ F1 Y F2.

The machine splits immediately into two copies of itself, which work exactly as N 1 and

N 2. It accepts a string if and only if at least one of the two main copies ends up in an

accepting state, i.e. in F1 or in F2, i.e. if and only if N 1 or N 2 accept it.

N

"

"

Concatenation Here we need an NFA N that accepts a word w if and only if w can be broken

into two pieces: a prefix and a suffix w “ wpws such that wp is accepted by N 1 and wsis accepted by N 2. We set q1 as the initial state and let the machine read the same way

N 1 would do. Any time that N 1 finds itself in an accepting state, we want N to non-

deterministically start reading as if it were N 2 but still remaining a copy of itself: so we

make it split any time it comes to some final state of N 1. The reason is that we want

to be able to check longer sub-strings as well, because it might be the case that the first

prefix that is found to be accepted by N 1 corresponds to a suffix that is rejected by N 2,

Recursivity 15

while there is a longer prefix which is also accepted by N 1 that yields a suffix which is this

time also accepted by N 2. Formally, we define ∆ by: pp, s, rq P ∆ if and only if one of the

following is true

(1) p, r P Q1

pp, s, rq P ∆1

(2) p, r P Q2

pp, s, rq P ∆2

(3) p P F1

s “ ε

r “ q2

The third condition guarantees the splitting. Finally, we set the accepting set to be F “ F2.

N

"

""

"

Star Here the machine N should be able to check if a word w can be broken into a finitely

many pieces w “ w1w2 ¨ ¨ ¨wk, each of them being accepted by N 1. So N has to read w1

as if it were N 1, and when it finds itself in an accepting state, it needs to start all over

again and read w2 and so on and so forth. The construction is similar to the one of the

concatenation, but since A˚1 contains the empty string, we want N to accept ε. So we just

add an initial state q0 which is also an accepting state, and from where the initial state of

N 1 is reached by an ε move. % 9

N

"

"

"

1.3 Regular Expressions

Definition 10 We say that R is a regular expression if R is of one the following form:


(1) a (for some a P Σ)

(2) ε

(3) H

(4) R1 YR2

(5) R1 ˝R2

(6) R1˚

where R1 and R2 are regular expressions.

The expression ε represents the language containing a single sequence, namely, the empty se-

quence, whereas H represents the language that doesn’t contain any sequence. Notice that

(1) R ˝ H “ H ˝R “ H (2) H˚ “ tεu.

Definition 11 Let R be a regular expression. We define by induction its associated language

LpRq as follows:

(1) Lpaq “ tau

(2) Lpεq “ tεu

(3) LpHq “ H

(4) LpR1 YR2q “ LpR1q Y LpR2q

(5) LpR1 ˝R2q “ LpR1q ˝ LpR2q

(6) LpR˚1q “ LpR1q˚.

Theorem 12 A language L is regular if and only if there exists a regular expression R such

that L “ LpRq.

Proof of Theorem 12:

(ñ) (1) Lpaq “ tau

(2) Lpεq “ tεu

(3) LpHq “ H

(4) LpR1 YR2q “ LpR1q Y LpR2q

(5) LpR1 ˝R2q “ LpR1q ˝ LpR2q

(6) LpR˚1q “ LpR1q˚

(ð) (1) We go from some n-states DFA to some n` 2-states Generalized-NFA:

(a) we add

(A) an initial state “s”

(B) an accepting state “a”

(C) a transition aεÝÑ q0

(D) a transition qεÝÑ a (each accepting state q ‰ a)

(b) we reduce the set of accepting states to tau.

Recursivity 17

(2) We go from some k`1`2-states Generalized-NFA 2 to some k`2-states Generalized-

NFA by removing one state from the original automaton: qrip R ts, au and for each

states qin R ta, qripu and qout R ts, qripu we set the new transition to be:

qinRinÑrip ˝ pRripÑripq

˚ ˝ RripÑout Y RinÑoutÝÝÝÝÝÝÝÝÝÝÝÝÝÝÝÝÝÝÝÝÝÝÝÝÝÝÝÝÝÝÑ qout

where RinÑrip, RripÑrip, RripÑout and RinÑout denote the following transitions:

(a) qinRinÑripÝÝÝÝÝÑ qrip

(b) qripRripÑripÝÝÝÝÝÑ qrip

(c) qripRripÑoutÝÝÝÝÝÝÑ qout

(d) qinRinÑoutÝÝÝÝÝÑ qout.

(3) We end up with a 2-states (“s” and “a”) Generalized-NFA with a single transition of

the form sRÝÑ a. The regular expression R gives the solution.

Example 13

(a)

1

0

0, 1(b)

1

0

0, 1"

"s

a

(c)

0"

s

a

1(0 [ 1)⇤

(d)

s

a

0⇤1(0 [ 1)⇤

An other example with an automaton a bit more complicated.

Example 14

(a)

2an NFA whose transitions are labelled with regular expressions.


1

0

0

0

1

1

(b)

1

0

0

0

1

1

"

"s a

"

(c)

0

1 "

s a"

00 [ 1

01 10 [ 0

11

(d)

s a

(10 [ 00)(00 [ 1)⇤01 [ 11

0(00 [ 1)⇤01 [ b

0(00 [ 1)⇤

(10 [ 0)(00 [ 1)⇤ [ "

(e)

s a

�0(00 [ 1)⇤01 [ b

��(10 [ 00)(00 [ 1)⇤01 [ 11

�⇤�(10 [ 0)(00 [ 1)⇤ [ "

�[�0(00 [ 1)⇤

�

% 12

Recursivity 19

1.4 Non-Regular Languages

Notice that any finite word on Σ can be coded by an integer, so that there are only ℵ0 many

regular languages. But there are 2ℵ0 many languages for there are as many as the number of

subsets of N. Hence most languages are not regular!

Theorem 15 (Pumping Lemma) If A is a regular language, then there is a number p (the

pumping length) where, if s is any sequence in A of length at least p, then s may be divided into

three pieces, s “ xyz, satisfying the following conditions:

(1) for each i ě 0, xyiz P A,

(2) |y| ą 0, and

(3) |xy| ď p

Proof of Theorem 15: Let A be any DFA such that LpAq “ A. Set p to be the number of states

of A. Let s be accepted by A. Then s may be broken into three pieces: s “ xyz. Such that the

path q0xÝÑ q never visits twice the same state. The path q

yÝÑ q visits twice the state q but none

of the others twice. This holds since for every word u of length at least p every path q1uÝÑ q” in

A visits at least twice the same state.

z

y

x

% 15

Example 16 The language t0m1m | n P Nu is not regular.

By contradiction, assume there exists some DFA A “ pQ,Σ, δ, q0, F q which recognises t0m1m |

n P Nu. We consider p “ |Q| the number of states of A. the word 0p1p is accepted by A. By

the previous Pumping Lemma there exist x, y and z such that 0p1p “ xyz and

(1) for each i ě 0, xyiz P t0m1m | n P Nu,

(2) |y| ą 0, and

(3) |xy| ď p


But since |xy| ď p, it turns out that xy P 0˚ and z P 0˚1˚. Therefore, for each integer i ą 1 we

have xyi P 0˚, hence xyiz contains too many 0’s compared to 1’s: a contradiction.

1.5 Pushdown Automata

Definition 17 A pushdown automaton (PDA) is a 6-tuple pQ,Σ,Γ, δ, q0, F q, where Q, Σ, Γ

and F are all finite sets, and

(1) Q is the set of states,

(2) Σ is the input alphabet,

(3) Γ is the stack alphabet,

(4) δ : Qˆ Σε ˆ Γε ÝÑ PpQˆ Γεq is the transition function3,


(6) F Ď Q is the set of accepting states.

Definition 18 A pushdown automaton M “ pQ,Σ,Γ, δ, q0, F q computes as follows. It accepts

input w if w can be written as w “ w1w2 . . . wm, where each wi P Σε and sequences of states

r0, r1, . . . , rm P Q and sequences s0, s1, . . . , sm P Γ˚ exist that satisfy the next three conditions.

The sequences si represent the sequence of stack contents that M has on the accepting branch of

the computation.

(1) r0 “ q0 and s0 “ ε. This condition testifies that M starts out properly: both in the initial

state and with an empty stack.

(2) For i “ 0, . . . ,m ´ 1, we have pri`1, bq P δpri, wi`1, aq, where si “ at and si`1 “ bt for

some a, b P Γε and t P Γ˚. This condition states that M moves properly according to the

state, stack, and next input symbol.

(3) rm P F . This condition states that an accepting state occurs right at the end of the reading

of the input.

(1) One step of a computation:

3For a deterministic version, replace PpQˆ Γεq by Qˆ Γε.

Recursivity 21

ri

aec

w1w2w3 · · · wi�1wiwi+1 · · · wm

ri+1ri

bec

w1w2w3 · · · wi�1wiwi+1 · · · wm

ri+1

(2) The special case where a “ ε and b P Γ (the PDA “pushes” b to the top of the stack)

ri

ec

w1w2w3 · · · wi�1wiwi+1 · · · wm

ri+1ri

bec

w1w2w3 · · · wi�1wiwi+1 · · · wm

ri+1

(3) The special case where a P Γ and b “ ε (the PDA “pops off” a from the top of the stack)

ri

aec

w1w2w3 · · · wi�1wiwi+1 · · · wm

ri+1ri

ec

w1w2w3 · · · wi�1wiwi+1 · · · wm

ri+1

Example 19 (1) The language t0m1m | n P Nu is recognised by a PDA.

1, 0 ! "

", "! ?0, "! 0

",? ! "1, 0 ! "

(2) The language t0i1j2k | i, j, k ě 0 and i “ j or i “ ku is recognizable by a PDA, however

it is not recognizable by a deterministic PDA.


1, 0 ! "

", "! ?

0, "! 0

",? ! ?

",? ! ?", "! "

", "!" ", "! "

1, "! " 2, 0 ! "

2, "! "

1.6 Context-Free Grammar

Definition 20 A context-free grammar is a 4-tuple pV,Σ, R, Sq, where

(1) V is a finite set whose elements are called variables,

(2) Σ is a finite set, disjoint from V . Its elements are called terminals,

(3) R is a finite set of rules. Each rule is a couple of the form pξ, uq where ξ P V and

u P pV Y Σq˚.

(4) S P V is the initial variable.

If u, v and w are sequences of variables and terminals, and A Ñ w is a rule of the grammar,

we say that uAv yields uwv (written uAv ñ uwv). We write u ñ˚ v if u “ v or if a sequence

u1, u2, . . . , uk exists for k ą 0 and

uñ u1 ñ u2 ñ . . .ñ uk ñ v.

The language generated by the grammar is tw P Σ˚ | S ñ˚ wu.

Example 21 Consider pV,Σ, R, Sq the context-free grammar where V “ tS,Bu, Σ “ t0, 1, 7u

and R is the following set of production rules:

˝ S ÝÑ 0S1 ˝ S ÝÑ B ˝ B ÝÑ 7

This grammar generates the language t0n71n | n P Nu.

Theorem 22 A language is recognised by a PDA if and only if it is context-free.

Recursivity 23


(ð) Get a context-free grammar. The Pushdown P works as follows:

(1) Places a marker symbol “K” and the start variable on the stack.

(2) Repeat:

(a) If the top stack is a variable A it selects non-deterministically one of the rules

for A and substitutes A by the string on the right hand side of the rule.

(b) If the top stack is a terminal symbol a, it reads the next input symbol from

the input and compares it to a. If they don’t match, rejects (for this branch of

non-deterministic). If they do, repeat.

(c) If the top of stack is the symbol “K” enters the accepting state (If a letter from

the input must be read, it rejects).

(ñ) We start from a PDA and construct P an equivalent one such that

(1) P has a single accepting state qacc.

(2) It empties its stack before accepting

(3) Each transition either pushes a symbol onto the stack or pops one off, but does not

do both at the same time so that the the content of the stack never stays put.

From P “ pQ,Σ,Γ, δ, q0, tqacc.uq we construct G.

(1) V “ tApq | p, q P Qu,

(2) Σ is unchanged,

(3) the start variable is Aq0, qacc..

(4) The set of rule R is:

(a) For each p, q, r, s P Q, t P Γ and a, b P Σε if δpp, a, εq contains pr, tq and δps, b, tq

contains pq, εq put the rule Apq Ñ a Arsb in R.

(b) For each p, q, r P Q put the rule Apq Ñ AprArq in R.

(c) For each p P Q put the rule App Ñ ε in R.

Why is the language recognised by P is the one derived by G?

(ñ) If w is accepted by P , then there exists a computation that accepts it. This compu-

tation goes from q0 ÝÑ qacc.. It determines one derivation.

(ð) Any successful derivation induces an accepting computation.

% 22

Every regular language is context-free. But many languages are neither regular nor context-free.


Theorem 23 (Pumping Lemma for Context-Free Languages) If A is a context-free lan-

guage, then there is a number p (the pumping length) where, if s is any sequence in A of length

at least p, then s may be divided into five pieces, s “ vwxyz, satisfying the following conditions:

(1) for each i ě 0, vwixyiz P A,

(2) |wy| ą 0, and

(3) |wxy| ď p

Proof of Theorem 23: See Theorem 2.19 in [43]. We first fix a grammar. Then we concentrate

on getting a derivation tree4 large enough so that there is one path – from the root to some leaf

– that visits twice the same variable T . For this, if k is the number of variables in the grammar,

we need a tree of height at least k ` 1. We take m to be the maximum number of symbols in

the right hand side of a rule 5, and take n “ maxp2,mq. Every word of height at least nk`1

that is generated by this grammar has a derivation tree with at least one branch whose length

is ě k ` 1. We set p “ nk`1.

Take any word u generated by this grammar such that |u| ď p holds. Consider the smallest –

in terms of nodes – derivation tree that produces u, and consider a node T which repeats only

once and such that there is no other variable that repeats in the subtree induced by this node.

The whole derivation tree is described below:

T

S

T

x y zwv

Notice that |wxy| ď p holds, because the subtree induced by T has never twice the same variable

(except for T itself which appears only twice). Hence every branch on this subtree has length

at most k ` 1, which guarantees that wxy has length at most p “ nk`1.

Notice also that |wy| ą 0 because otherwise, we would have w “ y “ ε. But then the derivation

tree below would also produce the same word which would contradict the minimality of the one

we chose.

4notice that in a derivation tree every leaf is a terminal symbol, and very other node is a variable.5k is the maximum number of immediate successors of a node in the derivation tree.

Recursivity 25

S

T

x

zv

We also clearly have, for each i ě 0, vwixyiz P A:

T

S

T

x y zwv

T

T

x yw

S

z

T

ywv

% 23

Example 24 The following language is not context-free:

tanbncn | n P Nu.

Towards a contradiction we assume that this language is context-free so that there exists some

integer p that verifies the conditions of Theorem 23. We consider the word u “ 0p1p2p P A. By

Theorem 23, there exist words v, w, x, y, z such that u “ vwxyz and

(1) vwixyiz P A (@i ě 0) (2) |wy| ą 0, and (3) |wxy| ď p

Since |wxy| ď p holds, this word cannot contain all three letters 0,1 and 2. We distinguish two

different cases:

(1) if wxy P 0˚1˚, then z P 1˚2˚. Therefore for each i ą 1 vwixyiz contains either more 0’s

than 2’s or 1’s than 2’s.

(2) if wxy P 1˚2˚, then v P 0˚1˚. Therefore for each i ą 1 vwixyiz contains either more 1’s

than 0’s or 2’s than 0’s.


Chapter 2

Turing Machines

A Turing Machine (TM) is a general model of computation introduced in 1936 by Alan Turing

[50]. It consist in an infinite tape and a tape head that can read, write and move around. It can

both read the content of the tape and write on it. The read-write head can move both to the

left and to the right. The tape is infinite. There are special states for rejecting and accepting

which both take immediate effect.

100 01 1

Control

t t t t t

2.1 Deterministic Turing Machines

Definition 25 A (deterministic) TM is a 7-tuple pQ,Σ,Γ, δ, q0, qacc., qrej.q where Q,Σ,Γ are all

finite sets and

(1) Q is the set of states,

(2) Σ is the alphabet not containing the blank symbol, \,

(3) Γ is the tape alphabet where \ P Γ and Σ Ď Γ

(4) δ : Qˆ Γ ÝÑ Qˆ Γˆ tL,Ru is the transition function

(5) q0 is the initial state

(6) qacc. is the accepting state

(7) qrej. is the rejecting state


Clearly qacc. and qrej. must be different states.

Notice that the head cannot move off the left hand end of the tape. If δ says so, it stays put. A

configuration of a TM is a snapshot: it consists in the actual control state (q), the position of the

head and what is written on the tape (w). To indicate the position of the head we consider the

word w0 which is located to the left of the head and slice the tape content w into the w0w1 “ w.

This means that the head is actually positioned on the first letter of w1. Strictly speaking the

content of the tape is an infinite word:

w \\\ . . . . . .\\ . . .

but we forget about the infinite suffix \\\ . . .. We then write w0qw1 to say that

˝ the tape content is w0w1 \\\ . . .

˝ the head is positioned on the first letter of w1 \\\ . . .

˝ the actual control state is q.

The initial configuration on input w P Σăω is q0w.

An halting configuration is

˝ either an accepting configuration of the form w0qacc.w1,

˝ or a rejecting configuration of the form w0qrej.w1.

Given any two configurations C,C 1 we write C ñ C 1 (for C yields C 1 in one step) if there exist

a, b, c P Γ, and u, v P Γ˚ such that

˝ either C “ uaqibv, C 1 “ uqjacv and δpqi, bq “ pqj , c, Lq,

˝ or C “ uaqibv, C 1 “ uacqjv and δpqi, bq “ pqj , c, Rq.

Definition 26 A TM accepts input w if there is a sequence of configuration C0, . . . , Ck such

that

(1) C0 “ q0w

(2) Ci yields Ci`1 (for any 0 ă i ă k)

(3) Ck is an accepting configuration.

Definition 27 The set of all words accepted by a TM M is the language it recognises:

LpMq “ tw P Σ˚ |M accepts wu.

Recursivity 29

Example 28 A Turing machine that recognises tw7w | w P t0, 1u˚u – where w is the mirror of

w (for instance 001011 “ 110100).

pQ,Σ,Γ, δ, q0, qacc., qrej.q where

(1) Q “ tq0, qremember 0 look for \ go right, qremember 1 look for \ go right, qwrite 0, qwrite 1,

qlook for \ go left, qstep rightu

(2) Σ “ t0, 1u

(3) Γ “ t0, 1,\u

(4) δ : Qˆ Γ ÝÑ Qˆ Γˆ tL,Ru is defined by

pq0,\q ÝÑ qacc.pq0, 0q ÝÑ pqremember 0 look for \ go right,\, Rq

pq0, 1q ÝÑ pqremember 1 look for \ go right,\, Rq

pqremember 0 look for \ go right,\q ÝÑ pqwrite 0,\, Lq

pqremember 0 look for \ go right, 0q ÝÑ pqremember 0 look for \ go right, 0, Rq


pqremember 1 look for \ go right,\q ÝÑ pqwrite 1,\, Lq



pqwrite 0,\q ÝÑ qrej.pqwrite 0, 0q ÝÑ pqlook for \ go left,\, Lq

pqwrite 0, 1q ÝÑ qrej.pqwrite 1,\q ÝÑ qrej.pqwrite 1, 0q ÝÑ qrej.pqwrite 1, 1q ÝÑ pqlook for \ go left,\, Lq

pqlook for \ go left,\q ÝÑ pqstep right,\, Rq

pqlook for \ go left, 0q ÝÑ pqlook for \ go left, 0, Lq

pqlook for \ go left, 1q ÝÑ pqlook for \ go left, 1, Lq

pqstep right,\q ÝÑ qacc.pqstep right, 0q ÝÑ pqremember 0 look for \ go right,\, Rq

pqstep right, 1q ÝÑ pqremember 1 look for \ go right,\, Rq


If we rename the states :

q0 ; q0

qremember 0 look for \ go right ; q1

qremember 1 look for \ go right ; q2

qwrite 0 ; q3

qwrite 1 ; q4

qlook for \ go left ; q5

qstep right ; q6

the transition function becomes:

pq0,\q ÝÑ qacc.pq0, 0q ÝÑ pq1,\, Rq

pq0, 1q ÝÑ pq2,\, Rq

pq1,\q ÝÑ pq3,\, Lq

pq1, 0q ÝÑ pq1, 0, Rq


pq2,\q ÝÑ pq4,\, Lq



pq3,\q ÝÑ qrej.pq3, 0q ÝÑ pq5,\, Lq

pq3, 1q ÝÑ qrej.pq4,\q ÝÑ qrej.pq4, 0q ÝÑ qrej.pq4, 1q ÝÑ pq5,\, Lq

pq5,\q ÝÑ pq6,\, Rq

pq5, 0q ÝÑ pq5, 0, Lq

pq5, 1q ÝÑ pq5, 1, Lq

pq6,\q ÝÑ qacc.pq6, 0q ÝÑ pq1,\, Rq

pq6, 1q ÝÑ pq2,\, Rq

Definition 29 A language L is Turing recognizable if there exists a TM M such that

L “ LpMq.

Proposition 30 Turing Machines with bi-infinite tapes are equivalent to Turing machines.

Proof of Proposition 30: Left as an exercise.

% 30

Proposition 31 2 stack Pushdown automata are equivalent to Turing machines.


% 31

Definition 32 A Decider is a TM that halts on all inputs.

Recursivity 31

Definition 33 A language is Turing decidable iff there exists a Decider that recognises it.

Turing recognizable is also called recursively enumerable (r.e. for short) and Decidable is also

called recursive.

Example 34 A Decider for tanbncn | n P wu:

˝ Scan the input from left to right to be sure that it is a member of a˚b˚c˚ and reject if it

isn’t.

˝ Return the head to the left and change one c into an x, then one b into x, then one a into

x. Go back to the first blank \.

Repeat again until the tape is only composed of x, in which case accept. Otherwise reject.

Definition 35 A k tape TM is the same as a TM except that is composed of k tapes: 1 ,. . . , k ,

with k independent heads so that the transition function becomes

δ : Qˆ Γk ÝÑ Qˆ Γk ˆ tL,Ruk

Notice that a configuration of a k-tape Turing machine is of the form

´

u1qv1 , u2qv2 , . . . . . . , ukqvk

¯

1 2 k.

Proposition 36 Given any TM there exist

(1) an equivalent TM with a bi-infinite tape,

(2) a multi-tape TM,

(3) a multi-tape with bi-infinite tapes TM.

Proof of Theorem 36: Left as an exercise. % 36

Theorem 37 Every multi-tape TM has an equivalent single tape TM.

Proof of Theorem 37: Let M be a multi-tape TM. We will describe a TM S that recognises

the same language. Let pw1, w2, . . . , wkq be the input of M on its k tapes. The corresponding

input of S will be 7w17w27 . . . 7wk7, where 7 does not belong to the alphabet of M. To simulate a

single move of M, S scans its tape from the first 7 which marks the left-hand end, to the k`1th

7 (which marks the right-hand end) replacing each letter a right after the 7 symbol (except for

the k ` 1th one) by a to indicate the position of the heads. Then S makes a second pass to


update the tapes according to M’s transition functions. If at any point S moves one of the

virtual heads to the right onto a 7, this action signifies that M has moved the corresponding

head onto the previously unread blank portion of that tape. So S writes a blank symbol on this

tape cell and shifts the tape contents from this cell until the rightmost 7, one unit to the right.

Then it continues the simulation as before.

00 01 t t

01 1 t t t

0 01 1

t t t t

]]]] 1 1

t

10 01 1

t

00 01 t t t10 bt

M

S% 37

2.2 Non-Deterministic Turing Machines

Definition 38 A non-deterministic TM (NTM) is the same as a deterministic TM except for

the transition function which is of the form:

δ : Qˆ Γ ÝÑ PpQˆ Γˆ tL,Ruq.

The computation of a (deterministic) TM is a sequence of configurations

C0 ùñ C1 ùñ . . . ùñ Ck ùñ . . .

that may be finite or infinite.

It accepts the input if this sequence is finite and the last configuration is an accepting one.

The computation of a non-deterministic TM is no more a sequence of configurations but a tree

whose nodes are configurations. This tree may have both infinite and finite branches. The

machine accepts the input if and only if there exists some branch that is finite and whose leaf is

an accepting configuration.

Recursivity 33

Theorem 39 For every NTM there exists a deterministic TM that recognises the same lan-

guage.


2434 2 24 t t t2

Mt t t t

t t t

t t

0 0

0 0

1

111

1

1 1 1 176

1

2

3

We consider a 3-tape ( 1 , 2 and 3 ) deterministic TM M to simulate a NTM N :

˝ (1) 1 is the input tape,

(2) 2 is the simulation tape, and

(3) 3 is the address tape.

˝ Initially, 1 contains the input w and 2 and 3 are empty.

˝ 1 always keeps the input w. So the content of 1 is never modified.

˝ 2 simulates N on one – initial segment of a – branch of its non-deterministic computation

tree.

˝ 3 contains a finite word which corresponds to a succession of non deterministic choices.

For instance the word 132 stands for: among the non-deterministic options choose the first

one for the first transition, the third one for the second and the second one for the third.

This means that we consider k P N to be

maxtCardpδpq, γqq | q P Q, γ P Γu

and for each | q P Q, γ P Γ we fix a total ordering of δpq, γq.

Words on 3 all belong to t1, 2, . . . , ku˚. Moreover, during the running time, the content

of 3 changes over and over again until the machine accepts. This series gives rise to an

enumeration of the infinite k-ary tree in a breadth-first search. This means it enumerates

all words in t1, 2, . . . , ku˚ along the following well-ordering:

u ă v ðñ

|u| ă |v|

or

|u| “ |v| and u ălexic. v


Which gives:

ε, 1, 2, . . . , k, 11, 12, . . . , 1k, 21, 22, . . . , 2k, . . . . . . , k1, k2, . . . , kk, 111, 112, . . . , 11k, . . . . . . . . . . . .

˝ At first, M Copies the content of 1 (= the input w) to 2 .

˝ It then uses 2 to simulate N with input w on the branch b of its non-deterministic

computation which is lodged on 3 . In case the word b does not correspond to a real

computation1 or if the simulation of N on 2 either reaches the rejecting state or does not

reach any halting state at all, then M erases completely 2 , replaces b on 3 with its the

immediate ă-successor, and starts all over again – by copying 1 on 2 and simulating Non 2 in accordance with the series of choices recorded on 3 .

% 39

Proposition 40

˝ Recursive languages are closed under union, intersection and complementation.

˝ Recursively enumerable languages are closed under union and intersection.


% 40

Definition 41 An enumerator is a TM. We say that it enumerates a language L if the result

of its computation (possibly infinite) is of the form

w0 \ w1 \ w2 \ . . .\ wn \ wn`1 \ . . . . . .

where twi | i P Nu “ L.

We say that a language L is “recursively enumerable” if there is an enumerator that enumerates

L.

Theorem 42 A language is Turing Recognizable if and only if it is recursively enumerable.


(ñ) from M we build E that enumerates LpMq. Fix a recursive enumeration psiqiPN of Σ˚.

(1) Repeat the following for i “ 1, 2, 3, . . .

(2) Run M for i steps on each input s1, s2, . . . , si

(3) If any computation accepts, print out the corresponding sj .

1this is the case for instance if from the initial configuration q0w there are only two control states non-

deterministically available, whereas the word on 3 reads 3 . . ..

Recursivity 35

(ð) From E we build M: on input w: run E , and every time E outputs some word v, find out

whether v “ w or not, and accept if they are the same.

% 42

Proposition 43 For any infinite L Ď Σ˚,

L is Turing decidable ðñ

$

’

’

’

’

’

’

’

’

’

’

’

’

’

’

’

&

’

’

’

’

’

’

’

’

’

’

’

’

’

’

’

%

there exits E an enumerator that prints out

u0 \ u1 \ u2 \ . . . . . .\ ui \ ui`1 \ . . . . . . . . .

such that

$

’

’

’

’

’

’

’

’

’

&

’

’

’

’

’

’

’

’

’

%

L “ tui | i P Nu

and

i ă j ùñ

$

&

%

|u| ă |v|

or

|u| “ |v| and u ălexic. v.


% 43

2.3 The Concept of Algorithm

In 1900, Hilbert gave a list of the main mathematical problems of the time [26, 27]. The 10th

one was the following: given a Diophantine equation with any number of unknown quantities,

and with rational integral numerical coefficients, can we derive a process according to which

it can be determined in a finite number of operations whether the equation admits a rational

integer solution? This corresponds to the intuitive notion of an algorithm. Proving that such an

algorithm does not exist requires a formal definition of the notion of “algorithm”. The “Church-

Turing thesis” states that the informal notion of an algorithm corresponds exactly to the notion

of a λ-calculus formula or equivalently to a Turing machine.

In 1970, Yuri Matijasevic proved2 that the 10th problem of Hilbert is undecidable [33]: assuming

that the notation P px1, . . . , xnq stands for a polynomial with integer coefficients, then there is

no decider for

tP px1, . . . , xnq | Dpa1, . . . anq P Nn P pa1, . . . , anq “ 0u.

Definition 44 A “coding” is a rule for converting a piece of information into another object.

Given any non empty sets E,F , a coding is a one-to-one (total) function

c : E1´1ÝÝÑ F.

2this is combined work of Martin Davis, Yuri Matiyasevich, Hilary Putnam and Julia Robinson


Example 45 E “ t0, 1u˚, F “ N and c : E1´1ÝÝÑ F is a coding defined by:

cpwq “ 1w2p= the word “1w” read in base 2q.

Notation 46 Given any Turing machine M, we write

˝ Mpwq Ó to say that the machine M stops on input w

‚ Mpwq Ó acc

.

means that M stops in an accepting configuration, and

‚ Mpwq Ó rej.

means that M stops in a rejecting configuration.

˝ Mpwq Ò to say that the machine M never stops on input w.

Definition 47 Given any two non-empty finite sets A,B, a partial function f : A˚ ÝÑ B˚ is

“Turing computable” if and only if there exists a Turing machine Mf such that

˝ on input w R dompfq: Mf pwq Ò, and

˝ on input w P dompfq: Mf pwq Ó acc

.

with the word “fpwq” on its tape.

Remarks 48

(1) Given any finite alphabet Σ, and any TM M whose alphabet is Σ, there exists a Turing

computable coding: c : Σ˚ ÝÑ t0, 1,\u˚ and a TM Mc with tape alphabet t0, 1,\u such

that M accepts w if and only if Mc accepts cpwq.

(2) Every regular language is decidable because a DFA is nothing but a deterministic TM that

always goes right.

(3) Every Context-free language is decidable, because any PDA can be easily simulated by

some equivalent non-deterministic TM.

(4) We have the following strict inclusions of languages.

Regular Ĺ Context-Free Ĺ Decidable

“

Recursive

Ĺ Turing Recognizable.

“

Recursively Enumerable

In computer science, a programming language is said be “Turing complete” or “universal” if

it can be used to simulate any single-tape Turing machine. Examples of Turing-complete pro-

gramming languages include:

Recursivity 37

˝ Ada

˝ C

˝ C++

˝ Common Lisp

˝ Java

˝ Lisp

˝ Pascal

˝ Prolog, etc.

2.4 Universal Turing Machine

If we compare a Turing Machine with a computer, on one hand the TM seems much better

because it can compute for ever without any chance to breakdown and it has an infinitely large

storage facility. But on the other hand, a TM seems to be more of a computer with a single

software program, whereas a computer can run different programs. A computer resemble more

of a Turing machine with finite capacity but, a Turing machine that we can modify by changing

its transition function – every program is like a new transition function for the machine.

How are we going to address this issue, since we claimed that a Turing machine is an abstract

model of computation ? This answer to this is the Universal Turing Machine. It is a machine

that can work just like any other machine provided that we feed it with the right code of the

machine.

We will employ Turing machines to obtain:

(1) a languages that is Turing recognizable but not decidable3,

(2) a language that is not Turing recognizable.

From now on, we only consider Turing Machines with fixed alphabets Σ “ t0, 1u,Γ “ t0, 1,\u.

Any such TM is of the form:

M “ xtq0, q1, . . . , qku, t0, 1u, t0, 1,\u, δ, q0, qacc., qrej.y

Where δ is the description of the transition function of M:

δ “ tpq3, 0, q1, 1, Rq, pq8, 1, q4, 0, Lq, pq3, 0, q3, 0, Lq, . . . . . .u

The description of such a machine is a finite sequence M over some finite alphabet A:!

x, y, q, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, \, L, R, t, u, p, q, ,)

“ A.

Since CardpAq ă 28 we can code any letter l P A by a sequence of eight 0’s and 1’s, i.e we take

any 1-1 mapping

C : A ÝÑ t0, 1ur8s

and we define a Turing computable coding

c : A˚ ÝÑ t0, 1u˚

by

cpa0 . . . apq “ Cpa0q Cpa1q Cpa2q . . .ˆCpapq.

3in other words: a non-recursive recursively enumerable language.


We denote by xMy the code of M, i.e.

xMy “ cpMq.

Clearly, the following language is decidable:

txMy : M is a TMu.

Proposition 49 (Universal Turing Machine) There exists a Turing machine4 U such that

on each input of the form vw P t0, 1u˚,

if v “ xMy for some Turing machine5 M, then U works as M on input w.

Notice that for any word u P t0, 1u˚, if there is a prefix of u which is the code of a Turing

Machine, then this prefix is unique 6. Therefore, in case a word u P t0, 1u˚ can be decomposed

into u “ xMyw for some Turing machine M, this decomposition is then unique.

This means for instance that on any input w:

˝ Upwq Ò if and only if Mpwq Ò;

˝ Upwq Ó rej.

with the word w1 on its tape if and only if Mpwq Ó rej.

with the word w1 on its tape;

˝ Upw1q Ó acc.

with w1 on its tape if and only if Mpwq Ó acc.

with w1 on its tape.

(1) on 1 the input xMyw is inserted. It will never be modified during the rest of the compu-

tation. Then U copies the code of

(a) the transition function of M – xδy – on 2 ;

(b) the initial state of M – xq0y – on 3 7;

(c) the accepting state of M – xqacc.y – on 4 8;

(d) the rejecting state of M – xqrej.y – on 5 9.

(2) It then uses 6 to simulate M on input w: for each step of M

(a) U reads a letter – say 0 – on 6 , and

(b) using the code of the actual state – say xq3y – on 3 , U looks in 2 for the code of the

corresponding transition – say xpq3, 0, q1, 1, Rqy – and then

4working on alphabets ΣU “ t0, 1u and ΓU “ t0, 1,\u5also working on alphabets ΣU “ t0, 1u and ΓU “ t0, 1,\u6this comes from the fact the last letter of a word that defines a TM is y. Therefore, reading u from left to

right by blocks of eight 0’s or 1’s, the first block that corresponds to xyy marks the end of the wanted prefix.7later on this tape will store the code of the actual state that M is on.8the content of 4 will never be modified in the future.9the content of 5 will never be modified in the future.

Recursivity 39

(c) U verifies that the code of the new state – here xq1y – is different from the content of

4 and 5 (otherwise if it corresponds to the content of 4 it means that it is xqacc.y,

and U accepts right away, and if it corresponds to the content of 5 it means it is

xqrej.y, in which case U rejects).

(d) If the new state is different from both qacc. and qrej. – in our example q1 is different

from both qacc. and qrej. – U replaces on 6 the letter it just read with the new one

– here it replaces 0 by 1 – and still on tape 6 it makes the move indicated – here it

goes right – and finally,

(e) U replaces on 3 the code of the old state by the new one – here it replaces xq3y by

xq1y.

U

0 01 1 1111 1 1 00 01 t t10 1

M

01 1 01 1

00 01 1 01 1

00 01 00 0 1

00 01 1 01 1

0

0 01 111 1 10

0 01 1 1111 1 1 00 01 t t10 10

4

5

6

1

2

3

% 49


2.5 The Halting Problem

Proposition 50 The following language is Turing recognizable but not decidable:

txMyw P t0, 1u˚ |M is a TM that accepts wu.

Proof of Proposition 50: Towards a contradiction we assume there exists a Decider D that

decides this language. We build a Turing machine H which works the following way:

on input w

˝ if D accepts ww, then H does not halt.

˝ if D rejects ww, then H accepts.

Notice that

H accepts xHy ðñ D rejects xHyxHy ðñ H does not accept xHy.

Or to say it differently

HpxHyq Ó acc.

ðñ DpxHyxHyq Ó rej.

ðñ HpxHyq Ò .

To see things slightly differently, since the machine H only stops when it accepts we can refor-

mulate the contradiction in

HpxHyq Óðñ HpxHyq Ò .

% 50

Proposition 51 The following language is Turing recognizable but not decidable:

txMy P t0, 1u˚ |Mpεq Óu.


% 51

Corollary 52 The following languages are not recursively enumerable:

(1) t0, 1u˚z

xMyw P t0, 1u˚ |Mpwq Ó acc

.

(

(2) t0, 1u˚z

xMy P t0, 1u˚ |Mpεq Ó(

(3)

xMyw P t0, 1u˚ |Mpwq Ó rej.

or Mpwq Ò(

(4)

xMy P t0, 1u˚ |Mpεq Ò(

.

Proof of Corollary 52: Left as an exercise.

% 52

Recursivity 41

2.6 Turing Machine with Oracle

A Turing machine with an oracle is one finite object (a Turing machine suitable for any oracle:

an almost regular 2-tape Turing Machine) plus one infinite object so that this TM can have

access to an infinite amount of information – which a normal one never does.

Definition 53

(1) An oracle is any subset O Ď N.

(2) An oracle-compatible-Turing machine (o-c-TM) is a 2-tape Turing machine similar to any

2-tape Turing machine except that it only reads but never writes on tape 2 :

O “ pQ,Σ,Γ, δ, q0, qacc., qrej.q

(3) An oracle-compatible-Turing machine O equipped with the oracle O, on input word w P Σ˚

(in short an oracle TM OO on word w P Σ˚) is nothing but the TM O whose initial

configuration is´

q0w , q0χO

¯

1 2

where χO P t0, 1uω is the infinite word

χOp0qχOp1qχOp2q . . . . . . χOpnqχOpn` 1q . . . . . . . . . . . .

defined by

χOpnq “

$

’

’

&

’

’

%

1 if n P O

and

0 if n R O.

This means that on tape 2 the whole characteristic function of the oracle is already available

once the machine starts. So that the machine is granted access to all of this ”external” infor-

mation: it knows which integers belong to O and which do not. For instance, in case O is the

set of all integers n such that:

(1) n reads “ 1xMyw ” in the decimal numeral system,

(2) Mpεq Ó;

then OO may be able to decide the Halting Problem. Of course this does not lead to a contra-

diction since there is no chance that such a Turing machine 10 ever sees its own code onto tape

2 (although the code of O – or the code of an equivalent TM – does show on 2 ).

10we are talking about OO and not just O!


OO

0 00 01 10

1 111 1 101

2

t t

00 000 0 0 00 0 0 0 0

Example 54 Let O Ď N be the set of all the codes of Turing machines that halt on the empty

input:

O “ t1xMy2P N |Mpεq Óu.

We describe an oracle-compatible-TM O that, once equipped with the oracle O, decides the

language

txMy P t0, 1u˚ |Mpεq Óu.

The machine OO works this way:

(1) on input w P t0, 1u˚, the TM OO checks whether w is the code of a TM

if it is not the case it rejects right away. Otherwise,

(2) it computes n “ 1xMy2, then checks on tape 2 whether χOpnq “ 1 – in which case it

accepts – or χOpnq “ 0 – in which case it rejects.

Notation 55

˝ Notice that the mapping f : t0, 1u˚ ÐÑ Nw ÞÝÑ 1w

2´ 1

is a bijection.

For any word w we write xwy for fpwq, and for any integer k we write xky for f´1k.

For instance x0010y “ 100102´ 1 “ 18 ´ 1 “ 17, and x12y “ 101 since 1101

2´ 1 “

p8` 4` 1q ´ 1 “ 13´ 1 “ 12.

˝ Given any language L Ď t0, 1u˚, we write OL Ď N for the set

OL “

!

xwy P N | w P L)

“

!

k P N | xky P L)

.

˝ Given any subset O Ď N, we write LpOq Ď t0, 1u˚ for the language

LpOq “!

w P t0, 1u˚ | xwy P O)

“

!

xky P t0, 1u˚ | k P O)

.

Recursivity 43

So OL is the oracle associated with the language L, and LpOq is the language associated with

the oracle O.

Notice that the oracle for the empty language is the empty set: OH “ H.

So, we have

(1) a coding for the Turing machines:

tx, y, q, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1,\, L,R, t, u, p, q, ,u˚ÐÑ t0, 1u˚

M ÞÝÑ xMy

(2) a coding for the words:

t0, 1u˚ ÐÑ N

w ÞÝÑ xwy

(3) a coding for the integers:

N ÐÑ t0, 1u˚

k ÞÝÑ xky

We will use the notation xMy instead of xxMyy which means we consider first the word in t0, 1u˚

that codes the Turing machine M, then the integer that codes this word. All we mean is that

xMy is an integer that codes the Turing machine M.

Proposition 56 Given any recursive language L Ď t0, 1u˚, and any oracle Turing machine

OOL:

˝ LpOOLq is recursively enumerable, and moreover

˝ if OOL is an oracle Decider11, then LpOOLq is recursive.


% 56

Definition 57 (Turing Reducibility) Given any A,B Ď N,

A is “Turing reducible” to B – denoted A ďT B – if there exists an o-c-TM OB which on empty

tape computes χA.

Proposition 58 Given any A,B Ď N, the following are equivalent:

(1) A is Turing reducible to B,

11meaning that OOL halts on every input.


(2) for every o-c-TM M, there exists an o-c-TM N such that L`

MA˘

“ L`

NB˘

.

Moreover, in case MA is an oracle Decider, we may ensure that NB be one too.

Proof of Proposition 58: Left as an easy exercise.

% 58

Notation 59 Given any A,B Ď N, we write

˝ A ďT B if A is Turing reducible to B;

˝ A ”T B if A ďT B and B ďT A;

˝ A ăT B if A ďT B but B ďT A.

Notice that we have

˝ A ďT A`

hence A ”T A˘

;

˝ pA ďT B and B ďT Cq ùñ A ďT C`

hence pA ”T B and B ”T Cq ùñ A ”T C˘

;

˝ A ”T B ðñ B ”T A.

So that ”T is an equivalence relation.

Examples 60 Given any language L Ď t0, 1u˚,

(1) OL ”T OLA “ NrOL,

(2) H ďT OL,

(3) L is recursive ðñ OL ”T H,

(4) L is not recursive ðñ H ăT OL.

(5) OL ”T OLpOLq holds since we have OL “ OLpOLq .

An equivalence class – rAs”T “ tB Ď N | B ”T Au for some A Ď N – is called a Turing degree.

The ordering on oracles induces an ordering on the set TD of all Turing degrees: given any

d, e P TD,

d ď e ðñ A ďT B holds for some A P d and some B P e

or equivalently

d ď e ðñ A ďT B holds for all A P d and all B P e

Recursivity 45

As usual, the notation

d ă e ðñ d ď e but e ę d

Examples 61 We list a few basic facts about Turing degrees.

(1) Given any d P TDcardpdq “ ℵ0.

The reason is that there are countably many Turing machines and always infinitely many

oracle that are Turing equivalent: for instance, given any A Ď N and any k P N form

Ak “ t2n | n P Au Y t2k ` 1u

We have both A ”T Ak (any k P N) and Ak ‰ Al (any k ‰ l P N).

(2) Given any set A Ď N the set

tB Ď N | B ďT Au

is countable for the reason that there are only countably many Turing machines.

(3) Given any d P TDcard

e P TD | e ď d(

ď ℵ0.

(4) Given any d P TDcard

e P TD | d ď e(

“ 2ℵ0 .

To see this, observe that if

d “ rAs”T

then given any B Ď N the set

A‘B “ t2n | n P Au Y t2n` 1 | n P Bu

satisfies

A ďT A‘B.

Moreover,

card

A‘B | B Ď N(

“ 2ℵ0 .

Since every Turing degree is countable, we obtain

card´

A‘B | B Ď N(

L

”T

¯

“ 2ℵ0

which gives the result.

(5) As Sacks showed in 1961 – see [23] p. 157 and also [40, 41] – the ordering pTD,ďq does

not have a familiar shape since every countable partial ordering pP,ďq can be embedded

into pTD,ďq.


Proposition 62

(1)!

LpOq Ď t0, 1u˚ | O ďT H)

is the class Rec. of all recursive languages.

(2)!

L Ď t0, 1u˚ | OL ďT Halt

)

Ľ R.E . (= the class of all r.e. languages)12.

(3)!

LpOq Ď t0, 1u˚ | O ďT Halt

)

Ľ R.E .

Where Halt stands for the set of codes of Turing machines that halt on the empty input:

Halt “ O xMyPt0,1u˚ | MpεqÓ

(

“

xMy P N | Mpεq Ó(

.


% 62

We now introduce an operation called the “jump” which shows that there is no maximum Turing

degree, since from any given oracle A it provides us with some oracle A1 that satisfies A ăT A1.

Definition 63 (jump operator) Given any subset A Ď N, the “jump” of A (denoted A1) is

A1 “ O xMyPt0,1u˚ | M an o-c-TM, MA

pεqÓ(

“

xMy P N | MApεq Ó(

.

Example 64

Halt ”T H1.

Proposition 65 For every A Ď N the set

A: “

α2pxMy, xwyq P N | MApwq Ó(

satisfies

A1 ”T A:.

12notice that the inclusion is strict since HAalt satisfies both HAalt ”T Halt and LpHAaltq R Rec.

Recursivity 47

¨

˚

˝

See page 60 for the definition of α2 : Nˆ N bij.ÐÝÑ N

px, yq ÞÝÑpx`yq¨px`y`1q

2 ` y.

˛

‹

‚

Proof of Proposition 65: Left as an easy exercise. % 65

Proposition 66 For every A Ď N,

A ăT A1.

Proof of Proposition 66: We decompose A ăT A1 into first A ďT A

1, then A1 ďT A.

(A ďT A1) We need to find an o-c-TM M that outputs χA while being equipped with the oracle

A1. To compute χApnq this machine proceeds as follows: it computes the code xN ny of any

o-c-TM N n that, no matter what its input w is, proceeds as follows when it is equipped

with the oracle O:

˝ if χOpnq “ 1, then N npwq Ó;

˝ if χOpnq “ 0, then N npwq Ò.

Then MA1 outputs

χApnq “ χA1pxN nyq.

(A1 ďT A) Towards a contradiction, we assume that A1 ďT A holds. Since A: ”T A1 we have

A: ďT A holds as well. So, there exists an o-c-TM N such that NA computes χA: .

We build an o-c-TM H such that HA on every input w P t0, 1u˚:

(1) computes k “ α2pxwy, xwyq, then

(2) by making use of N as a subprogram, computes the value χA:pkq, then

˝ if χA:pkq “ 0, then HApwq Ó;

˝ if χA:pkq “ 1, then HApwq Ò.

We obtain the following contradiction:

HApxHyq Ó ðñ α2pxHy, xHyq R A: ðñ HApxHyq Ò .

Below we show a picture that illustrates this diagonal argument that we have just used. If

pMiqiPN is a enumeration of all the oracle-compatible Turing machines, then we made sure

that the machine H we built is none of them by ensuring that for each i P N, there exists


an input word (its own code xMiy) such that HA has a completely different behaviour

than MAi on this word.

MA0 MA

1 MA2 MA

3 MA4 MA

5 MAn

xM0y 0 1 1 0 1 0 . . . 0 . . .

xM1y 1 1 1 0 0 0 . . . 0 . . .

xM2y 1 0 1 0 0 0 . . . 1 . . .

xM3y 0 0 1 0 1 0 . . . 0 . . .

xM4y 0 1 0 1 1 1 . . . 0 . . .

xM5y 1 1 0 0 0 0 . . . 0 . . .

......

......

......

......

xMny 1 0 0 0 1 1 . . . 1 . . ....

......

......

......

...

Diagonal argument: swap 0’s and 1’s on the diagonal.

% 66

Corollary 67 The following strict ordering between jumps is satisfied:

H ăT H1 ăT H

2 ăT . . . ăT H

nhkkikkj

2 ¨ ¨ ¨ 1 ăT H

n`1hkkikkj

2 ¨ ¨ ¨ 1 ăT . . . ăT H

ωhkkikkj

2 ¨ ¨ ¨ ăT H

ω`1hkkikkj

2 ¨ ¨ ¨ 1 ăT . . . .

where

k P H

ωhkkikkj

2 ¨ ¨ ¨ 1 ðñ

$

’

’

’

’

&

’

’

’

’

%

k “ pn`mqpn`m`1q2 `m

and

m P H

nhkkikkj

2 ¨ ¨ ¨ 1 .

Proof of Corollary 67: Let us use the notations Hpnq for H

nhkkikkj

2 ¨ ¨ ¨ 1 and Hpωq for H

ωhkkikkj

2 ¨ ¨ ¨ .

Recursivity 49

The only thing one needs to prove is that

Hpnq ăT Hpωq

holds for every integer n.

Hpnq ďW Hpωq is almost immediate, since it is straightforward to build an o-c-TM On that

outputs χHpnq when it is equipped with the oracle Hpωq since

χHpnqpmq “ χHpωq

ˆ

pn`mqpn`m` 1q

2`m

˙

.

Hpωq ďW Hpnq it is enough to proceed by contradiction and show that

Hpωq ďW Hpnq

would imply

Hpn`1q ďW Hpnq.

% 67

Iterating the jump operator into the transfinite

Notice that if for every limit countable ordinal λ we fix some bijection

fλ : N ÐÑ λˆ Nk ÞÝÑ pα,mq

we may then define an uncountable sequence of jumps`

Hpαq˘

αăω1by ordinal induction:

˝ Hp0q “ H

˝ Hpα`1q “ Hpαq1

˝ Hpλq “ tfλpα,mq P N | m P Hpαqu.

It is immediate to see that the sequence`

Hpαq˘

αăω1is strictly ăT -increasing, or in other words

Hpαq ăT Hpβq

holds for every α ă β ă ω1.


Chapter 3

Recursive Functions

The whole chapter is highly inspired by Rene Cori and Daniel Lascar book’s book: “Math-

ematical Logic, Part 2, Recursion Theory, Godel Theorems, Set Theory, Model Theory” [9].

Recursive functions are functions from Np to N. We will show that they have a strong relation

with the Turing computable ones.

We define the set of recursive functions by induction. For this purpose, for any integer p, we

denote by NpNpq the set of all mappings of the form Np ÝÑ N. Notice that Np is a notation for

the set of all mappings ti P N | i ă pu ÝÑ N. When p “ 0, the set ti P N | i ă pu becomes

ti P N | i ă 0u “ H. Thus the set N0 only contains one element: the empty function whose

graph is H. Therefore the set of all mappings of the form N0 ÝÑ N contains all mappings that

assign one integer to the empty function:

N0 ÝÑ N “"

f : tHu ÝÑ NH ÝÑ n

ˇ

ˇ

ˇ

ˇ

n P N*

.

So, as may be expected, mappings in NpN0q are identified with elements of N.

3.1 Primitive Recursive Functions

Definition 68

projection: If i is any integer such that 1 ď i ď p holds, the ith projection πpi is the function

of NpNpq defined by

πpi px1, . . . , xpq “ xi

successor: S P NN is the successor function1.

1Spnq “ n` 1.


composition: Given f1, . . . , fn P NpNpq and g P NpNnq, the composition h “ gpf1, . . . , fnq P NpN

pq

is defined by

hpx1, . . . , xpq “ g`

f1px1, . . . , xpq, . . . , fnpx1, . . . , xpq˘

We often make use the notation ÝÑx for px1, . . . , xpq so that for instance g`

f1pÝÑx q, . . . , fnpÝÑx q

˘

stands for g`

f1px1, . . . , xpq, . . . , fnpx1, . . . , xpq˘

.

recursion: Given g P NpNpq and h P NpNp`2q, there exists a unique f P NpNp`1q such that for allÝÑx P Np and y P N satisfies

(1) fpÝÑx , 0q “ gpÝÑx q

(2) fpÝÑx , y ` 1q “ h`

ÝÑx , y, fpÝÑx , yq˘

We say f is defined by recursion on both g (for the initial step) and h (for the successor

steps).

Definition 69 The set of primitive recursive (Prim. Rec.) functions is the least that

(1) contains:

(a) All constants Np ÝÑ N (all i P NpNpq s.t. ipÝÑx q “ i – any i, p P N).

(b) All projections πpi (any p P N, any 1 ď i ď p)

(c) The successor function S P NN.

(2) and is closed under

(a) composition

(b) recursion

We set up these functions in a hierarchy pRnqnPN:

(1) R0 is the set of all functions in (1)(a),(1)(b) and (1)(c).

(2) Rn`1 is the closure2 of Rn under (2)(a) and (2)(b).

Clearly

R “ď

nPNRn.

Example 70

2Rn`1 “ Rn Y th obtained by composition on the basis of functions in Rnu Y th obtained by induction on the

basis of functions in Rnu.

Recursivity 53

(1) Addition: px, yq ÝÑ x` y

We have:"

x` 0 “ x

x` py ` 1q “ px` yq ` 1.(3.1)

Formally:#

addpx, 0q “ π11pxq

addpx, y ` 1q “ S´

π33

`

x, y, addpx, yq˘

¯

.(3.2)

(2) Multiplication: px, yq ÝÑ x ¨ y

We have"

x ¨ 0 “ 0

x ¨ py ` 1q “ x ¨ y ` x.(3.3)

Formally:

#

multpx, 0q “ 0pxq

multpx, y ` 1q “ add´

π33

`

x, y,multpx, yq˘

, π31

`

x, y,multpx, yq˘

¯

.(3.4)

(3) Exponentiation: x ÝÑ nx

We have"

n0 “ 1

nx`1 “ nx ¨ n.(3.5)

Formally:#

expnp0q “ 1

expnpx` 1q “ mult´

π22

`

x, expnpxq˘

, n¯

.(3.6)

(4) Factorial: x ÝÑ x!

We have"

0! “ 1

px` 1q! “ x! ¨ px` 1q.(3.7)

Formally:$

&

%

factp0q “ 1

factpx` 1q “ mult

ˆ

π22

`

x, factpxq˘

, S´

π21

`

x, factpxq˘

¯

˙

.(3.8)


Example 71 We define 9 P NpN2q by

"

x 9 y “ x´ y if x ą y,

“ 0 otherwise.

We show that 9 P NpN2q belongs to Prim. Rec. by first showing that : x ÝÑ x 9 1 belongs to

Prim. Rec."

0 9 1 “ 0

px` 1q 9 1 “ x(3.9)

Formally:"

0 9 1 “ 0pxq

px` 1q 9 1 “ π21

`

x, x 9 1˘ (3.10)

"

x 9 0 “ x

x 9 py ` 1q “`

x 9 y˘

9 1(3.11)

Formally:#

x 9 0 “ π11pxq

x 9 py ` 1q “´

π33

`

x, y, x 9 y˘

¯

9 1(3.12)

Definition 72 A set A Ď Np is primitive recursive (Prim. Rec.) if its characteristic function

(χA P NpNpq) is primitive recursive.

Example 73

(1) The set H is Prim. Rec. since χH “ 0 is Prim. Rec.

(2) The set N is Prim. Rec. since χN “ 1 is Prim. Rec.

(3) The set ăN“ tpx, yq | x ă yu is Prim. Rec. χăNpx, yq “ 1 9 p1 9 py 9 xqq.

On computable and partial functions

Definition 74

(1) pdomf , fq is a partial function Np ÝÑ N if f is a mapping domf ÝÑ N where domf Ď Np.

(2) pdomf , fq is a total function Np ÝÑ N if domf “ Np holds.

Recursivity 55

We say that f is undefined on x – or fpxq is undefined – if x R domf .

Notation 75 We write f P NpdomĎNpq for pdomf , fq is a partial function Np ÝÑ N whose

domain is domf .

Notice that given any two partial functions f, g P NpdomĎNpq,

f “ g ðñ

$

&

%

domf “ domg

and

@x fpxq “ gpxq.

Definition 76 A partial function f P NpdomĎNpq is “Turing Computable” (TC) if there exists

a TM M such that on input ÝÑx “ px1, . . . , xpq:

(1) if fpÝÑx q is not defined, then MpÝÑx q Ò;

(2) if pÝÑx q P domf , MpÝÑx q Ó with fpÝÑx q written on its tape.

Proposition 77 Given any partial function f P NpdomĎNpq,

f is Turing Computable ðñ Gf “ `

ÝÑx , fpxq˘

| ÝÑx P domf

(

is Turing Recognizable.

Proof of Proposition 77:

(ñ) From M that computes f it is immediate to build N that recognises Gf . On input pÝÑx , yq

it simulates M if MpÝÑx , yq Ó acc.

it compares fpxq with y and if fpxq “ y it accepts, otherwise

it rejects.

(ð) From N that recognises Gf , we build M that computes f as follows, on input ÝÑx repeatedly

for i “ 1, 2, 3, . . . it recursively simulates N on pÝÑx , 0q, pÝÑx , 1q, pÝÑx , 2q, . . . , pÝÑx , iq for i many

steps. If N accepts pÝÑx , nq, then M prints out the value n.

% 77

Corollary 78 Given any function f : Np ÝÑ N,

f is both total and Turing Computable ùñ Gf is recursive (decidable).

Proof of Corollary 78: Left as an immediate exercise.

% 78


f P NpNpq f P NpdomĎNpq

Gf Ď Np`1recursive

||

decidable

rec. enum.

||

Turing rec.

Figure 3.1: Relations between Turing computable functions and their graphs

Remark 79 All functions in R0 are total and Turing computable. By induction on n, it is easy

to show that all functions in Rn are also total and Turing computable. Therefore

˝ All Prim. Rec. functions are total and Turing computable.

˝ All graphs of Prim. Rec. functions are recursive

Even though the class of all Prim. Rec. functions is included in the class of total and turing

computable functions, the inverse inclusion does not hold3.

3.2 Variable Substitution

Proposition 80 (PPPrim. RRRec. closed under variable substitution)

If f P NpNpq, is Prim. Rec., then given any σ : t1, . . . , pu ÝÑ t1, . . . , pu, the function g P NpNpqdefined by

gpx1, . . . , xpq “ fpxσp1q, . . . , xσppqq

is also Prim. Rec.

Proof of Proposition 80: The result is proved for all g P Rn by induction on n. Left as an

exercise.

% 80

Proposition 81 If A Ď Nk is Prim. Rec. and f1, . . . , fn P NpNkq are Prim. Rec. then

tÝÑx P Np | pf1pÝÑx q, . . . , fnpÝÑx qq P Au

is Prim. Rec.3see exercise on the Ackermann function A P NpN

2q defined by

Apm,nq “

$

&

%

n` 1 if m “ 0,

Apm´ 1, 1q if m ą 0 and n “ 0,

A`

m´ 1, Apm,n´ 1q˘

if m ą 0 and n ą 0.

Apm,nq is fast growing ; for instance 2 ¨ 1019728ă Ap4, 2q ă 3 ¨ 1019728.

Recursivity 57

Proof of Proposition 81: Set B “ tÝÑx P Np | pf1pÝÑx q, . . . , fnpÝÑx qq P Au. We have

χBpÝÑx q “ χA

´


¯

.

% 81

Example 82 If f, g P NpNpq are Prim. Rec., then the following sets are also Prim. Rec.:

(1) tÝÑx | fpÝÑx q ą gpÝÑx qu (2) tÝÑx | fpÝÑx q “ gpÝÑx qu (3) tÝÑx | fpÝÑx q ă gpÝÑx qu.

Proposition 83 If A,B Ď Np are Prim. Rec. then A Y B,A X B,A r B, A∆B and Np r A

are all Prim. Rec.


χAYB “ 1 9

`

1 9 pχA ` χBq˘

χAXB “ χA ¨ χBχArB “ χA ¨

`

1 9 χB˘

χA∆B “`

1 9 χA ¨ χB˘

¨

ˆ

1 9

´

`

1 9 χA˘

¨`

1 9 χB˘

¯

˙

χAc “ 1 9 χA.

% 83

Proposition 84 (Case study) If f1, . . . , fn`1 P NpNpq and A1, . . . , An P Np are all Prim. Rec.,

then g P NpNpq defined by:

gpÝÑx q “

$

’

’

’

’

’

’

’

’

’

’

&

’

’

’

’

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’

%

f1pÝÑx q if ÝÑx P A1

f2pÝÑx q if ÝÑx P A2 rA1

f3pÝÑx q if ÝÑx P A3 r pA1 YA2q...

......

fipÝÑx q if ÝÑx P Ai r pA1 YA2 Y . . .YAi´1q...

......

fn`1pÝÑx q if ÝÑx R pA1 Y . . .YAnq

is also Prim. Rec.

Proof of Proposition 84: g “ f1 ¨ χA1 ` f2 ¨ χpA2rA1q ` . . .` fn`1 ¨ χÀ1YA2Y...YAn

˘A . % 84

Corollary 85 sup(x1, . . . , xn) and inf(x1, . . . , xn) are Prim. Rec.


Proof of Corollary 85: Left as an exercise.

% 85

Proposition 86 f P NpNp`1q is Prim. Rec., then g, h P NpNpq below are Prim. Rec.:

gpx1, . . . , xpq “

t“yÿ

t“0

fpx1, . . . , xp, tq,

hpx1, . . . , xpq “

t“yź

t“0

fpx1, . . . , xp, tq.

Proof of Corollary 86: Left as an exercise(both are easily defined by induction).

% 86

3.3 Bounded Minimisation and Bounded Quantification

Proposition 87 (Bounded minimisation) If A Ď Np`1 is Prim. Rec., then f P NpNp`1q

defined below is also Prim. Rec.:

fpÝÑx , zq “

"

0 if @t ď z pÝÑx , tq R A,

the least t ď z such that pÝÑx , tq P A otherwise.

fpÝÑx , zq is denoted by µt ď z pÝÑx , tq P A.

Proof of Proposition 87: f is defined by:$

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fpÝÑx , 0q “ 0

fpÝÑx , z ` 1q “

$

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%

fpÝÑx , zq ify“zř

y“0χApÝÑx , yq ě 1

z ` 1 ify“zř

y“0χApÝÑx , yq “ 0 and pÝÑx , z ` 1q P A

0 ify“z`1ř

y“0χApÝÑx , yq “ 0.

% 87

Proposition 88 (PPPrim. RRRec. closed under bounded quantification) The set of all Prim. Rec.predicates is closed under bounded quantification: i.e. If A Ď Np`1 is Prim. Rec., then

˝ tpÝÑx , zq : Dt ď z pÝÑx , tq P Au ˝ tpÝÑx , zq : @t ď z pÝÑx , tq P Au

are both Prim. Rec.

Proof of Proposition 88: Set

Recursivity 59

˝ B “ tpÝÑx , zq : Dt ď z pÝÑx , tq P Au, ˝ C “ tpÝÑx , zq : @t ď z pÝÑx , tq P Au.

We have

˝ χBpÝÑx , zq “ 1 9 p1 9

t“zř

t“0χApÝÑx , tqq, ˝ χCpÝÑx , zq “

t“zś

t“0χApÝÑx , tqq.

% 88

Example 89

˝ t2n : n P Nu is Prim. Rec. It is defined by recursion"

χp0q “ 0

χpn`1q “ 1 9 χpnq

˝ The mapping P NpN2q px, yq ÝÑ

„

x

y

defined below is Prim. Rec.:$

’

&

’

%

„

x

y

“ 0 if y “ 0

“ integer part ofx

yotherwise.

Formally$

&

%

„

x

y

“ 0 if y “ 0

“ µt ď x y ¨ pt` 1q ą x otherwise.

˝

px, yq | y divides x(

P Prim. Rec.:

χpx, yq “ 1 9

˜

x 9

ˆ

y ¨

„

x

y

˙

¸

.

˝ Prime “ tx P N | x is a prime numberu P Prim. Rec.:

x P Prime ðñ

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x ą 1

and

@y ď x

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y “ 1

or

y “ x

or

y does not divide x.


˝ Π : N ÝÑ N defined by Π pnq “ n` 1th prime number P Prim. Rec.."

Π p0q “ 2

Π pn` 1q “ µz ď pΠ pnq!` 1q z ą Π pnq and z P Prime.

3.4 Coding Sequences of Integers

We define Prim. Rec. functions that allow to treat finite sequences of integers as integers.

Every sequence xx1, . . . , xpy will be “coded” by a single integer αppx1, . . . , xpq. And from this

single integer αppx1, . . . , xpq one will be able to recover the elements of the original sequence by

having Prim. Rec. functions βip that satisfy

βip

´

αppx1, . . . , xpq¯

“ xi.

Proposition 90 For every non-zero p P N there exists Prim. Rec. functions β1p , β

2p , . . . , β

pp P NN

and αp P NpNpq such that

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αp : Np 1´1 and ontoÐÝÝÝÝÝÝÝÝÑ N

and

α´1p pxq “

`

β1ppxq, . . . , β

pppxq

˘

.

Proof of Proposition 90: We start by defining α1 “ β11 “ id. Then we move on to

α2px, yq “px` yq ¨ px` y ` 1q

2` y.

This is obtained by looking at the following picture and noticing that

(1) α2px, yq “ α2px` y, 0q ` y, and

(2) α2px` y, 0q “ 1 ` 2 ` ¨ ¨ ¨ ` px` yq

“ 12

˜ 1

`

x` y

`

2

`

x` y ´ 1

` ¨ ¨ ¨ `

x` y

`

1

¸

.

Recursivity 61

0,4

0,3

0,2

0,0

0,1

1,3

1,2

1,0

1,1

3,2

3,0

3,1

2,3

2,2

2,0

2,1

4,0

4,1

5,0

0 1

2

5

9

14

4

8

13

3

7

12

18

6

11

17

10

16

15

19

We have

(1) β12pnq “ µx ď n Dt ď n α2px, tq “ n

(2) β22pnq “ µy ď n Dt ď n α2pt, yq “ n.

Then we define αp`1, β1p`1, β

2p`1, . . . . . . , β

p´1p`1 , β

pp`1 and βp`1

p`1 by induction on p P N:

˝ αp`1px1, . . . , xp, xp`1q “ αp`

x1, . . . , xp´1, α2pxp, xp`1q˘

˝ β1p`1 “ β1

p ;

˝ β2p`1 “ β2

p ;

...

˝ βp´1p`1 “ βp´1

p ;

˝ βpp`1 “ β12 ˝ β

pp ;

˝ βp`1p`1 “ β2

2 ˝ βpp .

% 90


Example 91 A different way of coding sequences of integers:

"

cpεq “ 1

cpx0, . . . , xpq “ Π p0qx0`1¨Π p1qx1`1

¨ ¨ ¨Π ppqxp`1.

From n P Nrt0u we recover the sequence xx0, . . . , xpy such that cpx0, . . . , xpq “ n by considering

the Prim. Rec. function d P NpN2q which yields the exponents of the prime numbers:

dpi, nq “ µx ď n Π piqx`1 does not divide n.

3.5 Partial Recursive Functions

We recall that

(1) pdomf , fq is a partial function Np ÝÑ N if f is a mapping domf ÝÑ N where domf Ď Np.

(2) pdomf , fq is a total function Np ÝÑ N if domf “ Np holds.

We say that f is undefined on x – or fpxq is undefined – if x R domf .

We use the notation f P NpdomĎNpq to signify that pdomf , fq is a partial function Np ÝÑ Nwhose domain is domf .

Notice that for any two partial functions f, g P NpdomĎNpq:

f “ g holds ðñ

$

&

%

domf “ domg

and

@x fpxq “ gpxq.

Definition 92 (composition) Given f1, . . . , fn P NpdomĎNpq and g P NpdomĎNnq, the composi-

tion h “ gpf1, . . . , fnq P NpNpq is defined by

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hpÝÑx q is undefined iff

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ÝÑx Rč

1ďiďn

domfi

or otherwise

`


˘

R domg.

hpÝÑx q is defined otherwise and hpÝÑx q “ g`


˘

.

“ g`

f1px1, . . . , xpq, . . . , fnpx1, . . . , xpq˘

Recursivity 63

Definition 93 (recursion) Given g P NpdomĎNpq and h P NpdomĎNp`2q, there exists a unique

f P NpdomĎNp`1q such that for all ÝÑx P Np and y P N:

(1)$

&

%

fpÝÑx , 0q is undefined if ÝÑx R domg

and

fpÝÑx , 0q is defined otherwise with fpÝÑx , 0q “ gpÝÑx q.

(2)$

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fpÝÑx , y ` 1q is undefined if

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pÝÑx , yq R domf

or

`


R domh.

and

otherwise fpÝÑx , y ` 1q is defined and fpÝÑx , y ` 1q “ h`


.

Definition 94 (minimization) Given f P NpdomĎNp`1q, we define g P NpdomĎNpq by:

gpÝÑx q “ µy fpÝÑx , yq “ 0.

Notice that

gpÝÑx q “ y ðñ

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@z ă y

$

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fpÝÑx , zq is defined!

and

fpÝÑx , zq ą 0

and

fpÝÑx , yq “ 0.

Definition 95 (partial recursive functions) The set of partial recursive (Part. Rec.) func-

tions is the least that

(1) contains:

(a) All constants Np ÝÑ N (all i P NpNpq s.t. ipÝÑx q “ i – any i, p P N).

(b) All projections πpi (any p P N, any 1 ď i ď p)

(c) The successor function S P NN.

(2) and is closed under


(a) composition

(b) recursion

(c) minimisation.

Our next goal is to show that a function f is in Part. Rec. if and only if it is Turing computable.

One direction is easy, the other one is more involved. One side effect of our proof will show that

every partial recursive function can be obtained by applying the minimisation at most once.

Lemma 96 Every partial recursive function is Turing computable.

Proof of Lemma 96: We need to show that given any p P N r t0u and any f P NpdomĎNpq there

exists some Turing machine M that computes f . This means that on input pn1, . . . , npq it

stops in configuration qacc.fpn1, . . . , npq if pn1, . . . , npq P domf , and it never stops otherwise. Of

course, we need to fix a certain representation of both integers and finite sequence of integers.

For simplicity, let us say that the integers are represented in base-ten and the sequences as

pn1, . . . , npq so that the input alphabet is

Σ “!

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, p, q, ,)

.

So, for instance a TM computes Add if on input word “p385, 218q” it returns the word “603”.

We do the proof by induction on the number of operation among

˝ composition ˝ recursion ˝ minimisation

that are necessary to obtain f P NpdomĎNpq on the basis of

˝ all constants ˝ all projections ˝ the successor function.

(1) It is quite obvious that

˝ if f P NpdomĎNpq is constant, then there exists some basic TM that computes it.

˝ Every projection πpi (any 1 ď i ď p) is also trivially computable.

˝ The successor function is clearly computable as well.

(2) (a) Assume f1, . . . , fn P NpdomĎNpq are computed respectively by Mf1 , . . . ,Mfn and g P

NpdomĎNnq is computed by Mg. Then f “ gpf1, . . . , fnq P NpNpq is computed by Mf

which works as follows:

on input ÝÑx “ pn1, . . . , npq:

successively for each i :“ 1, . . . ,n Mf simulates Mfi on input ÝÑx , if MfipÝÑx q Ó a

cc.

with some output mi it stores mi.

(In case all simulation of machines Mf1 , . . . ,Mfn do stop) Mf finally simulates Mg

on input pm1, . . . ,mnq.

It is clear that if either

Recursivity 65

(A) ÝÑx Rč

1ďiďn

domfi or (B)´


¯

R domg,

then Mf pÝÑx q Ò. In the opposite case, Mf p

ÝÑx q Ó acc.

with the right answer.

(b) Assume g P NpdomĎNpq and h P NpdomĎNp`2q are computed respectively by Mg and

Mh. Then f defined by recursion:

(A) fpÝÑx , 0q “ gpÝÑx q

(B) fpÝÑx , y ` 1q “ h`


is computed by Mf which works as follows:

˝ on input pÝÑx , 0q it simply simulates Mg on input ÝÑx , and

˝ on input pÝÑx , n ` 1q Mf first simulates Mg on input ÝÑx which gives4 fpÝÑx , 0q.

Then

recursively for i :“ 0, . . . ,n Mf simulates Mh on`

ÝÑx , i, fpÝÑx , iq˘

which yields5

fpÝÑx , i` 1q.

It is clear that Mf brings the result if and only if every step fpÝÑx , iq (i :“ 0, . . . , n`1)

is defined. Otherwise it simply never stops.

(c) Assume g P NpdomĎNp`1q is computed by Mg. We design Mf that on input ÝÑx

computes

µy gpÝÑx , yq “ 0.

set i :“ 0 Mf simulates Mg on input pÝÑx , iq. If Mg stops and outputs the value of

gpÝÑx , iq,

˝ if gpÝÑx , iq “ 0 Mf stops and outputs i,

˝ if gpÝÑx , iq ‰ 0, Mf starts over again with i :“ i` 1.

Notice that Mf stops and outputs n “ µy gpÝÑx , yq “ 0 if and only if

˝ @i ď n pÝÑx , iq P domg,

˝ @i ă n gpÝÑx , iq ą 0 and

˝ gpÝÑx , nq “ 0.

% 96

Lemma 97 Every Turing computable partial function f P NpdomĎNpq is Part. Rec.

Proof of Lemma 97: We show an even stronger result: given any Turing Machine M and any

recursive coding of words on the tape alphabet Γ we show that the partial function Σ˚ ÝÑ Γ˚

which maps v P Σ˚ to w P Γ˚ if and only if the Turing machine from the initial configuration

4in case MgpÝÑx q Ó a

cc.

.

5in case Mh

`

ÝÑx , i, fpÝÑx , iq˘

Ó acc.

.


q0v stops in some configuration w0qacc.w1 with w0w1 “ w is partial recursive in the code. This

means the function f 1 P NpdomĎN1q defined by

#

f 1`

codepvq˘

is undefined if Mpvq Ò or Mpvq Ó rej.

;

f 1`

codepvq˘

“ codepw0w1q if M Ó acc.

in config. w0qacc.w1.

We first choose a coding of the configurations of M: We assume

˝ Σ “ t1, . . . , k ´ 1u and Γ “ t0, . . . , k ´ 1u with k ą 1 and necessarily 0 “ \.

˝ Q “ tq0, . . . , qmu with q0, q1, q2 being respectively the initial state, the rejecting state and

the accepting state.

The coding of a word w “ a0 . . . an that we choose is

xa0 . . . any “ÿ

0ďiďn

ai ¨ ki.

This coding is not injective (this will not matter for our purpose) for any two word which differ

by the tailing blanks in their prefix will have the same encoding:

xa0 . . . any “ xa0 . . . an\y.

But on the other hand, this encoding is surjective.

A configuration w0qrw1 of the Turing machine will be coded by

xw0qrw1y “ α4pr, xw0y, xw1y, |w0|q.

For instance, that the initial configuration of the Turing machine with input w is:

xq0wy “ α4p0, 0, xwy, 0q

“ α2

´

0, α2

`

0, α2pxwy, 0q˘

¯

“ α2

´

0, α2

`

0, xwypxwy`1q2

˘

¯

“ α2

´

0,

`

xwypxwy`1q2

˘`

xwypxwy`1q2

`1˘

2

¯

“

¨

˝

`

xwypxwy`1q2

˘`

xwypxwy`1q2 `1

˘

2

˛

‚¨

¨

˝

`

xwypxwy`1q2

˘`

xwypxwy`1q2 `1

˘

2`1

˛

‚

2 .

To say that w is an input word is to say that w P Σ˚ (we will identify words of the form w P Σ˚

with words of the form w\ . . .\ since our coding will not be able to distinguish them6 and also

because the input word really is the infinite word w \ . . .\ . . .).

6remember that this is the reason why we chose 0 for the coding of the blank symbol.

Recursivity 67

So we see that a word w “ a0 . . . an P Γ˚ is an input word if for no i ă n we have both ai “ \ and

ai`1 ‰ \. This means that xa0 . . . any “ a0 ¨k0à1 ¨k

1` . . .àn ¨kn satisfies ai “ 0 ñ ai`1 “ 0

(any i ă n). With our coding, we recover the coefficient ai as

ai “

„

xa0 . . . any

ki

9

ˆ„

xa0 . . . any

ki`1

¨ k

˙

.

Therefore the set InputM of all the codes of input words of M is Prim. Rec.:

χInputMpmq “ 1 if @i ď m”m

ki

ı

9

´” m

ki`1

ı

¨ k¯

“ 0 ùñ

” m

ki`1

ı

9

´” m

ki`2

ı

¨ k¯

ą 0

“ 0 otherwise

Since the coding of words that we choose is surjective, it comes that a configuration C is an

initial configuration if and only if there exists m P InputM

xCy “

ˆ`

mpm`1q2

˘`

mpm`1q2

`1˘

2

˙

¨

ˆ`

mpm`1q2

˘`

mpm`1q2

`1˘

2 ` 1

˙

2.

Therefore the set InitM of all the codes of initial configurations of M is Prim. Rec.

χInitMpcq “

$

&

%

1 if Dm ď c`

χInputMpmq “ 1 ^ α4p0, 0,m, 0q “ c˘

,

0 otherwise.

We now describe the transition from a given configuration C to the next configuration C 1

(C ñ C 1), in other words, we analyse the we obtain the code of C 1 on the basis of both the code

of C and the transition function δ.

A transition yields a move of the head either to the right or to the left: δpqr, alq “ pqr1 , al1 , Rq or

δpqr, alq “ pqr1 , al1 , Lq. We need to consider differently these two forms, together with differen-

tiating also whether the head can or cannot move to the left when the transition function says

so7.

When δpqr,alq “ pqr1 ,al1 ,Rq, we have C ñ C 1 is w0qrw1 ñ w10qr1w11 with

(1) w10 “ w0al1 (2) |w10| “ 1` |w0| (3) alw11 “ w1.

so that

(1) xw10y “ xw0y` l1 ¨ k|w0| “ β24pxCyq ` l1 ¨ kβ

44pxCyq

7this means whether or not the head is already in position 0 and the transition is of the form

δpqr, alq “ pqr1 , al1 , Lq.


(2) |w10| “ |w0| ` 1 “ β44pxCyq ` 1

(3) alw11 “ w1 so that xw11y “

„

β34pxCyq

k

.

So, we obtain:

if

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β14pxCyq “ r

and

β34pxCyq 9

ˆ

k ¨

„

β34pxCyq

k

˙

“ l,

then

xC 1y “ α4

ˆ

r1, β24pxCyq ` l1 ¨ kβ

44pxCyq,

„

β34pxCyq

k

, β44pxCyq ` 1

˙

.

When δpqr,alq “ pqr1 ,al1 ,Lq, there are two different cases depending on whether the head can

move to the left or not.

if w0 “ ε, then we have C ñ C 1 is w0qrw1 ñ w10qr1w11 with

(1) w10 “ ε

(2) |w10| “ |w0| “ |ε|

(3) w11 “ al1v and w1 “ alv for some word v.

So that we get

(1) xw10y “ xεy “ 0

(2) |w10| “ |ε| “ 0

(3) xw11y “ l1 `

„

β34pxCyq

k

¨ k.

So, all in all we obtain:

Recursivity 69

if

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β14pxCyq “ r

and

β34pxCyq 9

ˆ

k ¨

„

β34pxCyq

k

˙

“ l,

and

β44pxCyq “ 0

then

xC 1y “ α4

ˆ

r1, 0, l1 `

„

β34pxCyq

k

¨ k, 0

˙

.

if w0 ‰ ε, then we have C ñ C 1 is w0qrw1 ñ w10qr1w11 with

(1) w10al “ w0

(2) |w10| “ |w0| ´ 1

(3) w11 “ al1w1.

so that

(1) xw10y “ β24pxCyq 9

„

β24pxCyq

kpβ44pxCyq 9 1q

¨ kpβ44pxCyq 9 1q

(2) |w10| “ β44pxCyq 9 1

(3) xw11y “ l1 ` β34pxCyq ¨ k.

So, we have found:

if

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β14pxCyq “ r

and

β34pxCyq 9

ˆ

k ¨

„

β34pxCyq

k

˙

“ l,

and

β44pxCyq ‰ 0


then

xC 1y “ α4

ˆ

r1, β24pxCyq 9

„

β24pxCyq

kpβ44pxCyq 9 1q

¨ kpβ44pxCyq 9 1q, l1 ` β3

4pxCyq ¨ k, β44pxCyq 9 1

˙

.

To wrap up everything that we did so far, we recursively define a mapping f : N2 ÝÑ N such

that if n codes a word on Σ – i.e. n “ xwy for some w P Σ˚ – then fpn, tq “ xCn,ty where Cn,tstands for the configuration that the Turing machine reaches after t-many steps from the initial

configuration q0w.

Since the machine stops if it reaches an accepting or a rejecting configuration, we will simply

assume that in any of these two cases the configuration of the Turing machines remains the

same: if Cn,t is either an accepting or a rejecting configuration, then Cn,t`x “ Cn,t holds for

every x P N.

We set δL and δR are the two following finite – hence Prim. Rec. – subsets of N4:

δL “!

pr, l, r1, l1q P t0, . . .mu ˆ t0 . . . k ´ 1u ˆ t0, . . .mu ˆ t0 . . . k ´ 1u δpr, lq “ pr1, l1, Lq)

and

δR “ t0, . . .mu ˆ t0 . . . k ´ 1u ˆ t0, . . .mu ˆ t0 . . . k ´ 1u r δL.

For our convenience we assume that Γ “ Σ Y t\u. This way the mapping f : N2 ÝÑ N we

currently construct is total (f P NpN2q). For readability we use the notation ki

for βi4pkq (any

k P N, i ď 4).

initial case

fpn, 0q “

$

’

&

’

%

1 if n R InputM

α4p0, 0, n, 0q if n P InputM

successor case

fpn, t`1q “

$

’

’

’

’

’

’

’

’

’

’

’

’

’

’

’

’

’

’

’

’

’

’

’

’

’

’

’

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’

’

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’

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’

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’

’

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’

’

%

fpn, tq if fpn, tq1“ 1 or fpn, tq

1“ 2;

α4

˜

r1, fpn, tq2` l1 ¨ kfpn,tq

4

,

«

fpn, tq3

k

ff

, fpn, tq4` 1

¸

if

˜

fpn, tq1, fpn, tq

39

˜

k ¨

«

fpn, tq3

k

ff¸

, r1, l1

¸

P δR;

α4

˜

r1, 0, l1 `

«

fpn, tq3

k

ff

¨ k, 0

¸

if

$

’

’

’

’

’

’

&

’

’

’

’

’

’

%

˜

fpn, tq1, fpn, tq

39

˜

k ¨

«

fpn, tq3

k

ff¸

, r1, l1

¸

P δL

and

fpn, tq4“ 0;

α4

˜

r1, fpn, tq2

9

«

fpn, tq2

k

´

fpn,tq4

9 1¯

ff

¨ k

´

fpn,tq4

9 1¯

, l1 ` fpn, tq3¨ k, fpn, tq

49 1

¸

if

$

’

’

’

’

’

’

&

’

’

’

’

’

’

%

˜

fpn, tq1, fpn, tq

39

˜

k ¨

«

fpn, tq3

k

ff¸

, r1, l1

¸

P δL

and

fpn, tq4‰ 0.


Notice that f P NpN2q is Prim. Rec. and that Mpwq Ó acc.

if and only if there exists some t P Nsuch that β1

4

`

fpxwy, tq˘

“ 2.

Moreover, in this case, we recover the code of the content of the tape wt from fpxwy, tq by

xwty “ β24

`

fpxwy, tq˘

` kβ44

`

fpxwy,tq˘

¨ β34

`

fpxwy, tq˘

.

Finally, we only need to fix both (1) a recursive representation of the natural numbers and (2)

what it means for a machine to compute a partial function.

(1) we fix our coding of integers: every integer n is coded by the word 11 . . . 1loomoon

n

. The function

x y : N ÝÑ Nn ÞÝÑ xny

such that xny “ÿ

iăn

1 ¨ ki (i.e. xny is the code of the word 11 . . . 1loomoon

n

) is Prim. Rec.:

$

&

%

x0y “ 0

xn` 1y “ xny` kn.

(2) We only consider Turing machines that on any input words of the form

nhkkikkj

11 . . . 1, provided

they reach an accepting configuration, they reach one of the form 11 . . . 1loomoon

n1

qacc..

We define the function fM P NpdomĎNrq that M computes by:

˝ fMpn1, . . . , nrq is undefined if on input

αrpn1,...,nrqhkkikkj

11 . . . 1 either M Ò or M Ó rej.

;

˝ fMpn1, . . . , nrq “ n if on input

αrpn1,...,nrqhkkikkj

11 . . . 1 M Ó acc.

in configuration 11 . . . 1loomoon

n

qacc..

Then the function leastM P NpdomĎNrq that picks the minimum number of steps – if any

– the machine takes before reaching an accepting configuration when starting from the

initial one q0 11 . . . . . . 1loooomoooon

αrpn1,...,nrq

is defined by

leastMpn1, . . . , nrq “ µt β14 ˝ f

`

xαrpn1, . . . , nrqy, t˘

“ 2.

It is undefined if the machine never halts or halts on the rejecting state.

Recursivity 73

At last, we are ready to provide the desired fM P NpdomĎNrq. We make use of the fact the

position of the head in an accepting configuration indicates precisely the number of 1’s

there are on the tape:

fMpn1, . . . , nrq “ β44 ˝ f

´

xαrpn1, . . . , nrqy, leastMpn1, . . . , nrq¯

.

% 97

Corollary 98

Every partial recursive function admits a construction that requires at most once minimisation.

Proof of Corollary 98: This is an immediate consequence of the whole proof of Lemma 97.

% 98

Theorem 99 For every k ą 0 and every f P NpdomĎNkq the following are equivalent

˝ f is Part. Rec.,

˝ f is Turing computable.

Proof of Theorem 99: Follows immediately from Lemmas 96 and 97. % 99


Part II

Arithmetic

Chapter 4

Representing Functions

4.1 Robinson Arithmetic

We start by describing a first order theory called Robinson Arithmetic. The language we consider

is LA “ t0, S,`, ü where

(1) 0 is a constant symbol,

(2) S is a unary function symbol1,

(3) ` and ¨ are binary function symbols2.

The axioms are

axiom 1. @x Sx ‰ 0

axiom 2. @x Dy px ‰ 0 Ñ Sy “ xq

axiom 3. @x @y pSx “ Sy Ñ x “ yq

axiom 4. @x x`0 “ x

axiom 5. @x @y´

x`Sy “ Spx`yq¯

axiom 6. @x x¨0 “ 0

axiom 7. @x @y´

x¨Sy “ pxÿq`x¯

1for any terms of LA t, we use the notation St instead of Sptq.2for any terms of LA t0, t1, we use the notation t0`t1 (respectively t0¨t1) instead of `pt0, t1q (respectively

¨pt0, t1q).


Example 100 The standard model of Robinson Arithmetic is

〈N, 0, S,`, ¨〉

where S is the successor function, ` is the usual addition, and ¨ is the customary multiplication.

By abuse of notation we identify N with the model 〈N, 0, S,`, ¨〉. So that, for instance, we will

write

N |ù φ

instead of the correct notation

〈N, 0, S,`, ¨〉 |ù φ.

Example 101 A very simple non standard model of Robinson arithmetic M in which

˝ SM admits a fixed point.

˝ ¨M is not commutative.

M “ 〈NY tau, 0MSM,`M, ¨M〉Where 0M “ 0 and the operations SM,`M, ¨M are defined the usual way on the integers.

i.e

˝ SM æ N “ S ˝ `M æ N “ ` ˝ ¨M æ N “ ¨

And when the unique non standard integer a is involved:

˝ SMa “ a

˝ a`Mα “ α`Ma “ a (any α P NY tau)

˝ a¨M0 “ 0

˝ α¨Ma “ a (any α P NY tau)

˝ a¨Mα “ a (any α P`

Nr t0u˘

Y tau)

We verify that

M |ù Rob.

axiom 1. since the only non standard integer verifies SMa “ a we have

M |ù @x Sx ‰ 0

Arithmetic 79

axiom 2. since every standard integer from 0 has a predecessor, and SMa “ a, so we have

M |ù @x Dy px ‰ 0 Ñ Sy “ xq

axiom 3. holds for standard integers, and for every n P N we have SMa ‰ SMn, thus

M |ù @x @y pSx “ Sy Ñ x “ yq

axiom 4. holds for standard integers, and we have a`M0 “ a, hence

M |ù @x x`0 “ x

axiom 5. if k, n P N, then k`MSMn “ SMpk`Mnq holds. If α P NY tau, we have

˝ a`MSMα “ a

˝ SMpa`Mαq “ SMa “ a

˝ α`MSMa “ α`Ma “ a

˝ SMpα`Maq “ SMa “ a

therefore we have

M |ù @x @y´

x`Sy “ Spx`yq¯

.

axiom 6. if α P NY tau, we have α¨M0 “ 0. Thus

M |ù @x x¨0 “ 0

axiom 7. if k, n P N and α P NY tau, then

˝ a¨MSMα “ a

˝ α¨MSMa “ a

˝ pa¨Mαq`Ma “ a

˝ pα¨Maq`Mα “ a

˝ k¨MSMn “ pk¨Mnq`Mk

so we have

M |ù @x @y´


.

Notation 102 For any integer n, we write n for the LA-term S . . . Sloomoon

n

0.

Example 103 Let us show that the following holds for all integers k:

Rob. $c @x`

0`x “ k ÝÑ x “ k˘

(4.1)


We will make use of the excluded middle:

$c @x`

x “ 0 _ x ‰ 0˘

and distinguish between the two cases.

The proof is by induction on k:

if k “ 0:

if x “ 0: since $c 0 “ 0 holds, we obtain

Rob. $c 0`0 “ 0 ÝÑ 0 “ 0

if x ‰ 0: by 2 @x Dy px ‰ 0 Ñ Sy “ xq it is enough to show

Rob. $c @y`

0`Sy “ 0 ÝÑ Sy “ 0˘

by 5 @x @y`

x`Sy “ Spx`yq˘

this comes down to establishing

Rob. $c @y`

Sp0`yq “ 0 ÝÑ Sy “ 0˘

which is immediate by 1 @x Sx ‰ 0 .

if k “ n` 1:

if x “ 0: we need to show

Rob. $c 0`0 “ n` 1 ÝÑ 0 “ n` 1.

By 4 @x x`0 “ x this comes down to showing

Rob. $c 0 “ n` 1 ÝÑ 0 “ n` 1.

because of 1 @x Sx ‰ 0 we have

Rob. $c 0 ‰ n` 1

which yields the result.

if x ‰ 0: by 2 @x Dy px ‰ 0 Ñ Sy “ xq it is enough to show

Rob. $c @y`

0`Sy “ n` 1 ÝÑ Sy “ n` 1˘

Arithmetic 81

by 5 @x @y`

x`Sy “ Spx`yq˘

this comes down to establishing

Rob. $c @y`

Sp0`yq “ n` 1 ÝÑ Sy “ n` 1˘

which is exactly

Rob. $c @y`

Sp0`yq “ Sn ÝÑ Sy “ Sn˘

by 3 @x @y pSx “ Sy Ñ x “ yq this amounts to showing

Rob. $c @y`

0`y “ n ÝÑ Sy “ Sn˘

.

The induction hypothesis gives

Rob. $c @y`

0`y “ n ÝÑ y “ n˘

from where we immediately get what we want.

Example 104 Let us show that the following holds for all integers k:

Rob. $c @x@y`

Sy`x “ k ÝÑ Spy`xq “ k˘

. (4.2)

We make use of the excluded middle:

$c @x`

x “ 0 _ x ‰ 0˘



Rob. $c @y`

Sy`0 “ k ÝÑ Spy`0q “ k˘

which is immediate by 4 @x x`0 “ x .

if x ‰ 0: it is enough to show

Rob. $c @y@z`

Sy`Sz “ k ÝÑ Spy`Szq “ k˘

.

The proof goes by induction on k:

if k “ 0: we need to show

Rob. $c @y@z`

Sy`Sz “ 0 ÝÑ Spy`Szq “ 0˘

.


By 5 @x @y`

x`Sy “ Spx`yq˘

this comes down to

Rob. $c @y@z`

SpSy`zq “ 0 ÝÑ Spy`Szq “ 0˘

which trivially holds by 1 @x Sx ‰ 0 .

if k “ n` 1: we need to show

Rob. $c @y@z`

Sy`Sz “ n` 1 ÝÑ Spy`Szq “ n` 1˘

.

By 5 @x @y`

x`Sy “ Spx`yq˘

this comes down to

Rob. $c @y@z`

SpSy`zq “ Sn ÝÑ Spy`Szq “ n` 1˘

.

By 3 @x @y pSx “ Sy Ñ x “ yq this amounts to proving

Rob. $c @y@z`

Sy`z “ n ÝÑ Spy`Szq “ n` 1˘

.

if z “ 0: we need to show

Rob. $c @y`

Sy`0 “ n ÝÑ Spy`S0q “ n` 1˘

.

By 4 @x x`0 “ x and 5 @x @y`

x`Sy “ Spx`yq˘

this comes down to

showing

Rob. $c @y`

Sy “ n ÝÑ SSy “ n` 1˘

.

which holds by definition.

if z ‰ 0: what we need to prove is equivalent to

Rob. $c @y@z1`

Sy`Sz1 “ n ÝÑ Spy`SSz1q “ n` 1˘

.

By 3 @x @y pSx “ Sy Ñ x “ yq this comes down to showing

Rob. $c @y@z1`

Sy`Sz1 “ n ÝÑ y`SSz1 “ n˘

.

The induction hypothesis yields

Rob. $c @y@z1`

Sy`Sz1 “ n ÝÑ Spy`Sz1q “ n˘

.

from where we easily get the result by 5 @x @y`

x`Sy “ Spx`yq˘

.

Arithmetic 83

4.2 Representable Functions

Definition 105 Let f P NpNnq and φpx0, x1, . . . , xnq be any LA-formula whose free variables are

among tx0, x1, . . . , xnu.

φpx0, x1, . . . , xnq represents the function f if for all i1, . . . , in P N

Rob. $c @x0

´

fpi1, . . . , inq “ x0 ÐÑ φpx0, i1, . . . , inq¯

.

Definition 106 Let A Ď Nn and φpx1, . . . , xnq be any LA-formula whose free variables are

among tx1, . . . , xnu.

φpx1, . . . , xnq represents the set A if for all i1, . . . , in P N we have:

˝ if pi1, . . . , inq P A, then Rob. $c φ`

i1, . . . , in˘

;

˝ if pi1, . . . , inq R A, then Rob. $c φ`

i1, . . . , in˘

.

Proposition 107 For any A Ď Nn,

A is representable if and only if χA is representable.


(ñ) If A is represented by φpx1, . . . , xnq, then χA is represented by`

x0 “ 1^ φpx1, . . . , xnq˘

_`

x0 “ 0^ φpx1, . . . , xnq˘

.

(ð) If χA is represented by φpx0, x1, . . . , xnq, then A is represented by

φpS0, x1, . . . , xnq.

% 107

Example 108 The constant function f P NpNnq defined by fpi1, . . . , inq “ k (any i1, . . . , in P N)

is represented by the following formula of the form φpx0, i1, . . . , inq:

x0 “ k.

It is enough to verify

Rob. $c @x0

´

fpi1, . . . , inq “ x0 ÐÑ x0 “ k¯

.

which is exactly

Rob. $c @x0

´

k “ x0 ÐÑ x0 “ k¯

.


Example 109 The projection πnj P NpNnqis represented by the formula:

x0 “ ij .


Rob. $c @x0

´

πnj pi1, . . . , inq “ x0 ÐÑ x0 “ ij

¯

.

i.e.

Rob. $c @x0

´

ij “ x0 ÐÑ x0 “ ij

¯

.

Example 110 The successor S P NN is represented by the formula:

x0 “ Sx1.


Rob. $c @x0

´

Spiq “ x0 ÐÑ x0 “ Si¯

.

i.e.

Rob. $c @x0

´

Si “ x0 ÐÑ x0 “ Si¯

.

Example 111 The addition ` P NpN2q is represented by the formula:

x0 “ x1`x2.


Rob. $c @x0

´

i1 ` i2 “ x0 ÐÑ x0 “ i1ì2

¯

(4.3)

The proof is by induction on i2:

i2 “ 0 because of 4 @x x`0 “ x we have

Rob. $c @x0

´

i1 “ x0 ÐÑ x0 “ i1`0¯

which is

Rob. $c @x0

´

i1 ` 0 “ x0 ÐÑ x0 “ i1`0¯

.

Arithmetic 85

i2 “ i` 1 by 5 @x @y`

x`Sy “ Spx`yq˘

we have

Rob. $c @x0

´

Spi1ìq “ x0 ÐÑ x0 “ i1`Si¯


Rob. $c @x0

´

i1 ` i “ x0 ÐÑ x0 “ i1ì¯

hence we obtain

Rob. $c @x0

´

Spi1 ` iq “ x0 ÐÑ x0 “ i1`Si¯

by the very definition of the terms involved we finally have

Rob. $c @x0

´

i1 ` pi` 1q “ x0 ÐÑ x0 “ i1`pi` 1q¯

.

Example 112 The multiplication ¨ P NpN2q is represented by the formula:

x0 “ x1¨x2.


Rob. $c @x0

´

i1 ¨ i2 “ x0 ÐÑ x0 “ i1ï2

¯

. (4.4)

The proof is by induction on i2:

i2 “ 0

because of 6 @x x¨0 “ 0 we have

Rob. $c @x0

´

0 “ x0 ÐÑ x0 “ i1¨0¯

which is

Rob. $c @x0

´

i1 ¨ 0 “ x0 ÐÑ x0 “ i1¨0¯

.

i2 “ i` 1

by 7 @x @y`

x¨Sy “ pxÿq`x˘

we have

Rob. $c @x0

´

i1¨Si “ x0 ÐÑ x0 “ pi1ïqì1

¯


which is exactly

Rob. $c @x0

´

i1¨pi` 1q “ x0 ÐÑ x0 “ pi1ïqì1

¯


Rob. $c @x0

´

i1 ¨ i “ x0 ÐÑ x0 “ i1ï¯

so we have

Rob. $c @x0

´

i1¨pi` 1q “ x0 ÐÑ x0 “ pi1 ¨ iqì1

¯

by 7 @x @y`


and (4.3) we have

Rob. $c @x0

´

pi1 ¨ iq ` i1 “ x0 ÐÑ x0 “ pi1 ¨ iqì1

¯

which is exactly

Rob. $c @x0

´

i1 ¨ pi` 1q “ x0 ÐÑ x0 “ pi1 ¨ iqì1

¯

and we finally obtain

Rob. $c @x0

´

i1 ¨ pi` 1q “ x0 ÐÑ x0 “ i1¨pi` 1q¯

.

Lemma 113 The set of representable functions is closed under composition.

Proof of Lemma 113: Assume f1, . . . , fn P NpNpq and g P NpNnq are represented respectively

by φf1px0, x1, . . . , xpq, . . . , φfnpx0, x1, . . . , xpq and φgpx0, x1, . . . , xnq. i.e. we have for all integers

i1, . . . , ip, k1, . . . , kn and 1 ď j ď n:

Rob. $c @x0

´

fjpi1, . . . , ipq “ x0 ÐÑ φfj px0, i1, . . . , ipq¯

and

Rob. $c @x0

´

gpk1, . . . , knq “ x0 ÐÑ φgpx0, k1, . . . , knq¯

.

The function

h “ gpf1, . . . , fnq P NpNpq

defined by

hpi1, . . . , ipq “ g`

f1pi1, . . . , ipq, . . . , fnpi1, . . . , ipq˘

Arithmetic 87

is represented by

φhpx0, x1, . . . , xpq “ Dy1 Dy2 . . . Dyn

´

ľ

1ďjďn

φfj pyj , x1, . . . , xpq ^ φgpx0, y1, . . . , ynq¯

.

Indeed, by the very definition of h for every i1, . . . , ip P N we have

$c @x0

ˆ

hpi1, . . . , ipq “ x0 ÐÑ Dy1 Dy2 . . . Dyn

´

ľ

1ďjďn

fjpi1, . . . , ipq “ yj ^ gpy1, . . . , ynq “ x0

¯

˙

.

Therefore

Rob. $c @x0

ˆ

hpi1, . . . , i1q “ x0 ÐÑ Dy1 Dy2 . . . Dyn

´

ľ

1ďjďn

φfj pyj , i1, . . . , i1q ^ φgpx0, y1, . . . , ynq¯

˙

.

% 113

We now turn to minimisation. We need to prove that if A Ď Nn`1 is representable and f P NpNnqis some total function defined by minimisation the following way:

fpi1, . . . , inq “ µk pk, i1, . . . , inq P A,

then f is representable.

This proof requires some good amount of preliminary work.

Example 114 We first notice that

˝ for all non-zero integer i the following holds

Rob. $c i ‰ 0. (4.5)

To see this, let i “ j ` 1, by the very definition of the terms involved we have

$c i “ Sj

hence by 1 @x Sx ‰ 0 we obtain

Rob. $c Sj ‰ 0


˝ for all integers i, j such that i ‰ j the following holds

Rob. $c i ‰ j. (4.6)

the proof is by induction on minti, ju:


minti, ju “ 0 : this is case 4.5 Rob. $c i ‰ 0 for all i P N, i ‰ 0 .

minti, ju ą 0 : set k ` 1 “ i and n` 1 “ j.

By 3 @x @y pSx “ Sy Ñ x “ yq we have

Rob. $c @x @y px ‰ y Ñ Sx ‰ Syq.

so that we easily obtain

Rob. $c k ‰ nÑ Sk ‰ Sn

which is precisely

Rob. $c k ‰ nÑ k ` 1 ‰ n` 1

i.e.

Rob. $c k ‰ nÑ i ‰ j.

By induction hypothesis we have

Rob. $c k ‰ n,

therefore by modus ponens we finally get

Rob. $c i ‰ j.

˝ The following holds

Rob. $c @x@y`

y ‰ 0 ÝÑ x`y ‰ 0˘

. (4.7)

By 2 @x Dy px ‰ 0 Ñ Sy “ xq 5 @x @y`

x`Sy “ Spx`yq˘

and 1 @x Sx ‰ 0

we obtain

Rob. $c @x@yDz´

y ‰ 0 ÝÑ py “ Sz ^ x`Sz “ Spx`zq ^ Spx`zq ‰ 0˘

¯

which immediately yields the result.

˝ The following holds

Rob. $c @x@y´

x`y “ 0 ÝÑ px “ 0 ^ y “ 0q¯

. (4.8)

By previous result 4.7 Rob. $c @x@y`

y ‰ 0 ÝÑ x`y ‰ 0˘

we see that

Rob. $c @x@y`

x`y “ 0 ÝÑ y “ 0˘

.

and by 4 @x x`0 “ x we obtain immediately the result.

Arithmetic 89

Notation 115 We introduce “ x ď z” to abbreviate the formula “ Dy y`x “ z”. We also

introduce “ x ă z” for the formula “ Dy`

y`x “ z ^ x ‰ z˘

”.

Example 116 We establish

Rob. $c @x x ă 0 (4.9)

We recall that x ă y stands for “ Dz`

z`x “ y ^ x ‰ y˘

”.

So we need to prove

Rob. $c @x Dz`

z`x “ 0 ^ x ‰ 0˘

which is

Rob. $c @x @z`

z`x ‰ 0 _ x “ 0˘

.

This is logically equivalent to

Rob. $c @x @z`

z`x “ 0 ÝÑ x “ 0˘

.

Which is also logically equivalent to 4.7 Rob. $c @x@y`

y ‰ 0 ÝÑ x`y ‰ 0˘

.

Example 117 For all integer n the following holds:

Rob. $c @x”

x ď nÐÑ`

x “ 0 _ x “ S0 _ . . . _ x “ n˘

ı

. (4.10)

The direction

Rob. $c @x´

`

x “ 0 _ x “ S0 _ . . . _ x “ n˘

ÝÑ x ď n¯

First, by making use of 4 @x x`0 “ x and 5 @x @y`

x`Sy “ Spx`yq˘

, the very def-

inition of k “ S . . . Sloomoon

k

0 and “ x ď z”:= “Dy`

y`x “ z ^ x ‰ z˘

”, it is straightforward to

establish by induction on n

Rob. $c @x”

`

x “ 0 _ x “ S0 _ . . . _ x “ n˘

ÝÑ Dy`

y`x “ n˘

ı

.

So it only remains to prove

Rob. $c @x”

x ď n ÝÑ`

x “ 0 _ x “ S0 _ . . . _ x “ n˘

ı

.

The proof is by induction on n:


n “ 0 : we need to show

Rob. $c @x`

x ď 0 ÝÑ x “ 0˘

which is

Rob. $c @x`

Dy y`x “ 0 ÝÑ x “ 0˘

i.e.

Rob. $c @x`

Dy y`x “ 0 _ x “ 0˘

i.e.

Rob. $c @x`

@y y`x ‰ 0 _ x “ 0˘

i.e.

Rob. $c @x@y`

y`x ‰ 0 _ x “ 0˘

i.e.

Rob. $c @x@y`

y`x “ 0 ÝÑ x “ 0˘

.

We easily obtain the result by 4.8 Rob. $c @x@y`

x`y “ 0 ÝÑ px “ 0 ^ y “ 0q˘

n “ k` 1 : we need to show

Rob. $c @x´

x ď k ` 1 ÝÑ`

x “ 0 _ x “ S0 _ . . . _ x “ k _ x “ k ` 1˘

¯

.

which really is

Rob. $c @x´

Dy y`x “ k ` 1 ÝÑ`

x “ 0 _ x “ S0 _ . . . _ x “ k _ x “ k ` 1˘

¯

.

i.e.

Rob. $c @x@y´

y`x “ k ` 1 ÝÑ`

x “ 0 _ x “ S0 _ . . . _ x “ k _ x “ k ` 1˘

¯

.


$c @y`

y “ 0 _ y ‰ 0˘

and distinguish between the two cases:

y “ 0: by 4.1 Rob. $c @x`

0`x “ k ÝÑ x “ k˘

we obtain

Rob. $c @x`

0`x “ k ` 1 ÝÑ x “ k ` 1˘

.

Arithmetic 91

y ‰ 0: by 2 @x Dy px ‰ 0 Ñ Sy “ xq what we need to show comes down to

Rob. $c @x@z´

Sz`x “ k ` 1 ÝÑ`

x “ 0 _ x “ S0 _ . . . _ x “ k _ x “ k ` 1˘

¯

.

by 4.2 Rob. $c @x@y`


we already know that

Rob. $c @x@z`

Sz`x “ k ÝÑ Spz`xq “ k˘

so that we need to prove

Rob. $c @x@z´

Spz`xq “ k ` 1 ÝÑ`

x “ 0 _ x “ 1 _ . . . _ x “ k _ x “ k ` 1˘

¯

.

By 3 @x @y pSx “ Sy Ñ x “ yq we only need to prove

Rob. $c @x@z´

z`x “ k ÝÑ`

x “ 0 _ x “ 1 _ . . . _ x “ k _ x “ k ` 1˘

¯

which is

Rob. $c @x´

Dz z`x “ k ÝÑ`

x “ 0 _ x “ 1 _ . . . _ x “ k _ x “ k ` 1˘

¯

i.e.

Rob. $c @x´

x ď k ÝÑ`

x “ 0 _ x “ 1 _ . . . _ x “ k _ x “ k ` 1˘

¯

.

By induction hypothesis gives

Rob. $c @x´

x ď k ÝÑ`

x “ 0 _ x “ 1 _ . . . _ x “ k˘

¯

so that we obtain the result very easily.

So we have proved the following two statements:

(1) Rob. $c @x@y´

y “ 0 ÝÑ`

x`y “ k ` 1 ÝÑ x “ k ` 1˘

¯

and

(2) Rob. $c @x@y

ˆ

y ‰ 0 ÝÑ´

x`y “ k ` 1 ÝÑ`

x “ 0_ x “ 1_ . . ._ x “ k˘

¯

˙

.

Therefore, by an immediate application of the excluded middle we have proved the result.


Example 118 For all integer n the following holds:

Rob. $c @x`

x ď n _ n ď x˘

. (4.11)

What we need to show is

Rob. $c @x`

Dy y`x “ n _ Dy y`n “ x˘

.


$c @x`

x “ 0 _ x ‰ 0˘



Rob. $c Dy y`0 “ n _ Dy y`n “ 0

which is immediate by 4 @x x`0 “ x .

if x ‰ 0: what wee need to prove comes down to

Rob. $c @z`

Dy y`Sz “ n _ Dy y`n “ Sz˘

.

The proof goes by induction on n:

if n “ 0 : we need to prove

Rob. $c @z`

Dy y`Sz “ 0 _ Dy y`0 “ Sz˘

.

By 4 @x x`0 “ x this comes down to

Rob. $c @z`

Dy y`Sz “ 0 _ Dy y “ Sz˘

which is immediate.

if n “ k` 1 : we need to prove

Rob. $c @z`

Dy y`Sz “ k ` 1 _ Dy y`pk ` 1q “ Sz˘

.

By 5 @x @y`

x`Sy “ Spx`yq˘

and 3 @x @y pSx “ Sy Ñ x “ yq this

amounts to proving

Rob. $c @z`

Dy y`z “ k _ Dy y`k “ z˘

which is exactly the induction hypothesis.

Arithmetic 93

We have already proved

p4.1q Rob. $c @x`

0`x “ k ÝÑ x “ k˘

p4.2q Rob. $c @x@y`


p4.3q Rob. $c @x0

`

i1 ` i2 “ x0 ÐÑ x0 “ i1ì2˘

p4.4q Rob. $c @x0

`

i1 ¨ i2 “ x0 ÐÑ x0 “ i1ï2˘

p4.5q Rob. $c i ‰ 0 (any i P N, i ‰ 0)

p4.6q Rob. $c i ‰ j (any i, j P N, i ‰ j)

p4.7q Rob. $c @x@y`

y ‰ 0 ÝÑ x`y ‰ 0˘

p4.8q Rob. $c @x@y`

x`y “ 0 ÝÑ px “ 0 ^ y “ 0q˘

p4.9q Rob. $c @x x ă 0

p4.10q Rob. $c @x`

x ď nÐÑ px “ 0 _ x “ S0 _ . . . _ x “ nq˘

p4.11q Rob. $c @x`

x ď n _ n ď x˘

Lemma 119 Let A Ď Nn`1 be representable. If the following function f P NpNnq is total, then

f is representable.

fpi1, . . . , inq “ µk pk, i1, . . . , inq P A,

Proof of Lemma 119: Assume φpx0, x1, . . . , xnq represents the set A. We claim the function f

is represented by the formula

φpx0, x1, . . . , xnq ^ @y ă x0 φpy, x1, . . . , xnq.

We need to show that for all i1, . . . , in P N

Rob. $c @x0

ˆ

fpi1, . . . , inq “ x0 ÐÑ

´

φpx0, i1, . . . , inq ^ @y ă x0 φpy, i1, . . . , inq¯

˙

.

(We writeÝÑi for i1, . . . , in and

ÝÑi for i1, . . . , in)

(ñ) (1) by the very definition of f and the fact that φpx0, x1, . . . , xnq represents A we trivially

have :

Rob. $c φpfpÝÑi q,ÝÑi q

¯

which is equivalent to

Rob. $c @x0

´

fpÝÑi q “ x0 ÝÑ φpx0,

ÝÑi q

¯

.


(2) To show

Rob. $c @x0

´

fpÝÑi q “ x0 ÝÑ @y ă x0 φpy,

ÝÑi q

¯

we show the equivalent

Rob. $c @y´

y ă fpÝÑi q ÝÑ φpy,

ÝÑi q

¯

We have two cases

if fpÝÑi q “ 0: we make use of 4.9 Rob. $c @x x ă 0 which settles this case.

if fpÝÑi q “ k` 1: by the very definition of f and the fact that φpx0, x1, . . . , xnq rep-

resents A we trivially have :

Rob. $c φp0,ÝÑi q ^ φpS0,

ÝÑi q ^ . . . ^ φpk,

ÝÑi q

which sums up to

Rob. $c @y´

`

y “ 0 _ y “ S0 _ . . . _ y “ k˘

ÝÑ φpy,ÝÑi q

¯

by 4.10 Rob. $c @x“

x ď nÐÑ px “ 0 _ x “ S0 _ . . . _ x “ nq‰

we ob-

tain

Rob. $c @y`

y ă k ` 1 ÝÑ py “ 0 _ y “ S0 _ . . . _ y “ kq˘

which yields

Rob. $c @y´

y ă k ` 1 ÝÑ φpy,ÝÑi q

¯

.

(ð) we need to show

Rob. $c @x0

ˆ

´

φpx0,ÝÑi q ^ @y ă x0 φpy,

ÝÑi q

¯

ÝÑ fpÝÑi q “ x0

˙

.

We prove the result by contraposition, which means we prove

Rob. $c @x0

ˆ

fpÝÑi q ‰ x0 ÝÑ

´


ÝÑi q

¯

˙

.

By 4.11 Rob. $c @x`

x ď n _ n ď x˘

we have

Rob. $c @x0

`

x0 ď fpÝÑi q _ fp

ÝÑi q ď x0

˘

Which is also

Rob. $c @x0

`

x0 ă fpÝÑi q _ fp

ÝÑi q ă x0 _ x0 “ fp

ÝÑi q

˘

Arithmetic 95

so that we only need to prove

Rob. $c @x0

ˆ

`

x0 ă fpÝÑi q ^ fp

ÝÑi q ă x0

˘

ÝÑ

´


ÝÑi q

¯

˙

.

We will successively prove

(1) Rob. $c @x0

ˆ

x0 ă fpÝÑi q ÝÑ

´


ÝÑi q

¯

˙

.

We have two cases

if fpÝÑi q “ 0: we make use of 4.9 Rob. $c @x x ă 0 which settles this case.

if fpÝÑi q “ k` 1: by 4.10 Rob. $c @x

“

x ď nÐÑ px “ 0 _ x “ S0 _ . . . _ x “ nq‰

Rob. $c @x0

`

x0 ă k ` 1 ÝÑ px0 “ 0 _ x0 “ S0 _ . . . _ x0 “ kq˘

and by the very definition of f :

Rob. $c φp0,ÝÑi q ^ φpS0,

ÝÑi q ^ . . . ^ φpk,

ÝÑi q

which gives

Rob. $c @x0

`

x0 ă k ` 1 ÝÑ φpx0,ÝÑi q

˘

which settles this case.

(2) Rob. $c @x0

ˆ

fpÝÑi q ă x0 ÝÑ

´


ÝÑi q

¯

˙

.

By the very definition of f :

Rob. $c φ`

fpÝÑi q,ÝÑi˘

Thus

Rob. $c @x0

´

fpÝÑi q ă x0 ÝÑ Dy ă x0 φpy,

ÝÑi q

¯

i.e.

Rob. $c @x0

´

fpÝÑi q ă x0 ÝÑ @y ă x0 φpy,

ÝÑi q

¯

which yields what we want.

(1) and(2) finish the proof.

% 119


Theorem 120 (Chinese Remainder Theorem) Suppose n0, n1, . . . , nk are positive integers

which are pairwise co-prime. Then, for any given sequence of integers a0, a1, . . . , ak there exists

an integer x solving the system of simultaneous congruences

$

’

’

’

&

’

’

’

%

x ” a0 mod n0

x ” a1 mod n1...

x ” ak mod nk.

Proof of Theorem 120: We set

α “ź

iďk

ni

and notice that for each i ď k the two integers ni and αni

are co-prime. By Bezout there exist

coefficients ci, di P Z such that

ci ¨ ni ` di ¨α

ni“ 1

if we set

ei “ di ¨α

ni

we see that

#

ei ” 1 mod ni

ei ” 0 mod nj (any j ‰ i)

It follows immediately that

β “ÿ

iďk

ai ¨ ei

is a solution to the system.

% 120

Lemma 121 (Godel β-function) There exists some function β P NpN3q which is both repre-

sentable and Prim. Rec. such that for all k P N and every sequence n0, n1, . . . , nk there exists

a, b P N such that

$

’

’

’

&

’

’

’

%

βp0, a, bq “ n0

βp1, a, bq “ n1...

βpk, a, bq “ nk.

Arithmetic 97

Proof of Lemma 121: The function is defined3 by

βpi, a, bq “ b 9

ˆ„

b

api` 1q ` 1

¨ papi` 1q ` 1q

˙

This shows that it is Prim. Rec.. To show that β is representable we consider the formula

x0 ă Spx2¨Sx1q ^ Dy ď x3

´

y¨Spx2¨Sx1q

¯

`x0 “ x3

To show that this formula represents the function β, we need to show that for all i1, i2, i3 P N

Rob. $c @x0

´

βpi1, i2, i3q “ x0 ÐÑ x0 ă Spi2¨Si1q ^ Dy ď i3`

y¨Spi2¨Si1q˘

`x0 “ i3

¯

which is left as a tedious but straightforward exercise.

Now given n0, n1, . . . , nk, in order to find a and b, we consider any integer m that satisfies both

(1) m ě k ` 1 (2) m! ě maxtn0, n1, . . . , nku.

We set a “ m! so that we make sure that a` 1, a ¨ 2` 1, . . . , a ¨ k` 1, a ¨ pk` 1q` 1 are co-prime.

To see this, we proceed by contradiction and consider there exists some prime number p that

divides both api` 1q ` 1 and apj ` 1q ` 1 for some 0 ď i ă j ď k. Then p also divides

apj ` 1q ` 1 ´`

api` 1q ` 1˘

“ apj ´ iq “ m!pj ´ iq

Since m ą pj ´ iq holds, p divides m! which contradicts p divides m!pi` 1q ` 1.

The Chinese Remainder Theorem (120) guarantees that there exists some integer b that satisfies

$

’

’

’

&

’

’

’

%

b ” n0 mod a` 1

b ” n1 mod a ¨ 2` 1...

b ” nk mod a ¨ pk ` 1q ` 1.

We chose m such that a “ m! ě maxtn0, n1, . . . , nku in order to insure ni ă api ` 1q ` 1 for

every integer i ď k. This makes certain that for each i ď k we have βpi, a, bq “ ni.

% 121

Lemma 122 If both functions g P NpNpq and h P NpNp`2q are representable, then the function

f P NpNp`1q defined by recursion below is also representable.$

&

%

fpÝÑx , 0q “ gpÝÑx q

fpÝÑx , y ` 1q “ h`


3βpi, a, bq is the remainder of the division of b by api` 1q ` 1.


Proof of Lemma 122: We let ÝÑx stand for x1, . . . , xp and assume g P NpNpq and h P NpNp`2q are

represented respectively by φgpx0,ÝÑx q and φhpx0,ÝÑx , xp`1, xp`2q.

We also consider the following formula that represents the β-function 4 :

φpx0, x1, x2, x3q :“ x0 ă Spx2¨Sx1q ^ Dy ď x3

´

y¨Spx2¨Sx1q

¯

`x0 “ x3

Instead of φpx0, x1, x2, x3q we prefer the formula φβpx0, x1, x2, x3q below which also obviously

represents β but in a strong way.

φβpx0, x1, x2, x3q :“ φpx0, x1, x2, x3q ^ @y ă x0 φpy, x1, x2, x3q

because for any integers i, n we have

Rob. $c @a@b@x0

”

“

φβpk, i, a, bq ^ φβpx0, i, a, bq‰

ÝÑ x0 “ kı

.

This holds because

Rob. $c x0

“

x0 ‰ k ÝÑ px0 ď k _ k ď x0q _ x0 ă k _ k ă x0

‰

and by 4.11 Rob. $c @x`

x ď n _ n ď x˘

, this comes down to

Rob. $c x0

“

x0 ‰ k ÝÑ x0 ă k _ k ă x0

‰

and by the definition of both φβ and k we have

(1) Rob. $c @a@b@x0

”

“

φβpk, i, a, bq ^ x0 ă k‰

ÝÑ φβpx0, i, a,@bqı

(2) Rob. $c @a@b@x0

”

“

φβpk, i, a, bq ^ k ă x0

‰

ÝÑ φβpx0, i, a,@bqı

.

which yields the following which is also logically equivalent to what we want:

Rob. $c @a@b@x0

”

“

x0 ‰ k ^ φβpk, i, a, bq‰

ÝÑ φβpx0, i, a, bqı

.

The formula the we choose to represent f is the following formula φf px0, x1, . . . , xp`1q. We use

the notation ÝÑx “ x1, . . . , xp so that φf px0, x1, . . . , xp`1q :“ φf px0,ÝÑx , xp`1q

4see the proof of Lemma 121.

Arithmetic 99

DaDb@i ď xp`1 Dy Dz

¨

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˝

i “ 0 ÝÑ φgpy,ÝÑx q

Ź

φβpy, i, a, bq

Ź

φhpz,ÝÑx , i, yq

Ź

φβpz, Si, a, bq

Ź

φβpx0, xp`1, a, bq

˛

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‚

In order to show that this formula φf px0,ÝÑx , xp`1q represents f , we need to prove that for all

integers i1, . . . , ip, ip`1 (we writeÝÑi for i1, . . . , ip) we have

Rob. $c @x0

¨

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˝

DaDb@i ď ip`1 Dy Dz

¨

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˝

i “ 0 ÝÑ φgpy,ÝÑi q

Ź

φβpy, i, a, bq

Ź

φhpz,ÝÑi , i, yq

Ź

φβpz, Si, a, bq

Ź

φβpx0, ip`1, a, bq

˛

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‚

ÐÑ fpÝÑi , ip`1q “ x0

˛

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‚

.

(ð) We first prove


Rob. $c @x0

¨

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˝

fpÝÑi , ip`1q “ x0 ÝÑ DaDb@i ď ip`1 Dy Dz

¨

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˝


Ź

φβpy, i, a, bq

Ź

φhpz,ÝÑi , i, yq

Ź

φβpz, Si, a, bq

Ź


˛

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‚

˛

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‚

.

We consider the sequence of integers fpÝÑi , 0q, . . . , fp

ÝÑi , ip`1q. Following the proof of

Lemma 121, we obtain two integers a and b to make the β-function work. Since the

formulas φβ, φg, φh respectively represent the functions β, g, h, we have

Rob. $c φβ`

gpÝÑi q, 0, a, b

˘

^ φg`

gpÝÑi q,ÝÑi˘

together with

Rob. $c φβ`

fpÝÑi , ip`1q, ip`1, a, b

˘

and for each integer n ă ip`1

Rob. $c φβ`

fpÝÑi , nq, n, a, b

˘

^ φh

´

fpÝÑi , n` 1q,

ÝÑi , n, fp

ÝÑi , nq

¯

^ φβ`

fpÝÑi , n` 1q, Sn, a, b

˘

.

hence we have

Arithmetic 101

Rob. $c @x0

¨

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˝

fpÝÑi , ip`1q “ x0 ÝÑ

ľ

kďip`1

¨

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˝

k“ 0 ÝÑ φgpgpÝÑi q,ÝÑi q

Ź

φβpfpÝÑi , kq, k, a, bq

Ź

φhpfpÝÑi , k ` 1q,

ÝÑi , k, fp

ÝÑi , kqq

Ź

φβpfpÝÑi , k ` 1q, Sk, a, bq

Ź


˛

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‚

˛

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‚

.

from which we logically derive

Rob. $c @x0

¨

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˝

fpÝÑi , ip`1q “ x0 ÝÑ

ľ

kďip`1

Dy Dz

¨

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˝

k“ 0 ÝÑ φgpy,ÝÑi q

Ź

φβpy, k, a, bq

Ź

φhpz,ÝÑi , k, yq

Ź

φβpz, Sk, a, bq

Ź


˛

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‚

˛

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‚

.

Furthermore, we know from 4.10 Rob. $c @x“

x ď nÐÑ px “ 0 _ x “ S0 _ . . . _ x “ nq‰


that

Rob. $c @i´

i ď ip`1 ÐÑ“

i “ 0 _ i “ 1 _ . . . _ i “ ip`1

‰

¯

.

Thus we obtain

Rob. $c @x0

¨

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˝

fpÝÑi , ip`1q “ x0 ÝÑ @i ď ip`1 Dy Dz

¨

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˝


Ź

φβpy, i, a, bq

Ź

φhpz,ÝÑi , i, yq

Ź

φβpz, Si, a, bq

Ź


˛

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‚

˛

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‚

and finally

Rob. $c @x0

¨

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˝

fpÝÑi , ip`1q “ x0 ÝÑ Da Db @i ď ip`1 Dy Dz

¨

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˝


Ź

φβpy, i, a, bq

Ź

φhpz,ÝÑi , i, yq

Ź

φβpz, Si, a, bq

Ź


˛

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‚

˛

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‚

which completes the first part of the proof.

Arithmetic 103

(ñ) We need to show

Rob. $c @x0

¨

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˝

DaDb@i ď ip`1 Dy Dz

»

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

–


Ź

φβpy, i, a, bq

Ź

φhpz,ÝÑi , i, yq

Ź

φβpz, Si, a, bq

Ź


fi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

fl

ÝÑ fpÝÑi , ip`1q “ x0

˛

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‚

.


Rob. $c @x0

¨

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˝

DaDb@i

»

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

–

i ď ip`1 ÝÑ Dy Dz

»

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

–


Ź

φβpy, i, a, bq

Ź

φhpz,ÝÑi , i, yq

Ź

φβpz, Si, a, bq

Ź


fi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

fl

fi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

fl


˛

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‚

.

By 4.10 Rob. $c @x“

x ď nÐÑ px “ 0 _ x “ S0 _ . . . _ x “ nq‰

this is equiva-


lent to

Rob. $c @x0

¨

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˝

DaDb@i

»

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

–

¨

˝

ł

kďip`1

i “ k

˛

‚ÝÑ Dy Dz

»

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

–


Ź

φβpy, i, a, bq

Ź

φhpz,ÝÑi , i, yq

Ź

φβpz, Si, a, bq

Ź


fi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

fl

fi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

fl


˛

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‚

.

which is equivalent to 5

Rob. $c @x0

¨

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˝

DaDb@i

»

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

–

ľ

kďip`1

¨

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˝

i “ k ÝÑ Dy Dz

»

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

–


Ź

φβpy, i, a, bq

Ź

φhpz,ÝÑi , i, yq

Ź

φβpz, Si, a, bq

Ź


fi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

fl

˛

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‚

fi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

fl


˛

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‚

.

which again is equivalent to

5we have pA_Bq ÝÑ C ” pA_Bq_C ” p A^ Bq_C ” p A_Cq^p B_Cq ” pA ÝÑ Cq^pB ÝÑ Cq.

Arithmetic 105

Rob. $c @x0

¨

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˝

DaDb

»

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

–

ľ

kďip`1

Dy Dz

»

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

–

k“ 0 ÝÑ φgpy,ÝÑi q

Ź

φβpy, k, a, bq

Ź

φhpz,ÝÑi , k, yq

Ź

φβpz, Sk, a, bq

Ź


fi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

fl

fi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

fl


˛

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‚

.


Rob. $c @x0

¨

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˝

DaDb . . . DykDzk . . .loooooomoooooon

kďip`1

»

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

–

ľ

kďip`1

»

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

–

k“ 0 ÝÑ φgpyk,ÝÑi q

Ź

φβpyk, k, a, bq

Ź

φhpzk,ÝÑi , k, ykq

Ź

φβpzk, Sk, a, bq

Ź


fi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

fl

fi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

fl


˛

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‚

.

and also to


Rob. $c @x0

¨

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˝

DaDb . . . DykDzk . . .loooooomoooooon

kďip`1

»

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

–

ľ

kďip`1

»

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

–

φgpy0,ÝÑi q

Ź

φβpyk, k, a, bq

Ź


Ź

φβpzk, Sk, a, bq

Ź


fi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

fl

fi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

fl


˛

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‚

.

and also to

Rob. $c @x0@a@b . . .@yk@zk . . .looooooomooooooon

kďip`1

¨

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˝

ľ

kďip`1

»

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

–

φgpy0,ÝÑi q

Ź

φβpyk, k, a, bq

Ź


Ź

φβpzk, Sk, a, bq

Ź


fi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

fl


˛

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‚

.

Finally, making use of the following three facts:

(1) φβ represents β in a strong way since we also have for all integers k, n we have

Rob. $c @a@b@x0

”

“

φβpn, k, a, bq ^ φβpx0, k, a, bq‰

ÝÑ x0 “ nı

Arithmetic 107

(2) φg represents g

(3) φh represents h

At last, by induction on ip`1 we show

Rob. $c @a@b . . .@yk@zk . . .looooooomooooooon

kďip`1

¨

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˝

ľ

kďip`1

»

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

–

φgpy0,ÝÑi q

Ź

φβpyk, k, a, bq

Ź


Ź

φβpzk, Sk, a, bq

fi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

fl

ÝÑľ

kďip`1

»

—

—

—

—

—

—

—

—

—

—

—

—

–

y0 “ gpÝÑi q

Ź

yk “ fpÝÑi , kq

Ź

zk “ fpÝÑi , k ` 1q

fi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

fl

˛

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‚

.

ip`1 “ 0: we only need to prove

Rob. $c @a@b@y0@z0

¨

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˝

»

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

–

φgpy0,ÝÑi q

Ź

φβpy0, 0, a, bq

Ź

φhpz0,ÝÑi , 0, y0q

Ź

φβpz0, S0, a, bq

fi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

fl

ÝÑ

»

—

—

—

—

—

—

—

—

—

—

—

—

–

y0 “ gpÝÑi q

Ź

y0 “ fpÝÑi , 0q

Ź

z0 “ fpÝÑi , 1q

fi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

fl

˛

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‚

which directly follows from the fact that φg and φh represent respectively g and h.

ip`1 “ n` 1: we assume


Rob. $c @a@b . . .@yk@zk . . .looooooomooooooon

kďn

¨

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˝

ľ

kďn

»

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

–

φgpy0,ÝÑi q

Ź

φβpyk, k, a, bq

Ź


Ź

φβpzk, Sk, a, bq

fi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

fl

ÝÑľ

kďn

»

—

—

—

—

—

—

—

—

—

—

—

—

–

y0 “ gpÝÑi q

Ź

yk “ fpÝÑi , kq

Ź

zk “ fpÝÑi , k ` 1q

fi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

fl

˛

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‚

.

We only need to show

Rob. $c @a@b@zn@yn`1@zn`1

¨

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˝

»

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

–

zn “ fpÝÑi , n` 1q

Ź

φβpzn, n` 1, a, bq

Ź

φβpyn`1, n` 1, a, bq

Ź

φhpzn`1,ÝÑi , n` 1, yn`1q

fi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

fl

ÝÑ

»

—

—

—

—

–

yn`1 “ fpÝÑi , n` 1q

Ź

zn`1 “ fpÝÑi , n` 2q

fi

ffi

ffi

ffi

ffi

fl

˛

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‚

.

which holds because:

(1) since φβ represents β in a strong way we have

Rob. $c @a@b@yn`1

”

“

φβpfpÝÑi , n` 1q, n` 1, a, bq ^ φβpyn`1, n` 1, a, bq

‰

ÝÑ yn`1 “ fpÝÑi , n` 1q

ı

(2) since φh represents h we have

Rob. $c @a@b@yn`1@zn`1

”

“

φhpzn`1,ÝÑi , n` 1, yn`1q ^ yn`1 “ fp

ÝÑi , n` 1q

‰

ÝÑ zn`1 “ fpÝÑi , n` 2q

ı

To sum up things, we have obtained

Arithmetic 109

Rob. $c @x0@a@b . . .@yk@zk . . .looooooomooooooon

kďip`1

¨

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˝

ľ

kďip`1

»

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

–

φgpy0,ÝÑi q

Ź

φβpyk, k, a, bq

Ź


Ź

φβpzk, Sk, a, bq

Ź


fi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

fl

ÝÑ

»

—

—

—

—

—

—

—

—

—

—

—

—

–

yip`1 “ fpÝÑi , ip`1q

Ź

φβpyip`1 , ip`1, a, bq

Ź


fi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

ffi

fl

˛

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‚

.

Once again, since φβ strongly represents β we have

Rob. $c @x0@a@b”

“

φβpfpÝÑi , ip`1q, ip`1, a, bq ^ φβpx0, ip`1, a, bq

‰

ÝÑ x0 “ fpÝÑi , n` 1q

ı

which finishes the proof.

% 122

Theorem 123 All total recursive functions are representable.

Proof of Theorem 123: An immediate consequence of Examples 108, 109 and 110 and Lemmas

113 , 119 and 122.

% 123


Chapter 5

Godel’s First IncompletenessTheorem

5.1 Godel Numbers

The idea is the following: intuitively, formulas from arithmetic talk about integers – no matter

whether these are standard or not – we ca turn them into formulas that talk about the arithmetic

itself by encoding formulas, proofs, etc. by integers. This way a formula φpxq may say something

like “x is the code of a closed formula from our language LA “ t0, S,`, ü” or ψpx, yq may

eventually say “x is the code of a closed formula θ from our language and y is the code of a

proof of θ in Robinson arithmetic”.

As always we start with the terms: given term t we write xty for its code.

Definition 124 (Godel numbering of the LA-terms) The Godel numbering of the terms

from the language LA “ t0, S,`, ü is

t “ 0 ù xty “ α3p0, 0, 0q

t “ xn ù xty “ α3pn` 1, 0, 0q

t “ St0 ù xty “ α3pxty, 0, 1q

t “ t0`t1 ù xty “ α3pxt0y, xt1y, 2q

t “ t0¨t1 ù xty “ α3pxt0y, xt1y, 3q

Lemma 125 The set T of all codes of terms from LA

T “

xty | t is a term from LA(

is Prim. Rec.


To almost immediately show this result we need to prove that some stronger form of construction

by recursion still produces Prim. Rec. functions.

Lemma 126 For all Prim. Rec. functions h P NpNn`p`1q, g P NpNnq, k1, . . . , kp P NN, such that

every integer y ą 0ľ

0ăiďp

kipyq ă y

the function f P NpNn`1q defined by

(1) fpÝÑx , 0q “ gpÝÑx q

(2) fpÝÑx , y ` 1q “ h`

fpÝÑx , k1pyqq, . . . , fpÝÑx , kppyqq,ÝÑx , y, fpÝÑx , yq˘

is also Prim. Rec.

First, notice that the Ackermann function A P NpN2q defined by

Apm,nq “

$

&

%

n` 1 if m “ 0,

Apm´ 1, 1q if m ą 0 and n “ 0,

A`

m´ 1, Apm,n´ 1q˘

if m ą 0 and n ą 0.

is not of this form for Apm,n´ 1q ă n certainly does not hold.

Proof of Lemma 126: The idea of the proof is to define by recursion a function that carries all

the datas fpÝÑx , zq for z ă y when defining fpÝÑx , yq. To achieve this, we make use of the two

Prim. Rec. functions c and d that we defined in Example 91:

(1) the function c : Năω ÝÑ N codes the finite sequences of integers and is defined by

"

cpεq “ 1

cpx0, . . . , xpq “ Π p0qx0`1¨Π p1qx1`1

¨ ¨ ¨Π ppqxp`1.

(2) And the function d P NpN2q that allows, from any integer n, to recover every element of

the sequence xx0, . . . , xpy that c encodes (i.e. cpx0, . . . , xpq “ n) by

dpi, nq “ µx ď n Π piqx`1 does not divide n.

We want to define some Prim. Rec. function θ P NpNn`1q such that

θpÝÑx , yq “ c`

fpÝÑx , 0q, fpÝÑx , 1q, . . . , fpÝÑx , yq˘

This is easily done by recursion:

(1) θpÝÑx , 0q “ 2gpÝÑx q`1

Arithmetic 113

(2) θpÝÑx , y ` 1q “ θpÝÑx , yq ¨Π py ` 1qh“

drk1pyqq,θpÝÑx ,yqs,...,drkppyqq,θpÝÑx ,yqs,ÝÑx ,y,fpÝÑx ,yq‰

`1.

Then, to show that f is also Prim. Rec., it only remains to set

fpÝÑx , yq “ d“

y, θpÝÑx , yq‰

9 1.

% 126

Proof of Lemma 125: Its characteristic function χT : N ÝÑ N is defined by:

if β33pkq “ 0 and β2

3pkq “ 0 and β13pkq “ 0 ù χT pkq “ 1


3pkq “ 0 and β13pkq ą 0 ù χT pkq “ 1


3pkq “ 0 and β13pkq ą 0 ù χT pkq “ χT ˝ β

13pkq

if β33pkq “ 2 ù χT pkq “ χT

`

β13pkq ¨ β

23pkq

˘

if β33pkq “ 3 ù χT pkq “ χT

`

β13pkq ¨ β

23pkq

˘

else ù χT pkq “ 0.

By Lemma 126 this definition by case study yields a Prim. Rec. function.

% 125

Definition 127 (Godel numbering of the LA-formulas) The Godel numbering of the LA-

formulas is

φ “ t0 “ t1 ù xφy “ α3pxt0y, xt1y, 4q

φ “ ψ ù xφy “ α3pxψy, 0, 5q

φ “ pφ0 ^ φ1q ù xφy “ α3pxφ0y, xφ1y, 6q

φ “ pφ0 _ φ1q ù xφy “ α3pxφ0y, xφ1y, 7q

φ “ pφ0 ÝÑ φ1q ù xφy “ α3pxφ0y, xφ1y, 8q

φ “ pφ0 ÐÑ φ1q ù xφy “ α3pxφ0y, xφ1y, 9q

φ “ @xn ψ ù xφy “ α3pxψy, n, 10q

φ “ Dxn ψ ù xφy “ α3pxψy, n, 11q.

Notice that for every formula φ, we have xφy ą 0.

Lemma 128 The set of all codes of formulas from LA

xφy | φ is a formula from LA(

is Prim. Rec.


Proof of Lemma 128: Its characteristic function χF : N ÝÑ N is defined by:

if β33pkq “ 4 ù χFpkq “ χT ˝ β

13pkq ¨ χT ˝ β

23pkq


3pkq “ 0 ù χFpkq “ χF ˝ β13pkq

if β33pkq “ 6 ù χFpkq “ χF

`

β13pkq ¨ β

23pkq

˘


`

β13pkq ¨ β

23pkq

˘


`

β13pkq ¨ β

23pkq

˘


`

β13pkq ¨ β

23pkq

˘


3pkq ą 0 ù χFpkq “ χF ˝ β13pkq


3pkq ą 0 ù χFpkq “ χF ˝ β13pkq

else ù χFpkq “ 0.


% 128

Lemma 129 The set of all codes of terms from LA that contain the variable xn

T3x “

pxty, nq | t is a term from LA and t contains xn(

is Prim. Rec.

Proof of Lemma 129: Its characteristic function χT 3xn: N2 ÝÑ N is defined by:


3pkq “ 0 and β13pkq “ n` 1 ù χT 3xn

pk, nq “ 1


3pkq “ 0 ù χT 3xnpk, nq “ χT 3xn

`

β13pkq, n

˘

if β33pkq “ 2 ù χT 3xn

pk, nq “ max´

χT 3xn

`

β13pkq, n

˘

, χT 3xn

`

β23pkq, n

˘

¯


pk, nq “ max´

χT 3xn

`

β13pkq, n

˘

, χT 3xn

`

β23pkq, n

˘

¯

else ù χT 3xnpk, nq “ 0.


% 129

Arithmetic 115

Lemma 130 The set of all codes of terms from LA that do not contain the variable xn

T7x “

pxty, nq | t is a term from LA and t does not contain xn(

is Prim. Rec.

Proof of Lemma 130: Its characteristic function χT 7xn: N2 ÝÑ N is defined by :


3pkq “ 0 and β13pkq ‰ n` 1 ù χT 7xn

pk, nq “ 1


3pkq “ 0 ù χT 7xnpk, nq “ χT 7xn

`

β13pkq, n

˘


pk, nq “ χT 7xn

`

β13pkq, n

˘

¨ χT 7xn

`

β23pkq, n

˘


pk, nq “ χT 7xn

`

β13pkq, n

˘

¨ χT 7xn

`

β23pkq, n

˘

else ù χT 7xnpk, nq “ 0.


% 130

Lemma 131 The set of all codes of formulas from LA that contain the variable xn

F3x “ `

xφy, n˘

| φ is a formula from LA and φ contains xn(

is Prim. Rec.

Proof of Lemma 131: Its characteristic function χF3xn: N2 ÝÑ N is defined by:

if β33pkq “ 4 ù χF3xn

pk, nq “ max´

χT 3xn

`

β13pkq, n

˘

, χT 3xn

`

β23pkq, n

˘

¯


3pkq “ 0 ù χF3xnpk, nq “ χF3xn

`

β13pkq, n

˘


pk, nq “ max´

χF3xn

`

β13pkq, n

˘

, χF3xn

`

β23pkq, n

˘

¯


pk, nq “ max´

χF3xn

`

β13pkq, n

˘

, χF3xn

`

β23pkq, n

˘

¯


pk, nq “ max´

χF3xn

`

β13pkq, n

˘

, χF3xn

`

β23pkq, n

˘

¯


pk, nq “ max´

χF3xn

`

β13pkq, n

˘

, χF3xn

`

β23pkq, n

˘

¯


pk, nq “ χF3xn

`

β13pkq, n

˘


pk, nq “ χF3xn

`

β13pkq, n

˘

else ù χF3xnpk, nq “ 0.



% 131

Lemma 132 The set of all codes of formulas from LA that do not contain the variable xn

F7x “ `

xφy, n˘

| φ is a formula from LA and φ does not contain xn(

is Prim. Rec.

Proof of Lemma 132: Its characteristic function χF7xn: N2 ÝÑ N is defined by :


pk, nq “ χT 7xn˝ β1

3pkq ¨ χT 7xn˝ β2

3pkq


3pkq “ 0 ù χF7xnpk, nq “ χF7xn

˝ β13pkq


pk, nq “ χF7xn˝ β1

3pkq ¨ χF7xn˝ β2

3pkq




3pkq




3pkq




3pkq



3pkq



3pkq

else ù χF7xnpk, nq “ 0.


% 132

Lemma 133 The set of all codes of formulas from LA that contain xn as a free variable

F3x free “ `

xφy, n˘

| φ is a formula from LA and xn is free in φ(

is Prim. Rec.

Proof of Lemma 133: Its characteristic function χF3x free: N2 ÝÑ N is defined by:

Arithmetic 117

if β33pkq “ 4 ù χF3x free

pk, nq “ max´

χF3x free

`

β13pkq, n

˘

, χF3x free

`

β23pkq, n

˘

¯


3pkq “ 0 ù χF3x freepk, nq “ χF3x free

`

β13pkq, n

˘


pk, nq “ max´

χF3x free

`

β13pkq, n

˘

, χF3x free

`

β23pkq, n

˘

¯


pk, nq “ max´

χF3x free

`

β13pkq, n

˘

, χF3x free

`

β23pkq, n

˘

¯


pk, nq “ max´

χF3x free

`

β13pkq, n

˘

, χF3x free

`

β23pkq, n

˘

¯


pk, nq “ max´

χF3x free

`

β13pkq, n

˘

, χF3x free

`

β23pkq, n

˘

¯


3pkq ‰ n ù χF3x freepk, nq “ χF3x free

`

β13pkq, n

˘


3pkq ‰ n ù χF3x freepk, nq “ χF3x free

`

β13pkq, n

˘

else ù χF3x freepk, nq “ 0.


% 133

Lemma 134 The set of all codes of formulas from LA that contain xn as a bound variable

F3x bound “ `

xφy, n˘

| φ is a formula from LA and xn is bound in φ(

is Prim. Rec.

Proof of Lemma 134: we have

F3x bound “ F3x r F3x free

% 134

Lemma 135 The set of all codes of closed formulas from LA

F3closed “

xφy | φ is a closed formula from LA(

is Prim. Rec.


Proof of Lemma 135: We have

k P F3closed ðñ k P F and @n ď k pk, nq R F3x free.

% 135

Lemma 136 The function STub. P NpN

3q defined below is Prim. Rec.

STub. pnu, nt, nq

$

&

%

xurt{xnsy if nu P T , nt P T and nu “ xuy, nt “ xty

0 otherwise .

Proof of Lemma 136: We first recall the definition of xty:

t “ 0 ù xty “ α3p0, 0, 0q

t “ xn ù xty “ α3pn` 1, 0, 0q

t “ St0 ù xty “ α3pxty, 0, 1q

t “ t0`t1 ù xty “ α3pxt0y, xt1y, 2q

t “ t0¨t1 ù xty “ α3pxt0y, xt1y, 3q

STub. P NpN

3q is defined by

STub. pnu, nt, nq “

$

’

’

’

’

’

’

’

’

’

’

&

’

’

’

’

’

’

’

’

’

’

%

0 if nu R T or nt R Tnt if nu, nt P T and β3

3pnuq “ 0 and β23pnuq “ 0 and β1

3pnuq “ n` 1

nu if nu, nt P T and β33pnuq “ 0 and β2

3pnuq “ 0 and β13pnuq ‰ n` 1

α3pSTub.

`

β13pnuq, nt, n

˘

, 0, 1q if nu, nt P T and β33pnuq “ 1 and β2

3pnuq “ 0 and β13pnuq P T

α3pSTub.

`

β13pnuq, nt, n

˘

,STub.

`

β23pnuq, nt, n

˘

, 2q if nu, nt P T and β33pnuq “ 2 and β2

3pnuq P T and β13pnuq P T

α3pSTub.

`

β13pnuq, nt, n

˘

,STub.

`

β23pnuq, nt, n

˘

, 3q if nu, nt P T and β33pnuq “ 3 and β2

3pnuq P T and β13pnuq P T

By Lemma 126 STub. is Prim. Rec.

% 136

Lemma 137 The function SFub. P NpN

3q defined below is Prim. Rec.

SFub. pnu, nt, nq “

$

&

%

xφrt{xnsy if nφ “ xφy P F , nt “ xty P T

0 otherwise .

Proof of Lemma 137: We first recall the definition of xφy:

Arithmetic 119

φ “ t0 “ t1 ù xφy “ α3pxt0y, xt1y, 4q

φ “ ψ ù xφy “ α3pxψy, 0, 5q

φ “ pφ0 ^ φ1q ù xφy “ α3pxφ0y, xφ1y, 6q

φ “ pφ0 _ φ1q ù xφy “ α3pxφ0y, xφ1y, 7q

φ “ pφ0 ÝÑ φ1q ù xφy “ α3pxφ0y, xφ1y, 8q

φ “ pφ0 ÐÑ φ1q ù xφy “ α3pxφ0y, xφ1y, 9q

φ “ @xn ψ ù xφy “ α3pxψy, n, 10q

φ “ Dxn ψ ù xφy “ α3pxψy, n, 11q.

SFub. P NpN

3q is defined by

SFub. pnφ, nt, nq “

$

’

’

’

’

’

’

’

’

’

’

’

’

’

’

’

’

’

’

’

’

’

’

’

’

&

’

’

’

’

’

’

’

’

’

’

’

’

’

’

’

’

’

’

’

’

’

’

’

’

%

0 if nφ R F or nt R Tα3pST

ub.

`

β13pnφq, nt, n

˘

,STub.

`

β23pnφq, nt, n

˘

, 4q if nφ P F and nt P T and β33pnφq “ 4

α3pSFub.

`

β13pnφq, nt, n

˘

, 0, 5q if nφ P F and nt P T and β33pnφq “ 5

α3pSFub.

`

β13pnφq, nt, n

˘

,SFub.

`

β23pnφq, nt, n

˘


α3pSFub.

`

β13pnφq, nt, n

˘

,SFub.

`

β23pnφq, nt, n

˘


α3pSFub.

`

β13pnφq, nt, n

˘

,SFub.

`

β23pnφq, nt, n

˘


α3pSFub.

`

β13pnφq, nt, n

˘

,SFub.

`

β23pnφq, nt, n

˘


α3pSFub.

`

β13pnφq, nt, n

˘

, β23pnφq, 10q if nφ P F and nt P T and β3

3pnφq “ 10 and β23pnφq ‰ n

α3pSFub.

`

β13pnφq, nt, n

˘

, β23pnφq, 11q if nφ P F and nt P T and β3

3pnφq “ 11 and β23pnφq ‰ n

β13pnφq if nφ P F and nt P T and β3

3pnφq “ 10 and β23pnφq “ n

β13pnφq if nφ P F and nt P T and β3

3pnφq “ 11 and β23pnφq “ n

By Lemma 126 SFub. is Prim. Rec.

% 137

We will now define a way of coding (finite) sets of formulas. We will note really encode the

set, but some finite sequence of formulas, because we will not care about the ordering of such a

sequence, even if what we really encode is the sequence, we will handle it as if it were a set.

Definition 138 (coding and decoding sequences) We define both x y : Năω ÝÑ N and

z y : N2 ÝÑ N by

"

xεy “ 0

xx0, . . . , xpy “ Π p0qx0 ¨Π p1qx1 ¨ ¨ ¨Π ppqxp .

Where Π piq enumerates the prime numbers1.

And

znyi “ µx ď n Π piqx`1 does not divide n.

1 Π p0q “ 2; Π p1q “ 3; Π p2q “ 5; etc.


Notice that for all i ď p we have zxx0, . . . , xpyyi “ xi. Furthermore, for every formula φ, the inte-

ger xφy is strictly positive. Therefore, given any sequence 〈x0, . . . , xp〉 P Năω if zxx0, . . . , xpyyi “ 0

then we know for sure that xi does not code a formula.

We will say that the integer 1 codes the empty set – which is also an empty set of formulas –

and another integer codes the set ∆ “ tφ0, φ1, . . . , φpu if this integer is of the form Π pi0qxφ0y

¨

Π pi1qxφ1y

¨ ¨ ¨Π pipqxφpy.

Definition 139 (Godel numbering of the LA-finite sets of formulas) The Godel number-

ing of any set ∆ “ tφ0, φ1, . . . , φpu of LA-formulas is any integer of the form#

xHy “ 1

x∆y “ Π pi0qxφ0y

¨Π pi1qxφ1y

¨ ¨ ¨Π pipqxφpy.

We denote CPfin.pFq the set of codes of finite sets of formulas:

CPfin.pFq “ tx∆y | ∆ is some finite set of LA formulasu.

Lemma 140 The set CPfin.pFq of codes of finite sets of formulas is Prim. Rec.

Proof of Lemma 140:

χCPfin.pFqpnq “

$

’

’

&

’

’

%

1 if n “ 1

1 if n ‰ 1 and @i ď n“

znyi ą 0 ÝÑ znyi P F‰

0 else

% 140

Lemma 141 There exist two Prim. Rec. functions Rem. : N2 ÝÑ N and Add. : N2 ÝÑ N such

that

Rem. pn,mq “

$

&

%

x∆Y tφuy if n “ xφy P F and m “ x∆y P CPfin.pFq

0 if n R F or m R CPfin.pFq

Add. pn,mq “

$

&

%

x∆ r tφuy if n “ xφy P F and m “ x∆y P CPfin.pFq


Arithmetic 121

Proof of Lemma 141:

Add. pn,mq “

$

&

%


m ¨Π`

µi ď m zmyi “ 0˘n

if xφy “ n P F and x∆y “ m P CPfin.pFq.

Rem. pn,mq “

$

’

’

’

’

&

’

’

’

’

%


m if n P F and m P CPfin.pFq and @i ď m zmyi ‰ n„

m

¨Π`

µi ď m zmyi “ n˘n

if n P F and m P CPfin.pFq and Di ď m zmyi “ n

% 141

Lemma 142 The following set is Prim. Rec.

Ins. “

!

pxφy, x∆yq P N2 | xφy P F , x∆y P CPfin.pFq and φ P ∆)

.

We will use a relation-like notation and write xφy Ins. x∆y instead of pxφy, x∆yq P Ins..

Proof of Lemma 142:

χIns.pn,mq “

$

’

’

’

’

&

’

’

’

’

%

1 if n P F and m P CPfin.pFq and Di ď m

$

’

’

&

’

’

%

Π piqn divides m

and

Π piqn`1 does not divides m

0 else.

% 142

Lemma 143 The following set is Prim. Rec.

Equ. “!

pxΓy, x∆yq P N2 | xΓy P CPfin.pFq, x∆y P CPfin.pFq and Γ “ ∆)

.

Notice that the equality “ Γ “ ∆” is between two sets, therefore it relies on extensionality. We

will use a relation-like notation and write xφy Equ. x∆y instead of pxφy, x∆yq P Equ..

Proof of Lemma 143: We have

χEqu.pn,mq “#

1 if n,m P CPfin.pFq and @i ď maxpn,mq`

i P F Ñ pi Ins. nØ i Ins. mq˘

0 else.

% 143


Lemma 144 There exists a Prim. Rec. function Union P NpN2q such that

Union pn,mq “

$

&

%

0 if n R CPfin.pFq or m R CPfin.pFq

xΓ1 Y∆1y if n “ xΓy P CPfin.pFq and m “ x∆y P CPfin.pFq and xΓ1 Y∆1y Equ. xΓY∆y

Proof of Lemma 144: We first define f P NpN3q by recursion:

$

&

%

fpγ, δ, 0q “ Π p0qzγy0¨Π p1qzδy0

fpγ, δ, n` 1q “ Π p2nqzγyn¨Π p2n` 1qzδyn

¨ fpγ, δ, nq.

Then we set

Union pγ, δq “ fpγ, δ,maxpγ, δqq.

% 144

Definition 145 (Godel numbering of the LA-sequents from sequent calculus) The Godel

numbering of any sequent Γ $ ∆ is

xΓ $ ∆y “ α2pxΓy, x∆yq

We denote SQ the set of codes of finite sets of formulas:

SQ “ txΓ $ ∆y | Γ, ∆ finite sets of LA formulasu.

Given any integer n we use the notation ln for β12pnq and rn for β2

2pnq. This way,

if n “ xΓ $ ∆y, then ln “ xΓy and rn “ x∆y.

Lemma 146 The set SQ of codes of sequents of sequent calculus is Prim. Rec.

Proof of Lemma 146:

χSQpnq “

$

&

%

1 if ln P CPfin.pFq and rn P CPfin.pFq

0 else

% 146

We will now denote AX the set of codes of axioms of sequent calculus which are not to be

mistaken for the axiom of Robinson arithmetic.

Arithmetic 123

Definition 147 (Godel numbering of the axioms of sequent calculus)

AX “

!

α2

`

2xφy, 2xφy˘

| xφy P F)

.

Lemma 148 The set AX of codes of axioms of sequent calculus is Prim. Rec.

Proof of Lemma 148:

χAX pnq “

$

&

%

1 if β12pnq “ β2

2pnq and zβ12pnqy

0 P F

0 else

5.2 Coding the Proofs

We recall that a proof in Sequent Calculus is a tree of the form

ax

@x pφÑ ψq $ @x pφÑ ψq@e

@x pφ ÝÑ ψq $ φry{xs Ñ ψry{xs

ax

@x φ $ @x φ@e

@x φ $ φry{xsÑ e

@x pφÑ ψq,@x φ $ ψry{xs@i

@x pφÑ ψq,@x φ $ @x ψÑ i

@x pφÑ ψq $ @x φÑ @x ψ

where the shape of the tree is controlled by the rules of Sequent Calculus.

We are now ready to define for each rule of the Sequent Calculus, a set of tuples of codes of

sequents that satisfy the property that the rule defines.

We will successively define

(1) ˝ Rax Ď N

(2) ˝ R^l1 Ď N2

˝ R^l2 Ď N2

˝ R l Ď N2

˝ R@l Ď N2

˝ RDl Ď N2

˝ R_r1 Ď N2

˝ R_r2 Ď N2

˝ R r Ď N2

˝ R@r Ď N2

˝ RDr Ď N2

˝ Rwknl Ď N2

˝ Rwknr Ď N2

˝ Rctr l&r Ď N2

˝ Rcut Ď N2

˝ RRep Ď N2

˝ RRef Ď N2

(3) ˝ R_l Ď N3 ˝ RÑlĎ N3 ˝ R^r Ď N3

and for each of them, the fact that it is Prim. Rec. will derive from its definition. We first recall

what the rules are.


Sequent Calculus

Axioms

ax

φ $ φ

Logical Rules

Γ, φ $ ∆^l1

Γ, φ^ ψ $ ∆

Γ, ψ $ ∆^l2

Γ, φ^ ψ $ ∆

Γ $ φ,∆ Γ $ ψ,∆^r

Γ $ φ^ ψ,∆

Γ, φ $ ∆ Γ, ψ $ ∆_l

Γ, φ_ ψ $ ∆

Γ $ φ,∆_r1

Γ $ φ_ ψ,∆

Γ $ ψ,∆_r2

Γ $ φ_ ψ,∆

Γ $ φ,∆ Γ, ψ $ ∆Ñl

Γ, φÑ ψ $ ∆

Γ, φ $ ψ,∆Ñr

Γ $ φÑ ψ,∆

Γ $ φ,∆ l

Γ, φ $ ∆

Γ, φ $ ∆ r

Γ $ φ,∆

Γ, φrt{xs $ ∆ 1

@l

Γ,@x φ $ ∆

Γ $ φry{xs,∆@r

Γ $ @x φ,∆ 2

Γ, φry{xs $ ∆Dl

Γ, Dx φ $ ∆ 2

Γ $ φrt{xs,∆1

Dr

Γ $ Dx φ,∆

Γ, t “ t $ ∆Ref

Γ $ ∆

Γ, t “ s, φrs{xs, φrt{xs $ ∆Rep

Γ, s “ t, φrt{xs $ ∆

Structural Rules

Γ $ ∆wknl

Γ, φ $ ∆Γ $ ∆

wknr

Γ $ φ,∆

Γ, φ, φ $ ∆ctrl

Γ, φ $ ∆

Γ $ φ, φ,∆ctrr

Γ $ φ,∆

Cut Rule

Γ $ φ,∆ Γ1, φ $ ∆1

cut

Γ,Γ1 $ ∆,∆1

1t a term2y with no free occurrence the sequent concluding the rule (not in Γ, Dx φ nor @x φ, nor ∆)

Arithmetic 125

ax

φ $ φ

U P Rax ðñ U P AX

........................................................................

Γ, φ $ ∆^l1

Γ, φ^ ψ $ ∆

pU,Dq P R^l1ðñ

$

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’

%

U P SQand

D P SQand

rU Equ. rDand

Dxφy ď lU Dxψy ď lD

¨

˚

˚

˝

xφy Ins.lU and xφ^ ψy Ins.

lD

and

Rem.

`

xφy, lU˘

Equ. Rem.

`

xφ^ ψy, lD˘

˛

‹

‹

‚

where

˝ “ xφ^ ψy” stands for “ α3pxφy, xψy, 6q”.

˝ “ Dxφy ď k θrxφy{yss” stands for “ Dn ď k`

n P F ^ θrn{ys˘

” and more generally

˝ “ Dxφ1y ď k1 . . . . . . Dxφny ď kn θrxφ1y{y1,...,xφny{yns” stands for

“ Dp ď αnpk1, . . . , knq`

ľ

iďn

`

βinppq P F ^ βinppq ď ki˘

^ θrβ1nppq{y1,...,βnnppq{yns

˘

”.

........................................................................


Γ, ψ $ ∆^l2

Γ, φ^ ψ $ ∆

pU,Dq P R^l2ðñ

$

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%

U P SQand

D P SQand

rU Equ. rDand

Dxψy ď lU Dxφy ď lD

¨

˚

˚

˝

xψy Ins.lU and xφ^ ψy Ins.

lD

and

Rem.

`

xψy, lU˘

Equ. Rem.

`

xφ^ ψy, lD˘

˛

‹

‹

‚

........................................................................

Γ, φ $ ∆ Γ, ψ $ ∆_l

Γ, φ_ ψ $ ∆

pUl, Ur, dq P R_lðñ

$

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%

Ul, Ur, D P SQand

rUl Equ. rUr Equ. rDand

Dxφy ď lUl Dxψy ď lUr

¨

˚

˚

˝

xφy Ins.lUl and xψy Ins.

lUr and xφ^ ψy Ins.lD

and

Rem.

`

xφy, lUl˘

Equ. Rem.

`

xψy, lUr˘

Equ. Rem.

`

xφ^ ψy, lD˘

˛

‹

‹

‚

........................................................................

Arithmetic 127

Γ $ φ,∆ Γ, ψ $ ∆Ñl

Γ, φÑ ψ $ ∆

pUl, Ur, Dq P RÑl

ðñ$

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%

Ul, Ur, D P SQand

rUr Equ. rDand

Dxφy ď rUl Dxψy ď lUr

¨

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˝

xφy Ins.rUl and xψy Ins.

lUr and xφ^ ψy Ins.lD

and

Rem. pxφy, rUlq Equ. rUrand

Rem.

`

xψy, lUr˘

Equ. lUl Equ. Rem.

`

xφÑ ψy, lD˘

and

Rem.

`

xψy, lUr˘

Equ. Rem.

`

xφÑ ψy, lD˘

˛

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‹

‚

˝ where “ A Equ. B Equ. C ” stands for “ A Equ. B and B Equ. C ”

........................................................................

Γ $ φ,∆ l

Γ, φ $ ∆

pU,Dq P R lðñ

$

’

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&

’

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’

’

’

’

%

U,D P SQand

Dxφy ď rU

¨

˚

˚

˝

xφy Ins.rU and x φy Ins.

lD

and

Rem. pxφy, rUq Equ. Rem.

`

x φy, lD˘

˛

‹

‹

‚

........................................................................


Γ, φrt{xns $ ∆@l

Γ,@xn φ $ ∆

pU,Dq P R@lðñ

$

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%

U,D P SQand

rU Equ. rDand

Dn ď lD Dx@xnφy ď lD Dxty ď lU

¨

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˝

¨

˚

˚

˚

˚

˚

˚

˚

˝

x@xnφy Ins.lD and

`

xφy, n˘

P F3x free

and

SFub. pxφy, xty, nq Ins.

lU

and

Rem.

`

SFub. pxφy, xty, nq, lU

˘

Equ. Rem.

`

x@xnφy, lD˘

˛

‹

‹

‹

‹

‹

‹

‹

‚

or

¨

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˝

x@xnφy Ins.lD

and`

xφy, n˘

P`

F7x Y F3x bound

˘

and

xφy Ins.lU

and


lU

and

Rem.

`

xφy, lU˘

Equ. Rem.

`

x@xnφy, lD˘

˛

‹

‹

‹

‹

‹

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‚

˛

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‹

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‹

‹

‚

˝ “ Dn ď rU Dx@xnφy ď rU . . .” stands for

“ Dn ď rU Dm ď rU´

m P F ^ β33pmq “ 10 ^ β2

3pmq “ n ^ β13pmq “ xφy ^ . . .

¯

”

˝ “ xφy stands for “ β13

`

x@xnφy˘

”

˝ “ Dxty ď lU . . .” stands for “ Dv ď lU`

v P T ^ . . .˘

”

........................................................................

Arithmetic 129

Γ, φrxk{xns $ ∆Dl

Γ, Dxn φ $ ∆ 2

pU,Dq P RDlðñ

$

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%

U,D P SQand

rU Equ. rDand

Dxxny ď lD Dxxky ď lU DxDxnφy ď lD Dxψy ď lU

¨

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

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˚

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˚

˚

˚

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˚

˚

˚

˚

˚

˝

¨

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˚

˝

xDxnφy Ins.lD and xψy Ins.

lU

and`

xψy, k˘

P F3x free and´

k ‰ nÑ`

xψy, n˘

P`

F7x Y F3x bound

˘

¯

and

α3pSFub. pxψy, xxny, kq , n, 11q “ xDxnφy

and

Rem.

`

xψy, lU˘

Equ. Rem.

`

xDxnφy, lD˘

and

@xθy ď lD´

xθy Ins.lD Ñ pxθy, kq P

`

F7xk Y F3xk bound

˘

¯

and

@xδy ď rD´

xδy Ins.rD Ñ pxδy, kq P

`

F7xk Y F3xk bound

˘

¯

˛

‹

‹

‹

‹

‹

‹

‹

‹

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‹

‹

‹

‹

‚

or

¨

˚

˚

˚

˚

˚

˚

˚

˝

xDxnφy Ins.lD and xψy Ins.

lU and

`

xψy, k˘

P`

F7xk Y F3xk bound

˘

and`

xψy, n˘

P`

F7x Y F3x bound

˘

and α3pxψy, n, 11q “ xDxnφy and

Rem.

`

xψy, lU˘

Equ. Rem.

`

xDxnφy, lD˘

˛

‹

‹

‹

‹

‹

‹

‹

‚

˛

‹

‹

‹

‹

‹

‹

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‹

‹

‹

‹

‹

‹

‹

‚

where



m P F ^ β33pmq “ 11 ^ β2

3pmq “ n ^ β13pmq “ xφy ^ . . .

¯

”


`

x@xnφy˘

”

˝ “ Dxxky ď lU . . .” stands for “ Dk ď lU`

xxky “ α3pk ` 1, 0, 0q ^ . . .˘

”

........................................................................

2xk has no free occurrence in Γ, Dxn φ and ∆


Γ, t “ t $ ∆Ref

Γ $ ∆

pU,Dq P RRef

ðñ$

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U,D P SQand

rU Equ. rDand

Dxty ď lU

¨

˚

˚

˝

xt “ ty Ins.lU

and

Rem.

`

xt “ ty, lU˘

Equ. lD

˛

‹

‹

‚

........................................................................

Γ, t “ s, φrs{xns, φrt{xns $ ∆Rep

Γ, s “ t, φrt{xns $ ∆

pU,Dq P RRep

ðñ$

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U,D P SQand

rU Equ. rDand

Dxty ď lU Dxsy ď lU Dn ď U Dxφy ď UU

¨

˚

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˚

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˚

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˚

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˚

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˚

˚

˚

˝

xt “ sy Ins.lU

and

SFub. pxφy, xsy, nq Ins.

lU and SFub. pxφy, xty, nq Ins.

lU

and

xs “ ty Ins.lD and SF

ub. pxφy, xty, nq Ins.lD

and

Rem.

`

SFub. pxφy, xty, nq,Rem.

`

xs “ ty, lD˘˘

Equ.Rem.

`

SFub. pxφy, xsy, nq,Rem.

`

SFub. pxφy, xty, nq,Rem.

`

xt “ sy, lU˘˘˘

˛

‹

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‹

‚

........................................................................

Arithmetic 131

Γ $ φ,∆ Γ $ ψ,∆^r

Γ $ φ^ ψ,∆

pUl, Ur, Dq P R^rðñ

$

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%

Ul, Ur, D P SQand

lUl Equ. lUr Equ. lDand

Dxφy ď rUl Dxψy ď rUr

¨

˚

˚

˝

xφy Ins.rUl and xψy Ins.

rUr and xφ^ ψy Ins.rD

and

Rem. pxφy, rUlq Equ. Rem. pxψy, rUrq Equ. Rem. pxφ^ ψy, rDq

˛

‹

‹

‚

........................................................................

Γ $ φ,∆_r1

Γ $ φ_ ψ,∆

pU,Dq P R_r1ðñ

$

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U,D P SQand

lU Equ. lDand

Dxφy ď rU Dxψy ď rD

¨

˚

˚

˝

xφy Ins.rU and xφ_ ψy Ins.

rD

and

Rem. pxφy, rUq Equ. Rem. pxφ_ ψy, rDq

˛

‹

‹

‚

........................................................................


Γ $ ψ,∆_r2

Γ $ φ_ ψ,∆

pU,Dq P R_r2ðñ

$

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U,D P SQand

lU Equ. lDand

Dxψy ď rU Dxφy ď rD

¨

˚

˚

˝

xψy Ins.rU and xφ_ ψy Ins.

rD

and

Rem. pxψy, rUq Equ. Rem. pxφ_ ψy, rDq

˛

‹

‹

‚

........................................................................

Γ, φ $ ψ,∆Ñr

Γ $ φÑ ψ,∆

pUl, Ur, Dq P RÑr

ðñ$

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Ul, Ur, D P SQand

Dxφy ď lUl Dxψy ď rUr

¨

˚

˚

˚

˚

˚

˚

˚

˚

˝

xφy Ins.lUl and xψy Ins.

rUr and xφÑ ψy Ins.rD

and

Rem. pxψy, rUrq Equ. Rem. pxφÑ ψy, rDq

and

Rem.

`

xφy, lUl˘

Equ. lD

˛

‹

‹

‹

‹

‹

‹

‹

‹

‚

........................................................................

Arithmetic 133

Γ, φ $ ∆ r

Γ $ φ,∆

pU,Dq P R rðñ

$

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%

U,D P SQand

Dxφy ď lU

¨

˚

˚

˚

˚

˚

˚

˚

˝

xφy Ins.lU and x φy Ins.

rD

and

Rem.

`

xφy, lU˘

Equ. lDand

Rem. px φy, rDq Equ. rU

˛

‹

‹

‹

‹

‹

‹

‹

‚

........................................................................


Γ $ φrxk{xns,∆@r

Γ $ @xn φ,∆2

pU,Dq P R@rðñ

$

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%

U,D P SQand

lU Equ. lDand

Dxxny ď rD Dxxky ď rU DxDxnφy ď rD Dxψy ď rU

¨

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˝

¨

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˚

˚

˚

˚

˚

˝

x@xnφy Ins.rD and xψy Ins.

rU

and`

xψy, k˘

P F3x free and´

k ‰ nÑ`

xψy, n˘

P`

F7x Y F3x bound

˘

¯

and

α3pSFub. pxψy, xxny, kq , n, 11q “ x@xnφy

and

Rem. pxψy, rUq Equ. Rem. px@xnφy, rDq

and

@xθy ď rD´

xθy Ins.rD Ñ pxθy, kq P

`

F7xk Y F3xk bound

˘

¯

and

@xδy ď rD´

xδy Ins.rD Ñ pxδy, kq P

`

F7xk Y F3xk bound

˘

¯

˛

‹

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‚

or

¨

˚

˚

˚

˚

˚

˚

˚

˝

x@xnφy Ins.rD and xψy Ins.

rU and

`

xψy, k˘

P`

F7xk Y F3xk bound

˘

and`

xψy, n˘

P`

F7x Y F3x bound

˘

and α3pxψy, n, 11q “ x@xnφy and

Rem. pxψy, rUq Equ. Rem. px@xnφy, rDq

˛

‹

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˛

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‹

‹

‚

where



m P F ^ β33pmq “ 10 ^ β2

3pmq “ n ^ β13pmq “ xφy ^ . . .

¯

”


`

x@xnφy˘

”

˝ “ Dxxky ď rU . . .” stands for “ Dk ď rU`

xxky “ α3pk ` 1, 0, 0q ^ . . .˘

”

........................................................................

2xk has no free occurrence in Γ, @xn φ and ∆

Arithmetic 135

Γ $ φrt{xns,∆Dr

Γ $ Dxn φ,∆

pU,Dq P RDrðñ

$

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U,D P SQand

lU Equ. lDand

Dn ď rD DxDxnφy ď rD Dxty ď rU

¨

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˝

¨

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˚

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˚

˚

˚

˝

xDxnφy Ins.rD and

`

xφy, n˘

P F3x free

and


rU

and

Rem. pSFub. pxφy, xty, nq, rUq Equ. Rem. pxDxnφy, rDq

˛

‹

‹

‹

‹

‹

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‹

‚

or

¨

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˚

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˚

˚

˚

˝

xDxnφy Ins.rD

and`

xφy, n˘

P`

F7x Y F3x bound

˘

and

xφy Ins.rU

and


rU

and

Rem. pxφy, rUq Equ. Rem. pxDxnφy, rDq

˛

‹

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˛

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‚

˝ “ Dn ď rU DxDxnφy ď rU . . .” stands for


m P F ^ β33pmq “ 11 ^ β2

3pmq “ n ^ β13pmq “ xφy ^ . . .

¯

”


`

xDxnφy˘

”

˝ “ Dxty ď rU . . .” stands for “ Dv ď rU`

v P T ^ . . .˘

”

........................................................................


Γ $ ∆wknl

Γ, φ $ ∆

pU,Dq P Rwknl

ðñ$

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%

U,D P SQand

rU Equ. rDand

Dxφy ď lD

¨

˚

˚

˝

xφy Ins.lD

and

Rem.

`

xφy, lD˘

Equ. lU

˛

‹

‹

‚

........................................................................

Γ $ ∆wknr

Γ $ φ,∆

pU,Dq P Rwknr

ðñ$

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%

U,D P SQand

lU Equ. lDand

Dxφy ď rD

¨

˚

˚

˝

xφy Ins.rD

and

Rem. pxφy, rDq Equ. rU

˛

‹

‹

‚

........................................................................

Arithmetic 137

Γ, φ, φ $ ∆ctrl

Γ, φ $ ∆

Γ $ φ, φ,∆ctrr

Γ $ φ,∆

pU,Dq P Rctr l&r ðñ

$

’

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&

’

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’

’

’

’

%

U,D P SQand

lU Equ. lDand

rU Equ. rD

........................................................................

Γ $ φ,∆ Γ1, φ $ ∆1

cut

Γ,Γ1 $ ∆,∆1

pUl, Ur, Dq P Rcut

ðñ$

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%

Ul, Ur, D P SQand

Dxφy ď rUl

¨

˚

˚

˚

˚

˚

˚

˚

˚

˝

xφy Ins.rUl and xφy Ins.

lUr

and

Union

`

Rem.

`

xφy, lUr˘

, lUl˘

Equ. lDand

Union pRem. pxφy, rUlq,rUrq Equ. rD

˛

‹

‹

‹

‹

‹

‹

‹

‹

‚

........................................................................


We write

$

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%

R0 “ Rax

R1 “

$

&

%

R^l1 YR^l2 YR l YR@l YRDl YR_r1YR_r2 YR r YR@r YRDr YRwknlY

Rwknr YRctr l&r YRRep YRRef YRcut

R2 “ R_l YRÑlYR^r

We say an integer codes a proof if it is of the form

α4pnode, left proof-tree, right proof-tree, arity of the ruleq.

Definition 149 The set Proofs of the codes of all possible proofs is defined by

k “ α4pn1, n2, n3, n4q P Proofsðñ

$

’

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&

’

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’

’

’

’

’

%

n4 “ 0 and n3 “ 0 and n2 “ 0 and n1 P R0

or

n4 “ 1 and n3 “ 0 and n2 P Proofs and`

β14pn2q, n1

˘

P R1

or

n4 “ 2 and n3 P Proofs and n2 P Proofs and`

β14pn3q, β

14pn2q, n1

˘

P R2.

Notation 150 Given any proof P we write xP y for the integer described above that codes this

proof.

Lemma 151 The set Proofs is Prim. Rec..

Proof of Lemma 151:

χProofspnq “

$

’

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&

’

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’

’

’

’

’

%

1 if β44pnq “ 0 and β3

4pnq “ 0 and β24pnq “ 0 and β1

4pnq P R0

1 if β44pnq “ 1 and β3

4pnq “ 0 and χProofs

`

β24pnq

˘

“ 1 and`

β14 ˝ β

24pnq, β

14pnq

˘

P R1

1 if β44pnq “ 2 and χProofs

`

β34pnq

˘

“ χProofs

`

β24pnq

˘

“ 1 and`

β14 ˝ β

34pnq, β

14 ˝ β

24pnq, β

14pnq

˘

P R2

0 otherwise.

Arithmetic 139

By Lemma 126 Proofs is Prim. Rec.% 151

5.3 Undecidability of Robinson Arithmetic

Definition 152

(1) A theory T is recursive if the following set is recursive:!

xφy | φ P T)

.

(2) A theory T is decidable if the following set is recursive:

thms pT q “!

xφy | T $c φ)

.

Informally, this means that a theory is decidable if one has an algorithm which on any input

that represents a formula φ stops and accepts if T proves φ, and stops and rejects if T does not

prove φ.

Theorem 153 Given any theory T , the set!

`

xP y, xφy˘

P N2 | P is a proof of T $c φ)

is

˝ primitive recursive if T is primitive recursive,

˝ recursive if T is recursive.

Proof of Theorem 153: First, P is proof that T $c φ if P is a proof-tree whose root is some

sequent “∆ $ φ” for some finite ∆ Ď T .

We recall Lemma 135 which stated that the following set is Prim. Rec.

F3closed “

xφy | φ is a closed formula from LA(

Second, let χT P N ÝÑ N be the characteristic function of T . i.e.

χT pnq “

#

1 if n “ xφy P T

0 otherwise.

The characteristic function of

A “!

`

xP y, xφy˘



is

χApn,mq “

$

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%

1 if

$

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’

%

n P Proofsand

m P Fand

@i ď lβ14pnq zlβ1

4pnqyi ‰ 0 ÝÑ χT

`

zlβ14pnqy

i˘

“ 1

and

@j ď rβ14pnq

`

zrβ14pnqy

j˘

“ m

0 otherwise.

We see that this function is primitive recursive if χT is primitive recursive, and total recursive

if χT is total recursive.

% 153

Proposition 154 Given any theory T ,

!

xψy | ψ P T)

is recursive ùñ

!

xφy | T $c φ)

is recursively enumerable.

Proof of Proposition 154: We set

A “!

`

xP y, xφy˘


and B “!

xφy | T $c φ)

.

By Theorem 153, the set A is recursive. Hence the function

χpart

B pnq “ 1 9

´

1 9

`

µk χApk, nq “ 1˘

¯

is also Part. Rec. and it satisfies n P B ðñ χpart

B pnq “ 1. % 154

We recall that a theory is complete if it is both consistent and satisfies for each formula φ either

T $c φ or T $c φ.

Corollary 155 Let T be any recursive theory.

If T is complete, then T is decidable.

Proof of Corollary 155: By Proposition 154 both sets!

xφy | T $c φ)

and!

x φy | T $c φ)

Arithmetic 141

are recursively enumerable. Since T is complete we have

!

xφy | T &c φ)

“

!

x φy | T $c φ)

Hence

Nr!

xφy | T $c φ)

“`

Nr F˘

Y

!

x φy | T $c φ)

is recursively enumerable, which yields the result. % 155

Theorem 156 Let T Ě Rob. be any theory,

T is consistent ðñ T is undecidable.


(ð) T inconsistent ñ T decidable is straightforward since in this case we have

!

xφy | T $c φ)

“ F .

(ñ) Towards a contradiction, we assume that T is decidable. We then consider

F3x0 !free“

xφy | φ is a formula whose only free variable is x0

(

.

Since we already know that the set

F3x free “ `

xφy, n˘

| φ is a formula from LA and xn is free in φ(

is Prim. Rec. (see Lemma 133) and we have

xφy P F3x0 !freeðñ

`

xφy, 0˘

P F3x free and @n ď xφy

´

n ‰ 0 Ñ`

xφy, n˘

R F3x free

¯

.

an immediate consequence is that F3x0 !freeis also Prim. Rec.

Then the set2!

`

xφy, n˘

| xφy P F3x0 !freeand T $c φrn{x0s

)

“!

`

xφy, n˘

| xφy P F3x0 !freeand SF

ub. pxφy, xny, 0q P

xψy | T $c ψ(

)

2we recall that on page 118 we defined SFub. pnu, nt, nq “

$

&

%

xφrt{xnsy if nφ “ xφy P F , nt “ xty P T

0 otherwise .


is recursive.

We then consider the following set

:Diag. “

"

k P N | pk, kq R!

`

xφy, n˘

| xφy P F3x0 !freeand T $c φrn{x0s

)

*

.

:Diag. is clearly recursive, therefore there exists some function φ:px0q that represents:Diag..

This means that for all k P N we have:

˝ k P:Diag. ùñ Rob. $c φrk{x0s

˝ k R:Diag. ùñ Rob. $c φrk{x0s.

It is enough to consider the closed formula φrrxφ:ys{x0s where rxφ:ys stands for the termxφ:y

hkkikkj

S . . . S 0.

As with the Halting problem where we asked the question whether our machine would stop

on its own code as input, here we ask the question whether or not T proves φ:rrxφ:ys{x0s.

This depends on whether xφ:y belongs to:Diag. or not.

˝ xφ:y P:Diag. ùñ Rob. $c φ:rrxφ:ys{x0s

ùñ T $c φ:rrxφ:ys{x0sùñ xφ:y R

:Diag.

˝ xφ:y R:Diag. ùñ Rob. $c φ:rrxφ:ys{x0s

ùñ T $c φ:rrxφ:ys{x0s.

Since T is consistent we cannot have both

T $c φ:rrxφ:ys{x0sand T $c φ:rrxφ:ys{x0s

.

Therefore, we have T &c φ:rrxφ:ys{x0swhich immediately implies xφ:y P

:Diag..

We obtain

xφ:y P:Diag. ðñ xφ:y R

:Diag..

This contradiction finishes the proof that T is undecidable.

% 156

This brings to the mind what we did in the proof of Proposition 66.

We propose again a picture that illustrates this diagonal argument. If pφiqiPN is a enumeration

of all the formulas with x0 as one and only free variable, we make sure to define a formula which

satisfies this requirement although it is none of them.

Arithmetic 143

φ0 φ1 φ2 φ3 φ4 φ5 φn

xφ0y 0 1 1 0 1 0 . . . 0 . . .

xφ1y 1 1 1 0 0 0 . . . 0 . . .

xφ2y 1 0 1 0 0 0 . . . 1 . . .

xφ3y 0 0 1 0 1 0 . . . 0 . . .

xφ4y 0 1 0 1 1 1 . . . 0 . . .

xφ5y 1 1 0 0 0 0 . . . 0 . . .

......

......

......

......

xφny 1 0 0 0 1 1 . . . 1 . . ....

......

......

......

...

There is a 1 on the array – for instance on row 3 and column 2 – if T $c φ2prxφ3ysq, and there

is a 0 – for instance on row 2 and column 5 – if T &c φ5prxφ2ysq.

Now if T is decidable, the whole array is decidable. This means there is a Decider that on any

input pn,mq accepts if there is a 1 on position pn,mq, and rejects if there is a 0. Furthermore,

for the whole array is decidable, its diagonal is also decidable. Hence the complement of the

diagonal is decidable as well. Finally, since all recursive sets are representable, the complement of

the diagonal is represented by some formula among the enumeration – say φn – which inevitably

stumbles on rxφnysq.


Theorem 157 (undecidability of first order logic) The set

!

xφy | $c φ)

.

is not recursive

Proof of Theorem 157: Since Rob. is a finite theory, we let φRob. be the conjunction of the seven

axioms from Rob.. For any formula ψ we have

Rob. $c ψ ðñ φRob. $c ψ ðñ $c φRob. Ñ ψ.

xψy P

!

xφy | Rob. $c φ)

ðñ xφRob. Ñ ψy P

!

xφy | $c φ)

.

Therefore, if the set of codes of universally valid formulas were decidable, then Robinson arith-

metic would also be decidable.

% 157

Theorem 158 (Godel’s first incompleteness theorem) Let T Ě Rob. be any theory both

consistent and recursive.

T is incomplete.

Proof of Theorem 158: By corollary 155 every recursive complete theory is decidable. By

Theorem 156 the theory T is undecidable.

% 158

Another way of stating this theorem that one encounters very frequently among the philosophy

community is the following:

There exists a true sentence that is not provable.

Or even,

There exists a sentence that is true although it is not provable.

“Provable” usually refers to Peano Arithmetic, and “true” means true in the standard model.

And very often people add

Taking this true sentence as an axiom inevitably yields another one that is not provable.

Of course the understatement hidden behind is “the complete theory of the standard model is

not recursive”.

Chapter 6

Godel’s Second IncompletenessTheorem

6.1 Peano Arithmetic and IΣ01

To prove Godel’s second incompleteness theorem we need to work in a theory slightly more

expressive than Rob. For this reason we introduce the theory of Peano arithmetic.

Peano is a theory based on the same language as Robinson arithmetic : LA “ t0, S,`, ü

Peano has infinitely many axioms:

axiom 1. @x Sx ‰ 0

axiom 2. @x Dy px ‰ 0 Ñ Sy “ xq

axiom 3. @x @y pSx “ Sy Ñ x “ yq

axiom 4. @x x`0 “ x

axiom 5. @x @y´

x`Sy “ Spx`yq¯

axiom 6. @x x¨0 “ 0

axiom 7. @x @y´


axiom schema (induction) @x0 @x1 . . . @xn

ˆ

´

φr0{x0s ^ @x0

`

φÑ φrSx0{x0s

˘

¯

Ñ @x0 φ

˙

(for any formula φrx0,x1,...,xns)1.

So we see that Peano is nothing but Rob. plus the induction schema for all formulas constructed

on the language of arithmetic. In fact we will not need to work within Peano but only a part of

1the notation φrx0,x1,...,xns means that the free variable of φ are all among x0, x1, . . . , xn.


it formed of Rob. plus the induction schema restricted to the sole Σ01-formulas (see next section).

This theory is called Rob.` IΣ01

Example 159 We saw that Rob. does not prove that the addition is commutative. We want

to prove here that within Rob. ` IΣ01 the addition becomes commutative. For this purpose we

will have to use several instances of the inductions schema.

(1) We first show that

Rob.` IΣ01 $c @x x`0 “ 0`x.

Indeed we have both

˝ $c 0`0 “ 0`0

˝ Rob. $c @x´

`

x`0 “ 0`x˘

Ñ`

Sx`0 “ 0`Sx˘

¯

because we have by

4 @x x`0 “ x and 5 @x @y`

x`Sy “ Spx`yq˘

Rob. $c Sx`0 “ Sx ^ 0`Sx “ Sp0`xq

hence

Rob., x`0 “ 0`x $c Sx`0 “ Sx ^ 0`Sx “ Spx`0q ^ Spx`0q “ Sx

So by applying the induction schema to the ∆00-formula x`0 “ 0`x we obtain the result.

(2) We then show that

Rob.` IΣ01 $c @x @y x`Sy “ Sx`y.

Indeed we have both

˝ Rob. $c @x x`S0 “ Sx`0 by

4 @x x`0 “ x and 5 @x @y`

x`Sy “ Spx`yq˘

˝ Rob. $c @x´

`

x`Sy “ Sx`y˘

Ñ`

x`SSy “ Sx`Sy˘

¯

because we have by

5 @x @y`

x`Sy “ Spx`yq˘

Rob. $c x`SSy “ Spx`Syq

hence

Rob., x`Sy “ Sx`y $c Spx`Syq “ SpSx`yq “ Sx`Sy.

Arithmetic 147

So by applying the induction schema to the ∆00-formula x`Sy “ Sx`y we obtain the

result.

(3) Finally we show that

Rob.` IΣ01 $c @x @y x`y “ y`x.

Indeed we have both

˝ Rob.` IΣ01 $c @x x`0 “ 0`x for this was what we established in case (1).

˝ Rob. $c @x´

`

x`y “ y`x˘

Ñ`

x`Sy “ Sy`x˘

¯

because by

5 @x @y`

x`Sy “ Spx`yq˘

we have

Rob., x`y “ y`x $c x`Sy “ Spx`yq “ Spy`xq “ y`Sx

hence by applying case (2) we obtain

Rob.` IΣ01, x`y “ y`x $c x`Sy “ Sy`x.

So, in the end, by applying the induction schema to the ∆00-formula x`y “ y`x we obtain

the result.

Example 160 We saw that Rob. proves that every integer (standard or non-standard) is always

comparable with any standard integer:

4.11 Rob. $c @x`

x ď n _ n ď x˘

Now we establish that Rob.` IΣ01 proves that any two integers are always comparable:

Rob.` IΣ01 $c @x @y

`

x ď y _ y ď x˘

.

We recall that x ď y stands for Dz z`x “ y. So we consider the instance of the axiom schema

for the Σ01-formula

φrx, ys :“ Dz z`x “ y _ Dz z`y “ x.

@y

¨

˚

˝

¨

˚

˝

`

Dz z`0 “ y _ Dz z`y “ 0˘

^

@x´

`

Dz z`x “ y _ Dz z`y “ x˘

Ñ`

Dz z`Sx “ y _ Dz z`y “ Sx˘

¯

˛

‹

‚

Ñ @x

¨

˝

Dz z`x “ y

_

Dz z`y “ x

˛

‚

˛

‹

‚


(1) By 4 @x x`0 “ x have

Rob. $c Dz z`0 “ y

which takes care of the first part: Dz z`0 “ y _ Dz z`y “ 0.

(2) For the second part we need to distinguish between two cases:

if Dz z`y “ x we distinguish between z “ 0 and z ‰ 0

if 0`y “ x by Example 159, we have

Rob.` IΣ01 $c 0`y “ xÑ x “ y

and

Rob.` IΣ01 $c x “ y Ñ S0`y “ 0`Sy “ Sy “ Sx

if Dz ‰ 0 z`x “ y then

Rob., Dz ‰ 0 z`x “ y $c Dz1 Sz1`x “ y

By Example 159, we obtain what we need:

Rob.` IΣ01, Dz ‰ 0 z`x “ y $c Dz

1 z1`Sx “ y

if Dz z`x “ y By 5 @x @y`

x`Sy “ Spx`yq˘

we have

Rob., z`x “ y $c Spz`yq “ z`Sy “ Sx

and by Example 159, we have

Rob.` IΣ01 $c z`Sy “ SxÑ Sz`y “ Sx

which gives the result we need:

Rob.` IΣ01, Dz z`x “ y $c Dz z`y “ Sx

So, in the end, by applying the induction schema to the Σ01-formula

φrx, ys :“ Dz z`x “ y _ Dz z`y “ x

we obtain the result.

Arithmetic 149

Example 161 We saw that Rob. does not prove that the addition is associative. We show here

that Rob.` IΣ01 proves that the addition is associative:

Rob.` IΣ01 $c @x @y @z px`yq`z “ x`py`zq.


Rob. $c @x @y px`yq`0 “ x`py`0q.

Indeed by 4 @x x`0 “ x we have

Rob. $c px`yq`0 “ x`y “ x`py`0q.


Rob. $c @x @y @z´

px`yq`z “ x`py`zq¯

Ñ

´

px`yq`Sz “ x`py`Szq¯

.

by 5 @x @y`

x`Sy “ Spx`yq˘

we have

Rob. $c px`yq`Sz “ S`

px`yq`z˘

and also

Rob. $c S`

x`py`zq˘

“ x`Spy`zq “ x`py`Szq

therefore we obtain

Rob., px`yq`z “ x`py`zq $c px`yq`Sz “ x`py`Szq.

(3) Finally, by applying the induction schema to the ∆00-formula px`yq`z “ x`py`zq we

obtain the result.

Example 162 We saw that Rob. does not prove that the multiplication is commutative. We

show here Rob.` IΣ01 proves that the addition is commutative.


Rob.` IΣ01 $c @x x¨0 “ 0¨x.

Indeed we have both


˝ $c 0¨0 “ 0¨0

˝ Rob. $c @x´

`

x¨0 “ 0¨x˘

Ñ`

Sx¨0 “ 0¨Sx˘

¯

because we have by

6 @x x¨0 “ 0 and 7 @x @y`


Rob. $c Sx¨0 “ 0 ^ 0¨Sx “ p0¨xq`0

hence

Rob., x¨0 “ 0¨x $c Sx¨0 “ 0 “ 0`0 “ p0¨xq`0.

So by applying the induction schema to the ∆00-formula x¨0 “ 0¨x we obtain the result.


Rob.` IΣ01 $c @x @y Sxÿ “ pxÿq`y.

Indeed we have both

˝ Rob. $c @x Sx¨0 “ px¨0q`0 by a simple application of

4 @x x`0 “ x and 6 @x x¨0 “ 0

˝ Rob. $c @x´

`

Sxÿ “ pxÿq`y˘

Ñ`

Sx¨Sy “ px¨Syq`Sy˘

¯

because we have by

7 @x @y`


Rob. $c Sx¨Sy “ pSxÿq`Sx

hence

Rob., Sxÿ “ pxÿq`y $c Sx¨Sy “`

pxÿq`y˘

`Sx.

but we also know that the addition is associative and commutative, thus we have

Rob.` IΣ01 $c xÿ`

`

y`Sx˘

“ xÿ``

y`Sx˘

“ xÿ``

Sy`x˘

and

Rob.` IΣ01 $c xÿ`

`

Sy`x˘

“ xÿ``

x`Sy˘

“`

xÿ`x˘

`Sy “ x¨Sy`Sy.

So by applying the induction schema to the ∆00-formula Sxÿ “ pxÿq`y we obtain the

result.

(3) Finally we show that

Rob.` IΣ01 $c @x @y xÿ “ y¨x.

Indeed we have both

Arithmetic 151

˝ Rob.` IΣ01 $c @x x¨0 “ 0¨x for this was what we established in case (1).

˝ Rob.` IΣ01 $c @x

´

`

xÿ “ y¨x˘

Ñ`

x¨Sy “ Sy¨x˘

¯

because by

7 @x @y`


we have

Rob., xÿ “ y¨x $c x¨Sy “ pxÿq`x “ py¨xq`x

by case (2) we have

Rob.` IΣ01 $c py¨xq`x “ Sy¨x

which leads to

Rob.` IΣ01, xÿ “ y¨x $c x¨Sy “ Sy¨x

So, in the end, by applying the induction schema to the ∆00-formula xÿ “ y¨x we obtain

the result.

Example 163 We show here that Rob. ` IΣ01 proves that the multiplication distributes over

the addition:

Rob.` IΣ01 $c @x @y @z x¨py`zq “ pxÿq`px¨zq.


Rob. $c @x @y x¨py`0q “ pxÿq`px¨0q.

which is immediate by

4 @x x`0 “ x and 6 @x x¨0 “ 0


Rob.` IΣ01 $c @x @y @z

´

x¨py`zq “ pxÿq`px¨zq¯

Ñ

´

x¨py`Szq “ pxÿq`px¨Szq¯

.

by 7 @x @y`


we have

Rob. $c pxÿq`px¨Szq “ pxÿq``

px¨zq`x˘

and

Rob.` IΣ01 $c pxÿq`

`

px¨zq`x˘

“`

pxÿq`px¨zq˘

`x


So that we obtain

Rob.` IΣ01, x¨py`zq “ pxÿq`px¨zq $c pxÿq`px¨Szq “

`

x¨py`zq˘

`x.

At last, by

5 @x @y`

x`Sy “ Spx`yq˘

and 7 @x @y`


we have

Rob. $c

`

x¨py`zq˘

`x “`

x¨Spy`zq˘

“`

x¨py`Szq˘

which terminates this proof.

(3) Finally, by applying the induction schema to the ∆00-formula x¨py`zq “ pxÿq`px¨zq we

obtain the result.

Example 164 We show that Rob.` IΣ01 proves that the multiplication is associative:

Rob.` IΣ01 $c @x @y @z pxÿq¨z “ x¨py¨zq.


Rob. $c @x @y pxÿq¨0 “ x¨py¨0q.

Indeed by 6 @x x¨0 “ 0 we have

Rob. $c pxÿq¨0 “ 0

and

Rob. $c x¨py¨0q “ x¨0 “ 0


Rob. $c @x @y @z`

pxÿq¨z “ x¨py¨zq˘

Ñ`

pxÿq¨Sz “ x¨py¨Szq˘

.

by 7 @x @y`


we have

Rob. $c x¨py¨Szq “ x¨`

py¨zq`y˘

and by Example 163 we also have

Rob.` IΣ01 $c x¨

`

py¨zq`y˘

“`

x¨py¨zq˘

`pxÿq

Arithmetic 153

so that we obtain

Rob.` IΣ01 , pxÿq¨z “ x¨py¨zq $c

`

x¨py¨zq˘

`pxÿq “`

pxÿq¨z˘

`pxÿq “ pxÿq¨Sz


(3) Finally, by applying the induction schema to the ∆00-formula pxÿq¨z “ x¨py¨zq we obtain

the result.

Proposition 165 The theory Rob.` IΣ01 proves that

(1) ` is commutative,

(2) ` is associative,

(3) ¨ is commutative,

(4) ¨ is associative, and

(5) ¨ distributes over `.


(1) “` is commutative” is Example 159,

(2) “` is associative” is Example 161,

(3) “¨ is commutative” is Example 162,

(4) “¨ is associative” is Example 164,

(5) “¨ distributes over `” is Example 163.

% 165

6.2 The Arithmetical Hierarchy

For the purpose of defining the arithmetical hierarchy we add a binary symbol “ ă” to our

language but essentially for the purpose of denoting bounded formulas such as Dy ď t φ and

@y ď t φ. In a sense, this differs from the use of this same symbol inside Robinson arithmetic

(see page 89) where it was an abbreviation for “ Dy`

y`x “ z ^ x ‰ z˘

”. For the reason that

in what follows we will have

˝ “ Dy`

y`x “ z ^ x ‰ z˘

” is a Σ01-formula, and

˝ “ Dx ď z x ‰ z” is a ∆00-formula.


We will be working with Rob.` IΣ01 so for every x and y we will have both

x ď y _ y ď x

and

Dz z`x “ y ðñ Dz x`z “ y.

Definition 166 (∆00-formulas) The set of ∆0

0-formulas is the least that

(1) contains all atomic formulas: t0 “ t1

(2) is closed under conjunctions, disjunctions and negations

(3) is closed under bounded quantifications

if φ P ∆00 and t is a term, then @x ă t φ and Dx ă t φ both belong to ∆0

0.

Definition 167 (arithmetical hierarchy) The hierarchy of formulas from arithmetic is de-

fined by induction on n P N:

(1) Σ00 “ Π0

0 “ ∆00

(2) Σ0n`1 is the set of all formulas of the form Dx1 . . . Dxk φ where φ P Π0

n.

(3) Π0n`1 is the set of all formulas of the form @x1 . . .@xkφ where φ P Σ0

n.

(4) ∆0n`1 “ Σ0

n`1 XΠ0n`1

Remark 168 All the classes defined above are closed under (finite) conjunctions and disjonc-

tions.

Example 169

(1) x0 ă Spx2¨Sx1q ÝÑ @y ď x3 y`x0 “ x3 P ∆00

(2) @x @y´


P Π01

(3) @x Dy px ‰ 0 Ñ Sy “ xq P Π02

Arithmetic 155

Proposition 170 Given any n P N and any Σ01-formula φ :“ Dx0 Dx1 . . . Dxnψ where ψ is ∆0

0,

there exists some ∆00-formula ψ1 such that

Rob. $c Dx0 Dx1 . . . Dxnψ ÐÑ Dxψ1.

Proof of Proposition 170: We set

ψ1 “ Dx0 ď x Dx1 ď x . . . Dxn ď x´

αn`1

`

x0, x1, . . . , xn˘

“ x ^ ψ¯

where αn`1

`

x0, x1, . . . , xn˘

“ x denotes the ∆00-formula defined by induction on n ą 0 by

˝ “α2

`

x0, x1

˘

“ x” is “Dy ď x px0`x1q¨px0`x1`S0q “ y`y ^ y`x1 “ x”

˝ “αn`1

`

x0, x1, . . . , xn˘

“ x” is “Dz ď x´

α2

`

x0, x1

˘

“ z ^ αn`

x0, x1, . . . , xn´1, z˘

“ x¯

”.

% 170

Proposition 171 Given any n, k1, . . . , kn P N and any Σ0n`1-formula and Π0

n`1-formula respec-

tively

φ :“ Dxn1 Dxn2 . . . Dx

nkn@x

n´11 . . .@xn´1

kn´1Dxn´2

1 . . . Dn´2kn´2

@xn´31 . . . . . . Qx1

1 . . . Qx1k1ψ

where Q is either @ or D depending on the parity of n, and ψ is ∆00, and

θ :“ @xn1 @xn2 . . .@x

nknDx

n´11 . . . Dxn´1

kn´1@xn´2

1 . . .@xn´2kn´2

Dxn´31 . . . . . . Qx1

1 . . . Qx1k1γ

where Q is either @ or D depending on the parity of n, and γ is ∆00,

there exists ∆00-formulas ψ1 and γ1 such that

˝ Rob. $c φ ÐÑ Dyn @yn´1 Dyn´2 . . . Qy1ψ1, and

˝ Rob. $c θ ÐÑ @yn Dyn´1 @yn´2 . . . Qy1γ1.

Proof of Proposition 171: This is an easy exercise based on the same idea as the one used to

prove Proposition 170.

% 171

From now on, Γ stands for any of the classes Σ0n`1, Π0

n`1, ∆0n (any n P N).

Proposition 172 Given any term t that does not contain the variable z, and any formula ψ,

˝ $c Dy ď x Dz ψ ÐÑ Dz Dy ď x ψ


˝ $c @y ď x @z ψ ÐÑ @z @y ď x ψ

Proof of Proposition 172: Immediate. % 172

Proposition 173 Given any n P N, any term t that does not contain the variable z, and any

formula ψ P Γ there exists ψ1 P Γ,

(1) Rob. $c Dy ď t @z ψ ÐÑ @z Dy ď t ψ1

(2) Rob. $c @y ď t Dz ψ ÐÑ Dz @y ď t ψ1.


(1) The idea is to have z encode a sequence of t many integers, and to consider all such

sequences. For this we set

ψ2 “ φβpx0, y, z1, z2q ^ ψrx0{zs

where φβpx0, y, z1, z2q stands for

x0 ă Spz1¨Syq ^ Dθ ď z2´

θ¨Spz1¨Syq¯

`x0 “ z2

so that we obtain

Rob. $c Dy ď t @z ψ ÐÑ @z1 @z2 Dy ď t ψ2

from where we apply Proposition 171 to get the result.

(2) Mutatis mutandis, the same idea works fine.

% 173

We are now able to state a stronger version of Theorem 123:

Theorem 174 Every total recursive function is representable by some Σ01-formula.

Proof of Theorem 174: It is enough to go through the proofs of Examples 108, 109, 110 and

Lemmas 113 , 119, 122, and notice that all formulas we defined were Σ01-formulas. % 174

Arithmetic 157

6.3 A first glance at Godel’s second incompleteness theorem

We first recall that by Theorem 153 the set below is

!

`

xP y, xφy˘


˝ primitive recursive if T is primitive recursive,

˝ recursive if T is recursive.

We consider any recursive theory T Ě Rob. and consider some Σ01-formula φ

proofTpx1, x2q which

represents the set above. This means that for all i1, i2 P N we have:

˝ if pi1, i2q P!

`

xP y, xφy˘


, then Rob. $c φproofT

pi1, i2q;

˝ if pi1, i2q R!

`

xP y, xφy˘


, then Rob. $c φproofTpi1, i2q.

so in particular if T is consistent, we have

P is a proof of T $c φ ðñ Rob. $c φproofT

prxP ys, rxφysq.

We consider the following primitive recursive function diag : N ÝÑ N.

diagpnq “

#

xφrrxφys{x0sy if n “ xφy P F3x0 !free

0 otherwise

together with any Σ01-formula φdiagpx0, x1q that represents diag. This means we have

for all n P NRob. $c @x0

´

diagpnq “ x0 ÐÑ φdiagpx0, nq¯

.

We define the Σ01-formula Ξ px0q by

Ξ px0q :“ Dx1 Dx2

`

φproofT

px1, x2q ^ φdiagpx2, x0q˘

Proposition 175 For every integer n we have

N |ù Ξ pnq ðñ Rob. $c Ξ pnq .


(1) if n “ xφy P F3x0 !freeand there is a proof P of T $c φrrxφys{x0s we have both

Rob. $c φdiagprxφrrxφys{x0sys, nq


and

Rob. $c φproofT

`

rxP ys, rxφrrxφys{x0sys˘

therefore

Rob. $c Dx1 Dx2

`

φproofT

px1, x2q ^ φdiagpx2, nq˘

which is

Rob. $c Ξ pnq .

(2) if n “ xφy P F3x0 !freeand there is no proof P of T $c φrrxφys{x0s we have for all proofs P

Rob. $c @x2

´

φdiagpx2, nq ÐÑ x2 “ rxφrrxφys{x0sys

¯

and

Rob. $c φproofT

`

rxP ys, rxφrrxφys{x0sys˘

and furthermore for every integer i

Rob. $c φproofT

`

i, rxφrrxφys{x0sys˘

therefore, since N |ù φRob., by the soundness theorem we have

Rob. &c Ξ pnq .

(3) if n R F3x0 !free, then for every integer i1,

Rob. $c φproofTpi1, diagpnqq

for the reason that for all integer i1

pi1, 0q R!

`

xP y, xφy˘


because 0 is never the code of a formula. Hence, by application of the soundness theorem

we have

Rob. &c Ξ pnq .

% 175

So to speak, N |ù Ξ pnq asserts that there exists a proof that there is a 1 on position pin, inq in

the array on page 143, where n is the integer that codes the formula φin .

We now consider the formula Ξ px0q – that we write Ξ – together with the term that represents

its code x Ξy and the term that represents the code of the formula Ξ px0q which “eats up” its

own code x Ξrrx Ξys{x0sy

Arithmetic 159

Claim 176

Rob. $c Ξrrx Ξys{x0s ÐÑ Dx1φproofT

`

x1, rx Ξrrx Ξys{x0sys˘

which is precisely

Rob. $c Dx1 Dx2

`

φproofT

px1, x2q ^ φdiagpx2, rx Ξysq˘

ÐÑ Dx1φproofT

`


.

Proof of Claim 176:

(ð) By the very definition of the function diag and that φdiag represents that function we have

Rob. $c φdiagprx Ξrrx Ξys{x0sys, rx Ξysq˘

thus

Rob. $c Dx1φproofT

`


ÝÑ Dx1 Dx2

`

φproofT


.

(ñ) Since φdiag represents the function diag we have

Rob. $c @x2

´

φdiagpx2, rx Ξysq˘

ÐÑ x2 “ rx Ξrrx Ξys{x0sys

¯

hence

Rob. $c

´

Dx1 Dx2

`

φproofT


ÝÑ x2 “ rx Ξrrx Ξys{x0sys

¯

therefore

Rob. $c Dx1 Dx2

`

φproofT


ÝÑ Dx1φproofT

`


.

% 176

Claim 177

T &c Ξrrx Ξys{x0s.

Proof of Claim 177: Towards a contradiction, we assume that

T $c Ξrrx Ξys{x0s.

It follows that there exists an integer xP y such that

`

xP y , rx Ξrrx Ξys{x0sys˘

P

!

`

xQy, xφy˘

P N2 | Q is a proof of T $c φ)

.

Therefore, since φproofT

represents the set above, we have

Rob. $c φproofT

`

rxP ys, rx Ξrrx Ξys{x0sys˘


and by Claim 176 we obtain

Rob. $c Ξrrx Ξys{x0s.

Since Rob. Ď T we obtain

T $c Ξrrx Ξys{x0s

which contradicts the fact that T is consistent for we obtain both

T $c Ξrrx Ξys{x0s and T $c Ξrrx Ξys{x0s.

% 177

Claim 178

Rob.

Ξrrx Ξys{x0s ÝÑ Dx1φproofRob.

`

x1, rxΞrrx Ξys{x0sys˘

ff

$c Ξrrx Ξys{x0s ÝÑ conspT q.

Where conspT q stands for the formula2

Dxφy`

Dx0 φproofTpx0, xφyq ^ Dx0 φproofT

px0, x φyq˘

Proof of Claim 178: From Claim 176 we obtain

Rob. $c Ξrrx Ξys{x0s Ñ Dx1φproofT

`


.

Thus we have both

˝ Ξrrx Ξys{x0s Ñ Dx1φproofRob.

`


$c Ξrrx Ξys{x0s Ñ Dx1φproofRob.

`


˝ Rob. $c Ξrrx Ξys{x0s Ñ Dx1φproofT

`


which leads to

Rob.

Ξrrx Ξys{x0s Ñ Dx1φproofRob.

`


+

$c Ξrrx Ξys{x0s Ñ

¨

˚

˚

˚

˝

Dx1 φproofRob.

`


Ź

Dx1 φproofT

`


˛

‹

‹

‹

‚

By the very definition3 of φproofT

and φproofRob.

we have

2We recall that we write Dxφy . . . for Dx`

φF pxq ^ . . ..3this means “if we choose wisely the Σ0

1-formulas φproofTand φproofRob.

that represent the two recursive sets

!

`

xP y, xφy˘

P N2| P is a proof of T $c φ

)

and!

`

xP y, xφy˘

P N2| P is a proof of Rob. $c φ

)

.

Arithmetic 161

˝ Rob. $c @x0 @x1

`

φproofRob.

px0, x1q ÝÑ φproofT

px0, x1q˘

.

Therefore we obtain

Rob.

Ξrrx Ξys{x0s Ñ Dx1φproofRob.

`


+

$c Ξrrx Ξys{x0s Ñ

¨

˚

˚

˚

˝

Dx1 φproofT

`


Ź

Dx1 φproofT

`


˛

‹

‹

‹

‚

which yields the result.

% 178

Lemma 179 Let T Ě Rob. be any consistent recursive theory.

If

T $c Ξrrx Ξys{x0s ÝÑ Dx1φproofRob.

`


,

then

T &c conspT q.

Proof of Lemma 179: Follows immediately from Claims 177 and 178.

% 179

So we are left with the problem of characterising the consistent theories that both extend Robin-

son arithmetic and prove this very strange formula: Ξrrx Ξys{x0s ÝÑ Dx1φproofRob.

`


.

Ultimately we will show that Rob.` IΣ01 is a good candidate. That is we will have

Rob.` IΣ01 $c Ξrrx Ξys{x0s ÝÑ Dx1φproofRob.

`


.

It can easily be seen that this formula is Σ01. We will prove a more general result.

6.4 The core of the proof

We are going to prove the following, and for this we will make use of the completeness theorem.

Lemma 180 Let φ be any closed Σ01-formula.

Rob.` IΣ01 $c φ ÝÑ Dx1φproofRob.

px1, xφyq .

First notice that this lemma is far more involved than

Lemma 181 Let φ be any closed Σ01-formula.

Rob.` IΣ01 $c φ ùñ Rob.` IΣ0

1 $c Dx1φproofRob.px1, xφyq .


Indeed, the difficult part in proving


px1, xφyq

is when Rob.` IΣ01 &c φ and even more precisely when N * φ. Because as we will see below,

for a closed Σ01-formula, as soon as N |ù φ holds, N |ù Dx1φproofRob.

px1, xφyq holds as well and

every model M of Rob.` IΣ01 satisfies M |ù Dx1φproofRob.

px1, xφyq.

So, to sum up what we’ve just said, we have

N |ù φ ùñ Rob.` IΣ01 $c φ ÝÑ Dx1φproofRob.

px1, xφyq .

Now assume that N * φ holds and consider any model M of Rob.` IΣ01 such that M |ù φ

holds. Since N * φ, we also have Rob.` IΣ01 &c φ. Now proving


px1, xφyq

amounts to proving

M |ù Dx1φproofRob.px1, xφyq .

So, there should be some (necessarily non standard !) integer θ that “codes” a “proof” of the

sequent Rob. $φ. Of course this will not be a proof in the usual sense of a finite tree, but rather

some “non-standard” object.

Definition 182 Let M and N be two models of Rob., such that M is a substructure 4 of N .

N is a final extension of Mðñ

for every a P |M| and b P |N | we have

(1) N |ù b ď a ùñ b P |M|

(2) b R |M| ùñ a ď b.

Lemma 183 If N is a model of Rob., then the substructure M whose domain is

|M| “ tnN | n P Nu

is isomorphic to the standard model N.

4M is a substructure of N iff

˝ |M| Ď |N |˝ for every constant symbol c: cM “ cN

˝ for every function symbol f whose arity is n: fM“ fN

æ |M|n

˝ for every relation symbol R whose arity is n: RM“ RN

X |M|n.

Arithmetic 163

Proof of Lemma 183: Left as a very easy exercise.

% 183

Lemma 184 Up to isomorphism, every model N of Rob. is a final extension of the standard

model N.

Proof of Lemma 184: The mapping f : N ÝÑ |N | defined by fpnq “ nN is an injective

homomorphism that satisfies for every n P N and b P |N |:

(1)

N |ù b ď n ùñ N |ù b “ 0 _ b “ S0 _ . . . _ b “ n ùñ f´1pbq P N.

This is by

4.10 Rob. $c @x“

x ď nÐÑ px “ 0 _ x “ S0 _ . . . _ x “ nq‰

.

(2)

b R f rNs ùñ n ď b.

This is by

4.11 Rob. $c @x`

x ď n _ n ď x˘

and

4.10 Rob. $c @x“

x ď nÐÑ px “ 0 _ x “ S0 _ . . . _ x “ nq‰

.

This shows that N is a final extension of the structure induced by f rNs which is isomorphic to

the standard model.

% 184

Proof of Lemma 181: We make use of the completeness theorem and of the fact that the

standard model (N) is a model of Rob.` IΣ01.

(ð) Since

Rob.` IΣ01 $c Dx1φproofRob.

px1, xφyq

holds, we also have

N |ù Dx1φproofRob.px1, xφyq .

Therefore, there exists some (standard integer) n that codes a proof of φ in Robinson

arithmetic. Such a proof is also some proof of φ in Rob. ` IΣ01, which witnesses that we

have

Rob.` IΣ01 $c φ.


(ñ) Since

Rob.` IΣ01 $c φ

we also have

N |ù φ.

By induction on the height of φ we show that

N |ù Dx1φproofRob.px1, xφyq .

We recall that since M |ù Rob. is satisfied, the model M is some final extension of a

substructure which is isomorphic to the standard model. Therefore we can easily prove by

induction5:

(1) for every closed terms t1 and t2

N |ù t1 “ t2 ñ N |ù Dx1φproofRob.px1, xt1 “ t2yq .

This is done by induction on the complexity of t1 and t2.

(2) for every closed ∆00-formula φ

N |ù φ ñ N |ù Dx1φproofRob.px1, xφyq .

This is done by induction on the complexity of φ with the help of

4.10 Rob. $c @x“

x ď nÐÑ px “ 0 _ x “ S0 _ . . . _ x “ nq‰

.

(3) for every closed Σ01-formula of the form Dx1 . . . Dxn φ where φ is some ∆0

0-formula,

N |ù Dx1 . . . Dxn φ ñ N |ù Dx1φproofRob.px1, xDx1 . . . Dxn φyq .

This holds for

N |ù Dx1 . . . Dxn φ ùñ for some k1 . . . kn P N, φrk1{x1,...,kn{xns holds in N

ùñ N |ù φrk1{x1,...,kn{xns

ùñ N |ù Dx1φproofRob.

`

x1, xφrk1{x1,...,kn{xnsy˘

ùñ N |ù Dx1φproofRob.px1, xDx1 . . . Dxn φyq

Finally, if N |ù Dx1φproofRob.px1, xφyq holds, then we use the fact that every model M of

Rob.` IΣ01 is a final extension of (a structure isomorphic to) N to ensure that

M |ù Dx1φproofRob.px1, xφyq

holds as well. Once again the proof is straightforward and goes by induction on the

complexity of φ.

5the whole proof involves many cases. It is tedious but straightforward.

Arithmetic 165

% 181

During the course of the proof, we have established:

Proposition 185 Let φ be any closed Σ01-formula.

N |ù φ ÐÑ Dx1φproofRob.px1, xφyq .

Proof of Proposition 185: we distinguish between the two directions of “ÐÑ”.

(ñ) was taken care of during the proof of Lemma 181.

(ð) from the standard integer that codes a proof of φ from Robinson arithmetic, we recover a

proof of Rob. $ φ, which shows that φ holds in all models of Rob., and in particular inside

the standard one (N).

% 185

This proposition yields an easy but amazing corollary.

Corollary 186 If the Goldbach conjecture6 is neither provable nor disprovable, then it holds

true in N.

Proof of Corollary 186: The Goldbach conjecture is some Π01 statement. If the negation of the

Goldbach conjecture were true in N, then by Proposition 185 it would be provable. % 186

Before we come to the proof of Lemma 180 – which will immediately yield Godel’s second

incompleteness theorem – we need to take care of some humongous preliminary work.

Lemma 187 Let trx1,...,xns be any LA-term (where LA “ t0, S,`, ü).

Rob.ÌΣ01 $c @x1 . . .@xn @xn`1

´

trx1,...,xns “ xn`1 ÝÑ Dx0 φproofRob.

`

x0, xtrx1,...,xns “ xn`1y˘

¯

.7

Proof of Lemma 187: We prove the result by induction on the height of the term trx1,...,xns.

htptq “ 0 we have three cases:

6Goldbach conjecture is: “every even integer strictly greater than 2 is the sum of two prime numbers”.7Where xtrx1,...,xns “ xn`1y stands for the formula xtrxk1

,...,xknvs“ xkn`1 yrxx1y{xk1

,...,xxny{xkn ,xxn`1y{xkn`1s

meaning that xtrxk1,...,xkn s

“ xkn`1 y is a formula whose n ` 1 (necessarily free) variables are

xk1 , . . . , xkn , xkn`1 and xtrx1,...,xns “ xn`1y is this term after the subsequent substitutions have taken place:

SFub.

´

. . .SFub.

´

SFub.

´

xtrxk1,...,xkn s

“ xkn`1 y, xx1y, k1

¯

, xx2y, k2

¯

, . . . , xxn`1y, kn`1

¯

.


t “ 0 we need to show

Rob.` IΣ01 $c @xn`1

´

0 “ xn`1 ÝÑ Dx0 φproofRob.px0, x0 “ xn`1yq

¯

which is


´

0 ‰ xn`1 _ Dx0 φproofRob.px0, x0 “ xn`1yq

¯

which comes down to proving

Rob.` IΣ01 $c Dx0 φproofRob.

px0, x0 “ 0yq

¯

.

The code of the following proof is just what is needed:

ax

0 “ 0 $ 0 “ 0Ref

$ 0 “ 0

t “ xn`1 we need to show


´

xn`1 “ xn`1 ÝÑ Dx0 φproofRob.px0, xxn`1 “ xn`1yq

¯

in which case the code of the following proof is what is needed

axxn`1 “ xn`1 $ xn`1 “ xn`1

Ref$ xn`1 “ xn`1

t “ xi pi ‰ n` 1q we need to show

Rob.` IΣ01 $c @xi @xn`1

´

xi “ xn`1 ÝÑ Dx0 φproofRob.px0, xxi “ xn`1yq

¯

which is also

Rob.` IΣ01 $c @xi @xn`1

´

xi ‰ xn`1 _ Dx0 φproofRob.px0, xxi “ xiyq

¯

which comes down to proving

Rob.` IΣ01 $c @xi Dx0 φproofRob.

px0, xxi “ xiyq .

in which case the code of the following proof is what is needed

axxi “ xi $ xi “ xi

Ref$ xi “ xi

htptq “ k` 1 we have three cases:

Arithmetic 167

t “ Su We need to show

Rob.ÌΣ01 $c @x1 . . .@xn @xn`1

´

Surx1,...,xns “ xn`1 ÝÑ Dx0 φproofRob.

`

x0, xSurx1,...,xns “ xn`1y˘

¯

.

We proceed by induction on xn`1, which means we need to show

(1) Rob.ÌΣ01 $c @x1 . . .@xn

´

Surx1,...,xns “ 0 ÝÑ Dx0 φproofRob.

`

x0, xSurx1,...,xns “ 0y˘

¯

.

The result follows immediately by 1 @x Sx ‰ 0 .

(2) And assuming that

Rob.ÌΣ01 $c @x1 . . .@xn

´


`


¯

holds, we need to show

Rob.ÌΣ01 $c @x1 . . .@xn

´

Surx1,...,xns “ Sxn`1 ÝÑ Dx0 φproofRob.

`

x0, xSurx1,...,xns “ Sxn`1y˘

¯

.

By 3 @x @y pSx “ Sy Ñ x “ yq together with the induction hypothesis we

obtain

Rob.ÌΣ01 $c @x1 . . .@xn

´

Surx1,...,xns “ Sxn`1 ÝÑ Dx0 φproofRob.

`

x0, xurx1,...,xns “ xn`1y˘

¯

.

We notice then that for any term a, b we have the following proof:ax.

Sa “ Sx rb{xs $ Sa “ Sx rb{xswknl

a “ b, Sa “ Sx rb{xs $ Sa “ Sx rb{xswknl

a “ b, Sa “ Sx rb{xs, Sa “ Sx ra{xs $ Sa “ Sx rb{xsRep

a “ b, Sa “ Sa $ Sa “ SbRef

a “ b $ Sa “ SbOr simply

ax.

Sa “ Sb $ Sa “ Sbwknl

a “ b, Sa “ Sb $ Sa “ Sbwknl

a “ b, Sa “ Sb, Sa “ Sa $ Sa “ SbRep


a “ b $ Sa “ SbThen, we consider one application of the cut rule to get a proof of

Rob. $ Sa “ Sb assuming a proof of Rob. $ a “ b

...Rob. $ a “ b

ax.

Sa “ Sb $ Sa “ Sbwknl

a “ b, Sa “ Sb $ Sa “ Sbwknl

a “ b, Sa “ Sb, Sa “ Sa $ Sa “ SbRep


a “ b $ Sa “ SbcutRob. $ Sa “ Sb


So, replacing a by urx1,...,xns, and b by xn`1 we obtain

...Rob. $ urx1,...,xns “ xn`1

ax.

Surx1,...,xns “ Sxn`1 $ Surx1,...,xns “ Sxn`1wknl

urx1,...,xns “ xn`1, Surx1,...,xns “ Sxn`1 $ Surx1,...,xns “ Sxn`1wknl

urx1,...,xns “ xn`1, Surx1,...,xns “ Sxn`1, Surx1,...,xns “ Surx1,...,xns $ Surx1,...,xns “ Sxn`1Rep

urx1,...,xns “ xn`1, Surx1,...,xns “ Surx1,...,xns $ Surx1,...,xns “ Sxn`1Ref

urx1,...,xns “ xn`1 $ Surx1,...,xns “ Sxn`1cutRob. $ Surx1,...,xns “ Sxn`1

So, from the code of a “proof” of Rob. $ urx1,...,xns “ xn`1 we easily obtain a

“proof” of Rob. $ Surx1,...,xns “ Sxn`1.

We then make use of the fact that urx1,...,xns “ xn`1 is ∆00 to obtain

Rob.ÌΣ01 $c @x1 . . .@xn`1

´


`


¯

.

t “ u`v We need to show

Rob.ÌΣ01 $c @x1 . . .@xn`1

´

pu`vqrx1,...,xns “ xn`1 Ñ Dx0 φproofRob.

`

x0, xpu`vqrx1,...,xns “ xn`1y˘

¯

.

The proof goes by induction on v.

v “ 0 We need to show

Rob.ÌΣ01 $c @x1 . . .@xn`1

´

pu`0qrx1,...,xns “ xn`1 Ñ Dx0 φproofRob.

`

x0, xpu`0qrx1,...,xns “ xn`1y˘

¯

Since by 4 @x x`0 “ x we have

Rob. $c @x1 . . .@xn`1 pu`0qrx1,...,xns “ urx1,...,xns

Since our proof is by induction on the complexity of the term t and u is less

complicated than t “ u`v, we have

Rob.ÌΣ01 $c @x1 . . .@xn`1

´

urx1,...,xns “ xn`1 Ñ Dx0 φproofRob.

`

x0, xurx1,...,xns “ xn`1y˘

¯

The “code” of the following “proof” yields the result.

...Rob. $ u “ xn`1

ax.

u`0 “ u $ u`0 “ u@l

@x0 x0`0 “ x0 $ u`0 “ u

ax.

u`0 “ xn`1 $ u`0 “ xn`1wknl

u “ xn`1, u`0 “ xn`1 $ u`0 “ xn`1wknl

u “ u`0, u “ xn`1, u`0 “ xn`1 $ u`0 “ xn`1Rep

u`0 “ u, u “ xn`1 $ u`0 “ xn`1cut

@x0 x0`0 “ x0, u “ xn`1 $ u`0 “ xn`1cutRob. $ u`0 “ xn`1

Arithmetic 169

v “ xi We need to show

Rob.ÌΣ01 $c @x1 . . .@xn`1

´

pu`xiqrx1,...,xns “ xn`1 Ñ Dx0 φproofRob.

`

x0, xpu`xiqrx1,...,xns “ xn`1y˘

¯

For this purpose we use the fact pu`xiqrx1,...,xns “ xn`1 is ∆00 and proceed by

induction on xn`1:

(1) The initial case is xi “ 0, which we already considered.

(2) Assuming

Rob.ÌΣ01 $c @x1 . . .@xn`1

´

pu`xiqrx1,...,xns “ xn`1 Ñ Dx0 φproofRob.

`


¯

we need to show

Rob.ÌΣ01 $c @x1 . . .@xn`1

´

pu`Sxiqrx1,...,xns “ xn`1 Ñ Dx0 φproofRob.

`

x0, xpu`Sxiqrx1,...,xns “ xn`1y˘

¯

Once again, we proceed by induction on xn`1:

(a) The initial case is xn`1 “ 0. We need to show

Rob.ÌΣ01 $c @x1 . . .@xn`1

´

pu`Sxiqrx1,...,xns “ 0 Ñ Dx0 φproofRob.

`

x0, xpu`Sxiqrx1,...,xns “ 0y˘

¯

which follows immediately by 5 @x @y`

x`Sy “ Spx`yq˘

and 1 @x Sx ‰ 0 .

(b) We assume

Rob.ÌΣ01 $c @x1 . . .@xn`1

´

pu`Sxiqrx1,...,xns “ xn`1 Ñ Dx0 φproofRob.

`

x0, xpu`Sxiqrx1,...,xns “ xn`1y˘

¯

and we need to show

Rob.ÌΣ01 $c @x1 . . .@xn`1

´

pu`Sxiqrx1,...,xns “ Sxn`1 Ñ Dx0 φproofRob.

`

x0, xpu`Sxiqrx1,...,xns “ Sxn`1y˘

¯

By 5 @x @y`

x`Sy “ Spx`yq˘

and 3 @x @y pSx “ Sy Ñ x “ yq

we have

Rob.ÌΣ01 $c @x1 . . .@xn`1

´

pu`Sxiqrx1,...,xns “ Sxn`1 Ñ pu`xiqrx1,...,xns “ xn`1

¯

By the previous induction hypothesis, we have

Rob.ÌΣ01 $c @x1 . . .@xn`1

´

pu`Sxiqrx1,...,xns “ Sxn`1 Ñ Dx0 φproofRob.

`


¯

The “code” of the “proof” on next page gives us what we need, which

terminates the proof by induction on xn`1.

So we obtain precisely the formula that we needed to complete the proof by

induction on xi.


Hence the whole result is proved.

v “ v0`v1 is left as a tedious exercise.

v “ v0¨v1 is left as a tedious exercise as well.

t “ u¨v We need to show

Rob.ÌΣ01 $c @x1 . . .@xn`1

´

pu¨vqrx1,...,xns “ xn`1 Ñ Dx0 φproofRob.

`

x0, xpu¨vqrx1,...,xns “ xn`1y˘

¯

.

The proof is similar to the case of the addition, and we leave it as a very tedious

exercise.

% 187

We obtain the wanted “proof” by making the following replacement in the “proof” below:

˝ a by urx1,...,xns ˝ b by xirx1,...,xns ˝ c by xn`1

ax.

a`Sb “ Spa`bq $ a`Sb “ Spa`bq@l

@x1 a`Sx1 “ Spa`x1q $ a`Sb “ Spa`bq@l

@x0 @x1 x0`Sx1 “ Spx0`x1q $ a`Sb “ Spa`bq

...Rob. $ a`b “ c

ax.

Spa`bq “ Sc $ Spa`bq “ Scwknl

pa`bq “ c, Spa`bq “ Sc $ Spa`bq “ Scwknl

pa`bq “ c, Spa`bq “ Sc, Spa`bq “ Spa`bq $ Spa`bq “ ScRep

pa`bq “ c, Spa`bq “ Spa`bq $ Spa`bq “ ScRef

pa`bq “ c $ Spa`bq “ Sccut

Rob. $ Spa`bq “ Sc

ax.

a`Sb “ Sc $ a`Sb “ Scwknl

a`Sb “ Sc, Spa`bq “ Sc $ a`Sb “ Scwknl

Spa`bq “ a`Sb, a`Sb “ Sc, Spa`bq “ Sc $ a`Sb “ ScRef

a`Sb “ Spa`bq, Spa`bq “ Sc $ a`Sb “ Sccut

Rob., a`Sb “ Spa`bq $ a`Sb “ SccutRob. $ a`Sb “ Sc


Lemma 188 The set of all formulas φ that satisfies

Rob.` IΣ01 $c @x1 . . .@xn

´

φrx1,...,xns ÝÑ Dx0 φproofRob.

`

x0, xφrx1,...,xnsy˘

¯

.

is closed under

(1) conjunction

(2) disjunction

(3) existential quantification

(4) bounded universal quantification.

Proof of Lemma 188:

conjunction if φ is ψ ^ θ, and

˝ Rob.` IΣ01 $c @x1 . . .@xn

´

ψrx1,...,xns Ñ Dx0 φproofRob.

`

x0, xψrx1,...,xnsy˘

¯

˝ Rob.` IΣ01 $c @x1 . . .@xn

´

θrx1,...,xns Ñ Dx0 φproofRob.

`

x0, xθrx1,...,xnsy˘

¯

.

We consider the “code” of the following “proof”:

...Rob. $ φ

...Rob. $ ψ

^rRob. $ φ^ ψ

which yields

˝ Rob.ÌΣ01 $c @x1 . . .@xn

´

pψ^θqrx1,...,xns Ñ Dx0 φproofRob.

`

x0, xpψ ^ θqrx1,...,xnsy˘

¯

.

disjunction this case is similar to the conjuction and left as an exercise.

existential quantification we consider Dxk φ, assuming that the following holds:

˝ Rob.` IΣ01 $c @x1 . . .@xn @xk

´

φrx1,...,xn,xks Ñ Dx0 φproofRob.

`

x0, xφrx1,...,xn,xksy˘

¯

Now we have

˝ $c @x1 . . .@xn @xk

´

φrx1,...,xn,xks Ñ Dxkφrx1,...,xns

¯

by considering the “code” of the following “proof”

...Rob. $ φrx1,...,xn,xks

DrRob. $ Dxk φrx1,...,xns

Arithmetic 173

we obtain

˝ Rob.` IΣ01 $c @x1 . . .@xn @xk

´


`

x0, xDxk φrx1,...,xnsy˘

¯

which immediately gives

˝ Rob.` IΣ01 $c @x1 . . .@xn

´

Dxk φrx1,...,xns Ñ Dx0 φproofRob.

`

x0, xDxk φrx1,...,xnsy˘

¯

.

bounded universal quantification we consider @xk ă xm φ, assuming that the following

holds:

˝ Rob.` IΣ01 $c @x1 . . .@xn @xk

´


`

x0, xφrx1,...,xn,xksy˘

¯

.

Strictly speaking, the formula @xk ă xm φ stands for @xk`

xk ă xm ÝÑ φ˘

which is

nothing but @xk`

Dy xk`y “ xm ÝÑ φ˘

and we have

˝ $c @xk`

Dy xk`y “ xm ÝÑ φ˘

ÐÑ @xk @y`

xk`y “ xm ÝÑ φ˘

We need to prove

˝ Rob.ÌΣ01 $c @x1 . . .@xn @xm

´

@xk ă xm φrx1,...,xns Ñ Dx0 φproofRob.

`

x0, x@xk ă xm φrx1,...,xnsy˘

¯

We prove the result by induction on xm:

(1) the initial case is

˝ Rob.ÌΣ01 $c @x1 . . .@xn

´

@xk ă 0 φrx1,...,xns Ñ Dx0 φproofRob.

`

x0, x@xk ă 0 φrx1,...,xnsy˘

¯

which is easy since the following holds.

˝ Rob.` IΣ01 $c @xk xk ă 0

(2) we assume

˝ Rob.ÌΣ01 $c @x1 . . .@xn @xm

´


`


¯

and we show

˝ Rob.ÌΣ01 $c @x1 . . .@xn @xm

´

@xk ă Sxm φrx1,...,xns Ñ Dx0 φproofRob.

`

x0, x@xk ă Sxm φrx1,...,xnsy˘

¯

Firstly, notice that we have

˝ Rob. $c @xk @xm

´

xk ă Sxm ÐÑ`

xk ă xm _ xk “ xm˘

¯

Secondly, we have both

˝ Rob.ÌΣ01 $c @x1 . . .@xn @xm

´

φrx1,...,xn,xms Ñ Dx0 φproofRob.

`

x0, xφrx1,...,xn,xmsy˘

¯

˝ Rob.ÌΣ01 $c @x1 . . .@xn @xm

´


`


¯

By mixing accordingly the two “proofs” we get what we need.

Finally, an instance of the induction schema yields the result.


% 188

We finally come to the proof of the main lemma which stated that


px1, xφyq

holds for every closed Σ01-formula.

Proof of Lemma 180: With Lemmas 187 and 188 we have almost everything we need. Because

we know that the result holds for the class that

˝ contains all atomic formulas

˝ and is closed under

(1) conjunction

(2) disjunction

(3) existential quantification

(4) bounded universal quantification.

Unfortunately, the class Σ01 contains all ∆0

0-formulas, and ∆00 is closed under negation. But we

see that if we can show that every formula which is the negations of an atomic formula belongs

to the class that Lemmas 187 and 188 provides, then we will have all Σ01-formulas.

Fortunately enough, given any formula of the form t ‰ u (where t and u are terms) satisfies

˝ Rob.` IΣ01 $c @x1 . . .@xn

´

t “ u ÐÑ`

Dx0 t`Sx0 “ u _ Dx0 u`Sx0 “ t˘

¯

.

Since the formula @x1 . . .@xn`


belongs to our class, we

have

˝ Rob.` IΣ01 $c @x1 . . .@xn

´

`


Ñ Dx0 φproofRob.

`

x0, x`


y˘

¯

then, from the “code” of a “proof” of`


we recover the

“code” of a “proof” of t ‰ u.

% 180

Theorem 189 (Godel’s second incompleteness theorem) Let T Ě Rob.ÌΣ01 be any con-

sistent recursive theory.

T &c conspT q.

Arithmetic 175

Proof of Theorem 189: Follows immediately from Lemmas 179 and 180. For the condition re-

quired by Lemma 179 on a theory T Ě Rob. to satisfy the conditions of the second incompleteness

theorem was that

T $c Ξrrx Ξys{x0s ÝÑ Dx1φproofRob.

`


.

And it turns out that the formula Ξrrx Ξys{x0s is both closed and Σ01 for it is the formula

Dx1 Dx2

`

φproofT


where both φproofT

px1, x2q and φdiagpx2, rx Ξysq are formulas that represent primitive recursive

relations, hence Σ01.

% 189


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