Recursion TheoryJoost J. Joosten

Institute for Logic Language and Computation

University of Amsterdam

Plantage Muidergracht 24

1018 TV Amsterdam

Room P 3.26, +31 20 5256095

jjoosten@phil.uu.nl

www.phil.uu.nl/∼jjoosten

Recursion Theory – p.1/9

Hamkin’s Course

Studying informational degrees via Turing Degrees

Recursion Theory – p.2/9

Hamkin’s Course

Studying informational degrees via Turing Degrees

What is the order and how complex is it?

Recursion Theory – p.2/9

Hamkin’s Course

Studying informational degrees via Turing Degrees

What is the order and how complex is it?

Computable tree with no computable branch

Recursion Theory – p.2/9

Hamkin’s Course

Studying informational degrees via Turing Degrees

What is the order and how complex is it?

Computable tree with no computable branch

Some computable model theory: e.g., does everycomputable consistent theory have a computablemodel?

Recursion Theory – p.2/9

Hamkin’s Course

Studying informational degrees via Turing Degrees

What is the order and how complex is it?

Computable tree with no computable branch

Some computable model theory: e.g., does everycomputable consistent theory have a computablemodel?

Infinite time Turing Machines

Recursion Theory – p.2/9

Gödel’s first

Gödel’s first incompleteness theorem can be seen as acorollary from computational complexity facts

Recursion Theory – p.3/9

Gödel’s first

Gödel’s first incompleteness theorem can be seen as acorollary from computational complexity facts

The second can not!

Recursion Theory – p.3/9

Gödel’s first

Gödel’s first incompleteness theorem can be seen as acorollary from computational complexity facts

The second can not!

Gödel 1: Any axiomatizable theory T which extends PAand which is

ω-consistent, any such theory is incomplete.

Recursion Theory – p.3/9

Gödel’s first

Gödel’s first incompleteness theorem can be seen as acorollary from computational complexity facts

The second can not!

Gödel 1: Any axiomatizable theory T which extends PA( or even a lot weaker is sufficient) and which isω-consistent, any such theory is incomplete.

Recursion Theory – p.3/9

Gödel’s first

Gödel’s first incompleteness theorem can be seen as acorollary from computational complexity facts

The second can not!

Gödel 1: Any axiomatizable theory T which extends PA( or even a lot weaker is sufficient) and which isω-consistent, any such theory is incomplete.

Incomplete: for some ψ we have that T 6⊢ ψ and T 6⊢ ¬ψ

Recursion Theory – p.3/9

Gödel’s first

Gödel’s first incompleteness theorem can be seen as acorollary from computational complexity facts

The second can not!

Gödel 1: Any axiomatizable theory T which extends PA( or even a lot weaker is sufficient) and which isω-consistent, any such theory is incomplete.

Incomplete: for some ψ we have that T 6⊢ ψ and T 6⊢ ¬ψ

Omega-inconsistent: for some ψ(x) we have

Recursion Theory – p.3/9

Gödel’s first

Gödel’s first incompleteness theorem can be seen as acorollary from computational complexity facts

The second can not!

Gödel 1: Any axiomatizable theory T which extends PA( or even a lot weaker is sufficient) and which isω-consistent, any such theory is incomplete.

Incomplete: for some ψ we have that T 6⊢ ψ and T 6⊢ ¬ψ

Omega-inconsistent: for some ψ(x) we haveT ⊢ ∃x ψ(x)

Recursion Theory – p.3/9

Gödel’s first

Gödel’s first incompleteness theorem can be seen as acorollary from computational complexity facts

The second can not!

Gödel 1: Any axiomatizable theory T which extends PA( or even a lot weaker is sufficient) and which isω-consistent, any such theory is incomplete.

Incomplete: for some ψ we have that T 6⊢ ψ and T 6⊢ ¬ψ

Omega-inconsistent: for some ψ(x) we haveT ⊢ ∃x ψ(x)

∀m T ⊢ ¬ψ(m)

Recursion Theory – p.3/9

Gödel’s first

Gödel’s first incompleteness theorem can be seen as acorollary from computational complexity facts

The second can not!

Gödel 1: Any axiomatizable theory T which extends PA( or even a lot weaker is sufficient) and which isω-consistent, any such theory is incomplete.

Incomplete: for some ψ we have that T 6⊢ ψ and T 6⊢ ¬ψ

Omega-inconsistent: for some ψ(x) we haveT ⊢ ∃x ψ(x)

∀m T ⊢ ¬ψ(m)

Omega-consistent is (not Omega-inconsistent)

Recursion Theory – p.3/9

Gödel’s first

Gödel’s first incompleteness theorem can be seen as acorollary from computational complexity facts

The second can not!

Gödel 1: Any axiomatizable theory T which extends PA( or even a lot weaker is sufficient) and which isω-consistent, any such theory is incomplete.

Incomplete: for some ψ we have that T 6⊢ ψ and T 6⊢ ¬ψ

Omega-inconsistent: for some ψ(x) we haveT ⊢ ∃x ψ(x)

∀m T ⊢ ¬ψ(m)

Omega-consistent is (not Omega-inconsistent)

Axiomatisable: the set of axioms is computable

Recursion Theory – p.3/9

Gödel’s first

Gödel’s first incompleteness theorem can be seen as acorollary from computational complexity facts

The second can not!

Gödel 1: Any axiomatizable theory T which extends PA( or even a lot weaker is sufficient) and which isω-consistent, any such theory is incomplete.

Incomplete: for some ψ we have that T 6⊢ ψ and T 6⊢ ¬ψ

Omega-inconsistent: for some ψ(x) we haveT ⊢ ∃x ψ(x)

∀m T ⊢ ¬ψ(m)

Omega-consistent is (not Omega-inconsistent)

Axiomatisable: the set of axioms is computable (c.e. issuff.)

Recursion Theory – p.3/9

Proving Gödel 1

It is a direct consequence of the Semi-RepresentabilityTheorem

Recursion Theory – p.4/9

Proving Gödel 1

It is a direct consequence of the Semi-RepresentabilityTheorem

A set S is semi-representable in a theory T wheneverthere is some formula ϕ(x) in the language of T suchthat

x ∈ S ⇔ T ⊢ ϕ(x).

Recursion Theory – p.4/9

Proving Gödel 1

It is a direct consequence of the Semi-RepresentabilityTheorem

A set S is semi-representable in a theory T wheneverthere is some formula ϕ(x) in the language of T suchthat

x ∈ S ⇔ T ⊢ ϕ(x).

SRT: The following are equivalent (provided PA isOmega-consistent)

Recursion Theory – p.4/9

Proving Gödel 1

It is a direct consequence of the Semi-RepresentabilityTheorem

A set S is semi-representable in a theory T wheneverthere is some formula ϕ(x) in the language of T suchthat

x ∈ S ⇔ T ⊢ ϕ(x).

SRT: The following are equivalent (provided PA isOmega-consistent)

S is c.e.

Recursion Theory – p.4/9

Proving Gödel 1

It is a direct consequence of the Semi-RepresentabilityTheorem

A set S is semi-representable in a theory T wheneverthere is some formula ϕ(x) in the language of T suchthat

x ∈ S ⇔ T ⊢ ϕ(x).

SRT: The following are equivalent (provided PA isOmega-consistent)

S is c.e.S is semi-representable in PA

Recursion Theory – p.4/9

Proving Gödel 1

It is a direct consequence of the Semi-RepresentabilityTheorem

A set S is semi-representable in a theory T wheneverthere is some formula ϕ(x) in the language of T suchthat

x ∈ S ⇔ T ⊢ ϕ(x).

SRT: The following are equivalent (provided PA isOmega-consistent)

S is c.e.S is semi-representable in PAS ≤m TPA

Recursion Theory – p.4/9

Proving Gödel 1

Corollary 1: theoremhood of PA is not computable

Recursion Theory – p.5/9

Proving Gödel 1

Corollary 1: theoremhood of PA is not computable(undecidable)

Corollary 2: PA is incomplete

Recursion Theory – p.5/9

Proving Gödel 1

Corollary 1: theoremhood of PA is not computable(undecidable)

Corollary 2: PA is incomplete

Proof: K is not semi-representable in PA, so certainlynot by ¬ϕ(x)

Recursion Theory – p.5/9

Proving Gödel 1

Corollary 1: theoremhood of PA is not computable(undecidable)

Corollary 2: PA is incomplete

Proof: K is not semi-representable in PA, so certainlynot by ¬ϕ(x)

where ϕ(x) semi-represents K

Recursion Theory – p.5/9

Proving Gödel 1

Corollary 1: theoremhood of PA is not computable(undecidable)

Corollary 2: PA is incomplete

Proof: K is not semi-representable in PA, so certainlynot by ¬ϕ(x)

where ϕ(x) semi-represents K

so, for some x ∈ K we have PA 0 ¬ϕ(x)

Recursion Theory – p.5/9

Proving Gödel 1

Corollary 1: theoremhood of PA is not computable(undecidable)

Corollary 2: PA is incomplete

Proof: K is not semi-representable in PA, so certainlynot by ¬ϕ(x)

where ϕ(x) semi-represents K

so, for some x ∈ K we have PA 0 ¬ϕ(x)

Natural incomplete sentences are hard to find

Recursion Theory – p.5/9

Proving Gödel 1

Corollary 1: theoremhood of PA is not computable(undecidable)

Corollary 2: PA is incomplete

Proof: K is not semi-representable in PA, so certainlynot by ¬ϕ(x)

where ϕ(x) semi-represents K

so, for some x ∈ K we have PA 0 ¬ϕ(x)

Natural incomplete sentences are hard to find

Goodstein’s sequences!

Recursion Theory – p.5/9

Proving Gödel 1

Corollary 1: theoremhood of PA is not computable(undecidable)

Corollary 2: PA is incomplete

Proof: K is not semi-representable in PA, so certainlynot by ¬ϕ(x)

where ϕ(x) semi-represents K

so, for some x ∈ K we have PA 0 ¬ϕ(x)

Natural incomplete sentences are hard to find

Goodstein’s sequences!

There is an interesting link from strange attractors inchaos theory to Goodstein’s process.

Recursion Theory – p.5/9

PA is creative

What does it mean for a theory to be creative?

Recursion Theory – p.6/9

PA is creative

What does it mean for a theory to be creative?

We know that K ≤m PA

Recursion Theory – p.6/9

PA is creative

What does it mean for a theory to be creative?

We know that K ≤m PA

Most other theories are also creative

Recursion Theory – p.6/9

PA is creative

What does it mean for a theory to be creative?

We know that K ≤m PA

Most other theories are also creative

Can we carve out a decidable piece out of PA?

Recursion Theory – p.6/9

PA is creative

What does it mean for a theory to be creative?

We know that K ≤m PA

Most other theories are also creative

Can we carve out a decidable piece out of PA?

Maybe a bit surprising, but NO, we can not

Recursion Theory – p.6/9

PA is creative

What does it mean for a theory to be creative?

We know that K ≤m PA

Most other theories are also creative

Can we carve out a decidable piece out of PA?

Maybe a bit surprising, but NO, we can not

Pure Predicate Calculus is undecidable

Recursion Theory – p.6/9

PA is creative

What does it mean for a theory to be creative?

We know that K ≤m PA

Most other theories are also creative

Can we carve out a decidable piece out of PA?

Maybe a bit surprising, but NO, we can not

Pure Predicate Calculus is undecidable

Can be done directly by coding the halting problem, wegive a shorter proof

Recursion Theory – p.6/9

Decidable theories

Not all theories are decidable

Recursion Theory – p.7/9

Decidable theories

Not all theories are decidable

Any complete axiomatisable theory is decidable

Recursion Theory – p.7/9

Decidable theories

Not all theories are decidable

Any complete axiomatisable theory is decidable

Funny little fact: if a theory is many-one reducible to asimple set, then that theory is decidable!

Recursion Theory – p.7/9

Decidable theories

Not all theories are decidable

Any complete axiomatisable theory is decidable

Funny little fact: if a theory is many-one reducible to asimple set, then that theory is decidable!

Vaught’s completeness test

Recursion Theory – p.7/9

Decidable theories

Not all theories are decidable

Any complete axiomatisable theory is decidable

Funny little fact: if a theory is many-one reducible to asimple set, then that theory is decidable!

Vaught’s completeness test

:if categorical in some cardinal, then decidable

Recursion Theory – p.7/9

Decidable theories

Not all theories are decidable

Any complete axiomatisable theory is decidable

Funny little fact: if a theory is many-one reducible to asimple set, then that theory is decidable!

Vaught’s completeness test

:if categorical in some cardinal, then decidable

Applications: dense linear ordering with no begin orend-points

Recursion Theory – p.7/9

Decidable theories

Not all theories are decidable

Any complete axiomatisable theory is decidable

Funny little fact: if a theory is many-one reducible to asimple set, then that theory is decidable!

Vaught’s completeness test

:if categorical in some cardinal, then decidable

Applications: dense linear ordering with no begin orend-points (Algebraicly closed fields of givencharacteristic)

Recursion Theory – p.7/9

PC is undecidable

Alonzo Church: PC is undecidable

Recursion Theory – p.8/9

PC is undecidable

Alonzo Church: PC is undecidable

Two main ingredients

Recursion Theory – p.8/9

PC is undecidable

Alonzo Church: PC is undecidable

Two main ingredients

(1) There is a finitely axiomatized creative theory(Raphael Robinson)

Recursion Theory – p.8/9

PC is undecidable

Alonzo Church: PC is undecidable

Two main ingredients

(1) There is a finitely axiomatized creative theory(Raphael Robinson)

(2) If T ′ is a finite extension of T , then

TT ′ ≤m TT

Recursion Theory – p.8/9

PC is undecidable

Alonzo Church: PC is undecidable

Two main ingredients

(1) There is a finitely axiomatized creative theory(Raphael Robinson)

(2) If T ′ is a finite extension of T , then

TT ′ ≤m TT

The proof of (2) is easy via the computable version ofthe deduction lemma

Recursion Theory – p.8/9

PC is undecidable

Alonzo Church: PC is undecidable

Two main ingredients

(1) There is a finitely axiomatized creative theory(Raphael Robinson)

(2) If T ′ is a finite extension of T , then

TT ′ ≤m TT

The proof of (2) is easy via the computable version ofthe deduction lemma

The proof of (1) is a bit more involved

Recursion Theory – p.8/9

Robinson’s Arithmetic

Q contains all the finite defining axioms of the symbols

Recursion Theory – p.9/9

Robinson’s Arithmetic

Q contains all the finite defining axioms of the symbols

Induction can be reduced to:

x 6= 0 → ∃y x = y′.

Recursion Theory – p.9/9

Robinson’s Arithmetic

Q contains all the finite defining axioms of the symbols

Induction can be reduced to:

x 6= 0 → ∃y x = y′.

One now can check that all the c.e. sets aresemi-representable in Q

Recursion Theory – p.9/9

Robinson’s Arithmetic

Q contains all the finite defining axioms of the symbols

Induction can be reduced to:

x 6= 0 → ∃y x = y′.

One now can check that all the c.e. sets aresemi-representable in Q (provided ω-consistency)

Recursion Theory – p.9/9

Robinson’s Arithmetic

Q contains all the finite defining axioms of the symbols

Induction can be reduced to:

x 6= 0 → ∃y x = y′.

One now can check that all the c.e. sets aresemi-representable in Q (provided ω-consistency)

Thus, we getK ≤m TQ ≤m TPC−

Recursion Theory – p.9/9

Robinson’s Arithmetic

Q contains all the finite defining axioms of the symbols

Induction can be reduced to:

x 6= 0 → ∃y x = y′.

One now can check that all the c.e. sets aresemi-representable in Q (provided ω-consistency)

Thus, we getK ≤m TQ ≤m TPC−

An easy lemma teaches us that (via the embedding)

TPC− ≤m TPC

Recursion Theory – p.9/9

Robinson’s Arithmetic

Q contains all the finite defining axioms of the symbols

Induction can be reduced to:

x 6= 0 → ∃y x = y′.

One now can check that all the c.e. sets aresemi-representable in Q (provided ω-consistency)

Thus, we getK ≤m TQ ≤m TPC−

An easy lemma teaches us that (via the embedding)

TPC− ≤m TPC

Thus we obtain that PC is creative!

Recursion Theory – p.9/9

Hamkin's CourseG"odel's firstProving G"odel 1Proving G"odel 1PA is creativeDecidable theoriesPC is undecidableRobinson's Arithmetic