Recursion Theory. Joost J. Joosten
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1 Recursion Theory Joost J. Joosten Institute for Logic Language and Computation University of Amsterdam Plantage Muidergracht TV Amsterdam Room P 3.26, jjoosten@phil.uu.nl jjoosten Recursion Theory p.1/9
2 Hamkin s Course Studying informational degrees via Turing Degrees Recursion Theory p.2/9
3 Hamkin s Course Studying informational degrees via Turing Degrees What is the order and how complex is it? Recursion Theory p.2/9
4 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
5 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 every computable consistent theory have a computable model? Recursion Theory p.2/9
6 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 every computable consistent theory have a computable model? Infinite time Turing Machines Recursion Theory p.2/9
7 Gödel s first Gödel s first incompleteness theorem can be seen as a corollary from computational complexity facts Recursion Theory p.3/9
8 Gödel s first Gödel s first incompleteness theorem can be seen as a corollary from computational complexity facts The second can not! Recursion Theory p.3/9
9 Gödel s first Gödel s first incompleteness theorem can be seen as a corollary from computational complexity facts The second can not! Gödel 1: Any axiomatizable theory T which extends PA and which is ω-consistent, any such theory is incomplete. Recursion Theory p.3/9
10 Gödel s first Gödel s first incompleteness theorem can be seen as a corollary 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 is ω-consistent, any such theory is incomplete. Recursion Theory p.3/9
11 Gödel s first Gödel s first incompleteness theorem can be seen as a corollary 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 is ω-consistent, any such theory is incomplete. Incomplete: for some ψ we have that T ψ and T ψ Recursion Theory p.3/9
12 Gödel s first Gödel s first incompleteness theorem can be seen as a corollary 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 is ω-consistent, any such theory is incomplete. Incomplete: for some ψ we have that T ψ and T ψ Omega-inconsistent: for some ψ(x) we have Recursion Theory p.3/9
13 Gödel s first Gödel s first incompleteness theorem can be seen as a corollary 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 is ω-consistent, any such theory is incomplete. Incomplete: for some ψ we have that T ψ and T ψ Omega-inconsistent: for some ψ(x) we have T x ψ(x) Recursion Theory p.3/9
14 Gödel s first Gödel s first incompleteness theorem can be seen as a corollary 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 is ω-consistent, any such theory is incomplete. Incomplete: for some ψ we have that T ψ and T ψ Omega-inconsistent: for some ψ(x) we have T x ψ(x) m T ψ(m) Recursion Theory p.3/9
15 Gödel s first Gödel s first incompleteness theorem can be seen as a corollary 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 is ω-consistent, any such theory is incomplete. Incomplete: for some ψ we have that T ψ and T ψ Omega-inconsistent: for some ψ(x) we have T x ψ(x) m T ψ(m) Omega-consistent is (not Omega-inconsistent) Recursion Theory p.3/9
16 Gödel s first Gödel s first incompleteness theorem can be seen as a corollary 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 is ω-consistent, any such theory is incomplete. Incomplete: for some ψ we have that T ψ and T ψ Omega-inconsistent: for some ψ(x) we have T 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
17 Gödel s first Gödel s first incompleteness theorem can be seen as a corollary 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 is ω-consistent, any such theory is incomplete. Incomplete: for some ψ we have that T ψ and T ψ Omega-inconsistent: for some ψ(x) we have T x ψ(x) m T ψ(m) Omega-consistent is (not Omega-inconsistent) Axiomatisable: the set of axioms is computable (c.e. is suff.) Recursion Theory p.3/9
18 Proving Gödel 1 It is a direct consequence of the Semi-Representability Theorem Recursion Theory p.4/9
19 Proving Gödel 1 It is a direct consequence of the Semi-Representability Theorem A set S is semi-representable in a theory T whenever there is some formula ϕ(x) in the language of T such that x S T ϕ(x). Recursion Theory p.4/9
20 Proving Gödel 1 It is a direct consequence of the Semi-Representability Theorem A set S is semi-representable in a theory T whenever there is some formula ϕ(x) in the language of T such that x S T ϕ(x). SRT: The following are equivalent (provided P A is Omega-consistent) Recursion Theory p.4/9
21 Proving Gödel 1 It is a direct consequence of the Semi-Representability Theorem A set S is semi-representable in a theory T whenever there is some formula ϕ(x) in the language of T such that x S T ϕ(x). SRT: The following are equivalent (provided P A is Omega-consistent) S is c.e. Recursion Theory p.4/9
22 Proving Gödel 1 It is a direct consequence of the Semi-Representability Theorem A set S is semi-representable in a theory T whenever there is some formula ϕ(x) in the language of T such that x S T ϕ(x). SRT: The following are equivalent (provided P A is Omega-consistent) S is c.e. S is semi-representable in PA Recursion Theory p.4/9
23 Proving Gödel 1 It is a direct consequence of the Semi-Representability Theorem A set S is semi-representable in a theory T whenever there is some formula ϕ(x) in the language of T such that x S T ϕ(x). SRT: The following are equivalent (provided P A is Omega-consistent) S is c.e. S is semi-representable in PA S m T PA Recursion Theory p.4/9
24 Proving Gödel 1 Corollary 1: theoremhood of PA is not computable Recursion Theory p.5/9
25 Proving Gödel 1 Corollary 1: theoremhood of PA is not computable (undecidable) Corollary 2: PA is incomplete Recursion Theory p.5/9
26 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 certainly not by ϕ(x) Recursion Theory p.5/9
27 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 certainly not by ϕ(x) where ϕ(x) semi-represents K Recursion Theory p.5/9
28 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 certainly not by ϕ(x) where ϕ(x) semi-represents K so, for some x K we have PA ϕ(x) Recursion Theory p.5/9
29 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 certainly not by ϕ(x) where ϕ(x) semi-represents K so, for some x K we have PA ϕ(x) Natural incomplete sentences are hard to find Recursion Theory p.5/9
30 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 certainly not by ϕ(x) where ϕ(x) semi-represents K so, for some x K we have PA ϕ(x) Natural incomplete sentences are hard to find Goodstein s sequences! Recursion Theory p.5/9
31 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 certainly not by ϕ(x) where ϕ(x) semi-represents K so, for some x K we have PA ϕ(x) Natural incomplete sentences are hard to find Goodstein s sequences! There is an interesting link from strange attractors in chaos theory to Goodstein s process. Recursion Theory p.5/9
32 PA is creative What does it mean for a theory to be creative? Recursion Theory p.6/9
33 PA is creative What does it mean for a theory to be creative? We know that K m PA Recursion Theory p.6/9
34 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
35 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
36 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
37 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
38 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, we give a shorter proof Recursion Theory p.6/9
39 Decidable theories Not all theories are decidable Recursion Theory p.7/9
40 Decidable theories Not all theories are decidable Any complete axiomatisable theory is decidable Recursion Theory p.7/9
41 Decidable theories Not all theories are decidable Any complete axiomatisable theory is decidable Funny little fact: if a theory is many-one reducible to a simple set, then that theory is decidable! Recursion Theory p.7/9
42 Decidable theories Not all theories are decidable Any complete axiomatisable theory is decidable Funny little fact: if a theory is many-one reducible to a simple set, then that theory is decidable! Vaught s completeness test Recursion Theory p.7/9
43 Decidable theories Not all theories are decidable Any complete axiomatisable theory is decidable Funny little fact: if a theory is many-one reducible to a simple set, then that theory is decidable! Vaught s completeness test :if categorical in some cardinal, then decidable Recursion Theory p.7/9
44 Decidable theories Not all theories are decidable Any complete axiomatisable theory is decidable Funny little fact: if a theory is many-one reducible to a simple set, then that theory is decidable! Vaught s completeness test :if categorical in some cardinal, then decidable Applications: dense linear ordering with no begin or end-points Recursion Theory p.7/9
45 Decidable theories Not all theories are decidable Any complete axiomatisable theory is decidable Funny little fact: if a theory is many-one reducible to a simple set, then that theory is decidable! Vaught s completeness test :if categorical in some cardinal, then decidable Applications: dense linear ordering with no begin or end-points (Algebraicly closed fields of given characteristic) Recursion Theory p.7/9
46 PC is undecidable Alonzo Church: PC is undecidable Recursion Theory p.8/9
47 PC is undecidable Alonzo Church: PC is undecidable Two main ingredients Recursion Theory p.8/9
48 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
49 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 T T m T T Recursion Theory p.8/9
50 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 T T m T T The proof of (2) is easy via the computable version of the deduction lemma Recursion Theory p.8/9
51 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 T T m T T The proof of (2) is easy via the computable version of the deduction lemma The proof of (1) is a bit more involved Recursion Theory p.8/9
52 Robinson s Arithmetic Q contains all the finite defining axioms of the symbols Recursion Theory p.9/9
53 Robinson s Arithmetic Q contains all the finite defining axioms of the symbols Induction can be reduced to: x 0 y x = y. Recursion Theory p.9/9
54 Robinson s Arithmetic Q contains all the finite defining axioms of the symbols Induction can be reduced to: x 0 y x = y. One now can check that all the c.e. sets are semi-representable in Q Recursion Theory p.9/9
55 Robinson s Arithmetic Q contains all the finite defining axioms of the symbols Induction can be reduced to: x 0 y x = y. One now can check that all the c.e. sets are semi-representable in Q (provided ω-consistency) Recursion Theory p.9/9
56 Robinson s Arithmetic Q contains all the finite defining axioms of the symbols Induction can be reduced to: x 0 y x = y. One now can check that all the c.e. sets are semi-representable in Q (provided ω-consistency) Thus, we get K m T Q m T PC Recursion Theory p.9/9
57 Robinson s Arithmetic Q contains all the finite defining axioms of the symbols Induction can be reduced to: x 0 y x = y. One now can check that all the c.e. sets are semi-representable in Q (provided ω-consistency) Thus, we get K m T Q m T PC An easy lemma teaches us that (via the embedding) T PC m T PC Recursion Theory p.9/9
58 Robinson s Arithmetic Q contains all the finite defining axioms of the symbols Induction can be reduced to: x 0 y x = y. One now can check that all the c.e. sets are semi-representable in Q (provided ω-consistency) Thus, we get K m T Q m T PC An easy lemma teaches us that (via the embedding) T PC m T PC Thus we obtain that PC is creative! Recursion Theory p.9/9
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