. How useful are random axioms? Laurent Bienvenu (CNRS & University of Paris 7) Andrei Romashchenko (University of Montpellier 2) Alexander Shen (University of Montpellier 2) Antoine Taveneaux (University of Paris 7) Stijn Vermeeren (University of Leeds) Séminaire Logique October 28, 2012
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Transcript
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How useful are random axioms?
Laurent Bienvenu (CNRS & University of Paris 7)Andrei Romashchenko (University of Montpellier 2)Alexander Shen (University of Montpellier 2)Antoine Taveneaux (University of Paris 7)Stijn Vermeeren (University of Leeds)
Séminaire Logique
October28, 2012
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1. Random axioms
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Peano arithmetic
In this talk, we work in the axiomatic system of Peano arithmetic, PA, over thelanguage (0, 1,+,×). Its standard model N is the set N of natural numbers,with standard addition and multiplication.
As it is well known, Gödel’s theorem applies to this theory: there are sentencesthat are true (in the standard model) but not provable.
PA ⊢ φ ⇒⇍ N |= φ
1. Random axioms 3/31
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Peano arithmetic
In this talk, we work in the axiomatic system of Peano arithmetic, PA, over thelanguage (0, 1,+,×). Its standard model N is the set N of natural numbers,with standard addition and multiplication.As it is well known, Gödel’s theorem applies to this theory: there are sentencesthat are true (in the standard model) but not provable.
PA ⊢ φ ⇒⇍ N |= φ
1. Random axioms 3/31
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Random axiomsNow imagine that a formula φ has the following properties:
1. PA ⊬ φ(n̄) for any n ≤ N
2. At least 99.9% of integers n ≤ N satisfy φ3. Item 2. is provable inside PA
Then, if we pick an integer r ≤ N at random and consider the theory
PA+φ(̄r)
with error probability at most 0.1% our new theory is coherent (assuming PA is)and true.
Now ifPA+φ(̄r) ⊢ ψ
we have (in some sense) “evidence” that ψ is likely to be true.
1. Random axioms 4/31
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Random axiomsNow imagine that a formula φ has the following properties:
1. PA ⊬ φ(n̄) for any n ≤ N2. At least 99.9% of integers n ≤ N satisfy φ
3. Item 2. is provable inside PA
Then, if we pick an integer r ≤ N at random and consider the theory
PA+φ(̄r)
with error probability at most 0.1% our new theory is coherent (assuming PA is)and true.
Now ifPA+φ(̄r) ⊢ ψ
we have (in some sense) “evidence” that ψ is likely to be true.
1. Random axioms 4/31
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Random axiomsNow imagine that a formula φ has the following properties:
1. PA ⊬ φ(n̄) for any n ≤ N2. At least 99.9% of integers n ≤ N satisfy φ3. Item 2. is provable inside PA
Then, if we pick an integer r ≤ N at random and consider the theory
PA+φ(̄r)
with error probability at most 0.1% our new theory is coherent (assuming PA is)and true.
Now ifPA+φ(̄r) ⊢ ψ
we have (in some sense) “evidence” that ψ is likely to be true.
1. Random axioms 4/31
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Random axiomsNow imagine that a formula φ has the following properties:
1. PA ⊬ φ(n̄) for any n ≤ N2. At least 99.9% of integers n ≤ N satisfy φ3. Item 2. is provable inside PA
Then, if we pick an integer r ≤ N at random and consider the theory
PA+φ(̄r)
with error probability at most 0.1% our new theory is coherent (assuming PA is)and true.
Now ifPA+φ(̄r) ⊢ ψ
we have (in some sense) “evidence” that ψ is likely to be true.
1. Random axioms 4/31
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Random axiomsNow imagine that a formula φ has the following properties:
1. PA ⊬ φ(n̄) for any n ≤ N2. At least 99.9% of integers n ≤ N satisfy φ3. Item 2. is provable inside PA
Then, if we pick an integer r ≤ N at random and consider the theory
PA+φ(̄r)
with error probability at most 0.1% our new theory is coherent (assuming PA is)and true.
Now ifPA+φ(̄r) ⊢ ψ
we have (in some sense) “evidence” that ψ is likely to be true.
1. Random axioms 4/31
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The canonical example
The canonical example:
C(x) ≥ L− 20 for binary strings x of length ≤ L
where C denotes Kolmogorov complexity (to which we will return shortly).
..Theorem (Chaitin)The above statement is true for a fraction ≥ 1 − 10−6 of strings x (and this wecan prove!), but is not provable in PA for any x (assuming L is large enough).
..Theorem (Chaitin)The above statement is true for a fraction ≥ 1 − 10−6 of strings x (and this wecan prove!), but is not provable in PA for any x (assuming L is large enough).
..Theorem (Chaitin)The above statement is true for a fraction ≥ 1 − 10−6 of strings x (and this wecan prove!), but is not provable in PA for any x (assuming L is large enough).
On this particular example, we could choose a binary string r of length L atrandom (say by flipping a coin), add to PA the new axiom C(r) ≥ L− 20 and tryto prove new statements.
1. Random axioms 5/31
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The canonical example
The canonical example:
C(x) ≥ L− 20 for binary strings x of length ≤ L
where C denotes Kolmogorov complexity (to which we will return shortly).
..Theorem (Chaitin)The above statement is true for a fraction ≥ 1 − 10−6 of strings x (and this wecan prove!), but is not provable in PA for any x (assuming L is large enough).
..Theorem (Chaitin)The above statement is true for a fraction ≥ 1 − 10−6 of strings x (and this wecan prove!), but is not provable in PA for any x (assuming L is large enough).
..Theorem (Chaitin)The above statement is true for a fraction ≥ 1 − 10−6 of strings x (and this wecan prove!), but is not provable in PA for any x (assuming L is large enough).
On this particular example, we could choose a binary string r of length L atrandom (say by flipping a coin), add to PA the new axiom C(r) ≥ L− 20 and tryto prove new statements.
1. Random axioms 5/31
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The canonical example
The canonical example:
C(x) ≥ L− 20 for binary strings x of length ≤ L
where C denotes Kolmogorov complexity (to which we will return shortly).
..Theorem (Chaitin)The above statement is true for a fraction ≥ 1 − 10−6 of strings x (and this wecan prove!), but is not provable in PA for any x (assuming L is large enough).
..Theorem (Chaitin)The above statement is true for a fraction ≥ 1 − 10−6 of strings x (and this wecan prove!), but is not provable in PA for any x (assuming L is large enough).
..Theorem (Chaitin)The above statement is true for a fraction ≥ 1 − 10−6 of strings x (and this wecan prove!), but is not provable in PA for any x (assuming L is large enough).
On this particular example, we could choose a binary string r of length L atrandom (say by flipping a coin), add to PA the new axiom C(r) ≥ L− 20 and tryto prove new statements.
1. Random axioms 5/31
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Probabilistic proofs (1)
This yields the natural idea of probabilistic proof.
Fix an error threshold δ in advance (margin of error that you find acceptable)which will be treated as a initial capital. Start with the theory T0 = PA.
1. Random axioms 6/31
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Probabilistic proofs (1)
This yields the natural idea of probabilistic proof.
Fix an error threshold δ in advance (margin of error that you find acceptable)which will be treated as a initial capital. Start with the theory T0 = PA.
1. Random axioms 6/31
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Probabilistic proofs (2)
At step n, we have two options:
• We can use the statements of our theory Tn to prove (in the normal way) anew fact ψ and take Tn+1 = Tn ∪ {ψ}
• Or, if we have in Tn a statement of type
#{n ∈ A | φ(n̄)} ≥ (1 − ε) · |A|
for some finite set A, we can choose some r ∈ A at random, pay ε fromour capital, and take Tn+1 = Tn ∪ {φ(̄r)} (unless our capital becomesnegative in which case this step is not allowed)
1. Random axioms 7/31
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Probabilistic proofs (2)
At step n, we have two options:
• We can use the statements of our theory Tn to prove (in the normal way) anew fact ψ and take Tn+1 = Tn ∪ {ψ}
• Or, if we have in Tn a statement of type
#{n ∈ A | φ(n̄)} ≥ (1 − ε) · |A|
for some finite set A, we can choose some r ∈ A at random, pay ε fromour capital, and take Tn+1 = Tn ∪ {φ(̄r)} (unless our capital becomesnegative in which case this step is not allowed)
1. Random axioms 7/31
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Probabilistic proofs (2)
At step n, we have two options:
• We can use the statements of our theory Tn to prove (in the normal way) anew fact ψ and take Tn+1 = Tn ∪ {ψ}
• Or, if we have in Tn a statement of type
#{n ∈ A | φ(n̄)} ≥ (1 − ε) · |A|
for some finite set A, we can choose some r ∈ A at random, pay ε fromour capital, and take Tn+1 = Tn ∪ {φ(̄r)} (unless our capital becomesnegative in which case this step is not allowed)
1. Random axioms 7/31
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Probabilistic proofs (3)
Such a proof scheme (or strategy) is probabilistic: if we run it twice we mightprove completely different statements!
Still, if a fixed statement ψ could be proven with high probability with thisscheme, our intuition tells us that ψ is very likely to be true. This raises twoquestions:
1. Is this intuition correct? i.e., does this proof scheme make sense?2. Is it useful? i.e., can we prove more things than we classically could?
1. Random axioms 8/31
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Probabilistic proofs (3)
Such a proof scheme (or strategy) is probabilistic: if we run it twice we mightprove completely different statements!
Still, if a fixed statement ψ could be proven with high probability with thisscheme, our intuition tells us that ψ is very likely to be true. This raises twoquestions:
1. Is this intuition correct? i.e., does this proof scheme make sense?2. Is it useful? i.e., can we prove more things than we classically could?
1. Random axioms 8/31
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Probabilistic proofs (3)
Such a proof scheme (or strategy) is probabilistic: if we run it twice we mightprove completely different statements!
Still, if a fixed statement ψ could be proven with high probability with thisscheme, our intuition tells us that ψ is very likely to be true. This raises twoquestions:
1. Is this intuition correct? i.e., does this proof scheme make sense?
2. Is it useful? i.e., can we prove more things than we classically could?
1. Random axioms 8/31
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Probabilistic proofs (3)
Such a proof scheme (or strategy) is probabilistic: if we run it twice we mightprove completely different statements!
Still, if a fixed statement ψ could be proven with high probability with thisscheme, our intuition tells us that ψ is very likely to be true. This raises twoquestions:
1. Is this intuition correct? i.e., does this proof scheme make sense?2. Is it useful? i.e., can we prove more things than we classically could?
1. Random axioms 8/31
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Probabilistic proofs (4)
The answer: Yes, it makes sense.
..TheoremIf a proof strategy with initial capital δ proves ψ with probability > δ, then ψ istrue.
..TheoremIf a proof strategy with initial capital δ proves ψ with probability > δ, then ψ istrue.
..TheoremIf a proof strategy with initial capital δ proves ψ with probability > δ, then ψ istrue.
No, it is not (really) useful.
..TheoremIf a proof strategy with initial capital δ proves ψ with probability > δ, then in factψ is already provable in PA.
..TheoremIf a proof strategy with initial capital δ proves ψ with probability > δ, then in factψ is already provable in PA.
..TheoremIf a proof strategy with initial capital δ proves ψ with probability > δ, then in factψ is already provable in PA.
1. Random axioms 9/31
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Probabilistic proofs (4)
The answer: Yes, it makes sense.
..TheoremIf a proof strategy with initial capital δ proves ψ with probability > δ, then ψ istrue.
..TheoremIf a proof strategy with initial capital δ proves ψ with probability > δ, then ψ istrue.
..TheoremIf a proof strategy with initial capital δ proves ψ with probability > δ, then ψ istrue.
No, it is not (really) useful.
..TheoremIf a proof strategy with initial capital δ proves ψ with probability > δ, then in factψ is already provable in PA.
..TheoremIf a proof strategy with initial capital δ proves ψ with probability > δ, then in factψ is already provable in PA.
..TheoremIf a proof strategy with initial capital δ proves ψ with probability > δ, then in factψ is already provable in PA.
1. Random axioms 9/31
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Probabilistic proofs (4)
The answer: Yes, it makes sense.
..TheoremIf a proof strategy with initial capital δ proves ψ with probability > δ, then ψ istrue.
..TheoremIf a proof strategy with initial capital δ proves ψ with probability > δ, then ψ istrue.
..TheoremIf a proof strategy with initial capital δ proves ψ with probability > δ, then ψ istrue.
No, it is not (really) useful.
..TheoremIf a proof strategy with initial capital δ proves ψ with probability > δ, then in factψ is already provable in PA.
..TheoremIf a proof strategy with initial capital δ proves ψ with probability > δ, then in factψ is already provable in PA.
..TheoremIf a proof strategy with initial capital δ proves ψ with probability > δ, then in factψ is already provable in PA.
1. Random axioms 9/31
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Probabilistic proofs (5)
Probabilistic proofs are still useful in the following sense: in the determinizationargument (transforming a probabilistic proof into a deterministic one), there is anexponential blow-up, i.e., the deterministic proof is in general exponentiallylonger than the probabilistic one.
In fact, this cannot be avoided:
..TheoremUnless NP=PSPACE, there are statements with polynomial-size probabilisticproofs and only exponential-size deterministic ones.
..TheoremUnless NP=PSPACE, there are statements with polynomial-size probabilisticproofs and only exponential-size deterministic ones.
..TheoremUnless NP=PSPACE, there are statements with polynomial-size probabilisticproofs and only exponential-size deterministic ones.
1. Random axioms 10/31
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Probabilistic proofs (5)
Probabilistic proofs are still useful in the following sense: in the determinizationargument (transforming a probabilistic proof into a deterministic one), there is anexponential blow-up, i.e., the deterministic proof is in general exponentiallylonger than the probabilistic one.
In fact, this cannot be avoided:
..TheoremUnless NP=PSPACE, there are statements with polynomial-size probabilisticproofs and only exponential-size deterministic ones.
..TheoremUnless NP=PSPACE, there are statements with polynomial-size probabilisticproofs and only exponential-size deterministic ones.
..TheoremUnless NP=PSPACE, there are statements with polynomial-size probabilisticproofs and only exponential-size deterministic ones.
1. Random axioms 10/31
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2. The axiomatic powerof Kolmogorov complexity
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Kolmogorov complexity (1)
The Kolmogorov complexity C(x) of a finite binary string x is the length of theshortest program (written in binary) that generates x. In a sense, it measuresthe amount of information contained in the string x.
For any given string x, we have
0 ≤ C(x) ≤ |x|+ O(1)
C(x) ≈ 0 meaning that x is simple (= far from random), while C(x) ≈ |x| meansthat x is quite random.
Most strings are quite random: there is only a fraction ≤ 2−c of strings of a givenlength such that C(x) ≤ |x|− c.
2. The axiomatic powerof Kolmogorov complexity 12/31
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Kolmogorov complexity (1)
The Kolmogorov complexity C(x) of a finite binary string x is the length of theshortest program (written in binary) that generates x. In a sense, it measuresthe amount of information contained in the string x.
For any given string x, we have
0 ≤ C(x) ≤ |x|+ O(1)
C(x) ≈ 0 meaning that x is simple (= far from random), while C(x) ≈ |x| meansthat x is quite random.
Most strings are quite random: there is only a fraction ≤ 2−c of strings of a givenlength such that C(x) ≤ |x|− c.
2. The axiomatic powerof Kolmogorov complexity 12/31
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Kolmogorov complexity (1)
The Kolmogorov complexity C(x) of a finite binary string x is the length of theshortest program (written in binary) that generates x. In a sense, it measuresthe amount of information contained in the string x.
For any given string x, we have
0 ≤ C(x) ≤ |x|+ O(1)
C(x) ≈ 0 meaning that x is simple (= far from random), while C(x) ≈ |x| meansthat x is quite random.
Most strings are quite random: there is only a fraction ≤ 2−c of strings of a givenlength such that C(x) ≤ |x|− c.
2. The axiomatic powerof Kolmogorov complexity 12/31
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Kolmogorov complexity (2)
Variation 1: K(x) is the prefix-free Kolmogorov complexity, i.e. the Kolmogorovcomplexity one gets if the set of valid programs is prefix-free.
Variation 2: C(x | y) and K(x | y) denote the conditional and prefix-freeconditional Kolmogorov complexity, i.e., the length of the shortest program thatproduces x given y as input.
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Kolmogorov complexity (2)
Variation 1: K(x) is the prefix-free Kolmogorov complexity, i.e. the Kolmogorovcomplexity one gets if the set of valid programs is prefix-free.
Variation 2: C(x | y) and K(x | y) denote the conditional and prefix-freeconditional Kolmogorov complexity, i.e., the length of the shortest program thatproduces x given y as input.
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Kolmogorov complexity (3)
Kolmogorov complexity is not a computable function. It is howeverupper-semicomputable: one can computably enumerate pairs (x, n) such thatC(x) ≤ n.
To summarize:• If C(x) ≤ n1 we will find out eventually• If C(x) ≥ n2 we might never know for sure
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Kolmogorov complexity (3)
Kolmogorov complexity is not a computable function. It is howeverupper-semicomputable: one can computably enumerate pairs (x, n) such thatC(x) ≤ n.
To summarize:• If C(x) ≤ n1 we will find out eventually• If C(x) ≥ n2 we might never know for sure
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Kolmogorov complexity (4)
Translation into the language of logic:
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Theorem (Chaitin)• If C(x) ≤ n1, then PA ⊢ “C(x) ≤ n1”• No matter what the actual value of C(x) is, PA ⊬ “C(x) ≥ n2” (provided n2 islarge enough).
..Theorem (Chaitin)• If C(x) ≤ n1, then PA ⊢ “C(x) ≤ n1”• No matter what the actual value of C(x) is, PA ⊬ “C(x) ≥ n2” (provided n2 islarge enough).
..Theorem (Chaitin)• If C(x) ≤ n1, then PA ⊢ “C(x) ≤ n1”• No matter what the actual value of C(x) is, PA ⊬ “C(x) ≥ n2” (provided n2 islarge enough).
Proof: Berry’s paradox → otherwise, given n, look for an xn such thatPA ⊢ “C(xn) ≥ n”. Since n is enough to describe xn, C(xn) ≤ O(log n), acontradiction.
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Kolmogorov complexity (4)
Translation into the language of logic:
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Theorem (Chaitin)• If C(x) ≤ n1, then PA ⊢ “C(x) ≤ n1”• No matter what the actual value of C(x) is, PA ⊬ “C(x) ≥ n2” (provided n2 islarge enough).
..Theorem (Chaitin)• If C(x) ≤ n1, then PA ⊢ “C(x) ≤ n1”• No matter what the actual value of C(x) is, PA ⊬ “C(x) ≥ n2” (provided n2 islarge enough).
..Theorem (Chaitin)• If C(x) ≤ n1, then PA ⊢ “C(x) ≤ n1”• No matter what the actual value of C(x) is, PA ⊬ “C(x) ≥ n2” (provided n2 islarge enough).
Proof: Berry’s paradox → otherwise, given n, look for an xn such thatPA ⊢ “C(xn) ≥ n”. Since n is enough to describe xn, C(xn) ≤ O(log n), acontradiction.
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Gödel’s theorem
In the previous argument, PA can be replaced by any stronger, recursive andcoherent theory. Thus, we have obtained a proof of Gödel’s first incompletenesstheorem using a Kolmogorov complexity interpretation of Berry’s paradox.
Note: one can also get a very elegant proof of Gödel’s secondincompleteness theorem via a Kolmogorov complexity interpretation of thesurprise examination paradox (Kritchman and Raz).
2. The axiomatic powerof Kolmogorov complexity 16/31
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Gödel’s theorem
In the previous argument, PA can be replaced by any stronger, recursive andcoherent theory. Thus, we have obtained a proof of Gödel’s first incompletenesstheorem using a Kolmogorov complexity interpretation of Berry’s paradox.
Note: one can also get a very elegant proof of Gödel’s secondincompleteness theorem via a Kolmogorov complexity interpretation of thesurprise examination paradox (Kritchman and Raz).
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Adding axioms about C (1)
Since PA does not prove them, a natural question:
How useful is the set C of true axioms of type “C(x) ≥ n”?
..TheoremThe statements provable from PA + C are exactly those of Π1-Th(N), i.e., thestatements φ that only contain ∀ as unbounded quantifiers and are true in N.
..TheoremThe statements provable from PA + C are exactly those of Π1-Th(N), i.e., thestatements φ that only contain ∀ as unbounded quantifiers and are true in N.
..TheoremThe statements provable from PA + C are exactly those of Π1-Th(N), i.e., thestatements φ that only contain ∀ as unbounded quantifiers and are true in N.
Proof. This is a direct translation into logic of the fact that if one can compute Cthen one can compute the halting problem ∅ ′.
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Adding axioms about C (1)
Since PA does not prove them, a natural question:
How useful is the set C of true axioms of type “C(x) ≥ n”?
..TheoremThe statements provable from PA + C are exactly those of Π1-Th(N), i.e., thestatements φ that only contain ∀ as unbounded quantifiers and are true in N.
..TheoremThe statements provable from PA + C are exactly those of Π1-Th(N), i.e., thestatements φ that only contain ∀ as unbounded quantifiers and are true in N.
..TheoremThe statements provable from PA + C are exactly those of Π1-Th(N), i.e., thestatements φ that only contain ∀ as unbounded quantifiers and are true in N.
Proof. This is a direct translation into logic of the fact that if one can compute Cthen one can compute the halting problem ∅ ′.
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Adding axioms about C (1)
Since PA does not prove them, a natural question:
How useful is the set C of true axioms of type “C(x) ≥ n”?
..TheoremThe statements provable from PA + C are exactly those of Π1-Th(N), i.e., thestatements φ that only contain ∀ as unbounded quantifiers and are true in N.
..TheoremThe statements provable from PA + C are exactly those of Π1-Th(N), i.e., thestatements φ that only contain ∀ as unbounded quantifiers and are true in N.
..TheoremThe statements provable from PA + C are exactly those of Π1-Th(N), i.e., thestatements φ that only contain ∀ as unbounded quantifiers and are true in N.
Proof. This is a direct translation into logic of the fact that if one can compute Cthen one can compute the halting problem ∅ ′.
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Adding axioms about C (1)
Interestingly, among the axioms “C(x) ≥ n”, some are more useful than others:
..
TheoremThere is a constant c and sequence of strings (xm) such that |xm| = m,C(xm) ≥m − c, and adding to PA all axioms “C(xm) ≥ m − c” proves all statements ofΠ1-Th(N).
..TheoremThere is a constant c and sequence of strings (xm) such that |xm| = m,C(xm) ≥m − c, and adding to PA all axioms “C(xm) ≥ m − c” proves all statements ofΠ1-Th(N).
..TheoremThere is a constant c and sequence of strings (xm) such that |xm| = m,C(xm) ≥m − c, and adding to PA all axioms “C(xm) ≥ m − c” proves all statements ofΠ1-Th(N).
On the other hand...
..
TheoremIf φ is a statement of Π1-Th(N) which is not provable in PA, it is possible toadd, for each length m, 99% of the true statements of type “C(y) ≥ m− c”, with|y| = m, and still be unable to prove φ.
..TheoremIf φ is a statement of Π1-Th(N) which is not provable in PA, it is possible toadd, for each length m, 99% of the true statements of type “C(y) ≥ m− c”, with|y| = m, and still be unable to prove φ.
..TheoremIf φ is a statement of Π1-Th(N) which is not provable in PA, it is possible toadd, for each length m, 99% of the true statements of type “C(y) ≥ m− c”, with|y| = m, and still be unable to prove φ.
2. The axiomatic powerof Kolmogorov complexity 18/31
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Adding axioms about C (1)
Interestingly, among the axioms “C(x) ≥ n”, some are more useful than others:
..
TheoremThere is a constant c and sequence of strings (xm) such that |xm| = m,C(xm) ≥m − c, and adding to PA all axioms “C(xm) ≥ m − c” proves all statements ofΠ1-Th(N).
..TheoremThere is a constant c and sequence of strings (xm) such that |xm| = m,C(xm) ≥m − c, and adding to PA all axioms “C(xm) ≥ m − c” proves all statements ofΠ1-Th(N).
..TheoremThere is a constant c and sequence of strings (xm) such that |xm| = m,C(xm) ≥m − c, and adding to PA all axioms “C(xm) ≥ m − c” proves all statements ofΠ1-Th(N).
On the other hand...
..
TheoremIf φ is a statement of Π1-Th(N) which is not provable in PA, it is possible toadd, for each length m, 99% of the true statements of type “C(y) ≥ m− c”, with|y| = m, and still be unable to prove φ.
..TheoremIf φ is a statement of Π1-Th(N) which is not provable in PA, it is possible toadd, for each length m, 99% of the true statements of type “C(y) ≥ m− c”, with|y| = m, and still be unable to prove φ.
..TheoremIf φ is a statement of Π1-Th(N) which is not provable in PA, it is possible toadd, for each length m, 99% of the true statements of type “C(y) ≥ m− c”, with|y| = m, and still be unable to prove φ.
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Adding axioms about C (2)
This raises the question:
..Open questionCharacterize the “useful” xn, i.e., those such that the collection “C(xn) ≥ n− c”is true and proves all of Π1-Th(N).
..Open questionCharacterize the “useful” xn, i.e., those such that the collection “C(xn) ≥ n− c”is true and proves all of Π1-Th(N).
..Open questionCharacterize the “useful” xn, i.e., those such that the collection “C(xn) ≥ n− c”is true and proves all of Π1-Th(N).
One last important point: axioms need to be precise to be useful.
..TheoremThere is no sequence (xn) such that C(xn) ≥ n − c and such that adding thecollection of axioms “C(xn) ≥ n/2 − c” proves all of Π1-Th(N).
..TheoremThere is no sequence (xn) such that C(xn) ≥ n − c and such that adding thecollection of axioms “C(xn) ≥ n/2 − c” proves all of Π1-Th(N).
..TheoremThere is no sequence (xn) such that C(xn) ≥ n − c and such that adding thecollection of axioms “C(xn) ≥ n/2 − c” proves all of Π1-Th(N).
2. The axiomatic powerof Kolmogorov complexity 19/31
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Adding axioms about C (2)
This raises the question:
..Open questionCharacterize the “useful” xn, i.e., those such that the collection “C(xn) ≥ n− c”is true and proves all of Π1-Th(N).
..Open questionCharacterize the “useful” xn, i.e., those such that the collection “C(xn) ≥ n− c”is true and proves all of Π1-Th(N).
..Open questionCharacterize the “useful” xn, i.e., those such that the collection “C(xn) ≥ n− c”is true and proves all of Π1-Th(N).
One last important point: axioms need to be precise to be useful.
..TheoremThere is no sequence (xn) such that C(xn) ≥ n − c and such that adding thecollection of axioms “C(xn) ≥ n/2 − c” proves all of Π1-Th(N).
..TheoremThere is no sequence (xn) such that C(xn) ≥ n − c and such that adding thecollection of axioms “C(xn) ≥ n/2 − c” proves all of Π1-Th(N).
..TheoremThere is no sequence (xn) such that C(xn) ≥ n − c and such that adding thecollection of axioms “C(xn) ≥ n/2 − c” proves all of Π1-Th(N).
2. The axiomatic powerof Kolmogorov complexity 19/31
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Adding axioms about C (2)
This raises the question:
..Open questionCharacterize the “useful” xn, i.e., those such that the collection “C(xn) ≥ n− c”is true and proves all of Π1-Th(N).
..Open questionCharacterize the “useful” xn, i.e., those such that the collection “C(xn) ≥ n− c”is true and proves all of Π1-Th(N).
..Open questionCharacterize the “useful” xn, i.e., those such that the collection “C(xn) ≥ n− c”is true and proves all of Π1-Th(N).
One last important point: axioms need to be precise to be useful.
..TheoremThere is no sequence (xn) such that C(xn) ≥ n − c and such that adding thecollection of axioms “C(xn) ≥ n/2 − c” proves all of Π1-Th(N).
..TheoremThere is no sequence (xn) such that C(xn) ≥ n − c and such that adding thecollection of axioms “C(xn) ≥ n/2 − c” proves all of Π1-Th(N).
..TheoremThere is no sequence (xn) such that C(xn) ≥ n − c and such that adding thecollection of axioms “C(xn) ≥ n/2 − c” proves all of Π1-Th(N).
2. The axiomatic powerof Kolmogorov complexity 19/31
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Adding axioms about C (2)
This raises the question:
..Open questionCharacterize the “useful” xn, i.e., those such that the collection “C(xn) ≥ n− c”is true and proves all of Π1-Th(N).
..Open questionCharacterize the “useful” xn, i.e., those such that the collection “C(xn) ≥ n− c”is true and proves all of Π1-Th(N).
..Open questionCharacterize the “useful” xn, i.e., those such that the collection “C(xn) ≥ n− c”is true and proves all of Π1-Th(N).
One last important point: axioms need to be precise to be useful.
..TheoremThere is no sequence (xn) such that C(xn) ≥ n − c and such that adding thecollection of axioms “C(xn) ≥ n/2 − c” proves all of Π1-Th(N).
..TheoremThere is no sequence (xn) such that C(xn) ≥ n − c and such that adding thecollection of axioms “C(xn) ≥ n/2 − c” proves all of Π1-Th(N).
..TheoremThere is no sequence (xn) such that C(xn) ≥ n − c and such that adding thecollection of axioms “C(xn) ≥ n/2 − c” proves all of Π1-Th(N).
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Adding axioms about C (3)
Another example: consider for a given k the set CCk of true axioms of type
“C(x | y) ≥ k”
..TheoremFor any k large enough, the statements provable fromPA+CCk are exactly thoseof Π1-Th(N).
..TheoremFor any k large enough, the statements provable fromPA+CCk are exactly thoseof Π1-Th(N).
..TheoremFor any k large enough, the statements provable fromPA+CCk are exactly thoseof Π1-Th(N).
Proof. We show that being able to decide the truth of “C(x | y) ≥ k” allows usto decide the halting problem, and translate the proof into logic.
2. The axiomatic powerof Kolmogorov complexity 20/31
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Adding axioms about C (3)
Another example: consider for a given k the set CCk of true axioms of type
“C(x | y) ≥ k”
..TheoremFor any k large enough, the statements provable fromPA+CCk are exactly thoseof Π1-Th(N).
..TheoremFor any k large enough, the statements provable fromPA+CCk are exactly thoseof Π1-Th(N).
..TheoremFor any k large enough, the statements provable fromPA+CCk are exactly thoseof Π1-Th(N).
Proof. We show that being able to decide the truth of “C(x | y) ≥ k” allows usto decide the halting problem, and translate the proof into logic.
2. The axiomatic powerof Kolmogorov complexity 20/31
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Adding axioms about C (3)
Another example: consider for a given k the set CCk of true axioms of type
“C(x | y) ≥ k”
..TheoremFor any k large enough, the statements provable fromPA+CCk are exactly thoseof Π1-Th(N).
..TheoremFor any k large enough, the statements provable fromPA+CCk are exactly thoseof Π1-Th(N).
..TheoremFor any k large enough, the statements provable fromPA+CCk are exactly thoseof Π1-Th(N).
Proof. We show that being able to decide the truth of “C(x | y) ≥ k” allows usto decide the halting problem, and translate the proof into logic.
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Adding axioms about C (4)
More interesting: the translation into logic does not necessarily go through!
..Theorem (BBFFGLMST)If one is given for each x a pair of values n1 and n2 such that C(x) = n1 orC(x) = n2, then one can compute ∅ ′.
..Theorem (BBFFGLMST)If one is given for each x a pair of values n1 and n2 such that C(x) = n1 orC(x) = n2, then one can compute ∅ ′.
..Theorem (BBFFGLMST)If one is given for each x a pair of values n1 and n2 such that C(x) = n1 orC(x) = n2, then one can compute ∅ ′.
But...
..
TheoremLetφ be any statement not provable in PA. Then it is possible to add for each xa true axiom of type “C(x) = n1 ∨ C(x) = n2” and still have a theory too weakto prove φ.
..TheoremLetφ be any statement not provable in PA. Then it is possible to add for each xa true axiom of type “C(x) = n1 ∨ C(x) = n2” and still have a theory too weakto prove φ.
..TheoremLetφ be any statement not provable in PA. Then it is possible to add for each xa true axiom of type “C(x) = n1 ∨ C(x) = n2” and still have a theory too weakto prove φ.
2. The axiomatic powerof Kolmogorov complexity 21/31
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Adding axioms about C (4)
More interesting: the translation into logic does not necessarily go through!
..Theorem (BBFFGLMST)If one is given for each x a pair of values n1 and n2 such that C(x) = n1 orC(x) = n2, then one can compute ∅ ′.
..Theorem (BBFFGLMST)If one is given for each x a pair of values n1 and n2 such that C(x) = n1 orC(x) = n2, then one can compute ∅ ′.
..Theorem (BBFFGLMST)If one is given for each x a pair of values n1 and n2 such that C(x) = n1 orC(x) = n2, then one can compute ∅ ′.
But...
..
TheoremLetφ be any statement not provable in PA. Then it is possible to add for each xa true axiom of type “C(x) = n1 ∨ C(x) = n2” and still have a theory too weakto prove φ.
..TheoremLetφ be any statement not provable in PA. Then it is possible to add for each xa true axiom of type “C(x) = n1 ∨ C(x) = n2” and still have a theory too weakto prove φ.
..TheoremLetφ be any statement not provable in PA. Then it is possible to add for each xa true axiom of type “C(x) = n1 ∨ C(x) = n2” and still have a theory too weakto prove φ.
2. The axiomatic powerof Kolmogorov complexity 21/31
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.
Adding axioms about C (4)
More interesting: the translation into logic does not necessarily go through!
..Theorem (BBFFGLMST)If one is given for each x a pair of values n1 and n2 such that C(x) = n1 orC(x) = n2, then one can compute ∅ ′.
..Theorem (BBFFGLMST)If one is given for each x a pair of values n1 and n2 such that C(x) = n1 orC(x) = n2, then one can compute ∅ ′.
..Theorem (BBFFGLMST)If one is given for each x a pair of values n1 and n2 such that C(x) = n1 orC(x) = n2, then one can compute ∅ ′.
But...
..
TheoremLetφ be any statement not provable in PA. Then it is possible to add for each xa true axiom of type “C(x) = n1 ∨ C(x) = n2” and still have a theory too weakto prove φ.
..TheoremLetφ be any statement not provable in PA. Then it is possible to add for each xa true axiom of type “C(x) = n1 ∨ C(x) = n2” and still have a theory too weakto prove φ.
..TheoremLetφ be any statement not provable in PA. Then it is possible to add for each xa true axiom of type “C(x) = n1 ∨ C(x) = n2” and still have a theory too weakto prove φ.
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..
3. Infinite binary sequences
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.
Random sequences
Kolmogorov complexity measures the amount randomness for finite binarysequences. For infinite binary sequences, it even allows us to provide an exactdefinition of randomness.
A sequence X = x1x2x3 . . . is said to be (Martin-Löf) random with constant c if
K(x1x2 ... xn) ≥ n− c
3. Infinite binary sequences 23/31
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Random sequences
Kolmogorov complexity measures the amount randomness for finite binarysequences. For infinite binary sequences, it even allows us to provide an exactdefinition of randomness.
A sequence X = x1x2x3 . . . is said to be (Martin-Löf) random with constant c if
K(x1x2 ... xn) ≥ n− c
3. Infinite binary sequences 23/31
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Randomness of a sequence, axiomatized (1)
Some random sequences do compute ∅ ′, such as Chaitin’s Omega number.
Could it then be that expressing the randomness of some sequences wouldprove all of Π1-Th(N)?
Interestingly, it depends on the way we express the randomness of X.
3. Infinite binary sequences 24/31
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Randomness of a sequence, axiomatized (1)
Some random sequences do compute ∅ ′, such as Chaitin’s Omega number.
Could it then be that expressing the randomness of some sequences wouldprove all of Π1-Th(N)?
Interestingly, it depends on the way we express the randomness of X.
3. Infinite binary sequences 24/31
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Randomness of a sequence, axiomatized (1)
Some random sequences do compute ∅ ′, such as Chaitin’s Omega number.
Could it then be that expressing the randomness of some sequences wouldprove all of Π1-Th(N)?
Interestingly, it depends on the way we express the randomness of X.
3. Infinite binary sequences 24/31
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Randomness of a sequence, axiomatized (2)
Given a sequence X random with constant c, the obvious way is to take the setMLRc(X) of all axioms of type
K(x1x2 ... xn) ≥ n− c
In that case, we cannot prove all of Π1-Th(N):
..
Theorem• If MLRc(X) is consistent, adding it to PA is never sufficient to proveΠ1-Th(N).• Moreover, MLRc(X) is asymptotically useless: if φ is not provable in PA,then PA+MLRc(X) does not prove φ for any c large enough.
..Theorem• If MLRc(X) is consistent, adding it to PA is never sufficient to proveΠ1-Th(N).• Moreover, MLRc(X) is asymptotically useless: if φ is not provable in PA,then PA+MLRc(X) does not prove φ for any c large enough.
..Theorem• If MLRc(X) is consistent, adding it to PA is never sufficient to proveΠ1-Th(N).• Moreover, MLRc(X) is asymptotically useless: if φ is not provable in PA,then PA+MLRc(X) does not prove φ for any c large enough.
3. Infinite binary sequences 25/31
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Randomness of a sequence, axiomatized (2)
Given a sequence X random with constant c, the obvious way is to take the setMLRc(X) of all axioms of type
K(x1x2 ... xn) ≥ n− c
In that case, we cannot prove all of Π1-Th(N):
..
Theorem• If MLRc(X) is consistent, adding it to PA is never sufficient to proveΠ1-Th(N).• Moreover, MLRc(X) is asymptotically useless: if φ is not provable in PA,then PA+MLRc(X) does not prove φ for any c large enough.
..Theorem• If MLRc(X) is consistent, adding it to PA is never sufficient to proveΠ1-Th(N).• Moreover, MLRc(X) is asymptotically useless: if φ is not provable in PA,then PA+MLRc(X) does not prove φ for any c large enough.
..Theorem• If MLRc(X) is consistent, adding it to PA is never sufficient to proveΠ1-Th(N).• Moreover, MLRc(X) is asymptotically useless: if φ is not provable in PA,then PA+MLRc(X) does not prove φ for any c large enough.
3. Infinite binary sequences 25/31
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Randomness of a sequence, axiomatized (3)
A very rough sketch of the proof. Suppose for the sake of contradiction thatMLRc(X) is true and proves Π1-Th(N).
We use an interesting back-and-forth between logic and computability.• Note that X can compute the set of axioms MLRc(X), hence X can
compute Π1-Th(N), thus X ≥T ∅ ′.• Now, consider the set of complete coherent extensions of PA, coded as
infinite binary sequences.• This forms a Π0
1 class P , so by the basis theorem for randomness, there isA ∈ P such that X is random relative to A.
3. Infinite binary sequences 26/31
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Randomness of a sequence, axiomatized (3)
A very rough sketch of the proof. Suppose for the sake of contradiction thatMLRc(X) is true and proves Π1-Th(N).
We use an interesting back-and-forth between logic and computability.
• Note that X can compute the set of axioms MLRc(X), hence X cancompute Π1-Th(N), thus X ≥T ∅ ′.
• Now, consider the set of complete coherent extensions of PA, coded asinfinite binary sequences.
• This forms a Π01 class P , so by the basis theorem for randomness, there is
A ∈ P such that X is random relative to A.
3. Infinite binary sequences 26/31
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Randomness of a sequence, axiomatized (3)
A very rough sketch of the proof. Suppose for the sake of contradiction thatMLRc(X) is true and proves Π1-Th(N).
We use an interesting back-and-forth between logic and computability.• Note that X can compute the set of axioms MLRc(X), hence X can
compute Π1-Th(N), thus X ≥T ∅ ′.
• Now, consider the set of complete coherent extensions of PA, coded asinfinite binary sequences.
• This forms a Π01 class P , so by the basis theorem for randomness, there is
A ∈ P such that X is random relative to A.
3. Infinite binary sequences 26/31
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.
Randomness of a sequence, axiomatized (3)
A very rough sketch of the proof. Suppose for the sake of contradiction thatMLRc(X) is true and proves Π1-Th(N).
We use an interesting back-and-forth between logic and computability.• Note that X can compute the set of axioms MLRc(X), hence X can
compute Π1-Th(N), thus X ≥T ∅ ′.• Now, consider the set of complete coherent extensions of PA, coded as
infinite binary sequences.
• This forms a Π01 class P , so by the basis theorem for randomness, there is
A ∈ P such that X is random relative to A.
3. Infinite binary sequences 26/31
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.
Randomness of a sequence, axiomatized (3)
A very rough sketch of the proof. Suppose for the sake of contradiction thatMLRc(X) is true and proves Π1-Th(N).
We use an interesting back-and-forth between logic and computability.• Note that X can compute the set of axioms MLRc(X), hence X can
compute Π1-Th(N), thus X ≥T ∅ ′.• Now, consider the set of complete coherent extensions of PA, coded as
infinite binary sequences.• This forms a Π0
1 class P , so by the basis theorem for randomness, there isA ∈ P such that X is random relative to A.
3. Infinite binary sequences 26/31
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Randomness of a sequence, axiomatized (4)
• Let T be the logical theory coded by A. Argue that X being A-randomimplies that X is seen as random in the theory T .
• Argue that therefore, in the theory T , the collection of axioms MLRc(X) is(almost) true.
• Thus the theory T proves Π1-Th(N).• Thus A ≥T ∅ ′.• Thus X is random relative to ∅ ′, a contradiction with X ≥T ∅ ′.
3. Infinite binary sequences 27/31
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Randomness of a sequence, axiomatized (4)
• Let T be the logical theory coded by A. Argue that X being A-randomimplies that X is seen as random in the theory T .
• Argue that therefore, in the theory T , the collection of axioms MLRc(X) is(almost) true.
• Thus the theory T proves Π1-Th(N).• Thus A ≥T ∅ ′.• Thus X is random relative to ∅ ′, a contradiction with X ≥T ∅ ′.
3. Infinite binary sequences 27/31
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.
Randomness of a sequence, axiomatized (4)
• Let T be the logical theory coded by A. Argue that X being A-randomimplies that X is seen as random in the theory T .
• Argue that therefore, in the theory T , the collection of axioms MLRc(X) is(almost) true.
• Thus the theory T proves Π1-Th(N).• Thus A ≥T ∅ ′.• Thus X is random relative to ∅ ′, a contradiction with X ≥T ∅ ′.
3. Infinite binary sequences 27/31
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.
Randomness of a sequence, axiomatized (4)
• Let T be the logical theory coded by A. Argue that X being A-randomimplies that X is seen as random in the theory T .
• Argue that therefore, in the theory T , the collection of axioms MLRc(X) is(almost) true.
• Thus the theory T proves Π1-Th(N).
• Thus A ≥T ∅ ′.• Thus X is random relative to ∅ ′, a contradiction with X ≥T ∅ ′.
3. Infinite binary sequences 27/31
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.
Randomness of a sequence, axiomatized (4)
• Let T be the logical theory coded by A. Argue that X being A-randomimplies that X is seen as random in the theory T .
• Argue that therefore, in the theory T , the collection of axioms MLRc(X) is(almost) true.
• Thus the theory T proves Π1-Th(N).• Thus A ≥T ∅ ′.
• Thus X is random relative to ∅ ′, a contradiction with X ≥T ∅ ′.
3. Infinite binary sequences 27/31
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.
Randomness of a sequence, axiomatized (4)
• Let T be the logical theory coded by A. Argue that X being A-randomimplies that X is seen as random in the theory T .
• Argue that therefore, in the theory T , the collection of axioms MLRc(X) is(almost) true.
• Thus the theory T proves Π1-Th(N).• Thus A ≥T ∅ ′.• Thus X is random relative to ∅ ′, a contradiction with X ≥T ∅ ′.
3. Infinite binary sequences 27/31
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Randomness of a sequence, axiomatized (5)
An alternative way: take the set MLR ′c(X) of all axioms of type
(∀N)(∃σ ≻ x1...xn) |σ| = N and K(σ) ≥ N− c
In this case, MLR ′c(X) may be powerful.
..
Theorem• There exists a consistent set of axioms MLR ′
c(X) such thatPA+MLR ′c(X)
proves all of Π1-Th(N) [take Chaitin’s Omega].• Still, MLR ′
c(X) is asymptotically useless as before.
..Theorem• There exists a consistent set of axioms MLR ′
c(X) such thatPA+MLR ′c(X)
proves all of Π1-Th(N) [take Chaitin’s Omega].• Still, MLR ′
c(X) is asymptotically useless as before.
..Theorem• There exists a consistent set of axioms MLR ′
c(X) such thatPA+MLR ′c(X)
proves all of Π1-Th(N) [take Chaitin’s Omega].• Still, MLR ′
c(X) is asymptotically useless as before.
3. Infinite binary sequences 28/31
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Randomness of a sequence, axiomatized (5)
An alternative way: take the set MLR ′c(X) of all axioms of type
(∀N)(∃σ ≻ x1...xn) |σ| = N and K(σ) ≥ N− c
In this case, MLR ′c(X) may be powerful.
..
Theorem• There exists a consistent set of axioms MLR ′
c(X) such thatPA+MLR ′c(X)
proves all of Π1-Th(N) [take Chaitin’s Omega].• Still, MLR ′
c(X) is asymptotically useless as before.
..Theorem• There exists a consistent set of axioms MLR ′
c(X) such thatPA+MLR ′c(X)
proves all of Π1-Th(N) [take Chaitin’s Omega].• Still, MLR ′
c(X) is asymptotically useless as before.
..Theorem• There exists a consistent set of axioms MLR ′
c(X) such thatPA+MLR ′c(X)
proves all of Π1-Th(N) [take Chaitin’s Omega].• Still, MLR ′
c(X) is asymptotically useless as before.
3. Infinite binary sequences 28/31
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Non-random sequencess can help too!
We finish on this intriguing fact: while MLRc(X) is never very useful for arandom X, there is a non-random sequence Z whose complexity gives usefulinformation:
..
TheoremThere is a sequence Z = z1z2... and constant c such that K(z1 ... zn) ≥ n − cfor infinitely many n, and such that if we take these infinitely many statements asaxioms, we can prove all of Π1-Th(N).
..TheoremThere is a sequence Z = z1z2... and constant c such that K(z1 ... zn) ≥ n − cfor infinitely many n, and such that if we take these infinitely many statements asaxioms, we can prove all of Π1-Th(N).
..TheoremThere is a sequence Z = z1z2... and constant c such that K(z1 ... zn) ≥ n − cfor infinitely many n, and such that if we take these infinitely many statements asaxioms, we can prove all of Π1-Th(N).
3. Infinite binary sequences 29/31
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Open questions
..Open questionCharacterize the “useful” xn, i.e., those such that the collection “C(xn) ≥ n− c”is true and proves all of Π1-Th(N).
..Open questionCharacterize the “useful” xn, i.e., those such that the collection “C(xn) ≥ n− c”is true and proves all of Π1-Th(N).
..Open questionCharacterize the “useful” xn, i.e., those such that the collection “C(xn) ≥ n− c”is true and proves all of Π1-Th(N).
..Open questionWhat axioms about Kolmogorov should we look at if we want to say somethingabout Πn-Th(N)?
..Open questionWhat axioms about Kolmogorov should we look at if we want to say somethingabout Πn-Th(N)?
..Open questionWhat axioms about Kolmogorov should we look at if we want to say somethingabout Πn-Th(N)?
3. Infinite binary sequences 30/31
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Open questions
..Open questionCharacterize the “useful” xn, i.e., those such that the collection “C(xn) ≥ n− c”is true and proves all of Π1-Th(N).
..Open questionCharacterize the “useful” xn, i.e., those such that the collection “C(xn) ≥ n− c”is true and proves all of Π1-Th(N).
..Open questionCharacterize the “useful” xn, i.e., those such that the collection “C(xn) ≥ n− c”is true and proves all of Π1-Th(N).
..Open questionWhat axioms about Kolmogorov should we look at if we want to say somethingabout Πn-Th(N)?
..Open questionWhat axioms about Kolmogorov should we look at if we want to say somethingabout Πn-Th(N)?
..Open questionWhat axioms about Kolmogorov should we look at if we want to say somethingabout Πn-Th(N)?
3. Infinite binary sequences 30/31
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Open questions
..Open questionCharacterize the “useful” xn, i.e., those such that the collection “C(xn) ≥ n− c”is true and proves all of Π1-Th(N).
..Open questionCharacterize the “useful” xn, i.e., those such that the collection “C(xn) ≥ n− c”is true and proves all of Π1-Th(N).
..Open questionCharacterize the “useful” xn, i.e., those such that the collection “C(xn) ≥ n− c”is true and proves all of Π1-Th(N).
..Open questionWhat axioms about Kolmogorov should we look at if we want to say somethingabout Πn-Th(N)?
..Open questionWhat axioms about Kolmogorov should we look at if we want to say somethingabout Πn-Th(N)?
..Open questionWhat axioms about Kolmogorov should we look at if we want to say somethingabout Πn-Th(N)?