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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.
2. The axiomatic powerof Kolmogorov complexity 13/31
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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 (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
2. The axiomatic powerof Kolmogorov complexity 14/31
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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.
2. The axiomatic powerof Kolmogorov complexity 15/31
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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.
2. The axiomatic powerof Kolmogorov complexity 15/31
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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).
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).
2. The axiomatic powerof Kolmogorov complexity 16/31
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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 ∅ ′.
2. The axiomatic powerof Kolmogorov complexity 17/31
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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 ∅ ′.
2. The axiomatic powerof Kolmogorov complexity 17/31
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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).
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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.
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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
.
.
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
.
.
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
.
.
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
.
.
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
.
.
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
.
.
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
.
.
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
.
.
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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.Thank You !
3. Infinite binary sequences 31/31