aa r X i v : . [ m a t h . L O ] J a n Randomising Realisability ⋆ Merlin Carl , Lorenzo Galeotti , and Robert Passmann , Europa-Universit¨at Flensburg, 24943 Flensburg, Germany Amsterdam University College, Postbus 94160, 1090 GD Amsterdam, TheNetherlands Institute for Logic, Language and Computation, Faculty of Science, University ofAmsterdam, P.O. Box 94242, 1090 GE Amsterdam, The Netherlands St John’s College, University of Cambridge, Cambridge CB2 1TP, England
Abstract.
We consider a randomised version of Kleene’s realisabilityinterpretation of intuitionistic arithmetic in which computability is re-placed with randomised computability with positive probability. In par-ticular, we show that (i) the set of randomly realisable statements isclosed under intuitionistic first-order logic, but (ii) different from theset of realisable statements, that (iii) ”realisability with probability 1”is the same as realisability and (iv) that the axioms of bounded Heyt-ing’s arithmetic are randomly realisable, but some instances of the fullinduction scheme fail to be randomly realisable.
Have you met skeptical Steve? Being even more skeptical than most mathemati-cians, he only believes what he actually sees. To convince him that there is an x such that A , you have to give him an example, together with evidence that A holds for that example. To convince him that A → B , you have to show hima method for turning evidence of A into evidence of B , and so on. Given thatSteve is “a man provided with paper, pencil, and rubber, and subject to strictdiscipline” [9], we can read “method” as “Turing program”, which leads us toKleene’s realisability interpretation of intuitionistic logic [4].Steve has a younger brother, pragmatical Per. Like Steve, Per is equippedwith paper and pencil; however, he also has a coin on his desk, which he is allowedto throw from time to time while performing computations. By his pragmaticalnature, he does not require being successful at obtaining evidence for a givenproposition A every time he gives it a try; he is quite happy when it works withprobability (1 − ) or so, which makes it highly unlikely to ever fail in hislifetime.Per wonders whether his pragmatism is more powerful than Steve’s method.After all, he knows about Sacks’s theorem [1, Corollary 8.12.2] that every func-tion f : ω → ω that is computable using coin throws with positive probability isrecursive. Can he find evidence for some claims where Steve fails? He also notices ⋆ The authors would like to thank Rosalie Iemhoff and Jaap van Oosten for discussionsabout the material included in this paper. Merlin Carl, Lorenzo Galeotti, and Robert Passmann that turning such “probabilistic evidence” for A into “probabilistic evidence” for B is a job considerably different (and potentially harder) than turning evidencefor A into evidence for B . Could it be that there are propositions whose truthSteve can see, but Per cannot? Although Per is skeptical, e.g., of the law ofthe excluded middle just like Steve, he is quite fond of the deduction rules ofintuitionistic logic; thus, he wonders whether the set of statements for which hecan obtain his “highly probably evidence” is closed under these.Steve is unhappy with his brother’s sloppiness. After all, even probability(1 − ) leaves a small, albeit nonzero, chance of getting things wrong. Hemight consider changing his mind if that chance was brought down to 0 bystrengthening Steve’s definition, demanding that the “probabilistic evidence”works with probability 1. However, he is only willing to give up absolute securityif that leads to evidence for more statements. Thus, he asks whether “probability1 evidence” is the same as “evidence”.These and other questions will be considered in this paper. To begin with, wewill model Per’s attitude formally, which gives us the concepts of µ -realisability and almost sure realisability . We will then show the following: There are state-ments that are µ -realisable, but not realisable (Theorem 14). The set of µ -realisable statements are closed under deduction in intuitionistic predicate cal-culus (Theorem 18); in a certain sense to be specified below, the law or excludedmiddle fails for µ -realisability (Lemma 16). The axioms of Heyting arithmeticexcept for the induction schema are µ -realised (Theorem 19); and there are in-stances of the induction schema that are not µ -realised (Theorem 20). Almostsure realisability is the same as realisability (Theorem 25). Realisability is one of the most common semantic tools for the study of con-structive theories and was introduced by Kleene in his seminal 1945 paper [4].In this work, Kleene connected intuitionistic arithmetic—nowadays called
Heyt-ing arithmetic —and recursive functions. The essential idea is that a statementis true if and only if there is a recursive function witnessing its truth. For moredetails on realisability, see also Troelstra’s 344 [8], and van Oosten’s paper [7]for an excellent historical survey of realisability. In particular, see [8, Definition3.2.2] for a definition of realisability in terms of recursive functions. In whatfollows, we denote this classical relation of realisability by ‘ (cid:13) ’.As mentioned in the introduction, we want to give pragmatic Per the abilityto throw coins while he tries to prove the truth of a statement. We will imple-ment this coin throwing by allowing Per to access an infinite binary sequence.Therefore, we will make use of the Lebesgue measure on Cantor space 2 ω . Fora full definition, see Kanamori’s section on ‘Measure and Category’ [2, Chapter0]. We denote the Lebesgue measure by µ . Recall that a set A is Lebesgue mea-surable if and only if there is a Borel set B such that the symmetric differenceof A and B is null. Given an element u of Cantor space we will denote by N u ↾ n the basic clopen set { v ∈ ω ; u ↾ n ⊂ v } where as usual u ↾ n is the prefix of u andomising Realisability 3 of length n , and u ↾ n ⊂ v if u ↾ n is a prefix of v . We recall that given a binarysequence of length n , we have that N s is measurable and µ (N s ) = n .We fix a computable enumeration ( p n ) n ∈ N of programs. Moreover, given aprogram p that uses an oracle and an element u ∈ ω we will denote by p u theprogram p where the oracle tape contains u at the beginning of the computation.Moreover, given n ∈ N we will denote by p ( n ) the program that for every oracle u ∈ ω returns p u ( n ).A sentence in the language of arithmetic is said to be ∆ if it does not containunbounded quantifiers. We will say that a sentence is a pretty Σ if it is ∆ or ofthe form Q Q . . . Q n ψ where ψ is ∆ and Q i is either an existential quantifieror a bounded universal quantifier for every 0 ≤ i ≤ n . Similarly, we will say thata sentence is a universal Π if it is ∆ or of the form Q Q . . . Q n ψ where ψ is ∆ and Q i is a universal quantifier for every 0 ≤ i ≤ n Throughout this paper, we fix codings for formulas and programs. In orderto simplify notation, we will use ϕ to refer to both the formula and its code, andsimilar for programs p . We end this section with some lemmas on realisabilityof pretty Σ and universal Π formulas. Lemma 1.
There is a program p that for every pretty Σ sentence ϕ does thefollowing: If ϕ is true then p ( ϕ ) halts and outputs a realiser of ϕ and otherwise,it diverges.Proof. First we define the program for ∆ formulas by recursion.(1) If ϕ is atomic, p first checks if ϕ is true. If so then p returns any naturalnumber, otherwise, it loops.(2) ϕ ≡ ψ ∧ ψ : the program p checks whether ψ and ψ are true. If bothcomputations are successful then p returns the code of a program q which returns p ( ψ ) on input 0 and p ( ψ ) on input 1.(3) ϕ ≡ ψ ∨ ψ the program p starts checking if at least one between ψ and ψ is true. If one of the two computations is successful then p returns the codeof a program q which returns p ( ψ i ) on input 1 and i on input 0 where i is thesmallest i such that ψ i is true. Otherwise the program loops.(4) ϕ ≡ ψ → ψ then p first checks whether ψ is true if not p returns 0otherwise returns a program that for every input returns p ( ψ ).(5) ϕ ≡ ∃ x < nψ then the program p checks if there is m < n such that ψ ( m )is true. If so then p returns the code of a program q which returns p ( ψ ( m )) oninput 0 and m on input 1 where m is the smallest natural number such that ψ ( m ) is true. Otherwise the program loops.(6) ϕ ≡ ∀ x < nψ then p checks in parallel the truth of all the instances of ψ ( m ) for m < n . If all of them are true then p returns the code of a program q which for all m ∈ N returns p ( ψ ( m )). Otherwise the program loops. Merlin Carl, Lorenzo Galeotti, and Robert Passmann
Now we extend the definition of p to pretty Σ sentences. Assume that ϕ isof the form Q . . . Q n ψ where ψ is ∆ and Q i is either an existential quantifieror a bounded universal quantifier for every 0 ≤ i ≤ n . We define p by recursionon n . Since the base case and the inductive step are essentially the same we willonly show the latter.Let n = m + 1 and ϕ ≡ Q Q . . . Q n ψ where ψ is ∆ . We assume that f isalready defined for Q . . . Q n ψ and need to show that we can extend it to ϕ . Wehave two cases(1) Q is a bounded quantifier. Then we repeat what we did in part (5) and(6) of this proof.(2) Q is an unbounded existential quantifier. Then the program p starts anunbounded search to find an i such that ψ ( i ) is true. If it finds it then p returnsthe code of a program q which returns p ( ψ ( i )) on input 0 and i on input 1 where i is the smallest natural number such that ψ ( i ) is true. Lemma 2.
There is a program p that for every universal Π sentence ϕ doesthe following: If ϕ is true then p ( ϕ ) halts and outputs a realiser of ϕ (we do notspecify a behaviour otherwise).Proof. Define p as in the proof of Lemma 1 for ∆ formulas. Then for formulasof the type ∀ xψ let p ( ψ ) be the code of the program that for all n runs p ( ψ ( n )). Lemma 3.
A pretty Σ sentence in the language of arithmetic is realised if andonly if it is true. The same result holds for universal Π sentences.Proof. The right-to-left direction follows from Lemma 1. The other direction isa straightforward induction on the complexity of ϕ . The proof for universal Π sentences is also an easy induction. Corollary 4.
There is a program p that for every pretty Σ sentence ϕ does thefollowing: If ϕ is true then p ( ϕ ) halts and outputs a realiser of ϕ and otherwise,it diverges. In this section we will introduce the notion of µ -realisability and prove the basicproperties of this relation. As we mentioned before, we will modify classicalrealisability in order to use realisers that can access an element of Cantor space.Then we will say that a sentence is randomly realised if for non-null set of oraclesin Cantor space the program does realise the sentence. Formally we define µ -realisability as follows: Definition 5 ( µ -Realisability). We define two relations (cid:13) O and (cid:13) µ by mutualrecursion. Let u ∈ ω , p be a program that uses an oracle, and ϕ be a sentencein the language of arithmetic. We define: andomising Realisability 5 ( p, u ) , (cid:13) O ⊥ ,2. ( p, u ) (cid:13) O n = m iff n = m ,3. ( p, u ) (cid:13) O ϕ ∧ ψ iff ( p u (0) , u ) (cid:13) O ϕ and ( p u (1) , u ) (cid:13) O ψ ,4. ( p, u ) (cid:13) O ϕ ∨ ψ iff we have p u (0) = 0 and ( p u (1) , u ) (cid:13) O ϕ or p u (0) = 1 and ( p u (1) , u ) (cid:13) O ψ ,5. ( p, u ) (cid:13) O ϕ → ψ iff for all s such that s (cid:13) µ ϕ , we have that p u ( s ) (cid:13) µ ψ ,6. ( p, u ) (cid:13) O ∃ x ϕ iff ( p u (0) , u ) (cid:13) O ϕ ( p u (1)) ,7. ( p, u ) (cid:13) O ∀ x ϕ iff for all n ∈ ω we have ( p u ( n ) , u ) (cid:13) O ϕ ( n ) ,For every program p that uses an oracle and every sentence ϕ in the languageof arithmetic, we will denote by C p,ϕ the set: { u ∈ ω ; ( p, u ) (cid:13) O ϕ } . Let ϕ be asentence in the language of arithmetic, r be a positive real number, and p be anatural number. We define p (cid:13) µ ϕ ≥ r as follows: p (cid:13) µ ϕ ≥ r iff µ ( C p,ϕ ) ≥ r. In this case we will say that p randomly realises (or µ -realises) ϕ with proba-bility at least r . We will say that ϕ is randomly realisable (or µ -realisable) withprobability at least r if and only if there is p such that p (cid:13) µ ϕ ≥ r . Moreover,we write p (cid:13) µ ϕ and say that p randomly realises (or µ -realises) ϕ if and onlyif p (cid:13) µ ϕ ≥ r for some r > . Finally, we will say that ϕ is randomly realisable(or µ -realisable) if sup { µ ( C p,ϕ ) ; p (cid:13) µ ϕ } = 1 . Why is it not possible to give a simpler definition of (cid:13) µ ? A natural at-tempt would be the following: p (cid:13) µ ϕ ≥ r ⇔ µ ( { u ; ( p, u ) (cid:13) Or ϕ } ) ≥ r where (cid:13) Or denotes oracle-realisability, obtained by replacing computability with com-putability relative to a fixed oracle in Kleene realisability. Unfortunately, it turnsout that this relation is not closed under modus ponens and the ∀ -GEN rule ofpredicate logic. Another natural approach is the one of §
6. We start our studyof µ -realisability by showing that the set of µ -realised sentences of arithmetic isconsistent. Lemma 6.
Let ϕ a sentence in the language of arithmetic. Then ϕ is µ -realisediff ¬ ϕ is not µ -realised.Proof. Assume that both p (cid:13) µ ϕ and q (cid:13) µ ¬ ϕ . Then for all u ∈ C q, ¬ ϕ we havethat q u ( p ) would be a realiser of ⊥ . But this is a contradiction.The following lemma has a crucial role in the theory of µ -realisability. Lemma 7 (Push Up Lemma).
Let ϕ be a sentence in the language of firstorder arithmetic and < r ≤ r ′ < be positive real numbers. Then ϕ is randomlyrealisable with probability at least r if and only if ϕ is randomly realisable withprobability at least r ′ .Proof. The right-to-left direction is trivial. For the left-to-right direction, let ϕ berandomly realisable with probability at least r . We will show that ϕ is randomlywith probability at least r ′ . Let p be a program such that µ ( C p,ϕ ) ≥ r >
0. Bythe Lebesgue Density Theorem [3, Exercise 17.9] there are u ∈ ω and n ∈ ω such that µ ( C p,ϕ ∩ N u ↾ n ) µ (N u ↾ n ) > r ′ . Now, let p ′ be the program that given an oracleruns p with oracle ( u ↾ n ) ◦ u . Note that µ ( C p ′ ,ϕ ) = µ ( C p,ϕ ∩ N u ↾ n ) µ (N u ↾ n ) > r ′ . Finally, Merlin Carl, Lorenzo Galeotti, and Robert Passmann it follows trivially by the definition that p ′ randomly realisable with probabilityat least r ′ as desired.By the Push Up Lemma, we can simplify our definition of µ -realisability. Corollary 8.
A sentence ϕ in the language of arithmetic is µ -realisable if andonly if there are < r ∈ R and p such that that µ -realises ϕ with probability atleast r . We conclude this section by proving some basic interactions between µ -realisability and the logical operators. Lemma 9.
For all programs p and sentences ϕ the set C p,ϕ is Borel. In partic-ular C p,ϕ is measurable.Proof. The proof is an induction on the complexity of ϕ . All the cases exceptimplication follow directly from the closure properties of the pointclass of Borelsets, see, e.g., [5, Theorem 1C.2]. Let us just prove the implication case. Let ϕ ≡ ψ → ψ and p be a program. For every program s let A s be 2 ω if s (cid:13) µ ψ and C p ( s ) ,ψ , otherwise. Then C p,ϕ = T s ∈ N A s . By inductive hypothesis we havethat A s is Borel for every s so C p,ϕ is a countable intersection of Borel sets,which is Borel. Corollary 10.
Let ψ and ψ be sentences and let ϕ be a formula. Then forevery p the following hold:1. p (cid:13) µ ψ ∧ ψ if and only if there are s and q such that s (cid:13) µ ψ and q (cid:13) µ ψ .2. p (cid:13) µ ψ ∨ ψ if and only if there is q such that q (cid:13) µ ψ or q (cid:13) µ ψ .3. If p (cid:13) µ ψ → ψ then p ( s ) (cid:13) µ ψ for all s such that s (cid:13) µ ψ .4. If p (cid:13) µ ∃ xϕ then there there is n ∈ N such that p (0) (cid:13) µ ϕ ( n ) .5. If p (cid:13) µ ∀ xϕ then for all n ∈ N we have p ( n ) (cid:13) µ ϕ ( n ) .Proof. Note that an case-by-case proof shows that for every n ∈ N , u ∈ ω , andformula ϕ we have that:( p ( n ) , u ) (cid:13) O ϕ iff ( p u ( n ) , u ) (cid:13) O ϕ. (1) First assume that p (cid:13) µ ψ ∧ ψ ≥ r . Then we have that µ ( C p,ψ ∧ ψ ) ≥ r and for all u ∈ C p,ψ ∧ ψ we have that ( p, u ) (cid:13) O ψ ∧ ψ . But then for all u ∈ C p,ψ ∧ ψ we have that ( p u (0) , u ) (cid:13) O ψ and ( p u (0) , u ) (cid:13) O ψ . So for every u ∈ C p,ψ ∧ ψ we have that ( p (0) , u ) (cid:13) O ψ and ( p (1) , u ) (cid:13) O ψ but then p (0) (cid:13) µ ψ ≥ r and p (1) (cid:13) µ ψ ≥ r . Let s = p (0) and q be p (1).Now assume that q (cid:13) µ ψ and s (cid:13) µ ψ . Therefore for all u ∈ C q,ψ and v ∈ C s,ψ we have that ( q, u ) (cid:13) O ψ and ( s, u ) (cid:13) O ψ . By Lemma 7 we canassume that s and q are such that µ ( C q,ψ ∩ C s,ψ ) > r . But then if we let p be the program that returns q on input 0 and s on input 1, we have that for all u ∈ C q,ψ ∩ C s,ψ we have that ( p (0) , u ) (cid:13) O ψ and ( p (1) , u ) (cid:13) O ψ . Therefore p (cid:13) µ ψ ∧ ψ as desired. andomising Realisability 7 (2) First assume that p (cid:13) µ ψ ∨ ψ . Then we have that µ ( C p,ψ ∨ ψ ) > u ∈ C p,ψ ∨ ψ we have that ( p, u ) (cid:13) O ψ ∨ ψ . But then for all u ∈ C p,ψ ∨ ψ we have that ( p u (1) , u ) (cid:13) O ψ or ( p u (1) , u ) (cid:13) O ψ . Moreover,note that by Lemma 9 we have that both C p (1) ,ψ and C p (1) ,ψ are measurablesets and since C p,ψ ∨ ψ ⊆ C p (1) ,ψ ∪ C p (1) ,ψ at least one of them is not null.Without loss of generality assume that µ ( C p (1) ,ψ ) >
0. So, ( p (1) , u ) (cid:13) O ψ forall u ∈ C p (1) ,ψ . Let q be the program that for every oracle returns p u (1) oninput 1 and 0 on input 0. Then trivially q (cid:13) µ ψ as desired.Now assume that q (cid:13) µ ψ , the same proof works in the case in which q (cid:13) µ ψ .Therefore for all u ∈ C q,ψ we have that ( q, u ) (cid:13) O ψ . Then if we let p be theprogram that returns q on input 1 and 0 on input 0, we have that for all u ∈ C q,ψ ( p (1) , u ) (cid:13) O ψ . Therefore p (cid:13) µ ψ ∨ ψ as desired.(3) Assume that p (cid:13) µ ψ → ψ . Then we have that µ ( C p,ψ → ψ ) > u ∈ C p,ψ → ψ we have that ( p, u ) (cid:13) O ψ → ψ . But then for all u ∈ C p,ψ → ψ and for every s such that s (cid:13) µ ψ we have p u ( s ) (cid:13) µ ψ . But for all s suchthat s (cid:13) µ ψ we have that ( p u ( s ) , u ) (cid:13) O ψ and therefore ( p ( s ) , u ) (cid:13) O ψ . So, p ( s ) (cid:13) µ ψ for all s such that s (cid:13) µ ψ as desired.(4) Assume that p (cid:13) µ ∃ xψ . Then we have that µ ( C p, ∃ xψ ) > u ∈ C p,ψ ∨ ψ we have that ( p, u ) (cid:13) O ∃ xψ . But then for all u ∈ C p, ∃ xψ we have that( p u (0) , u ) (cid:13) O ψ ( p u (1)). Moreover, note that by Lemma 9 we have that for every n ∈ N the set C p (0) ,ψ ( n ) is measurable and since C p, ∃ xψ ⊆ S n ∈ N C p (0) ,ψ ( n ) thereis n ∈ N such that µ ( C p (0) ,ψ ( n ) ) >
0. So, ( p (0) , u ) (cid:13) O ψ ( n ) for all u ∈ C p (0) ,ψ ( n ) and therefore p (0) (cid:13) µ ψ ( n ) as desired.(4) Assume that p (cid:13) µ ∀ xψ . Then we have that µ ( C p, ∀ xψ ) > u ∈ C p, ∀ xψ we have that ( p, u ) (cid:13) O ∀ xψ . But then for all u ∈ C p, ∃ xψ and every n ∈ N we have that ( p u ( n ) , u ) (cid:13) O ψ ( n ). So, ( p ( n ) , u ) (cid:13) O ψ ( n ) and p ( n ) (cid:13) µ ψ ( n )for all n ∈ N . Corollary 11.
Let ψ and ψ be sentences and let ϕ be a formula. Then forevery p the following hold:1. 1 p (cid:13) µ ψ ∧ ψ ≥ if and only if p (0) (cid:13) µ ψ ≥ and p (1) (cid:13) µ ψ ≥ .2. If p (1) (cid:13) µ ψ ≥ or p (1) (cid:13) µ ψ ≥ then p (cid:13) µ ψ ∨ ψ ≥ .3. If p ( s ) (cid:13) µ ψ ≥ for all s such that s (cid:13) µ ψ then p (cid:13) µ ψ → ψ ≥ .4. If there there is n ∈ N such that p ( n ) (cid:13) µ ϕ ( n ) ≥ then p (cid:13) µ ∃ xϕ ≥ .5. If all n ∈ N we have p ( n ) (cid:13) µ ϕ ( n ) ≥ then p (cid:13) µ ∀ xϕ ≥ .Proof. The proof is an easy modification of the proof of Corollary 10.
In this section, we will study the relationship between classical and randomrealisability. In particular we will show that the two notion do not coincide.We start by proving that classical realisability and µ -realisability agree onpretty Σ sentences and that therefore µ -realisability for pretty Σ sentencescoincides with truth. Merlin Carl, Lorenzo Galeotti, and Robert Passmann
Theorem 12.
Let ϕ be a pretty Σ sentence in the language of arithmetic.Then, there are two computable functions P µ and P − µ such that for every p ,(i) p (cid:13) ϕ implies P µ ( p, ϕ ) (cid:13) µ ϕ ≥ , and (ii) p (cid:13) µ ϕ implies P − µ ( p, ϕ ) (cid:13) ϕ .Therefore a pretty Σ formula is true if and only if it is µ -realised. The sameresult holds for universal Π sentences.Proof. We define P µ and P − µ by recursion on ϕ and will prove that they havethe desired properties. We first define P µ and P − µ on ∆ formulas.(1) ϕ is atomic: in this case realisability and µ -realisability have the samerealisers. So we can just let P µ and P − µ be the identity on atomic formulas.(2) If ϕ ≡ ψ ∧ ψ where ψ and ψ are ∆ .First assume that p (cid:13) µ ψ ∧ ψ . Then, µ ( C p,ϕ ) > u ∈ C p,ϕ we have ( p u (0) , u ) (cid:13) O ψ and ( p u (1) , u ) (cid:13) O ψ . Let p ( i ) be the program thatfor every oracle u just returns p u ( i ) for i ∈ { , } . Then for i ∈ { , } we have p ( i ) (cid:13) µ ψ i . Let P − µ ( p, ϕ ) be the program that given input i computes g ( p ( i ) , ψ i ).By inductive hypothesis we have that P − µ ( p ( i ) , ψ i ) (cid:13) ψ i for every i ∈ { , } andtherefore P − µ ( p, ϕ ) realises ϕ as desired.On the other hand let p (cid:13) ψ ∧ ψ . Then p ( i ) (cid:13) ψ i for every i ∈ { , } . Let P µ ( p, ϕ ) be the program that ignores the oracle and for every i ∈ { , } returns P µ ( p ( i ) , ψ i ). By inductive hypothesis we have that P µ ( p ( i ) , ψ i ) (cid:13) µ ψ i ≥
1. Notethat µ ( C P µ ( p (0) ,ψ ) ,ψ ∩ C P µ ( p (1) ,ψ ) ,ψ ) = 1 and that for all u ∈ C P µ ( p (0) ,ψ ) ,ψ ∩ C P µ ( p (1) ,ψ ) ,ψ we have that ( P µ ( p (0) , ψ ) , u ) (cid:13) O ψ and ( P µ ( p (1) , ψ ) , u ) (cid:13) O ψ .But then, since P µ ( p, ϕ ) u ( i ) = P µ ( p ( i ) , ψ i ) for every oracle u and every i ∈ { , } ,we have that P µ ( p, ϕ ) (cid:13) µ ψ ∧ ψ ≥ ϕ ≡ ψ ∨ ψ where ψ and ψ are ∆ .First assume that p (cid:13) µ ψ ∨ ψ . Then, µ ( C p,ϕ ) > u ∈ C p,ϕ we have that ( p u (1) , u ) (cid:13) O ψ p u (0) . Let p (1) be the program that for every oracle u returns p u (1). Then there is i ∈ { , } such that p (1) (cid:13) µ ψ i by the proof ofCorollary 10. Let P − µ ( p, ϕ ) be the program that for does the following: starts byrunning in parallel two instances of the program of Corollary 4, one with input ψ and one with input ψ . By inductive hypothesis note that at least one of thetwo instances will halt. Let i ∈ { , } be such that the ψ i instance halted first.Then, if the input is 0, the program halts with output P − µ ( p (0) , ψ i ), and if theinput is 1 the program returns i .Now let p (cid:13) ψ ∨ ψ where ψ and ψ are ∆ . Then p (1) (cid:13) ψ p (0) . Let P µ ( p, ϕ )be the program that if the input is 0 halts with output P µ ( p (1) , ψ p (0) ), and ifthe input is 1 the program returns p (0).By inductive hypothesis we have that P µ ( p (1) , ψ p (0) ) (cid:13) µ ψ p (0) ≥
1. But then,by the proof of Corollary 11 since for every u , P µ ( p, ϕ ) u (1) = P µ ( p (1) , ψ p (0) ),and P µ ( p, ϕ )(0) = p (0), we have P µ ( p, ϕ ) (cid:13) µ ψ ∨ ψ ≥ ϕ ≡ ψ → ψ where ψ and ψ are ∆ . First assume that p (cid:13) µ ψ → ψ .Let P − µ ( p, ϕ ) be the program that does the following: for every input returns andomising Realisability 9 the code of the instance of the program in Corollary 4 with input ψ . Note that,if ψ is realisable, then by inductive hypothesis is µ -realisable, by assumptions ψ is µ -realisable and by inductive hypothesis ψ is realisable. In this case forevery s we have that P − µ ( p, ϕ )( s ) (cid:13) ψ and therefore P − µ ( p, ϕ ) is a realiser of ψ . On the other hand, if ψ is not realisable, then any natural number realises ϕ , so P − µ ( p, ϕ ) is again a realiser of ϕ .Now let p (cid:13) ψ → ψ where ψ and ψ are ∆ . Then for every realiser s of ψ we have that p ( s ) (cid:13) ψ . Let P µ ( p, ϕ ) be the code of the program thatfor every input s and every oracle returns P µ ( p ( P − µ ( s, ψ )) , ψ ). By inductivehypothesis if s is a µ -realiser of ψ , so P − µ ( s, ψ ) is a realiser of s . By assumption p ( P − µ ( s, ψ )) is a realiser of ψ , and again by inductive hypothesis we havethat P µ ( p ( P − µ ( s, ψ ) , ψ ) (cid:13) µ ψ ≥
1. But then by Corollary 11 we have that P µ ( p, ϕ ) (cid:13) µ ϕ ≥ Σ formulas.(6) If ϕ ≡ ∃ xψ where ψ is pretty Σ . First assume that p (cid:13) µ ∃ xψ . Then,by Corollary 10 there must be n ∈ N such that p (0) (cid:13) µ ψ ( n ) therefore, byinductive hypothesis, ψ ( n ) is realised. Let P − µ ( p, ϕ ) be the program that doesthe following: run in parallel all the instances of the program of Corollary 4 withinput ψ ( n ) with n ∈ N . By inductive hypothesis note that one of these instancesmust halt. Let i ∈ N be the least such that the ψ ( i ) instance halts. Then, ifthe input is 0, the program returns P − µ ( p (0) , ψ ( i )) and if it is 1, the programreturns i .Note that, by inductive hypothesis, the program halts and returns a realiserof ϕ , as desired.Now assume that p (cid:13) ∃ xψ . Then p (0) (cid:13) ψ ( p (1)). Let f ( p, ϕ ) be the programthat returns P µ ( p (0) , ψ ( p (1))) if the input is 0 and p (1) if the input is 1. Byinductive hypothesis P µ ( p (0) , ψ ( p (1))) (cid:13) µ ψ ( p (1)) ≥
1. But then by Corollary 11since for all u we have P µ ( p, ϕ ) u (0) = f ( p (0) , ψ ( p (1))) and P µ ( p, ϕ ) u (0) = p (1),we have that P µ ( p, ϕ ) (cid:13) µ ∃ xψ ≥ ϕ ≡ ∀ x < nψ , where ψ is pretty Σ . First assume that p (cid:13) µ ∀ x < nψ .Then, by Corollary 10, for every natural number m < n and for every program s we have that p ( m ) (cid:13) µ ψ ( m ) and by inductive hypothesis ψ ( m ) is realised.Let P − µ ( p, ϕ ) be the program that for every m returns a program that for everyinput if m < n returns P − µ ( p ( m ) , ψ ( m )) and returns 0 otherwise. For all m < n ,by inductive hypothesis we have that P − µ ( p ( m ) , ψ ( m )) (cid:13) ψ ( m ) and therefore P − µ ( p, ϕ ) is a realiser of ϕ as desired.Now assume that p (cid:13) ∀ x < nψ . Then for all m < n and s we have that p ( m )( s ) (cid:13) ψ ( m ). Let P µ ( p, ϕ ) be the program that ignores the oracle andfor every m returns a program that given q as input, if m < n then returns P µ ( p ( m )( q ) , ψ ) otherwise returns 0. Now note that for every m < n and for every program q we have that P µ ( p ( m )( q ) , ψ ) (cid:13) µ ψ ( m ) ≥
1. But then for every m < n , every q , and every u ∈ ω we have that ( P µ ( p, ϕ ) u ( m )) u ( q ) (cid:13) µ ψ ≥ P µ ( p, ϕ ) (cid:13) µ ϕ ≥ ∆ case to universal Π formulas.(8) If ϕ ≡ ∀ xψ where ψ is universal Π . First assume that p (cid:13) µ ∀ xψ . Let P − µ ( p, ϕ ) be the program that for all n runs P µ ( P − µ ( p ( n ) , ψ ( n ))). By Corollary10 and the inductive hypothesis P − µ ( p ( n ) , ψ ( n )) is a realiser of ψ ( n ). Therefore P − µ ( p, ϕ ) is a realiser of ∀ xψ as desired.Now assume that p (cid:13) ∀ xψ . Let P µ ( p, ϕ ) be the program that for every n andfor every oracle returns P µ ( p ( n ) , ψ ( n )). Once more by inductive hypothesis forall n and all µ ( C P µ ( p ( n ) ,ψ ( n )) ,ψ ( n ) ) = 1 but then µ ( T n ∈ N C P µ ( p ( n ) ,ψ ( n )) ,ψ ( n ) ) = 1and P µ ( p, ϕ ) (cid:13) µ ϕ ≥ Corollary 13.
Let ϕ be any false pretty Σ sentence in the language of arith-metic. Then ( p, u ) (cid:13) O ϕ → ⊥ and p (cid:13) µ ϕ → ⊥ ≥ for every p and u . The sameholds for universal Π formulas.Proof. By Theorem 12 every µ -realisable pretty Σ (universal Π ) sentence ϕ istrue. Therefore ϕ cannot be µ -realised and every program is going to µ -realise ϕ → ⊥ , which means that for all p and for all u we have p (cid:13) µ ϕ → ⊥ and( p, u ) (cid:13) O ϕ → ⊥ as desired.We are now ready to prove the main result of this section, namely that µ -realisability and classical realisability do no coincide. This result is surprisinggiven that by Sacks’s theorem [1, Corollary 8.12.2] functions that are computablewith a non-null set of oracles are computable by a classical Turing machine. Theorem 14.
There is a sentence ϕ in the language of arithmetic that is ran-domly realisable but not realisable.Proof. Let ϕ be the sentence “For all k there is n such that for all ℓ the executionof p k ( k ) does not stop in at most ℓ steps or p k ( k ) = n ” and let ψ ( k ) be thesentence “There is n such that for all ℓ the execution of p k ( k ) does not stop inat most ℓ steps or p k ( k ) = n ”.A classical realiser for ϕ would be a program that computes a total functionthat, for every code k of a program, returns a natural number which is notthe output of p k ( k ). By diagonalization, such a program cannot exists: If p k was such a program, then it would follow that for every n ∈ ω we have that p k ( k ) = n ⇔ p k ( k ) = n .Now we want to show that ϕ is randomly realisable.Fix any realiser s . Let p be the program that given an oracle u ∈ ω , a naturalnumber k , and i ∈ { , } does the following : Let p u ( k )( i ) = u ↾ ( k + 1) if i = 1 Here, we do not distinguish between the finite sequence u ↾ ( k + 1) and the naturalnumber coding it.andomising Realisability 11 and p u ( k )( i ) = p ′ if i = 0 where p ′ is the program that ignores the oracle anddoes the following:On input ℓ , p ′ checks whether p k ( k ) stops in ℓ steps. If not, then p ′ ( ℓ ) is thecode of a program that returns 0 on input 0 and s on. input 1. Otherwise p ′ ( ℓ )is the code of a program that returns 1 on input 0 and on input 1 looks for an µ -realiser of the ∆ formula expressing the fact that “ p k ( k ) = u ↾ ( k + 1)” byrunning the algorithms in Lemma 3 and Theorem 12.Now, for every k ∈ ω and u ∈ ω we have two cases: p k ( k ) does not halt: then we have that p u ( k )(1) = u ↾ ( k +1) and p u ( k )(0) = p ′ .Since p k ( k ) does not halt, we have p ′ ( ℓ )(0) = 1 and p ′ ( ℓ )(1) = s for every ℓ .Moreover, by Corollary 13 ( s, u ) (cid:13) O “ p k ( k ) does not halt in ℓ steps” and there-fore ( p u ( k ) , u ) (cid:13) O ψ ( k ). p k ( k ) halts: then we have that p u ( k )(1) = u ↾ ( k + 1) and p u ( k )(0) = p ′ . Let ℓ be such that p k ( k ) halts in at most ℓ steps. Then, p ′ ( ℓ )(0) = 0. Moreover, notethat if the output of p k ( k ) is not the same as the first k bits of the oracle then( p ′ ( ℓ ) u (1) , u ) (cid:13) O u ↾ ( k + 1) = p k ( k ).We only need to show that µ ( C p,ϕ ) >
0. To see this, it is enough to note thatthe set of u such that p k ( k ) = u ↾ ( k + 1) has measure ≥ − ( k +1) . Therefore, µ ( C p,ϕ ) = Q k ∈ N (1 − ( k +1) ) > Corollary 15.
There is a sentence in the language of arithmetic which is real-isable but not randomly realisable.Proof (Corollary 15).
It is enough to consider the sentence ϕ → ⊥ where ϕ isthe sentence in the proof of Theorem 14. The sentence is trivially realised since ϕ is not realised. Moreover, the sentence is not µ -realised since ϕ is µ -realisedand ⊥ is not µ -realised. In this section, we study the logic and arithmetic of µ -realisability. We firstobserve that, in a certain sense, the Law of Excluded Middle is not µ -realisable. Lemma 16.
There is ϕ such that ∀ x ( ϕ ( x ) ∨ ¬ ϕ ( x )) is not µ -realisable.Proof. Let ϕ ( x ) be the formula expressing the fact that the program p x ( x ) halts.Assume that ∀ x ( ϕ ( x ) ∨ ¬ ϕ ( x )) is randomly realised. Then, there is a program p such that p (cid:13) µ ∀ x ( ϕ ( x ) ∨ ¬ ϕ ( x )). Therefore, p computes the halting problemfor a set of oracles of measure >
0. But this directly contradicts Sacks’ theorem[1, Corollary 8.12.2].We now show that µ -realisability is preserved by the inference rules of first-order intuitionistic proof calculus.First, we need to fix what it means for ϕ to be µ -realizable when x occursfreely in ϕ : This is defined to mean the same as the µ -realisability of ∀ xϕ . Definition 17 (Intuitionistic Calculus).
Inference rules are:
MP : from ϕ and ϕ → ψ infer ψ ∀ − GEN : from ψ → ϕ infer ψ → ( ∀ x ϕ ) , if x is not free in ψ. ∃ − GEN : from ϕ → ψ infer ( ∃ x ϕ ) → ψ , if x is not free in ψ. The axioms are
THEN − ϕ → ( χ → ϕ )THEN − ϕ → ( χ → ψ )) → (( ϕ → χ ) → ( ϕ → ψ ))AND − ϕ ∧ χ → ϕ AND − ϕ ∧ χ → χ AND − ϕ → ( χ → ( ϕ ∧ χ ))OR − ϕ → ϕ ∨ χ OR − χ → ϕ ∨ χ OR − ϕ → ψ ) → (( χ → ψ ) → (( ϕ ∨ χ ) → ψ ))FALSE : ⊥ → ϕ PRED − ∀ x ϕ ( x )) → ϕ ( t ) , if the term t is free for substitutionfor the variable x in ϕ PRED − ϕ ( t ) → ( ∃ x ϕ ( x )) , with the same restriction as for PRED − .Proof (Theorem 18). We show that (i) all instantiations of the axioms of in-tuitionistic first-order calculus are µ -realisable and (ii) the set of µ -realizablestatements is closed under modus ponens, ∀ -GEN and ∃ -GEN.We start with (i).THEN-1: A µ -realiser for an instance of ϕ → ( χ → ϕ ) needs to turn anygiven µ -realiser r for ϕ into one for χ → ϕ . The µ -realiser for χ → ϕ works bysimply returning r for any input.THEN-2: Here, we are given a µ -realiser r for ϕ → ( χ → ψ ) and our goal isto turn any µ -realiser p for ( ϕ → χ ) into a µ -realiser q for ϕ → ψ . Given r and p , q works as follows: Given a µ -realiser s for ϕ , first use r to compute from s a µ -realiser t for χ → ψ with positive probability; moreover, use p to computefrom s a µ -realiser for u χ with positive probability. Then apply t to u .AND-1 works by projecting the µ -realiser for ϕ ∧ χ to the first component,AND-2 by projecting to the second component.AND-3: We need to turn any µ -realiser p for ϕ into a µ -realiser q for χ → ( ϕ ∧ χ )) with positive probability. Let p be given. Also, let a µ -realiser r for χ begiven. Now, q works as follows: For a given oracle x , let x = x ⊕ x , where, forreal numbers a and b , a ⊕ b denotes the join of a and b , i.e., 2 i ∈ a ⊕ b iff i ∈ a and 2 i + 1 ∈ a ⊕ b iff i ∈ b . Now ( q x (0) , x ) runs ( p x (0) , x ) while ( q x (1)(0) , x )runs ( r x , x ).OR-1 works by, given a µ -realiser r for ϕ , sending 0 to 0 and 1 to r , OR-2by sending 0 to 1 and 1 to r . andomising Realisability 13 OR-3: We need to turn any µ -realiser p for ϕ → ψ into one for (( χ → ψ ) → (( ϕ ∨ χ ) → ψ )) with positive probability. Let q be a µ -realiser for χ → ψ , andlet r be a µ -realiser for ϕ ∨ χ . Now, the sets S , S of oracles relative to which r realizes ϕ or χ , respectively, are measurable, and as their union has positivemeasure, at least one of the sets S and S has positive measure. Thus, for apositive measure set S of oracles u , r u (0) will terminate with output i ∈ { , } such that S i has positive measure, so that ( r u (1) , u ) will be an O -realiser of χ (if i = 0) or ψ (if i = 1), respectively. Let us denote by r (1) the program that,on oracle u , runs the program with index r u (1) in the oracle u . With positiveprobability, r (1) will be an O -realiser of ϕ (if i = 0) or ψ (if i = 1). Now weproceeds as follows: Given u , first compute r u (0). If this is 0, apply p to r (1).If it is 1, apply q to r (1). With positive probability, it then happens that p isapplied to a µ -realiser of ϕ or that q is applied to a µ -realiser of χ . In bothcases, we obtain a µ -realiser of ψ . Thus, we obtain a µ -realiser of ψ with positiveprobability, as desired.FALSE is µ -realized by any program, as ⊥ does not have µ -realisers.PRED-1: Here, t will just be a natural number. Let r be a µ -realiser for ∀ xϕ ( x ). Let an oracle u be given, and suppose that r works for u (i.e., ( r, u ) (cid:13) O ∀ xϕ ( x )), which happens for all u from a set of positive measure. For each such u , ( r u ( t ) , u ) will be an O -realiser for ϕ ( t ) by definition. Thus, the program r ( t )that, for given u , runs the program with index r u ( t ) in the oracle u is a µ -realiserfor ϕ ( t ).PRED-2: Let r be a µ -realiser for ϕ ( t ). Then a µ -realiser p for ∃ xϕ ( x ) worksby letting p u (1) output t and letting p u (0) output r for every u .Now for (ii).(1) (MP) If ϕ and ϕ → ψ are µ -realizable, then so is ψ .Suppose that p µ -realizes ϕ and that q ϕ -realizes ϕ → ψ . Pick a real number x such that ( q, x ) realizes ϕ → ψ and run q x ( p ). By definition, the output is a µ -realiser for ψ .(2) ∀ -GENLet p be a µ -realiser for ψ → ϕ , and let q µ -realize ψ → ( ∀ xϕ ), where x does not occur freely in ψ but (possibly) in ϕ . If x does not occur freely in ψ ,the claim is trivial since then ∀ xϕ is µ -realizable if and only if ϕ is. We aregiven n ∈ ω , our goal is to compute a realiser for ϕ ( n ). Pick some oracle y suchthat ( p, y ) realizes ψ → ϕ . Note that this means that ( p, y ) computes a realiserfor ψ → ϕ ( n ) from any given n ∈ ω . Now run this realiser in the input n ; bydefinition, the output will be a µ -realiser for ϕ ( n ), as desired.(3) ∃ -GENLet p be a µ -realiser for ϕ → ψ and let q be a µ -realiser for ( ∃ xϕ ) → ψ ,where x is not free in ψ . Pick oracles y and z such that ( q, y ) realizes ( ∃ xϕ ) → ψ and ( p, z ) realizes ϕ → ψ . Thus, ( q y (0) , y ) realizes ϕ ( n ), where n is the outputof q y (1). Recall that p is a µ -realiser for ∀ x ( ϕ → ψ ). Thus, p z ( n ) is the index ofa program r that turns µ -realisers for ϕ ( n ) into µ -realisers for ψ . Consequently,running p z ( n ) on the input q (1) yields a µ -realiser for ψ . Theorem 18 (Soundness).
The set of µ -realizable statements is closed underthe rules of intuitionistic first-order calculus. It is a classical result that the axioms of Heyting Aritmetic are realisable,see [6, Theorem 1]. We show that only a fragment of HA is µ -realisability. LetHA − denote the axioms of Peano arithmetic without the induction schema. Asusual, Heyting arithmetic
HA is the theory obtained from adding the inductionschema to HA − . We say that a set of formulas Γ is µ -realised if ϕ is µ -realisedfor all ϕ ∈ Γ .Since all the axioms except for the induction schema are universal Π state-ments, it follows by Theorem 12 that the axioms of HA − are all µ -realised. Theorem 19.
The set HA − is µ -realised. Contrary to the classical case the induction schema fails for µ -realisability. Theorem 20.
The induction schema is not µ -realised.Proof. Let ϕ ( x ) be the formula expressing the fact that “Every program withcode i < x halts or does not halt”. By the proof of Lemma 16, ϕ is not µ -realisable.On the other hand, a µ -realiser p ( n ) for ϕ ( n ) is given by a program thatdoes the following: for every i < n , p returns a program that if the i th elementof the oracle is 1 returns 1 on input 0 and any number on input 1. While if the i th element of the oracle is 0 the program returns 0 on input 0 and on input 1starts building a realiser of “the program i halts” using the algorithm in Lemma4; if it finds one, it runs the algorithm in Theorem 12 to compute the desired µ -realiser.It is not hard to see that the algorithm works with probability n . Thus, torealize the implication ϕ ( n ) → ϕ ( n + 1), we can ignore the µ -realiser for ϕ ( n )and just output p ( n ). So the premise of the instance of the induction schema is µ -realised, while the conclusion is not.Note that the proof of Theorem 20 heavily relies on the fact that the definitionof µ -realisability does not require any relationship between the measures of theset of oracles realising the antecedent of an implication and the set of oraclesrealising the consequent. We think that a modification of this definition couldlead to a notion of probabilistic realisability that realises the induction schema.Even though the axiom schema of induction is not µ -realisable, one can provethat all ∆ -instances of the schema are realisable. Indeed, by Theorem 12 andthe fact that if ϕ is a ∆ formula then ∀ xϕ ( x, ¯ y ) is a universal Π formula, wehave the following: Corollary 21.
The set HA − together with the induction schema restricted to ∆ formulas is µ -realisable. andomising Realisability 15 In this section, we will consider other natural definitions of realisability arisingfrom notions of big sets of oracles on the real numbers. More specifically, wewill consider “almost sure realisability,” “comeagre realisability,” “interval-freerealisability,” and “positive measure realisability.” It will turn out, however,that the first three are equivalent to standard realisability, while the final onecoincides with truth. We begin with the following general definition.
Definition 22.
Let F be a family of subsets of Cantor space ω . We then define F -realisability recursively as follows:1. p (cid:13) F ⊥ never,2. p (cid:13) F n = m if and only if n = m ,3. p (cid:13) F ψ ∧ ψ if and only if p ( i ) (cid:13) F ψ i for i < ,4. p (cid:13) F ψ ∨ ψ if and only if there is some O ∈ F and some i < such thatfor every u ∈ O , we have p u (0) = i and p u (1) (cid:13) F ψ i ,5. p (cid:13) F ϕ → ψ if and only if there is a set O ∈ F , such that for every u ∈ O and s (cid:13) F ϕ , we have p u ( s ) (cid:13) F ψ ,6. p (cid:13) F ∃ xϕ if and only if there is some O ∈ F , such that there is some n forall u ∈ O such that p u (0) = n and p u (1) (cid:13) ϕ ( n ) ,7. p (cid:13) F ∀ xϕ if and only if there is a set O ∈ F , such that for every u ∈ O and n ∈ N we have p u ( n ) (cid:13) F ϕ ( n ) . From this definition, we derive the following notions of realisability: Let F if be the family of co-interval-free subsets of the Cantor space, i.e. X ∈ F cif if andonly if X ∈ ω and there is no open interval I such that I ⊆ ω \ X , and (cid:13) cif denotes F cif -realisability. Let C be the family of comeagre subsets of the Cantorspace, then let (cid:13) C denote C -realisability. Let F =1 be the family of subsets of theCantor space that are of measure 1, and let (cid:13) =1 denote F =1 -realisability. Let F > be the family of subsets of the Cantor space of positive measure, and (cid:13) > denotes F > -realisability. As before, we will write (cid:13) F ϕ if and only if there issome realiser p such that p (cid:13) F ϕ .In what follows we will make use of the bounded exhaustive search with p ( n ),i.e. the following procedure. Given a program p (and possibly some input n ), dothe following successively for all k ∈ ω . Enumerate all 0-1-strings of length k .For each of these strings s , do the following: Run p s ( n ) for k many steps. If thecomputation does not halt within that time (which implies in particular that atmost the first k many bits of the oracle were requested), continue with the next s (if there is one, otherwise with ( k + 1)). If the computation halts with output x within that time, then the search terminates with output x .The crucial property of this procedure, which is also contained in the proofidea of Sacks’ theorem [1, Corollary 8.12.2], is the following: Lemma 23.
Let G ⊆ ω , n ∈ ω and let p be a program. Suppose that there is aset S ⊆ ω such that ω \ S is interval-free and p u ( n ) terminates for all u ∈ S with output k ∈ G . Then the bounded exhaustive search with p ( n ) will terminatewith output k ∈ G . Proof.
Note that for every n and u ∈ S we have that p u ( n ) terminates withoutput in G . So there is a finite initial segment s of u such that p s ( n ) terminateswith output p u ( n ). So, the bounded exhaustive search will halt.Now, note that if the search halts on the string s with output k ∈ ω , but k / ∈ G , then p x ( n ) ↓ k for all u ∈ N s . But then, N s ⊆ ω \ S which contradictsthe fact that 2 ω \ S is interval free. Lemma 24.
Let X ⊆ ω be a subset of Cantor space. If µ ( X ) = 0 or X ismeagre, then X is interval-free.Proof. The first statement follows trivially from the fact that every non-emptyopen interval has positive measure. For the second statement, recall that meagresets have empty interiour by the Baire Category Theorem (cf. [2, Theorem 0.11])and therefore contain no intervals.
Theorem 25.
Let F be a family of subsets of Cantor space such that every X ∈F is co-interval-free. There are programs P F and P − F such that the followingholds for all statements ϕ : (i) if p (cid:13) ϕ , then P F ( p, ϕ ) (cid:13) F ϕ , (ii) if p (cid:13) F ϕ , then P − F ( p, ϕ ) (cid:13) ϕ . Consequently, ϕ is realisable if and only if it is F -realisable, and (cid:13) , (cid:13) cif , (cid:13) C , and (cid:13) =1 coincide.Proof. We show both statements by simultaneous induction on the complexityof ϕ and simultaneously define P F and P − F by recursion on ϕ .(1) ϕ is t = t or t = t . In this case, F -realisers and realisers are the same,so the statement is trivial: P F and P − F just return the first component.(2) ϕ is ψ ∧ ψ .Let r = ( r , r ) be a realiser for ϕ such that r i realises ψ i for i <
2. Byinduction hypothesis, P F ( r i , ψ i ) will return an F -realiser for ψ i . Hence, P F ( r, ϕ )is the program that outputs P F ( r i , ψ i ) on input i . We obtain P − F in exactly thesame way.(3) ϕ is ψ ∨ ψ .Let r be a realiser for ϕ , i.e. r (0) returns some i < r (1) (cid:13) ψ i . Byinduction hypothesis, we have that P F ( r (1) , ψ i ) (cid:13) F ψ i . Hence, P F ( r, ϕ ) is theprogram that returns i on input 0 and P F ( r (1) , ψ i ) on input 1.Conversely, let r be an F -realiser for ϕ . Then there are some i < O ∈ F such that for all u ∈ O , r u (0) = i and r u (1) (cid:13) F ψ i . Hence, let P − F ( r, ϕ )be the program that executes a bounded exhaustive search with r u (0), whichterminates by Lemma 23 in some i <
2, and then returns i on input 0, and P − F ( r, ψ i ) on input 1. Then P − F ( r ) (cid:13) ϕ .(4) ϕ is ψ → ψ .Let r (cid:13) ϕ . Then r is a program that, given a realiser r (cid:13) ψ , returns arealiser r (cid:13) ψ . Let r ′ (cid:13) F ψ . By induction hypothesis, P − F ( r ′ , ψ ) (cid:13) ψ .Hence, r ( P − F ( r ′ , ψ )) (cid:13) ψ and P F ( r ( P − F ( r ′ , ψ )) , ψ ) (cid:13) F ψ . Therefore, let andomising Realisability 17 P F ( r, ϕ ) be the program that takes a realiser r ′ (cid:13) ψ as input and returns P F ( r ( P − F ( r ′ , ψ )) , ψ ).The proof for the other direction is symmetric by exchanging the roles of P F and P − F .(5) ϕ is ∃ xψ ( x ).Let r (cid:13) ∃ xψ ( x ). Then r (0) = n and r (1) (cid:13) ψ ( n ). By induction hypothesis, itfollows that P F ( r (1) , ψ ) (cid:13) F ψ ( n ). So let P F ( r, ϕ ) be the program that output n on input 0, and P F ( r (1) , ψ ) on input 1. Then, P F ( r, ϕ ) (cid:13) F ϕ .Conversely, let r (cid:13) F ∃ xψ ( x ). Then there is some O ∈ F and n ∈ ω such that r u (0) = n and p u (1) (cid:13) F ψ ( n ). By induction hypothesis, P − F ( p u (1) , ψ ) (cid:13) ψ ( n ).Define P − F ( r, ϕ ) to be the following program: First, start a bounded exhaustivesearch with r (0). By Lemma 23 this search must terminate with output n . Return n on input 0, and return P − F ( r u (1) , ψ ) on input 1. Then P − F ( r, ϕ ) (cid:13) ∃ xψ ( x ).(6) ϕ is ∀ xψ ( x ).Let r (cid:13) ∀ xψ ( x ). Then r ( n ) (cid:13) ψ ( n ) for every n ∈ ω . Let P F ( r, ϕ ) be theprogram that, given n ∈ ω , returns P F ( r ( n ) , ψ ). With the induction hypothesis,it follows that P F ( r, ϕ ) (cid:13) F ϕ .Conversely, let r (cid:13) F ∃ xψ ( x ). Then there is some O ∈ F such that for every u ∈ O and n ∈ N we have that r u ( n ) (cid:13) ψ ( n ). Define P − F ( r, ϕ ) to be thefollowing program: Start a bounded exhaustive search with r ( n ). By Lemma 23,this search will terminate with r ′ (cid:13) F ψ ( n ). Then return P − F ( r ′ , ψ ), which, byinduction hypothesis, is a realiser of ψ ( n ). Hence, P − F ( r, ϕ ) (cid:13) ψ . Theorem 26.
Let ϕ be a formula. Then (cid:13) > ϕ if and only if ϕ is true.Proof. The proof is an induction on the complexity of ϕ .(1) If ϕ is atomic the statement follows by the definitions.(2) Assume that ϕ ≡ ψ ∧ ψ .If ϕ is true then by inductive hypothesis there are p and q such that p F > -realises ψ and q F > -realises q . Let s be a sequence which starts with a code of p followed by a marker and by a code for q followed by a second marker. Then let t be the program that on input 0 returns the content of the oracle up to the firstmarker and on input 1 returns the content of the oracle between the first andthe second marker. Note that for all u ∈ N s , r u (0) (cid:13) > ψ and r u (1) (cid:13) > ψ .So, r (cid:13) > ϕ as desired.On the other hand if ϕ is F > -realised then by definition both ψ and ψ are F > -realised and the statement follows by the inductive hypothesis.(3) Assume that ϕ ≡ ψ ∨ ψ .If ϕ is true then by inductive hypothesis there is p such that p u (1) (cid:13) > ψ p u (0) for every u in some positive measure set O . Let s be a sequence which starts with p u (0) followed by a code for p u (1) followed by a marker. Then let q be theprogram that on input 0 returns the content of the first bit of the oracle andon input 1 returns the content of the oracle from the second bit to the marker.Note that for all u ∈ N s , q u (1) (cid:13) > ψ q u (0) . So, q (cid:13) > ϕ as desired.On the other hand if ϕ is F > -realised then by definition at least one between ψ and ψ is F > -realised and the statement follows by the inductive hypothesis.(4) Assume that ϕ ≡ ψ → ψ .Assume that ϕ is true. Then either ψ is true or ψ is false. If ψ is false thenby inductive hypothesis is not F > -realised and therefore any natural numberwill F > -realise ϕ . If ψ is true, then by inductive hypothesis is F > -realised bysome program p . Let s be the sequence that starts with a code of p followed by amarker. Let q the program that for every n and every oracle returns the contentof the oracle up to the first occurrence of the marker. Then for all u ∈ N s andfor every n we have that q u ( n ) (cid:13) > ψ . So, q (cid:13) > ϕ as desired.On the other hand if ϕ is F > -realised by some program p . If ψ is true thenit is F > -realised by some program q . Then there is a non-null set O such thatfor all u ∈ O we have that p u ( q ) (cid:13) > ψ . But then by inductive hypothesis ψ must be true.(5) Assume that ϕ ≡ ∃ xψ .Assume that ϕ is true. Then for some n ∈ N we have that ψ ( n ) is true.By inductive hypothesis there is p which F > -realises ψ ( n ). Let s be a sequencestarting with a code for n followed by a marker and then by the code of p followedby a marker. Let q be the program that on input 0 returns the content of theoracle up to the first marker, and on input 1 returns the content of the oraclebetween the first and second marker. Then for all u ∈ N s and for every n wehave that q u (1) (cid:13) > ψ ( q u (0)). So, q (cid:13) > ϕ as desired.On the other hand if ϕ is F > -realised by some program p . Then there is anon-null set O such that for all u ∈ O we have that p u (1) (cid:13) > ψ ( p u (0)). Butthen by inductive hypothesis ψ must be true.(6) Assume that ϕ ≡ ∀ xψ .Assume that ϕ is true. Then for all n ∈ N we have that ψ ( n ) is true. Withoutloss of generality we can assume that the main operator of ψ is not a universalquantifier, the proof can be easily modified otherwise. Let q be the program thatignores the oracle and depending on the main connective of ψ does the following: – if ψ is atomic q is just the constant function 1; – if ψ is ψ ∧ ψ then q ( n ) is the program r from the proof of case (2); – if ψ is ψ ∨ ψ then q ( n ) is the program q from the proof of case (3); – if ψ is ψ → ψ then q ( n ) is the program q from the proof of case (4); – if ψ is ∃ xψ then q ( n ) is the program q from the proof of case (5); andomising Realisability 19 By inductive hypothesis and by (2), (3), (4), and (5) of this proof we have thatfor all u ∈ ω and for every n we have that q u ( n ) (cid:13) > ψ ( n ). So, q (cid:13) > ϕ asdesired.On the other hand if ϕ is F > -realised by some program p . Then there is anon-null set O such that for all u ∈ O we have that p u ( n ) (cid:13) > ψ ( n ). But thenby inductive hypothesis ϕ must be true. References
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