Variations on a complex theme

Variations on a complex theme

Newton C. A. da Costa Francisco Antonio Doria

Variations on a complex theme

Klam & LabFuzzy 2015

c N. C. A. da Costa and F. A. Doria, March 2015.

Working notes; not to be quoted. This is a fully revised version of the Barebones set of notes, which were circulated by the authors among friends from 1998 until 2004. (The last version had the indexing number 16.5.) We stress: this is a set of working notes, with incomplete and underdeveloped ideas, sketchy discussions and proofs. We throw everything in here, with the hope that somehow the proverbial golden nugget will be found hidden within our messy thoughts. . . March 18, 2015.

N. C. A. da Costa and F. A. Doria are researchers at the Advanced Research Group at the Production Engineering Program, COPPE – UFRJ , and members of the Brazilian Academy of Philosophy.

on the Arnol’d Problems in 1994, and decided to try our hand at computational complexity issues. We surely knew that one must always approach a difficult question in a ludic, playful way; we work on it to enjoy ourselves, like someone who is amusing himself or herself by trying to solve a messy crossword problem. And nothing more: we just wanted to experience the sheer pleasure of solving a difficult, beautiful puzzle. We had at first no idea of how to tackle a problem like the P vs. NP question; we had only a hunch, that it could be independent of some strong axiomatic system, out of the weird relativization results by Baker, Gill and Solovay in 1975. We were then surprised when in mid–1996 Kreisel sent us an email — totally out of the blue — with the sketch of a research program that, if fulfilled, could prove the desired independence. Somehow Kreisel had heard of our efforts, and decided to give us a helping hand.

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E HAD FINISHED OUR WORK

What he said: suppose that we are dealing with independence with respect to ZFC: • Kreisel first told us that if P < NP is independent of a consistent ZFC, then P < NP holds true, provided that ZFC has a model with standard arithmetic. (The proof is easy, and circulates as folklore.) • He then noticed that the counterexample function to P = NP grows in its peaks beyond any ZFC–provably total recursive function. He never suggested any specific way to prove it, and it took us quite a few months to settle the matter. (Roughly, the counterexample function lists all first instances of failures of poly machines to output a satisfying line of values for the Satisfiability problem in Boolean algebra.) As Kreisel had anticipateed, it is a monster, a Busy Beaver like function. This is the essence of Kreisel’s program. In 2000 Marcel Guillaume offered to help us. Guillaume has an efficient, bulldozer– like style that helped us a lot with our imprecisions and loose ends [27, 28]. Also Leo Gordeev guided us with many criticisms and a counterexample to a conjecture we’d formulated. This set of notes collects some published results and lots of unpublished stuff. It also presents lots of unfinished business, sketchy, vague ideas. Now, if you want to go straight to the best result we’ve got, take a look at Proposition 7.137, p. 56.

Introduction. HE KEY FACT in the present discussion arises out of Proposition 7.130. In a nutshell: the function that lists first instances of all counterexamples to P = NP grows in its peaks at least as fast as the Busy Beaver function. Let’s add some more detail to our picture. ¨ (For notation see the numbered sections.) Enumerate all Godel numbers (supposed onto ω) for Turing machines. Define a function f such that:

T

• If n codes a poly machine (a Turing machine which is time–polynomial in the length of the input, binarily coded), then f (n) = the first instance x ∈ S AT, if any, which is input to {n} and fails to output a line of truth– values that satisfies x, +1. (x is of course binarily coded.) • If n doesn’t code a poly machine, then f (n) = 0. f is the counterexample function to P = NP. It is a monstrous, noncomputable, object. As it stands, we don’t even know whether it is total or not — since if P = NP, for some n, f (n) will be undefined. Moreover, since the set of poly machines isn’t even recursively enumerable, f cannot be given by an algorithm. And finally, as can be seen from Proposition 7.130, the “peaks” in f overtake all na¨ıvely total recursive functions (the total recursive functions in the standard model for arithmetic). So, we are here in the worst of worlds: we have a function that grows at least as fast as the Busy Beaver, is intractable, and doesn’t tell us whether it is total or not. However we can change it from a lion into a pussycat (or at least we can hope to): there is a well–known set of poly machines, the BGS set [1] which is recursive, so that we can obtain a (recursive or partial recursive) f over it which adequately represents f , because (i) every machine in the BGS set is a poly machine, and (ii) given any poly machine there is a machine in BGS that computes the same function. 9

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But now f over BGS becomes too tame, and its rate of growth isn’t transparent at all. So, we cannot argue for f as we did for f , and neither can we easily see whether it is provably total in some theory or not. The middle–ground is covered by what we have called the “exotic formulation” fF (or rather, as we shall see, the manifold exotic formulations), the first one being noted as before), which allows us to at least show that fF grows faster (over its domain) than any (similarly restricted) total recursive function dominated by F, which is taken to be a strictly increasing na¨ıvely total recursive function. The problem that arises is: if now F isn’t provably total in some theory S — for S see the table in the next page — then the equivalence

[f is total] ↔ [fF is total] is independent of S ! However it holds of the standard model for arithmetic. That’s our starting point. The present set of notes can be seen as variations on f as a Leitmotiv. We notice that a good deal of what is known about this problem — like the fact that f has such a monstrous growth rate; lots of similar interesting, basic facts, circulate as unpublished folklore. Still another example of folklore– like stuff is the fact that if (the formal arithmetic Π2 sentence) [ P < NP] is independent of ZFC, then it will hold true of the standard model for arithmetic, if ZFC has a model with a standard arithmetic portion. That’s a very important fact, as it implies that independence entails the truth of P < NP.

A first tentative plot. • We know that if [ P = NP] holds true of standard arithmetic, then it can be proved in theory PA + ξ, where ξ is a true Π1 sentence. • Also, if both [ P = NP] and [ P < NP] turn out to be independent of a theory S that includes PA, has a recursively enumerable set of theorems, has a first–order classical language and has a model with standard arithmetic (is arithmetically sound), then [ P < NP] is true of standard arithmetic. • We know that S + [ P < NP] ↔ [ P < NP]g is arithmetically sound, for g with an infinite domain and increasing over its domain. • We obtain a (partial recursive) G so that both S and S + [ P < NP] ↔ [ P < NP]G do not prove it to be total. G is strictly increasing over its domain which is infinite. • What happens if S ` [ P < NP], S + [ P < NP] ↔ [ P < NP]G has the same provably total recursive functions as S? • As a consequence, both S and S + [ P < NP] ↔ [ P < NP]G do not prove [ P < NP]G , since that sentence could prove [G is total]. • Then should follow the consistency of S + [ P < NP] ↔ [ P < NP]G + [ P = NP]G , and that of S + [ P = NP]. ¨ • We can thus show that if S proves [ P = NP], the set K of Godel numbers of the algorithms that settle all of S AT is proved by S (or by S + ξ, where ξ is a Π1 sentence) to be recursive. • Therefore the function that equals the counterexample function to [ P = NP] for all those n 6∈ K, and is zero for n ∈ K is recursive and provably total in S + ξ, and a fortiori in S. • But that function cannot be provably total in S, as its exotic G–version grows faster than any S–provably total recursive function in the standard model, and S + [ P < NP] ↔ [ P < NP]G makes both versions equivalent. • So, S cannot prove [ P = NP]. • Therefore S + [ P < NP] is consistent, and arithmetically sound. • That is, S + Shoenfield’s ω–rule ` [ P < NP]. Does that proceed? Is there an alternate route?

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1. Summary.

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the main ideas behind our chief argument. They essentially arise iut of what we have called “the exotic formulation for the P < NP conjecture”:

E SKETCH HERE

• The intuitive version of the hypothesis P < NP or P 6= NP can be formalized1 as an arithmetical sentence [ P < NP] of the form

∀ x ∈ ω ∃y ∈ ω A( x, y), where A is primitive recursive. This is a Π2 sentence. Its negation [ P = NP] ↔Def ¬[ P < NP] is a Σ2 sentence. • For a theory S which includes Peano Arithmetic, is based on a classical first–order language, and has a recursively enumerable set of theorems (and moreover has a model with standard arithmetic), for the (partial recursive) function {e A }( x ) = miny A( x, y): S ` ∀ x ∈ ω ∃y ∈ ω A( x, y) if and only if S ` ∀ x ∈ ω ∃z ∈ ω T (e A , x, z). That is, S proves that sentence if and only if S proves {e A } to be total. (Here T is Kleene’s predicate [36]) 1 There

are alternative, nonequivalent, and reasonable formalizations. See below.

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• We will only consider theories as S described above. PA and ZFC fit into the picture. (Such theories were first considered in the metamathematics of computer science by Hartmanis and Hopcroft in 1977 [29].) • Moreover, if S proves that {e A } isn’t dominated by all S–provably total recursive functions, then S 6` ∀ x ∈ ω ∃y ∈ ω A( x, y). (That means, {e A } overtakes all S–provably total recursive functions in the standard model for arithmetic in S.) • So, our goal is to study the behavior of {e A } (and of related functions) for P < NP, as they may tell us whether [ P < NP] is provable in S or not. Remark 1.1 Another formalization would be: • ∃y ∈ ω A(0, y). • ∃y ∈ ω A(1, y). • ∃y ∈ ω A(2, y). • .... Clearly ∀ x ∈ ω ∃y ∈ ω A( x, y) implies that alternative characterization, while the inverse implication isn’t in general valid.

A venerable line of research That approach has been followed before by DeMillo and Lipton, O’Donnell, Fortune and Leivant, Kowalczyk, Maté [14, 15, 22, 32, 33, 37, 41, 42]; we must thank E. Bir who uncovered and communicated to us the preceding results along this line of research [11].Those results go as follows: • For some fragments F of arithmetic, if F is consistent then so is F + [ P = NP]. • For Peano Arithmetic and beyond, the results are: given some nonintuitive, usually complicated, metamathematical condition, then S + [ P = NP] is consistent. The authors’ results in 2003 [7] were (stated for S): • We introduced a Π2 sentence [ P < NP]F called the “exotic formalization.” It is equivalent to [ P < NP] in the standard model, but has a nontrivial behavior within S. ([ P < NP] ↔ [ P < NP]F is actually independent of S.) • S 6` [ P < NP]F and if S is consistent, then so is S + [ P = NP]F .

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da Costa, Doria • Both S + [ P < NP] ↔ [ P < NP]F and S + [ P < NP] → Consis (S) are simply consistent theories, but one has no information (in the 2003 paper) on the consistency of S + [ P < NP] ↔ [ P < NP]F + [ P = NP]F , from which we would be able to get the consistency of S + [ P = NP]. • However there is a partial result that can be stated as: if S + [ P = NP]F is ω–consistent, then S + [ P = NP] is consistent.

We notice that it is certainly not obvious why S + [ P = NP]F should be ω– consistent, even if we restrict the requirement of ω–consistency to PA–provable sentences, which would be enough for our purposes. We explore here several — not always equivalent — exotic formulations.

Some notation We write P < NP and P = NP for the intuitive versions of the corresponding hypotheses. [ P < NP] and [ P = NP] are the Π2 and Σ2 sentences that formalize them. The set of all Boolean expressions in cnf can be coded by binary words in several reasonable ways, and those binary words can in turn be coded by the natural numbers ω in a 1–1 way; the set of satisfiable Boolean expressions in cnf, S AT ⊂ ω is a primitive recursive subset of ω, and so it can be given a primitive recursive coding which is 1–1 and onto ω; it is also a strict subset, as there are unsatisfiable Boolean expressions in cnf, namely the totally false expressions such as p ∧ ¬ p.

2. The exotic version. HE EXOTIC FORMALIZATION for P < NP is na¨ıvely the same as the (usual) standard formalization, but cannot (in general) be proved equivalent to the latter within even strong systems such as ZFC. Let tm ( x ) be the (primitive recursive function that gives the) operation time of {m} over an input x of length | x |. Recall that the operation time of a Turing machine is given as follows: if {m} stops over an input x, then:

T

tm ( x ) = | x | + [number of cycles of the machine until it stops]. tm is primitive recursive and can in fact be defined out of Kleene’s T predicate. Definition 2.2 (S TANDARD FORMALIZATION FOR P = NP.)

[ P = NP] ↔Def ∃m, a ∈ ω ∀ x ∈ ω [(tm ( x ) ≤ | x | a + a) ∧ R( x, m)]. R( x, y) is a polynomial predicate; as its interpretation we can say that it formalizes a kind of “verifying machine” that checks whether or not x is satisfied by the output of {m}. There is an equivalent formalization for [ P = NP] where one uses Kleene’s T predicate to get the time measure tm ; see Lemma F.3. Definition 2.3 [ P < NP] ↔Def ¬[ P = NP]. Now suppose that {ef } = f is total recursive and strictly increasing: Remark 2.4 The na¨ıve version for the exotic formalization is:

[ P = NP]f ↔ ∃m ∈ ω, a ∀ x ∈ ω [(tm ( x ) ≤ | x |f (a) + f ( a)) ∧ R( x, m)]. 15

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However as we will soon see, there is no reason why we should ask that f be total; on the contrary, there will be interesting situations where such a function may be partial and yet provide a reasonable exotic formalization for P < NP [28]. So, for the next definitions and results let f be in general a (possibly partial) recursive function which is strictly increasing over its domain, and let ef be the ¨ Godel number of an algorithm that computes f. Let p(hef , b, ci, x1 , x2 , . . . , xk ) be an universal Diophantine polynomial with parameters ef , b, c; that polynomial has integer roots if and only if {e f }(b) = c. We may if needed suppose that polynomial to be ≥ 0. We omit the “∈ ω” in the quantifiers, since they all refer to natural numbers. Definition 2.5 Mf ( x, y) ↔Def ∃ x1 , . . . , xk [ p(hef , x, yi, x1 , . . . , xk ) = 0]. Actually Mf ( x, y) stands for Mef ( x, y), or better, M(ef , x, y), as dependence is ¨ on the Godel number ef . Definition 2.6 ¬ Q(m, a, x ) ↔Def [(tm ( x ) ≤ | x | a + a) → ¬ R( x, m)]. Proposition 2.7 ( STANDARD FORMALIZATION , AGAIN .)

[ P < NP] ↔ ∀m, a ∃ x ¬ Q(m, a, x ). Definition 2.8 ¬ Qf (m, a, x ) ↔Def ∃ a0 [ Mf ( a, a0 ) ∧ ¬ Q(m, a0 , x )]. Remark 2.9 We will sometimes write ¬ Q(m, f ( a), x ) for ¬ Qf (m, a, x ), whenever f is total. Definition 2.10 (E XOTIC FORMALIZATION .)

[ P < NP]f ↔Def ∀m, a ∃ x ¬ Qf (m, a, x ). Notice that again this is a Π2 arithmetic sentence:

∀m, a ∃ x, a0 , x1 , . . . , xk {[ p(hef , a, a0 i, . . . , x1 , . . . , xk ) = 0] ∧ ¬ Q(m, a0 , x )}. (Recall that Q is primitive recursive.) Definition 2.11 [ P = NP]f ↔Def ¬[ P < NP]f .

In a nutshell. We will use through this paper a consistent theory S with the following characteristics: Remark 2.12 We require of S that: • Its underlying language is the first–order classical predicate calculus. • It has a recursively enumerable set of theorems. • It includes PA (Peano Arithmetic) in the following sense: it includes the language and proves all theorems of PA. • It has a model with standard arithmetic. So, for the next definitions and results let g be in general a (possibly partial) recursive function which is strictly increasing over its ¨ domain, and let eg be the Godel number of an algorithm that computes g. Let p(heg , b, ci, x1 , x2 , . . . , xk ) be an universal Diophantine polynomial with parameters eg , b, c; that polynomial has integer roots if and only if {e g }(b) = c. We may if needed suppose that polynomial to be ≥ 0. We omit the “∈ ω” in the quantifiers, since they all refer to natural numbers. Definition 2.13 Mg ( x, y) ↔Def ∃ x1 , . . . , xk [ p(heg , x, yi, x1 , . . . , xk ) = 0]. Actually Mg ( x, y) stands for Meg ( x, y), or better, M(eg , x, y), ¨ as dependence is on the Godel number eg . For R a polynomial predicate: Definition 2.14 ¬ Q(m, a, x ) ¬ R( x, m)].

↔Def

[(tm ( x ) ≤

| x | a + a) →

Definition 2.15 [ P < NP] ↔ ∀m, a ∃ x ¬ Q(m, a, x ). Definition 2.16 ¬ Qg (m, a, x ) ↔Def ∃ a0 [ Mg ( a, a0 ) ∧ ¬ Q(m, a0 , x )].

We will sometimes write ¬ Q(m, g( a), x ) for ¬ Qg (m, a, x ), whenever g is S–provably total. Definition 2.17 (E XOTIC FORMALIZATION .)

[ P < NP]g ↔Def ∀m, a ∃ x ¬ Qg (m, a, x ). Notice that this is a Π2 arithmetic sentence. Definition 2.18 [ P = NP]g ↔Def ¬[ P < NP]g .

3. Fast–growing functions.

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a well–known recursive function that is diagonalized over all S–provably total recursive functions. We note it F (and sometimes FS ). See [7] and [34], pp. 51–52 on it.

E USE HERE

Remark 3.19 For each n, F(n) = maxk≤n ({e}(k)) + 1, that is is the sup of those {e}(k) such that: 1. k ≤ n. 2. dPrS (d∀ x ∃z T (e, x, z)e)e ≤ n. ¨ PrS (dξ e) means, there is a proof of ξ in S, where dξ e means: the Godel ¨ number of ξ. So dPrS (dξ e)e means: “the Godel number of sentence ‘there is a proof of ξ in S.’ ” Condition 2 above translates as: there is a proof of [{e} is ¨ total] in S whose Godel number is ≤ n. Proposition 3.20 We can explicitly compute a Gödel number eF so that {eF } = F. Proposition 3.21 If S is consistent then S 6` ∀m∃n [{eF }(m) = n]. Remark 3.22 Notice that [ P < NP]F ↔ ∀m, a ∃ x ¬ QF (m, a, x ). Lemma 3.23 If I ⊆ ω is infinite and 0 ∈ I, then: ZFC ` {[∀m ∀ a ∈ I ∃ x ¬ Q(m, a, x )] → [∀m ∀ a ∈ ω ∃ x ¬ Q(m, a, x )]}. The meaning of this result is: as long as we have an infinite succession of ever larger bounds that make the Turing machines polynomial, our standard 18

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definitions hold. The size of the intermediate gaps between each pair of bounds doesn’t matter. However we again stress that there is in general no provable equivalence in S between [ P < NP] and [ P < NP]F (see [7]): Proposition 3.24 S ` [ P < NP]F ↔ {[F is total ] ∧ [ P < NP]}. Sketch of proof : For the formal, computational proof see [7]. It is easy to see that: S + [ P < NP] + [F is total] ` [ P < NP]F . For the converse, the fact that the exotic counterexample function fF (see below after Lemma 3.25) is total implies that [ P < NP] and [F is total] hold at the same time (we will require Lemma 3.23 here). We quote a result that follows from the above due to its importance: Lemma 3.25 S ` [ P < NP]F → [F is total.]. Remark 3.26 The formal proof is in [19]. The following argument clarifies the meaning of the lemma and gives an informal proof for it: let: fF (hm, ai) = min [¬ Q(m, F( a), x )], x

where we can look at F as a (partial) recursive function. (The brackets h. . . , . . .i note the usual 1–1 pairing function.) Now if fF is total, then F( a) has to be defined for all values of the argument a, that is, F must be total. The function fF is the so–called exotic counterexample function to [ P = NP]F . We can similarly define a standard counterexample function : f (hm, ai) = min [¬ Q(m, a, x )]. x

As S ` [F is total] ↔ [S is Σ1 –sound], and also as S ` [S is Σ1 –sound] → Consis (S), S ` [F is total] → Consis (S). (For the preceding discussion see Corollary 5.68.) Then: Lemma 3.27 S ` [ P < NP]F → Consis (S). Proposition 3.28 If S is consistent, then S doesn’t prove [ P < NP]F . Proof : S ` [[ P < NP]F → (F is total)]. (Lemma 3.25.) So, S cannot prove [ P < NP]F . Corollary 3.29 [ P = NP]F is consistent with S. If N |= S and makes it arithmetically sound, that is, N has a standard arithmetic part for the arithmetic in S : Proposition 3.30 N |= S + [ P < NP] ↔ [ P < NP]F .

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da Costa, Doria Proof : See Proposition 5.71.

Proposition 3.31 [ P < NP] ↔ [ P < NP]F is independent of S. Proof : S doesn’t prove that equivalence due to Proposition 3.24, since otherwise it would prove [F is total.]. On the other hand, consistency of: S + [ P < NP] ↔ [ P < NP]F follows from Propositions 3.30 and 5.71 below. Remark 3.32 Now one interesting question is, does theory: S + [ P < NP] ↔ [ P < NP]F prove [ P < NP]F ? What do you think about it? Proposition 3.33 1. For f as above, S ` [ P < NP] ↔ [ P < NP]f . 2. For F as above, [ P < NP] ↔ [ P < NP]F is independent of S. Proof :Recall that f is S–provably total recursive. Again S doesn’t prove that result due to Proposition 3.24, since otherwise it would prove [F is total.]. And on the other hand, consistency of: S + [ P < NP] ↔ [ P < NP]F follows from the fact that it trivially holds in the standard model for arithmetic. The first equivalence holds since S ` [ P < NP]f ↔ {[f is total ] ∧ [ P < NP]} and f is provably total. Remark 3.34 We have already noticed that we can write, if convenient,

[ P < NP] ↔Def ∀m, a ∃ x ¬ Q(m, f ( a), x ). Remark 3.35 If the sentence ¬ [F is total] holds in some model for our theory S, the fact that we have in that model ∃ x (F(0) = x ), ∃ x (F(1) = x ), ∃ x (F(2) = x ), . . . , together with ¬ [F is total] shows that: if ¬ [F is total] holds, then S + ¬ [F is total] is ω–inconsistent. More precisely: if some theory S0 proves ¬ [F is total], then it is ω–inconsistent. So, if some theory S0 is ω– consistent, then [F is total] is consistent with it. Thus a partial result: Proposition 3.36 If S + [ P = NP]F is ω–consistent, then S + [ P = NP] is consistent. So far we have no evidence that such a theory is in fact ω–consistent.

4. A blueprint.

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ATHEMATICS IS SOMETIMES full of unexpected turns, like a good mys-

tery novel. The Busy Beaver game looks na¨ıve enough, and yet it leads us to an incredibly fast–growing, uncomputable function [44]. Or, who would expect that questions about finite functions would depend [24] on the existence of large cardinals? Sometimes mathematics behaves in a rather predictable, reasonable way. When one goes from real numbers to complex numbers to quaternions and then to the Cayley numbers, each step in that succession is marked (as it should be) by a slight modification of previously known properties. Yet in many cases the generalization of properties leads us into the unexpected: a contemporary example can be found in the mathematics of gauge fields. Given an Abelian gauge field, that field is always (locally, at least) derived from a gauge potential (a connection form) which is unique modulo gauge transformations. However when we deal with non–Abelian gauge fields, we see that the same field can be derived from physically different gauge potentials, that is, potentials that aren’t related by a gauge transformation, not even locally [16, 17, 46]. Just by looking at the Abelian case we would never expect such a novel phenomenon to creep up in the non–Abelian case. Moreover this precise phenomenon is quite surprisingly related to a kind of transformation first used by Einstein in his theory of the asymmetric field [16, 20]. Can we say that we are dealing here with singularities in a kind of space of mathematical theories? We have in fact proposed this picture: we described a coding of formal systems by dynamical systems that bifurcate: for instance, if a given (undecidable) sentence holds, the system bifurcates; if not, it doesn’t. Thus undecidable sentences are coded by singularities in those systems [10]. However the actual situation may be in fact much more complicated than 21

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an opposition between singular and regular points: mathematics is the realm of the unexpected.

The P =?NP question Let’s consider a final example, which will lead us into these notes’ main topic. That example stems from the P =?NP question. The P =?NP question starts from a very simple, rather obvious question that we can formulate as a short tale: Mrs. H. is a gentle and able lady who has long been the secretary of a large university department. Every semester Mrs. H. is confronted with the following problem: there are courses to be taught, professors to be distributed among different classes of students, large and small classes, and a shortage of classrooms. She fixes a minimum acceptable level of overlap among classes and students and sets down in a tentative way to get the best possible schedule given that minimum desired overlap. It’s a tiresome task, and in most cases, when there are many new professors or when the dean changes the classroom allocation system, Mrs. H. quite frequently has to check nearly all conceivable schedulings before she is able to reach a conclusion. In despair she asks a professor whom she knows has a degree in math: “tell me, can’t you find in your math a fast way of scheduling our classes with a minimum level of overlap among them?” Mrs. H. unknowingly asks about the P =?NP question. She is able to understand its basic contours: there are questions for which it is easy to check whether a given arrangement of data fits in as a solution; however for the general case there are no known shortcuts in order to reach a solution. If you can guess the answer, it is easy to verify it. If you have to work it out, you’ll most likely have to sweat your shirt until you reach an adequate answer. Most likely: for one always wonders whether there is some general way to get a quick solution. Who knows? Are there general shortcuts available? This is the whole point—in a nutshell; and this is the main query in the P =?NP problem. P = NP is, roughly, “there is a time–polynomial Turing machine that inputs each instance of a problem in a NP class and that correctly guesses (outputs) a solution for that instance.” The negation P < NP is, “for each polynomial ¨ machine of Godel number m there is an instance x such that machine m guesses wrongly about x.” The counterexample function f (m) is the function that enumerates each first instance x where a polynomial algorithm fails; it is recursive on the set of all polynomial machines and P < NP holds if and only if f is total. So, one of the possible approaches to this major problem is to focus on the counterexample function f . How are we to understand its properties? Perhaps

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by analogy: we use a tamer function g that more or less looks like f , and try to infer the properties of f from those of g. The chief property we want to consider is whether f can be proved to be total in, say, Peano Arithmetic (PA), for in that case: If the counterexample function f is bounded by a prescribed total recursive function—for Peano Arithmetic, the fast–growing function Fe0 — then P < NP will be provable in Peano Arithmetic. For an adequately conceived g, it should be easy to check the desired property for g, and it should now be just a small step from g to f . (For Fe0 see [49] and references therein; for the counterexample function see [18].) Then our main query is: is g bounded by such a function Fe0 ? If not, where do the properties of g will lead us concerning f ?

Functions g0 and g∗ Remark 4.37 Let’s motivate g∗ . Consider the following: g0 (m) = min( x m < 2x ) x

(1)

(we require here and in all such functions that x ≥ 2 to avoid trivialities) for non–negative integers m. This function is trivially recursive and total, and bounded by the exponential 2x itself. Now suppose that we have defined some (necessarily nonrecursive) enumeration for all time–polynomial Turing machines (the poly machines). This enumeration is noted, Pm , m ∈ ω; we note P the set of all polymachines. The following function: g∗ (m) = min[Pm ( x ) < 2x ] (2) x

looks very similar to (1). In fact we have only changed the monomial x m for the output Pm ( x ) of a poly machine, and that output is necessarily bounded in its length by some polynomial, that is, there is a positive integer k so that

|Pm ( x )| < | x |k . Therefore g∗ is also na¨ıvely total. g∗ is usually presented as a kind of analogue of g0 in eq. (1), and it is argued that, since g0 is bounded by an exponential, the same should be true of g∗ in eq. (2). Yes, a moment’s thought shows that this should be the case. . . Wait a moment—is it really so? Let’s ponder it: the relation between exponent k and machine index m is possibly quite complicated—and index m roams over a nonrecursive set. The analogy starts to break down here. Yes, the analogy breaks down right at the beginning. Can we pass over those differences and still have a rather “tame” behavior for g∗ as the one exhibited by g0 ?

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We show in this paper how wildly different is the behavior of g∗ when compared to g0 , despite the fact that the two functions are apparently very similar. The whole point, as we will see, is the relation between machine index m and the exponent k for the bounding polynomial.

Analogies between f and g∗ The main analogy between the counterexample function f and g∗ is: both are defined over the set of all poly machines P and their values are given by the application of the min operator to a recursive predicate on P that depends on the index m of a poly machine Pm (for the case of f see [18]). Of course the main difference between both functions is that g∗ is intuitively total, while the whole point of the P =?NP question is whether f is total or not.

Summary of the section g∗ is part of the folklore around the P =?NP question. As noticed, it is sometimes used as a simile for the behavior of the counterexample function to the P = NP (no question–mark here!) hypothesis. We consider its behavior in detail in the present section in order to show how wildly different is the behavior of g∗ or its modification g (see below) when compared to that of g0 . This paper shows that this function g∗ isn’t well–behaved at all. In fact, given a natural version of it defined over all Turing machines, the modified g∗ (noted g) will oscillate in a wild way, and in its “ups” it will tower over all total recursive functions, so that the only immediate bounds we can find for it are in the realm of those functions which bound all total recursive functions. One of those upper bounds is a generalized Busy Beaver function, as we show in one of the next sections of this text.

Preliminary comments ¨ Remark 4.38 If M is the set of all Turing machines given by their Godel numbers, the subset of all poly machines, P ⊂ M, isn’t recursive. If we consider a function g given by: • g(m) = minx [Mm ( x ) < 2x ], for m in P , • g(m) = 0, otherwise, x ≥ 2, we intuitively see that g is total. We consider g instead of g∗ .

Goals Our goal in the present chapter is to discuss some properties of g. We will try to understand its relation to the set of all total recursive functions; everything

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proceeds within in an intuitive framework, where we will be able to compare it to a generalization of the Busy Beaver function [44]. As noticed above, functions g∗ and g are part of the folklore around the P =?NP question; g∗ was suggested to the authors by F. Cucker and, independently, by W. Mitchell.

Conventions on Turing and `–machines We will use here what we call `–machines, for “labeled” or “parametric” Turing machines; they are defined in this section. They are defined out of the usual Turing machines. They are simply Turing machines of which we keep track with the help of a tag given by a parameter. Remark 4.39 Suppose given the canonical enumeration of binary words ∅, 0, 1, 00, 01, 10, 11, 000, 001, . . . which code the empty word and the integers; they correspond to 0, 1, 2, . . . . We take this correspondence to be fixed for the rest of this text.

Turing machines Remark 4.40 We describe the behavior of the (standard) Turing machines we deal here: 1. Turing machines are defined over the set A2∗ of finite words on the binary alphabet A2 = {0, 1}. 2. Each machine has n ≥ 0 states s0 , s1 , . . . , sn−1 , where s0 is the final state. (The machine stops when it moves to s0 .) We allow for a machine with 0 states and an empty table; it is discussed below in Remark 4.44. 3. The machine’s head roams over a two–sided infinite tape. 4. Machines input a single binary word and either never stop or stop when the tape has a finite, and possibly empty set of binary words on it. 5. The machine’s output word will be the one over which the head rests if and when s0 is reached. (If the head lies on a blank square, then we take the output word to be the empty word, that is, 0.) Remark 4.41 The Turing machine inputs a binary string b x c and (if it stops over b x c) outputs a binary string byc. The corresponding recursive function inputs the numeral x and outputs the numeral y. Whenever it is clear from context, we write x for both the binary sequence and the numeral.

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Remark 4.42 We will use upper case and lower case sans serif letters (such as M,. . . ) for Turing machines (and also for `–machines). If Mn is a Turing machine ¨ of Godel number n, its input–output relation is noted Mn ( x ) = y. Remark 4.43 Turing machines are given by tables. We can write the tables as code lines ξ, ξ 0 , . . ., separated by blanks t, such as ξ t ξ 0 t . . . t ξ 00 . Let Ξ be one such set of code lines separated by blanks. Let Ξ0 be obtained out of Ξ by a permutation of the lines ξ, ξ 0 , . . .. Both Ξ and Ξ0 are seen as different machines that compute the same algorithmic function, that is, in this case, if f Ξ ( f Ξ0 ) is computed by Ξ (Ξ0 ), then f Ξ = f Ξ0 . Remark 4.44 We define the empty or trivial machine to be the Turing machine with an empty table; we take it to be the simplest example of the identity machine, again by definition. (See also Remark 4.54.)

`–machines Remark 4.45 We describe here the `–machines. Consider a two–tape Turing machine [31] where the input is written over tape 1, while tape 2 comes with a possibly empty binary string n. n is the machine’s label, or parameter. • One easily sees that each Turing machine Mm is simulated by infinitely many `–machines: write an arbitrary n on tape 2 and write all instructions for Mm solely for tape 1. • For the converse, if Mm is an arbitrary Turing machine, and if (by an abuse of notation) τ is the machine that polynomially computes the onto pairing function ω × ω → ω, then the set of all coupled machines Mm ◦ τ (n, x ), all n, m, represents the set of `–machines. • Moreover, to avoid a pre–fixed label or parameter, we can use the family of constant Turing machines in that print n over any input, and couple it to the arrangement above, that is Mm ◦ τ ◦ in (in order to generate n), to obtain a set of emulated `–machines. ¨ ¨ The Godel number k = c(m, #τ, n) (# . . . denotes the Godel number of . . .) is given by a primitive recursive c [40]. From that we have: Definition 4.46 An `–machine is a pair hn, Mm i, where n ∈ ω and Mm is a 2–tape Turing machine of Gödel number m, with n written on tape 2. Remark 4.47 From now on whenever we write “machine” we will mean “`– machine,” unless otherwise stated. Remark 4.48 So, `–machines are just a convenient bookkeeping device to keep track of families of Turing machines which are labeled by some parameter. However, under the guise of `–machines they allow us the simple result stated in Lemma 4.51.

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Godel ¨ numbering (We use here Kleene’s “method of digits,” without however “closing the gaps” to squeeze out the junk; see [36], p. 178.) Definition 4.49 Each `–machine is coded by the pair hn, mi, where both n, m range over the whole of ω. Remark 4.50 We refer to hn, mi as the Gödel number of the `–machine, or `–Gödel number, or `–index. In order to compute some values for g as we do here one needs to know the `–index for some poly machines which are given as `–machines. A lemma on Godel ¨ numbers ¨ In order to compute pieces of g one needs to know the Godel numbers for some polynomial Turing machines. We use the following result: Lemma 4.51 Let Mhn,mi , n ∈ ω, m fixed, be a family of `–machines. We can explicitly construct an `–Gödel numbering for the set of all `–machines so that the `–indexes for the Mhn,mi , n ∈ ω, m fixed, are given by a linear function N (n) = an + b. Moreover there is a primitive recursive map from that `–Gödel numbering to standard Gödel numberings. Therefore, if k is such a map, the induced N 0 (n) = k ◦ N (n) is primitive recursive. Remark 4.52 Recall that the length | x | of a finite string x is the number of letters in x. Proof of Lemma 4.51 : The idea goes as follows: 1547, 2547, 3547, . . . are in arithmetic progression with r = 1000 = 103 = 10length(547) . We extend this to the proof of our lemma. • Recall the ordering of binary words as described in Remark 4.39. • `–machines are fully characterized by pairs hn, mi, where n is the parameter and m the 2–tape Turing machine’s code. • If xn and xm are the corresponding binary words, code pair hn, mi as the word xn 10xc m , where if k 0 1 2 , xm = xm xm xm . . . xm

then

k k 0 0 1 1 xc m = xm xm xm xm . . . xm xm .

• So, given an arbitrary binary word, there is a simple recursive procedure to decide whether it codes an `–machine hn, mi or not: go to the rightmost end and (by going backwards) see if we have a duplicated word that ends in the sequence 10. If so, we have a coded binary word for one of our pairs.

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da Costa, Doria • Map the in–between mumbo–jumbo onto the trivial `–machine. • This coding is certainly recursive and 1–1 onto the natural numbers. • Finally, if | xm | is the length of xm , then | xc m | = 2| xm | + 2, and therefore, between `–machine hn, mi and `–machine hn + 1, mi there are 22| xm |+2 binary words.

¨ Follows the lemma, and the Godel number N (n) of the `–machines just described is given by an arithmetic progression N (n), of ratio 22| xm |+2 . From Remark 4.45 one immediately sees that there are primitive recursive ¨ maps that lead from that coding to more usual Godel numberings, such as the ¨ original one used by Godel, or Kleene’s “method of digits,” so that N 0 (n) is also primitive recursive. ¨ Remark 4.53 So, it is enough to know that the Godel numbers N for the families of machines we will be using here are related to the machine’s “tag” or parameter by N = q(n), where q is primitive recursive, that is, Ackermann’s function Fω dominates q. The construction we just described amounts to a particular case of the s–m– n theorem. We can directly use it.

Quasi–trivial machines We use `–machines; therefore here “machine” stands for `–machine. Remark 4.54 Again one of the trivial `–machines is the one with the empty table; more formally, `–machine h0, 0i. We will now consider a set of, say, nearly trivial machines — we call them “quasi–trivial machines.” They are all polynomial. Recall that the operation time of a Turing machine is given as follows: if M stops over an input x, then the operation time over x, tM = | x | + number of cycles of the machine until it stops. Example 4.55 • First trivial machine. Note it O. O inputs x and stops. tO = | x | + moves to halting state + stops. So, operation time of O has a linear bound. • Second trivial machine. Call it O0 . It inputs x, always outputs 0 (zero) and stops. Again operation time of O0 has a linear bound.

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• Quasi–trivial machines. A quasi–trivial machine Q operates as follows: for x ≤ x0 , x0 a constant value, Q = R, R an arbitrary total machine. For x > x0 , Q = O or O0 . This machine has also a linear bound. We will use several `–families of quasi–trivial machines. Please allow for some abuse of language here; we need it in order to avoid cumbersome nota¨ tions. (For instance, the machines defined below depend on the Godel numbers of the machines that appear as their subroutines.) Remark 4.56 Now let H be any fast–growing, superexponential total machine. Let H0 be another such machine. Form the following family QH(n) of quasi– trivial `–machines with subroutines H and H0 : 1. H(n) = k(n), all n, is the way we introduce the parameter in the family. 0

2. If x ≤ k(n), QH,H ,n ( x ) = H0 ( x ); 0

3. If x > k(n), QH,H ,n ( x ) = 0. For function g in Remark 4.38: Proposition 4.57 If N (n) is the Gödel number of an `–machine as in Remark 7.126, then g( N (n)) = k(n) + 1 = H(n) + 1. Proof : Use Lemma D.5. Recall that N (n) = an + b, a and b constants that ¨ depend on the actual algorithms (i.e., in the Godel numbers) for H, H0 and Q.

A domination lemma Recall: Definition 4.58 For f , g : ω → ω, f dominates g ↔Def ∃y ∀ x ( x > y → f ( x ) ≥ g( x )). We write f g for f dominates g. Our goal here is to prove the following result: Proposition 4.59 For no total recursive function h does h g. Proof : Suppose that there is a total recursive function h such that h g. Remark 4.60 Given such a function h, obtain another total recursive function h0 which satisfies: 1. h0 is strictly increasing.

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da Costa, Doria 2. For n > n0 , h0 (n) > h( ba n − 1b ), for integer values of the argument of g. That is, h0 ( an + b) > h(n).

¨ Constants a, b are from the Godel numbers of the quasi–trivial machines described in Remark 7.126. Lemma 4.61 Given a total recursive h, there is a total recursive h0 that satisfies the conditions in Remark 4.60. Proof : Given h, obtain out of that total recursive function a strictly increasing total recursive h∗ . Then if, for instance, Fω is Ackermann’s function, h0 = h∗ ◦ Fω will do. 0

0

¨ From Lemma 4.51, we have that the Godel numbers #[Qh ,K,n ] of the Qh ,K,n 0 ,K,n h are given by #[Q ] = an + b, a, b ∈ ω. Therefore, g( an + b) = h0 (n) + 1. From Lemma D.5, Remark 4.60 and Lemma 4.61 we conclude our argument. If we make explicit the computations, for q(n) = an + b (as the argument holds for any strictly increasing primitive recursive q): g(q(n)) = h0 (n) + 1 = h∗ (Fω (n)) + 1, and h∗ (Fω (n)) > h∗ (q(n)). For N = q(n) (see Remark 4.53), g( N ) > h∗ ( N ) ≥ h( N ), all N. Therefore no such h can dominate g. Remark 4.62 No need to emphasize that we could have used ordinary Turing machines, as at the end of Remark 4.45; then N (n) would be primitive recursive on n.

An extended Busy Beaver function So g overtakes all total recursive functions. Can we put a ceiling to it? Yes: and that ceiling is a kind of generalized Busy Beaver function (which depends on g itself). We now sketch an argument that shows that this generalized Busy Beaver function dominates g. The Busy Beaver function B can be easily and intuitively defined. If N is the number of states of a given Turing machine, then B( N ) is the biggest number of 1s a N–state Turing machine prints over a tape filled with 0s as its input. Remark 4.63 We are going to generalize that function to B0 as follows:

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• First, notice that there is a simple relation between a Turing machine’s table and the number [30] of its states. More precisely, from that reference we see that a N–state Turing machine with a binary alphabet has a table given by a N 0 = (3N + 1) × 4 matrix. ¨ So we can define a primitive recursive relation f from the Godel numbers ¨ to the set of states of Turing machines so that if Godel numbers m0 > m, then states f (m0 ) ≥ f (m). • We can therefore see the Busy Beaver function B as a monotonic nonde¨ creasing function of Turing machine Godel numbers. ¨ • Now for each Godel number m for a polynomial Turing machine, g(m) is effectively computed. • We will define B0 (m) as: the maximum number of 1s printed by a N (m)– state Turing machine over a tape with only 0s and over a tape with all inputs < g(m) + 1. • Clearly B0 ( N ) ≥ B( N ) and B0 ( N (m)) = B( N (m)) if the Turing machine ¨ with Godel number m isn’t a polynomial machine. Proposition 4.64 For all m, B0 (m) ≥ B(m). Proposition 4.65 B0 dominates g. Remark 4.66 We conjecture that g is of the same order of growth as the Busy Beaver function itself.

5. Σ1–soundness and F. show that F sort of codes, or represents, deep facts about the structure of the axiomatic system where it is defined. Namely, if it is total, then the axiomatic system T w.r.t. which F is defined, is consistent. We also show that F is total if and only if T is Σ1 –sound.

T

HE NEXT RESULTS

Remark 5.67 The local reflection principle Rfn( T ) for theory T is: PrT (dφe) → φ, (that is, if there is a proof for φ, we actually find it in T); and the (first) uniform reflection principle, RFN( T ) is:

[∀ x PrT (dφ( x˙ )e)] → [∀ x φ( x )], all φ with only x free, and x˙ standing for the x that can be represented by actual constants in T. This means that once one can list instances φ(0), φ(1), . . . (which are derivable due to the first supposition in the Reflection Principle) for all nameable n, then a restricted application of the ω–rule leads to the principle. For Σ1 –soundness one restricts φ to all ∃ x φ( x ), φ primitive recursive; the corresponding restricted reflection principle is noted RFNΣ1 ( T ). For T as PA, ZFC, and first–order extensions as considered here: Corollary 5.68 T ` RFNΣ1 ( T ) → Consis( T ). Proof : • Suppose T ` ¬(0 = 0). 32

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• Therefore, T ` PrT [¬(0 = 0)]. • Given Σ1 –soundness, as this is a trivial Σ1 sentence, PrT [¬(0 = 0)] → [¬(0 = 0)]. • By contraposition,

{¬[¬(0 = 0)]} → ¬PrT [¬(0 = 0)]. • Or, (0 = 0) → ¬PrT [¬(0 = 0)]. • However, T ` (0 = 0). Then follows: • ¬PrT [¬(0 = 0)]. A contradiction. (This proof is due to N. C. A. da Costa.)

Σ1 –soundness is equivalent to [F is total] We argue here for ZFC, since we want to have as much “elbow room” as possible. However we could have argued for any theory like the one we have denoted by S. If we allow for the abuse of language that subsumes the infinitely many sentences of the Reflection Principle in our formulation as a single one in the statement of our result: Lemma 5.69 ZFC ` [F is total ] ↔ [ ZFC is Σ1 –sound ]. Proof : Recall that [F is total] ↔ ∀ x ∃z T (eF , x, z), where — here — T is ¨ Kleene’s T predicate and eF is a Godel number for F.

(⇐). We first prove: assuming RFNΣ1 (ZFC), then ∀ x ∃z T (eF , x, z). Given the (recursively enumerable) infinite set of conditions RFNΣ1 (ZFC), for T we get: ˙ z)e)] → [∀ x ∃z T (eF , x, z)]. [∀ x PrZFC (d∃z T (eF , x, Now, for each ZFC constant n˙ we have that: ˙ z)]. ZFC ` [∃z T (eF , n, ˙ Therefore we conThen there are proofs of each of these sentences, for each n. ˙ z)e)], as x˙ only ranges over the constants. From clude: [∀ x PrZFC (d∃z T (eF , x, the corresponding restriction of RFNΣ1 (ZFC) we conclude that:

[∀ x ∃z T (eF , x, z)] holds, by modus ponens.

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(⇒). For the converse: given ∀ x ∃z T (eF , x, z), we deduce RFNΣ1 (ZFC). (We will have to deduce each instance of the Reflection Principle, for each 1–variable ∃z ψ(z, x ), ψ primitive recursive — for the meaning of z see below.) Recall that given each e we can explicitly construct a Diophantine polynomial pe (h x, yi, z1 , z2 , . . . , zm ) so that:

[∀ x ∃z T (e, x, z)] ↔ [∀ x ∃y, z1 , . . . , zm pe (h x, yiz1 , . . . , zm ) = 0]. (Of course [∃z1 , . . . zm pe (h x, yiz1 , . . . , zm ) = 0] ↔ [y = {e}( x )]. Since we aren’t using an universal equation, m may depend on e.) We will abbreviate the z1 , . . . , zm by z. 1. Now, if [F is total], then, for each n ∈ ω we have that F(n) is the sup of all {e}(k) so that k ≤ n and:

dPrZFC (d∀ x ∃z T (e, x, z)e)e ≤ n. ¨ 2. Since n is explicitly given, it is a bound on the Godel number of the proof. Therefore we can also obtain a n0 > n so that dPrZFC (d∃z T (e, x, z)e)e ≤ n0 . 3. Or, for another n00 , dPrZFC (d∃y, z pe (h x, yi, z) = 0e)e ≤ n00 . This follows from: • Every recursive function {e}(n) = m can be represented by a predicate Fe (n, m). (The algorithm to produce Fe given e is in Machtey and Young [40] p. 126 ff.) • Given Fe (n, m) we can use the procedure described in Davis’ paper on Hilbert’s 10th Problem [13] to get a polynomial pe out of Fe . 4. Since n00 is explicitly given, we can then recover proofs in ZFC of:

[∃y, z pe (h x, yi, z) = 0], all e under the specified conditions. 5. We then establish that ZFC proves, for all such e, PrZFC (d∃y, z pe (h x, yi, z) = 0e) → [∃y, z pe (h x, yiz) = 0]. 6. We now add the universal quantifier for x, and as it distributes over →,

[∀ x PrZFC (d∃y, z pe (h x, yi, z) = 0e)] → [∀ x ∃y, z pe (h x, yiz) = 0].

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7. This will of course also hold for (due to the implication’s properties): ˙ yi, z) = 0e)] → [∀ x ∃y, z pe (h x, yiz) = 0]. [∀ x PrZFC (d∃y, z pe (h x, 8. Finally the ∃wpe provide an enumeration of all Σ1 relations in ZFC of interest for Σ1 –soundness. Notice that in ZFC, f and F recursive,

[f is total] ↔ {[F is total] → [F dominates f ]}. 9. To show that the enumeration is exhaustive: suppose that for some p.r. ψ one has: ˙ y)), ∀ x PrZFC (∃y ψ( x, and that moreover the following sentence is proved: ˙ y))] → ∀ x ∃y ψ( x, y). [∀ x PrZFC (∃y ψ( x, 10. Put fψ ( x ) = miny ψ( x, y). Then:

[∀ x ∃y (fψ ( x ) = y)] ↔ [∀ x ∃y ψ( x, y)]. 11. That is, fψ is ZFC–provably total recursive, and therefore falls into the preceding case. Remark 5.70 The preceding result shows the essential inner connections between the inability of an axiomatic theory like S to “see” the totality of functions like F and beyond, and the usual formal sentences that assert the consistency of S itself. That is to say, the obstructions we have to face when trying to prove the totality of F in S have to do with the impossibility of proving the consistency of S itself with its own tools. The proof of the preceding result for PA can be found in the original Paris– Harrington paper [43].

Main theorem, first version The main result is first proved within informal arithmetic, that is, it holds true of the standard model for arithmetic; it generalizes a result proved in [7]: Proposition 5.71 If g is partial recursive, has an infinite domain, and is increasing over its domain, then [ P = NP]g ↔ [ P = NP]. Proof : It is enough to consider the bounding polynomial term. • [⇒]. There is a ∈ Domain g so that [tm ( x ) ≤ | x |g(a) + g( a)]. As g has an infinite domain and is increasing, we have that for any a in the domain, 0 there is g( a) = a0 . Thus the bounding term becomes [tm ( x ) ≤ | x | a + a0 ].

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• [⇐]. There is a0 ∈ ω so that [tm ( x ) ≤ | x | a + a0 ]. Since g is increasing and has an infinite domain, there is a ∈ Domain g so that a0 ≤ g( a). Therefore there exists a ∈ ω such that [tm ( x ) ≤ | x |g(a) + g( a)]. The argument above only requires that g have an infinite domain, and be increasing on its domain. This fact was pointed out to the authors by M. Guillaume [28, 35, 36]. From Proposition 5.71, for g as above: Proposition 5.72 If S has a model N with standard arithmetic, that is, if S is arithmetically sound, then Sg = S + [ P < NP]g ↔ [ P < NP] is consistent and holds of N. We should add here: Proposition 5.73 The following two assertions are equivalent in S: 1. [ P < NP]g ↔ [ P < NP]. 2. [g has an infinite domain and is increasing]. Proof : For (2) ⇒ (1), see Proposition 5.71. For (1) ⇒ (2), it suffices to show that if Domain g is finite, or nonincreasing, or both, then one cannot establish the desired equivalence. Remark 5.74 Do we have a function G such that it has an infinite domain, grows — over its domain — as fast as function F in Definition 6.107, but such that the formalized sentence [G is total] is undecidable in a reasonably useful SG ?

Function F and function G Remark 5.75 We will obtain a function G whose use here was implied by Guillaume [27, 28]. Notice that given function F as above in Definition 6.107, we can see that [F is total] is independent of S, but the structure of that function for the nonstandard models where ¬[F is total] holds isn’t clear at all. We immediately get an ω–inconsistency result, for S + ¬[F is total] + ∃y (F(0) = y) + ∃y (F(1) = y) + . . . is a consistent theory. Thus Σ1 –unsoundness (or equivalently ¬[F is total] implies ω–inconsistency. However is there a partial recursive function G like the one implied [27, 28] in Guillaume’s notes? We answer that question in the affirmative through the following: Proposition 5.76 There is a partial recursive function G so that: 1. If N |= S and has standard arithmetic, then N |= F = G.

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2. S + [F is total ] ` [Domain G is infinite and G is increasing]. 3. S + [F is total ] 6` [G is total ] and S + [F is total ] 6` ¬[G is total ]. 4. If S(α) , α > 1 an ordinal, is in the Turing–Feferman hierarchy over S extended by Σ1 –soundness reflection principles, then we can choose α as high as we wish in that hierarchy ≤ ω, so that S(α) ` [G is total ], but such that S( β) 6` [G is total ], β < α. Proof : As we have supposed that S has a model N with standard arithmetic, then so does theory S0 = S + [F is total]: • We can explicitly obtain a Diophantine polynomial p( x1 , . . . , xk ) so that: 1. S0 6` [∀ x1 , . . . p( x1 , . . .) > 0] while ∀ x1 , . . . p( x1 , . . .) > 0 holds of N. 2. S0 6` [∃ x1 , . . . p( x1 , . . .) = 0] and for some model M with nonstandard arithmetic, M |= ∃ x1 , . . . p( x1 , . . .) = 0. • Define: ζ = a1 , given that for some nonstandard model M, M |= [ p( a1 , a2 , . . . , ak ) = 0]. (We may wish to impose some uniqueness condition on h a1 , . . . , ak i.) By construction, for all models of S0 , [F is total]. Remark 5.77 We now informally describe an algorithm for the function we are looking for. Let h be such that S proves h to be total and strictly increasing. Put, for G: • If m < ζ, G(m) = F(m). • If m > ζ and n = h(m), then G(n) is undefined. • If m > ζ and n 6= h(m), then G(n) = F(n). So G will always have an infinite domain whenever [F is total] holds. And due to the dependence of G on ζ, it cannot be proved total even in a strong theory such as S + [F is total], that is, S + [S is Σ1 –sound]. For the last assertion, given the hierarchy over S plus Σ1 –sound reflection principles, it suffices to choose an adequate p so that ∀ x, . . . p( x, . . .) > 0 is only proved by S(α) , but not by any S( β) , β < α ≤ ω. (For the Turing–Feferman theorem see [2, 21, 23].) We can give a formal expression for G with the help of the ι symbol, which we then add to our formal background — S and the required extensions. • We can write down the algorithm for a Turing machine that never stops over any input n. Let {e0 } be that machine.

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da Costa, Doria • Consider h as in Remark 5.77. h is a S–provably total recursive function. Then write: H(n) = ι x {[( x = F(n)) ∧ (n 6∈ Image(h))]∨

∨ [( x = {e0 }(n)) ∧ (n ∈ Image(h))]}. Therefore S proves that such a function equals F (which is total for all models of S), but for the values of h, where it is undefined. As the image of h doesn’t exhaust all of ω, its complement is infinite, and H will have an infinite domain. • Now consider the first Diophantine polynomial p above together with ζ = a1 . If we write ∀ x p > 0 for [∀ x1 , . . . p( x1 , . . .) > 0], and similarly ∃ x p = 0, then consider the next expression that can be seen to be in the language of S (possibly extended for definitions): G(n) = ι x {[(∀ x p > 0) ∧ ( x = F(n))]∨

∨[(∃ x p = 0) ∧ [(n < ζ ) ∧ ( x = F(n))]∨ ∨ [(n ≥ ζ ) ∧ ( x = H(n))]]}. ζ is as above. This construction originates in the (symbolic) form (1 − β)X + βX0 , which we used in [5]; for the expression above see the construction in [8], p. 34. It is also akin to the construction in Kreisel’s Lemma.

Results for G We will proceed with G as we did with F in [7], as G is F over the standard model, like F it is undecidable in S whether it is total or not; the difference between both is that is proved (in an extended S) that G has an infinite domain and is increasing over its domain. Proposition 5.78 If S is consistent, then: 1. S ` [ P < NP]G ↔ [G is total ] ∧ [ P < NP]. 2. S ` [ P < NP]G → [G is total ]. 3. S 6` [ P < NP]G . 4. S ` [G is total ] → [F is total ], while the converse doesn’t hold. 5. S ` [G is total ] → Consis (S). Proof : The first two assertions are proved as in the cases for [ P < NP]F . For the third assertion see Proposition 5.76. Assertion 4 is a strict implication, since there will be models where [F is total] holds together with ¬ [G is total]. Finally assertion 5 is a consequence of S ` [F is total] → Consis (S). The same argument that we have used in [7] applies because [ P < NP]G has the same form as [ P < NP]F ; formal expressions that give F and G only differ in the corresponding M term; see the reference.

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A possible conclusion of the argument? Remark 5.79 Put: • S∗ = S + [ P < NP] ↔ [ P < NP]G , while: • S0 = S + [F is total ]. We review our results up to this point. We start from: If S is arithmetically sound, then so is S + [ P < NP] ↔ [ P < NP]g , where g has an infinite domain and is increasing over it. We also have that S proves the equivalence between [ P < NP] ↔ [ P < NP]g and [domain of g is infinite and g is increasing], but cannot prove each isolated clause on both sides of the equivalence, as [ P < NP] ↔ [ P < NP]G is independent of S. Given that, we introduce partial recursive function G. Notice ¨ that its Godel number can be explicitly computed, but it has the property that S0 = S + [F is total] neither proves [G is total] nor its negation. Now does S + [F is total] + [ P < NP] ↔ [ P < NP]F prove [G is total] ? No, it doesn’t, for the new condition only implies that domain of G is infinite and that G increases over its domain, and for a model with a strictly partial G, out of that function (in the model) we can establish [ P < NP] ↔ [ P < NP]G , since G is increasing and has an infinite domain. So, it doesn’t prove [G is total], and we conclude our argument. Proposition 5.80 For G as above: 1. S∗ ` [G has an infinite domain and is increasing ]. 2. S∗ 6` [G is total ] and S∗ 6` ¬[G is total ]. Proposition 5.81 S + [ P < NP]G ↔ [ P < NP] 6` [ P < NP]G . Proof : From S ` [ P < NP]G → [G is total].

Kreisel’s lemma Kreisel’s Lemma has a direct bearing on several results about the present matter [39]; it is nearly a folklore–like result, referred to by several authors. A summary proof also appears in a 1983 in a paper by Fortune et al. [22]. Kreisel’s Lemma gives a characterization for the set of S–provably total recursive functions, for S a theory like the ones we deal with in this paper: • S has a language based on first order classical predicate logic. • S has a recursively enumerable set of theorems. • S includes arithmetic; more precisely, Peano Arithmetic (PA) has an interpretation in S and has a model with standard arithmetic for that interpretation of PA.

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Essentially, what happens to the set of S–provably total recursive functions if we add true arithmetic sentences as new axioms to S? Let S be a theory as above. Let S1 be S plus all true Π1 sentences; S1 , we note, has a nonrecursive set of theorems. Then: Lemma 5.82 The S–provably total recursive functions and the S1 –provably total recursive functions coincide. Proof : Let S be a theory as just described. Let S1 be S plus all true Π1 sentences. Then: • The S–total recursive functions in S are in those of S1 , of course. • For the converse, consider the following: – If f is S–provably total, that property can be written as: S ` ∀ x ∃z T (ef , x, z), or, for Af ( x, y) ↔ (f ( x ) = y): S ` ∀ x ∃y Af ( x, y). – If f is provably total in S, then f 0 (m) = minx [ Af (m, x ) ∨ ∃ x P( x )] is also S–provably total. (Here P is arbitrary and primitive recursive.) Just compute: step 1, check for x = 0 to satisfy Af , step 2, check for x = 0 to satisfy ∃ x P( x ), . . . . The first x to satisfy one of them is output. • Now, notice that if ∀ x Q( x ) is true, again Q arbitrary and primitive recursive, then: S1 ` min [ Af (m, x ) ∨ ∃ x ¬ Q( x )] = f (m). x

That is, f and

f0

are both provably total and the same function in S1 .

• Let f be S1 –provably total recursive. Then it is thus proved by a finite subset of the axioms of S plus a finite set of Π1 sentences. • The conjunction of those Π1 sentences is also Π1 and is noted ∀ xQ( x ). • Then if S1 proves that f is total, it follows that S ` ∀ x Q( x ) → [f is total]. • Or S ` [f is total] ∨ ∃ x ¬ Q( x ). • Now define function f ∗ as below. Given n: – Compute f (n). – Simultaneously input 0, 1, 2, . . . to ¬ Q( x ) to check for ∃ x ¬ Q( x ). – If and when one of the two alternatives gives an output, write that output as the value f ∗ (n).

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• This is the same as f 0 above. Notice that f ∗ is total in S: at least one of the alternatives will give an output. Moreover, since ∀ x Q( x ) is true, the second alternative never holds, and f ∗ simulates f. We are done. Remark 5.83 Notice that the result also holds for Π1 sentences of the form ∀ x ∃y ≤ g( x ) Q( x, y), where g is S–provably total recursive. Independence results from some theory S that remain unaffected when we add all true Π1 sentences to S are called “nonteratological” by Kreisel.

More stuff Recall: Definition 5.84 If S is a first–order classical theory that includes arithmetic and has a model N |= S with a standard arithmetic fragment, then S is arithmetically sound. Go back to the proof of Proposition 5.71 to see that it suffices to suppose that f has an infinite domain and is increasing in order to establish the required equivalence. Therefore, if we have no further information, we see that S + [ P < NP] ↔ [ P < NP]g cannot prove that g is total, any such g. However we give below a stricter proof of that fact. We restate Proposition 5.72: Proposition 5.85 If S has a model N with standard arithmetic, that is, if S is arithmetically sound, then S + [ P < NP]g ↔ [ P < NP] is consistent and holds of N. Let’s now argue by contradiction. What happens if we do prove [ P < NP] in S? That is to say, what happens if the intuitive result contradicts the formal one, namely S 6` [ P < NP]F and S 6` [ P < NP]G , while formally we have S ` [ P < NP] ?

6. Good algorithms. that settles all of S AT (if there are such algorithms is a good algorithm. We will later see that there are approximately good algorithms and almost–everywhere good algorithms. Our goal here is to show:

A

NY POLY ALGORITHM

Lemma 6.86 S proves the (formal version of the) sentence:

[The set of Gödel numbers of good algorithms is recursive.]. We note the set of good algorithms K ⊂ ω. Of course, if K = ∅, then the lemma holds. We now consider the nontrivial case.

Preliminary definitions Recall that S is arithmetically sound if it has a model with a standard arithmetic part. From the arithmetic soundness of S, for a model for it with standard arithmetic (which will always be noted) N: Definition 6.87 Let theory S be such that N |= S, and such that we can construct a function FS for it as in Remark 6.107. Then S is unbounded if for any f so that N |= [f is strictly increasing total recursive], then S + [FS dominates f ] is consistent and holds of N. The idea is that we can take the bounding fast–growing function FS as fast– growing as possible. Remark 6.88 Again: from here on we will consider theory S, but now with an extra condition: 42

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• S includes Peano Arithmetic. • S is arithmetically sound, and model N makes it so. • S has a recursively enumerable set of theorems. • S is based on a first–order classical language. • S isn’t unbounded. We notice that both PA and ZFC satisfy the above requirements.

Preliminary results We get: Lemma 6.89

[ P = NP] ↔ ∃e, a ∀ x ∃z ≤ (| x | a + a) [ T (e, x, z) ∧ R( x, U ( T (e, x, z))]. Define: Remark 6.90 For our purposes, we specifically consider in this section Π1 sentences of the form: 1. ∀ x P, P primitive recursive. 2. ∀ x ∃y ≤ g( x ) P, where g, P are primitive recursive. Then:

∀ x ∃z ≤ (| x | a + a) [ T (e, x, z) ∧ R( x, U ( T (e, x, z))]

is a Π1 sentence, if a ∈ ω is seen as a constant. Hypothesis 6.91 We suppose that S ` [ P = NP]. Corollary 6.92 N |= [ P = NP]. Moreover, for integers e, a: Corollary 6.93 N |= ∀ x ∃z ≤ (| x | a + a) [ T (e, x, z) ∧ R( x, U ( T (e, x, z))]. Now let PA1 be the (nonrecursive) theory that consists of PA plus all true Π1 sentences. Then clearly: Lemma 6.94 If there are constants e, a so that

∀ x ∃z ≤ (| x | a + a) [ T (e, x, z) ∧ R( x, U ( T (e, x, z))] holds true of the standard model for arithmetic, then: PA1 ` [∀ x ∃z ≤ (| x | a + a) [ T (e, x, z) ∧ R( x, U ( T (e, x, z))]].

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Corollary 6.95 If there are constants e, a so that

∀ x ∃z ≤ (| x | a + a) [ T (e, x, z) ∧ R( x, U ( T (e, x, z))] holds true of the standard model for arithmetic, then: PA1 ` [ P = NP]. Proof : For we have:

[∀ x ∃z ≤ (| x | a + a) [ T (e, x, z) ∧ R( x, U ( T (e, x, z))]] → → [∃e, a ∀ x ∃z ≤ (| x | a + a) [ T (e, x, z) ∧ R( x, U ( T (e, x, z))]]. Then by detachment we get [ P = NP].

Conclusion of the argument From the arithmetic soundness of S, for a model for S with standard arithmetic N: Corollary 6.96 If Hypothesis F.5 holds, then N |= [ P = NP]. Now notice: • Suppose S ` [ P = NP]. • Then N |= [ P = NP]. • Therefore it is true that there is at least a (standard) pair e, a ∈ ω so that {e} with bound | x | a + a settles all of S AT. • Let PA1 be Peano Arithmetic plus all true Π1 arithmetic sentences. • Since PA1 ` [ P = NP], then either PA proves it or, at worst, PA0 = PA + ∀ x ∃z ≤ (| x | a + a) [ T (e, x, z) ∧ R( x, U ( T (e, x, z))] trivially proves it. Both are arithmetically sound theories with a recursively enumerable set of theorems. • Notice that both PA and PA0 have the same provably total recursive functions, and therefore have (in that sense) the same provability strength. This stems from Kreisel’s Lemma [11, 22]. • Then if: K (e, a) ↔Def [∀ x ∃z ≤ (| x | a + a) [ T (e, x, z) ∧ R( x, U ( T (e, x, z))], then for infinitely many numerals e, a, the sentences K (e, a) will be theorems of PA0 .

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• Moreover, for numerals e, a, the set of all PA0 –provable K (e, a) is a recursively enumerable set coded by the pair he, ai. This pair is the BGS code for poly machines. (A BGS pair [1] can be interpreted as a Turing machine {e} coupled to a clock that stops its operation on binary input x before | x | a + a cycles. Every BGS pair then codes a poly machine, and every poly machine can be represented as a BGS pair. Finally, if e runs over all of ω, then the BGS pairs run over ω × ω × ω.) • Now suppose that there is he, ai that settles S AT, that is to say, there is some Turing machine {e} with polynomial bound | x | a + a so that N |= K (e, a), but such that K (e, a) doesn’t come up among the theorems of PA0 . • Note that machine P0m . Then there is a machine M so that, for a machine that comes up in the theorem’s enumeration, Pm , we have: P0m = M ◦ P0m . (Of course M must be total and polynomial.) • We then have a proof that P0m is in the enumeration of theorems: – K (e, a) is a theorem, that is, Pm settles all of S AT. – Then there is a poly Turing machine M so that P0m = M ◦ Pm . – Therefore P0m settles all of S AT, that is, for m0 = he0 , a0 i, K (e0 , a0 ) is a theorem. We have proved: Proposition 6.97 Given Hypothesis F.5, the nonempty set K of all BGS–coded he, ai poly Turing machines so that K (e, a) holds true of the standard model for arithmetic is recursive. Proof : For it is recursively enumerable, and its complement (the set of counterexamples) is also recursively enumerable, as it is the image of a partial recursive function. Proposition 6.98 We can effectively compute the Gödel number of an algorithm that recursively enumerates the K (e, a). Proof : Begin a listing of all integers, 0, 1, 2, . . .. Input k ∈ ω: ¨ • We first check whether it is the Godel number of a proof in PA0 . If not, go to k + 1. If it is, go to the next step. • Check whether the last code in the proof is a code for K (e, a), for some pair e, a. • If not, go to k + 1. If so, output e, a and go to k + 1. ¨ Let eK be that Godel number.

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Corollary 6.99 Given Hypothesis F.5, for the nonempty set K, we have that T ` [K ⊂ ω is recursive.]. Proof : By the representation theorem, given Hypothesis F.5, as we can com¨ pute a Godel number for the algorithmic procedure that recursively enumerates K, we can represent it in PA, and therefore also in S; by definition it is (PA)–provably recursively enumerable, and as the complement of a recursively enumerable set it is recursive in PA and therefore in S. More precisely: • If {eK } is the algorithm that enumerates K as described above, then we can explicitly construct a Diophantine polynomial pK (y, x1 , . . . , xk ) so that it holds of a model N with standard arithmetic that:

[{eK }( a) = b] ↔ ∃ x1 , . . . , xk pK (h a, bi, x1 , . . . , xk ) = 0. • We then have that {eK }( a) = b if and only if: S ` ∃ x1 , . . . , xk pK (h a, bi, x1 , . . . , xk ) = 0. • Then (with perhaps some abuse of language): S proves:

[{y ∈ ω : ∃ x1 , . . . , xk pK (y, x1 , . . . , xk ) = 0} is recursively enumerable]. • As its complement is also a recursively enumerable set — it is the set of values of the counterexample function to [ P = NP] — we have that S proves that K is recursive. The immediate consequence is, Proposition 6.100 If S ` [ P = NP], then the following function f ∗ : 1. If n ∈ K, then f ∗ (n) = 0. 2. If n 6∈ K, then f ∗ (n) = the counterexample function to [ P = NP]. is provably total and recursive in S. Remark 6.101 We can give an expression for that function with the help of the ι–symbol added to the language S, if we also suppose that ∈ has been added to that language: f ∗ (n) = ι x {[( x = 0) ∧ (n ∈ K )] ∨ [( x = f (n)) ∧ (n 6∈ K )]. Here f is the counterexample function to [ P = NP]. Now, for the function f ∗ , notice that h∗ = f ∗ |ω −K , where | A denotes restriction to A:

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Proposition 6.102 If [ P = NP] holds, since f ∗ is recursive and proved total by S, then so is h∗ (over ω − K). Since ω − K is proved recursive by S, we can recursively renumber the domain of h∗ in order to obtain h which is recursive and provably total in T. We can do the same for the exotic formulations and obtain hF and hG [9]. Then: Proposition 6.103 1. S ` [fF is total ] → [F is total ]. 2. S ` [fG is total ] → [G is total ] 3. S ` [(f ∗ )F is total ] → [F is total ]. 4. S ` [(f ∗ )G is total ] → [G is total ]. Proof : The first two implications are simply [ P < NP]F → [F is total ] and [ P < NP]G → [G is total ]. To prove the third and fourth consider the restriction (f ∗ )F |ω −K , and correspondingly for G. Notice that (f ∗ )F , restrictions, etc, are expressed with the predicate ¬ Q(. . . , F( a), . . .), or equivalently with a function composition (f ∗ )F ( a) = f ∗ [F( a)]. The desired results follow. Remark 6.104 To conclude one has to establish the equivalence:

[(f ∗ )G is total ] ↔ [(f ∗ ) is total ]. (On that equivalence see also [9].) That equivalence is [ P = NP] → [ P = NP]G . As we have supposed that S ` [ P = NP], S + [ P = NP] + [ P = NP] ↔ [ P = NP]G is consistent and holds of a model for S with standard arithmetic part. So, the results in Section 3 apply here. We can summarize here our argument in the next proposition: Proposition 6.105 S ` [ P = NP] → [f ∗ is total ]. Proof : The preceding discussion arises from S + [ P = NP] ` [f ∗ is total ], and we supposed that S ` [ P = NP], a hypothesis that reasserts the consistency of S + [ P = NP]. Remark 6.106 We have also shown that if [ P = NP] is true of the standard model for arithmetic, then PA0 proves it. Therefore, if PA0 doesn’t prove [ P = NP], then [ P = NP] doesn’t hold true of the standard integers. Or, if either PA0 proves [ P < NP], or if [ P < NP] is independent of PA0 , then [ P < NP] holds true of the standard model for arithmetic. That is to say, we conclude that if

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we show [ P < NP] and [ P = NP] to be independent of PA0 , then we show that [ P < NP] is true. One final observation: since K is proved recursive by PA0 , which is a theory that has the same provably total recursive functions as PA, then its characteristic function is primitive recursive on some function in the Kreisel hierarchy [38] < e0 . We can improve on that result by choosing a weaker arithmetic theory that can still represent intuitive computations; see on these matters [4].

Functions F and f Function f will be any S–provably total, strictly increasing, recursive function. Recall that for such an f, S ` ∀ x ∃y [f ( x ) = y]. Also, we use here a well–known recursive function that is diagonalized over all S–provably total recursive functions. We note it F, or sometimes FS . See [7] and [34], pp. 51–52 on it. Remark 6.107 For each n, F(n) = maxk≤n ({e}(k)) + 1, that is is the sup of those {e}(k) such that: 1. k ≤ n. 2. dPrS (d∀ x ∃z T (e, x, z)e)e ≤ n. ¨ PrS (dξ e) means, there is a proof of ξ in S, where dξ e means: the Godel ¨ number of ξ. So dPrS (dξ e)e means: “the Godel number of sentence ‘there is a proof of ξ in S.’ ” Condition 2 above translates as: there is a proof of [{e} is ¨ total] in S whose Godel number is ≤ n. Proposition 6.108 We can explicitly compute a Gödel number eF so that {eF } = F. Proposition 6.109 If S is consistent then: S 6` ∀m∃n [{eF }(m) = n]. Notice that [ P < NP]F ↔ ∀m, a ∃ x ¬ QF (m, a, x ). Remark 6.110 Functions such as F are prominent in the study of transfinite progressions of theories [2, 21, 23]. The idea of such functions goes back to Kleene [35, 36] and was first thoroughly explored by Kreisel [38]. Recall the following crucial fact: Lemma 6.111 If I ⊆ ω is infinite and 0 ∈ I, then: S ` {[∀m ∀ a ∈ I ∃ x ¬ Q(m, a, x )] →

→ [∀m ∀ a ∈ ω ∃ x ¬ Q(m, a, x )]}.

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The meaning of this result is: as long as we have an infinite succession of ever larger bounds that make the Turing machines polynomial, our standard definitions hold. The size of the intermediate gaps between each pair of bounds doesn’t matter. However notice that there is in general no equivalence in S between [ P < NP] and [ P < NP]F (see [7]): Proposition 6.112 S ` [ P < NP]F ↔ {[F is total ] ∧ [ P < NP]}. Sketch of proof : For the formal, computational proof see [7]. It is easy to see that: S + [ P < NP] + [F is total] ` [ P < NP]F . For the converse, the fact that the exotic counterexample function fF (see below after Lemma 6.114) is total implies that [ P < NP] and [F is total] hold at the same time (we will require Lemma 6.111 here). Remark 6.113 Of course, if f is provably total in our theory, then S ` [ P < NP] ↔ [ P < NP]f . We again refer to a result that follows from the above due to it looking tantalizingly close to very interesting stuff: Lemma 6.114 S ` [ P < NP]F → [F is total.]. Remark 6.115 Formal proof is in [7]. The following argument clarifies the meaning of the lemma and gives an informal proof for it: let: f(c,F) (hm, ai) = min [¬ Q(m, F( a), x )], x

where we can look at F as a (partial) recursive function. (The brackets h. . . , . . .i note the usual 1–1 pairing function.) Now if fF is total, then F( a) has to be defined for all values of the argument a, that is, F must be total. The function f(c,F) is the so–called exotic counterexample function to [ P = NP]F . We can similarly define a standard counterexample function : fc (hm, ai) = min [¬ Q(m, a, x )]. x

Remark 6.116 Always keep in mind that the counterexample function is provably total in our theory S iff S proves P < NP. Again recall: as S ` [F is total] ↔ [S is Σ1 –sound] (see [2] on that equivalence), and also as S ` [S is Σ1 –sound] → Consis (S), S ` [F is total] → Consis (S). Then: Lemma 6.117 S ` [ P < NP]F → Consis (S). Proof,: For the proof see the discussion in sections 4 and 5.

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Proposition 6.118 If S is consistent, then S doesn’t prove [ P < NP]F . Proof : S ` [[ P < NP]F → (F is total)]. (Lemma 6.114.) So, S cannot prove [ P < NP]F . Corollary 6.119 [ P = NP]F is consistent with S. However this is not the whole story, far from it, as F may be partial in the theory (which will then only have nonstandard models) while the counterexample function is total, so that [ P < NP] holds. If N |= S and makes it arithmetically sound, that is, N has a standard arithmetic part for the arithmetic in S : Proposition 6.120 N |= S + [ P < NP] ↔ [ P < NP]F . Proof : F is total in the standard model for arithmetic. Proposition 6.121 1. For f as above, S ` [ P < NP] ↔ [ P < NP]f . 2. For F as above, [ P < NP] ↔ [ P < NP]F is independent of S. Proof : S doesn’t prove that equivalence due to Proposition 6.112, since otherwise it would prove [F is total.]. On the other hand, consistency of: S + [ P < NP] ↔ [ P < NP]F follows from the fact that it trivially holds in the standard model for arithmetic. Equivalence holds since S ` [ P < NP]f ↔ {[f is total ] ∧ [ P < NP]} and f is provably total. Remark 6.122 We can take, if convenient,

[ P < NP] ↔Def ∀m, a ∃ x ¬ Q(m, f ( a), x ).

Recall that if the sentence ¬ [F is total] holds in some model for our theory S, the fact that we have in that model ∃ x (F(0) = x ), ∃ x (F(1) = x ), ∃ x (F(2) = x ), . . . , together with ¬ [F is total] shows that: if ¬ [F is total] holds, then S + ¬ [F is total] is ω–inconsistent. More precisely: if some theory S0 proves ¬ [F is total], then it is ω–inconsistent. So, if some theory S0 is ω– consistent, then [F is total] is consistent with it. Thus again the partial result: Proposition 6.123 If S + [ P = NP]F is ω–consistent, then S + [ P = NP] is consistent.

7. Function f . It is fast–growing and non computable; in fact it grows in its peaks at least as fast as the Busy Beaver function. The full counterexample function f is defined as follows; let ω code an enumeration of the Turing machines. Similarly code by a standard code SAT onto ω:

W

E NOW CONSIDER the full counterexample function.

• If n ∈ ω isn’t a poly machine, f (n) = 0. • If n ∈ ω codes a poly machine: – f (n) = first instance x of SAT so that the machine fails to output a satisfying line for x, plus 1, that is, f (n) = x + 1. – Otherwise f (n) is undefined, that is, if P = NP holds for n, f (n) = undefined. As defined, f is non computable. It will also turn out to be at least as fast growing as the Busy Beaver function, since in its peaks it tops all intuitively total recursive functions. The idea in the proof goes as follows: ¨ • Use the s–m–n theorem to obtain Godel numbers for an infinite family of “quasi–trivial machines” — soon to be defined. The table for those Turing machines involves very large numbers, and the goal is to get a compact code for that value in each quasi–trivial machine so that their ¨ Godel numbers are in a sequence g(0), g(1), g(2), . . ., where g is primitive recursive. 51

52

da Costa, Doria • Then add the required clocks as in the BGS sequence of poly machines, ¨ and get the Godel numbers for the pairs machine + clock. We can embed the sequence we obtain into the sequence of all Turing machines. • Notice that the subsets of poly machines we are dealing with are (intuitive) recursive subsets of the set of all Turing machines. More precisely: if we formalize everything in some theory S, then the formalized version ¨ of the sentence “the set of Godel numbers for these quasi–trivial Turing ¨ machines is a recursive subset of the set of Godel numbers for Turing machines” holds of the standard model for arithmetic in S, and vice versa. However S may not be able to prove or disprove that assertion, that is to say, it will be formally independent of S. • We can thus define the counterexample functions over the desired set(s) of poly machines, and compare them to fast–growing total recursive functions over similar restrictions. Recall:

Definition 7.124 For f , g : ω → ω, f dominates g ↔Def ∃y ∀ x ( x > y → f ( x ) ≥ g( x )). We write f g for f dominates g.

Quasi–trivial machines Recall that the operation time of a Turing machine is given as follows: if M stops over an input x, then the operation time over x, tM = | x | + number of cycles of the machine until it stops. Example 7.125 • First trivial machine. Note it O. O inputs x and stops. tO = | x | + moves to halting state + stops. So, operation time of O has a linear bound. • Second trivial machine. Call it O0 . It inputs x, always outputs 0 (zero) and stops. Again operation time of O0 has a linear bound. • Quasi–trivial machines. A quasi–trivial machine Q operates as follows: for x ≤ x0 , x0 a constant value, Q = R, R an arbitrary total machine. For x > x0 , Q = O or O0 . This machine has also a linear bound.

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Remark 7.126 Now let H be any fast–growing, superexponential total machine. Let H0 be a total Turing machine. Form the following family Q... of quasi–trivial Turing machines with subroutines H and H0 : 0

1. If x ≤ H(n), QH,H ,n ( x ) = H0 ( x ); 0

2. If x > H(n), QH,H ,n ( x ) = 0. 0

Proposition 7.127 There is a family Rg(n,|H|,|H0 |) ( x ) = QH,H ,n ( x ), where g is primitive recursive, and |H|, |H0 | denotes the Gödel number of H and of H0 . Proof : By the composition theorem and the s–m–n theorem. Remark 7.128 We are interested in quasi–trivial machines where H0 = T, the standard truth–table exponential algorithm for SAT. We first give a result for the counterexample function when defined over all Turing machines (with the extra condition that the counterexample function = 0 if Mm isn’t a poly machine). We have: Proposition 7.129 If N (n) = g(n) is the Gödel number of a quasi–trivial machine as in Remark 7.126, then f ( N (n)) = H(n) + 1. Proof : Use the machines in Proposition 7.127 and Remark 7.128.

Proof of non–domination Our goal here is to prove the following result: Proposition 7.130 For no total recursive function h does h f . Proof : Suppose that there is a total recursive function h such that h f . Notice: • Given such a function h, we can obtain another total recursive function h0 which satisfies: 1. h0 is strictly increasing. 2. For n > n0 , h0 (n) > h(g(n)), with g as in Prop. 3. • Given a total recursive h, there is a total recursive h0 that satisfies the previous conditions. For given h, we obtain out of that total recursive function by the usual constructions a strictly increasing total recursive h∗ . Then if, for instance, Fω is Ackermann’s function, h0 = h∗ ◦ Fω will do. (The idea is that Fω dominates all primitive recursive functions, and therefore h∗ composed with it dominates g(n).)

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¨ We have that the Godel numbers of the quasi–trivial machines Q are given by g(n). Choose adequate quasi–trivial machines, so that f (g(n)) = h0 (n) + 1, from Proposition 3. From Remark 4.60 and Lemma 4.61 we conclude our argument. If we make explicit the computations, for g(n) (as the argument holds for any strictly increasing primitive recursive g): f (g(n)) = h0 (n) + 1 = h∗ (Fω (n)) + 1, and

h∗ (Fω (n)) > h∗ (g(n)).

For N = g(n),

f ( N ) > h∗ ( N ) ≥ h( N ), all N.

Therefore no such h can dominate f . Corollary 7.131 No total recursive function dominates f .

BGS–like sets We use here the BGS [1] set of poly machines:

hMm , | x | a + a i, where we couple a Turing machine Mm to a clock regulated by the polynomial | x | a + a, that is, it stops Mm after | x | a + a steps in the operation over x, where x is the machine’s binary input and | x | its bit–length. A more general machine–clock couple will also be dealt with here:

hMm , | x |(a) + f ( a)i 7→ Mc(m,|f |,a) , ¨ Its Godel number is given by c(m, |f |, a), with c primitive recursive by the s–m– n theorem. Remark 7.132 Notice that we can have c such that, for parameters a, b, if a < b, then c(. . . a . . .) < c(. . . b . . .). P < NP is given by a Π2 arithmetic sentence, that is, a sentence of the form “for every x there is an y so that P( x, y),” where P( x, y) is a very simple kind of relation.2 Now given a theory S with enough arithmetic in it, S proves a Π2 sentence ξ if and only if the associated Skolem function fξ is proved to be total recursive by S. For P < NP, the Skolem function is what we have been calling the counterexample function. However there are infinitely many counterexample functions we may consider, an embarras de choix, as they say in French. Why is it so? For many adequate, reasonable theories S, we can build a recursive (computable) scale of 2 It

is a primitive recursive predicate.

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functions3 f0 , f1 , . . . , fk , . . . with an infinite set of S–provably total recursive functions so that f0 is dominated by f1 which is then dominated by f2 , . . . , and so on. Given each function fk , we can form a BGS–like set BGSk , where clocks in the time–polynomial Turing machines are bounded by a polynomial:

| x | fk ( n ) + fk ( n ) , where | x | denotes the length of the binary input x to the machine. We can then consider the recursive set: [ BGSk k

of all such sets. Each BGSk contains representatives of all poly machines (time polynomial Turing machines). Now, what happens if: • There is a function g which is total provably recursive in S and which dominates all segments fk of counterexample functions over each BGSk ? • There is no such an g, but there are functions gk which dominate each particular fk , while the sequence g0 , g1 , . . . is unbounded in S, that is, grows as the sequence F0 , F1 , . . . in S ? We will take a look into these queries.

Exotic BGSF machines Now let F be a fast growing, intuitively total, algorithmic function. We consider exotic BGSF machines, that is, poly machines coded by the pairs hm, ai, which code Turing machines Mm with bounds | x |F(a) + F( a). Since the bounding clock is also a Turing machine, now coupled to Mm , there is a primitive recursive map c so that: hMm , | x |F(a) + F( a)i 7→ Mc(m,|F|,a) , where Mc(m,|F|,a) is a poly machine within the sequence of all Turing machines. We similarly obtain a g as above, and follows: Proposition 7.133 Given the counterexample function fk defined over the BGSk –machines, for no ZFC–provable total recursive h does h fk . ¨ Proof : As in Proposition 7.130; use Godel number coding primitive recur¨ sive function c to give the Godel numbers of the quasi–trivial machines we use in the proof. 3 We

thank L. Gordeev for this construction.

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Remark 7.134 Notice that we have a perfectly reasonable formalization for our big question: k

[ P < NP]k ↔ [ P < NP]f . Also, S ` [ P < NP]k ↔ [ fck is total]. So our analysis will give estimates for the growth rate of each counterexample function fck . Remark 7.135 The previous statements together with Remark 6.122 has interesting consequences, which we will briefly pursue below. For the proof of the proposition choose a BGSk so that fk dominates all strictly increasing fast growing provably total recursive functions that eventually appear in the proof. We can state, for total fck : Proposition 7.136 For each j there is a k, k > j + 1, so that S proves the sentence “fk doesn’t dominate the BGSk counterexample function fck .” However we cannot conclude that “for all j, we have that...” since that would imply that S proves “for all j, f j is total” as a scholium, which cannot be done (as that is equivalent to “FS is total,” which again cannot be proved in S). What can be concluded: let S0 be the theory S + FS is total. Then: Proposition 7.137 If S is consistent and if fck is total in a model with standard arithmetic for each k, then S0 proves: there is no proof of the totality of fck , any k, in S. Proof : See below the discussion. Remark 7.138 Notice that: • S0 ` ∀k ([ P < NP]k ↔ [fck is total]), while S cannot prove it. • S0 ` ∀k ([ P < NP]k ↔ [ P < NP]) while again S cannot prove it. • S0 is S + [S is Σ1 sound]. Remark 7.139 That means that we can conclude: S0 proves that, for every k, S cannot prove [ P < NP]k . Now : does the [ P < NP]k adequately translate our main question? Remark 7.140 Notice that theory S + FS is total is S + S is Σ1 –sound. This will have further consequences.

Properties of the counterexample function f and the infinitely many fc are very peculiar objects. They are fractal–like in the following sense: the essential data about NP–complete questions is reproduced mirror–like in each of the f (or over each BGSk ). The different BGSk are distributed over the set of all Turing machines by the primitive recursive function c(m, k, a). Also we cannot argue within S that for all k, fk dominates . . . , as that would imply the totality of the recursive function FS .

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A few intuitions It is interesting to keep in mind a picture of these objects. First notice that the BGS and BGSk machines are interspersed among the Turing machines. The ¨ quasi–trivial Turing machines have their Godel numbers given by the primitive recursive function c(k, n) — we forget about he other parameters — where: • k refers to f k and to BGSk as already explained; • n is the argument in f k (n). So fast–growing function f k is sort of cloned among the values of the BGSk counterexample function while slightly slowed down by c. (Recall that c is primitive recursive, and cannot compete in growth power with the f k . Function f k compresses what might be a very large number into a small ¨ code given by the Godel number of gk and by n (recall that the length of f k (n) is the order of log f k (n)). The effect is that all functions f j , j < k embedded into the k–counterexample function via our quasi–trivial machines keep their fast– growing properties and allow us to prove that the counterexample function is fast–growing in its peaks for BGSk . For j > k the growth power of f k doesn’t compensate the length of the parameters in the bounding polynomial that regulates the coupled clock in the BGSk machines. ¨ Finally while j < k, the compressed Godel numbers of the quasi–trivial mak chines — they depend on the exponent and constant of the polynomial xf (n) + f k (n) which regulates the clock — grow much slower that the growth rate of the counterexample function over these quasi–trivial machines (depending on f j ) and so their fast growing properties come out clearly.

8 Acknowledgments This paper was supported in part by CNPq, Philosophy Section. It is part of the research efforts of the Advanced Studies Group, Production Engineering Program, at C OPPE –UFRJ and of the Logic Group, HCTE – UFRJ. We thank Profs. R. Bartholo, C. A. Cosenza, S. Fuks (in memoriam), S. Jurkiewicz, R. Kubrusly, and F. Zamberlan for support.

A few half–baked intuitions A Does a large cardinal imply P < NP ? We consider the theory S0 as above. Waving hands a lot. . . Is there a large cardinal involved? Let’s consider the following (again very very sketchy) argument: • Notice that [ P < NP] is independent of ZFC if and only if, for the so– called counterexample function f to [ P = NP], [f is total] is independent of ZFC. • Form the theory ZFC∗ = ZFC + [f is total.] • If [ P < NP] ↔ ∀m ∃ x ¬ Q(m, x ), then for ZFC∗ :

[ P < NP] ↔ ∀m ∃ x ≤ f (m) [¬ Q(m, x )]. That is a Π01 sentence, and so it is true of the standard model of arithmetic (within an adequate model for ZFC∗ ). • Moreover, ZFC∗ implies the consistency of ZFC, as we have that: ZFC ` [f is total]→[F is total] and ZFC∗ ` [f is total]. F is of course as in [ P = NP]F . • Now: is there some large cardinal involved? That is to say, do we have that theory: ZFC + [Some large cardinal hypothesis ] ` [ P < NP ] ? • The idea here is as follows: we can give a meaning to 2ℵ0 . Then, in a nutshell: we know that the counterexample function f is highly oscillating, but (over all Turing machines) it overtakes all total recursive functions. Consider its increasing envelope, f ∗ (n) = supn0 ≤n f (n). Clearly f ∗ ≥ f. We can certainly give some set–theoretic meaning to f ∗ (ℵ0 ). Would that be an inaccessible cardinal? And, can we arrange things so that [f ∗ (ℵ0 ) inaccessible ]→[f is total ] ? 58

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More on inaccessible cardinals We can perhaps do better. Recall that a strongly inaccessible cardinal λ satisfies: 1. λ > ω. 2. If α is a cardinal and α < λ, then 2α < λ. 3. For every family β i , i ∈ ι, ι < λ, and for each i, β i < λ, then supi ( β i ) < λ. Let Consis( T ) mean the (usual) sentence that asserts that theory T is consistent. Let Card(λ) mean that λ is a cardinal, and let SInacT (λ) mean that λ is strongly inacessible for theory T. Finally let F be the function that appears in the exotic dormulation: Proposition A.1 There is a λF so that:

(ZFC + [F is total ]) ` Card(λ) ∧ SInacZFC λF ). Tentative sketch of proof, to be further developed : 1. As we suppose that Consis(ZFC + [F is total]) holds, then it has a model M. 2. Now, ZFC + [F is total ] ` Consis (ZFC). 3. It is a theorem of ZFC that: Consis(ZFC) ↔ ∃ x [ x |= ZFC]. (We can also take this as a definition for Consis( T ). For details see K. Kunen, Set Theory, North–Holland (1983), p. 144, and references therein.) 4. Given that: ZFC + [F is total ] ` Consis (ZFC), there is a set x ∈ M that is a model for ZFC. 5. Write Vλ = M − x. 6. Since Vλ is nonempty and as the axiom of choice holds, there are ordinals in it. 7. Therefore there is at least a cardinal in Vλ . 8. Pick up the smallest of such cardinals; note it λ: (a) One easily sees that for each cardinal α ∈ V, λ is different from 2α . (b) Also for each sequence β i etc., λ is different from supi β i . (Both conditions hold because if not, λ would be in V.)

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da Costa, Doria 9. Finally for all cardinals α ∈ V, λ > α. For if not, there would be a β ∈ V, and λ < β, and λ would be in V. This also means that V is in fact a set, V.

This argument doesn’t show that existence of this particular inaccessible cardinal λ proves P < NP; it only shows (or purports to show) that our extended theory ZFC + [F is total] implies the existence of an inaccessible cardinal. One must now show that f ∗ (ℵ0 ) can be interpreted as an inacessible cardinal, and that it then implies P < NP. We would then have λ ≤ f ∗ (ℵ0 ).

B Proof strength and Kolmogorov–Chaitin complexity Remark B.1 Let us be given a theory S (see the specifications above, in Remark 2.12) which moreover has in its alphabet an infinite sequence of terms x1 , x2 , x3 , . . . that stand for variables and let us be given a sequence of independent Π1 sentences

∀ x j P ( x j ), ∀ x k P ( x k ), ∀ x m P ( x m ), . . . , ∀ xr P ( xr ), where the sequence of positive integers j, k, m, . . . , r is arbitrarily long and has a characteristic function (when seen as a subset of the positive integers) of maximum KC–complexity. Then for theory S and S0 = S + ∀ x j P( x j ) + ∀ xk P( xk ) + ∀ xm P( xm ) + . . . + ∀ xr P( xr ) hold the following: • Both S and S0 have the same proof strength (this follows from Kreisel’s Lemma, that adding independent Π1 true sentences to a theory S does not change its proof strength). • KS–complexity of S0 may be made arbitrarily higher than that of S (as any binary coding for the axioms of S0 will include a coding for the characteristic function of j, k, m, . . . , r).

61

C The halting function and forcing We can obtain an expression for a ‘faceless’ object within the language of ZFC [6]. (A ‘faceless’ object is an object such that all its properties are undecidable within the formal theory.) Our construction goes as follows: let Q( x, a1 , a2 , . . . , an ) be a predicate on the fixed parameters a1 , . . .. Think of that predicate as, say, the characterization of the set of all hamiltonians on all symplectic manifolds. An object that satisfies Q is a Q–object. In our example, a Q–object is simply an arbitrary hamiltonian. Suppose given an enumeration of all the predicates Pk in T. Moreover we suppose that: 1. For ξ i ∈ L T , T ` Q(ξ i ). 2. For ξ i , ξ j , i 6= j, ∈ L T , T ` Pk (ξ i ) and T ` ¬ Pk (ξ j ).. 3. Out of that we list all nontrivial predicates Pk0 that apply to Q–defined objects. So, out of all formal expressions ζ i , i ∈ ω, that provably satisfy Q, we pick up those couples ζ i , ζ j where P can be proved to be satisfied for a member of the couple and is provably denied for the other member of the couple. Out of that we can obtain in T a recursive enumeration of pairs of expressions ξ 2i , ξ 2i+1 , i a natural number, that represent different objects and such that: 1. Both T ` Q(ξ 2i ) and T ` Q(ξ 2i+1 ). 2. T ` Pi (ξ 2i ) and T ` ¬ Pi (ξ 2i+1 ), where the Pi are nontrivial predicates (relative to Q) in T that range over Q– objects. Then: Proposition C.1 Within T: ∗ of expressions for Q–objects • Undecidability. There is a countable family ζ m in T such that there is no general algorithm to decide, for any nontrivial Q– property Pk in T whether that expression satisfies (or doesn’t satisfy) Pk .

• Incompleteness. There is a Q–object all whose nontrivial Q–properties cannot be proved within T. Sketch of the proof of the proposition : (Details in [6].) We prove the second assertion; a proof of the first assertion is similarly obtained. We must construct an expression for a problem that satisfies our hypotheses. We proceed in a stepwise manner: 62

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• Let ξ 1 , ξ 2 , ξ 3 , . . . , be an infinite countable sequence of mutually indepen¨ ¨ dent Godel sentences in T. Let m1 , m2 , . . . , be the corresponding Godel numbers. • Form the θ1 = θ (m1 ), θ2 = θ (m2 ), . . . . • Define: e0j = 1 − θ j , e1j = θ j , all j ∈ ω. • Let τn be a variable that ranges over all 2n binary sequences of length n. Code those by ordered n–tuples of zeroes and ones and establish a map f between those n–bit binary sequences and all n–factor products 0 n e1α e2α . . . enα , so that in the i–th position 0(1) 7→ ei0 (ei1 ). Given τn , the associated product is f (τn ). Notice that given a specific model for T, all such sequences equal 0 but for a single one, that equals 1. The actual nonzero sequence depends on which model is chosen. • Order the predicates P1 , P2 , . . . in some fixed way. We will convene that T ` Pi (ζ 2i−1 ) while T ` ¬ Pi (ζ 2i ). • Now list all finite binary sequences and select from those a countably infinite set of mutually incompatible sequences (incompatible sequences are those that do not have common extensions) of growing length (such as, say, 1, 01, 001, 0001, . . . ). Given each binary sequence it is either in our previously chosen set of incompatible sequences or will be the extension of some previously chosen sequence. Note the incompatible set {τ1 , τ2 , . . .}; if τj is any sequence, note τj0 , τj00 , . . . , its extensions. (A general extension is noted τjk .) The expression we require is: ζ

= ζ 1 [(1/2) f (τ1 ) + (1/2)2 ∑ ∗ f (τ10 ) + . . .] +

+ ζ 2 [(1/2) f (τ2 ) + (1/2)2 ∑ ∗ f (τ20 ) + . . .] + . . .

(3)

(∑∗ denotes sum over all extensions of equal length in the factors e.) Equation (3) can obviously be given a more compact (albeit less legible) expression in LT . We have T ` “ζ is an expression for a Q–object,” while none of the remaining properties Pi for Q–objects can be verified. (The first assertion is proved in a similar vein. Only we choose indepen¨ dent undecidable functions θk (m) instead of independent Godel numbers—for instance, we take a sequence of functions θ, θ 0 , θ 00 , . . . , of progressively higher arithmetic degrees; see [6].) Thus ζ in equation (3) is a ‘faceless’ object. The theory is blind to it; we can only prove that ζ is a Q–object, and nothing more. Now the intuition behind

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set–theoretically generic objects is a similar one: set–theoretically generic objects are faceless, with no properties beyond those that are ‘forced’ upon them by the forcing conditions. Therefore: can we go from ζ to a generic set?

D From ζ to a generic set Let the τi be as in the preceding Section. For i 6= j, τi ⊥τj . Let V be a model for ZFC, and let B = RO(2ω ) be the regular open algebra of 2ω [3]. (See that reference for notation.) Form the Boolean extension V B . Then define: Definition D.1 1. Bk = { g ∈ 2ω : τk ⊂ g}. 2. u(nˆ ) = Bn , for n ∈ ω. Remark D.2 Notice that the coefficients in equation (3) are a coding for the Bk ; for instance: [(1/2) f (τ1 ) + (1/2)2 ∑ ∗ f (τ10 ) + . . .] 7→ B1 . Those coefficients are therefore a coding for the Cohen ‘name’ of the generic set in V. Then: ˆ Lemma D.3 V B |= u ⊂ ω. Proof :

ku ⊂ ωˆ k B =

^

[u(nˆ ) → knˆ ∈ ωˆ k B ].

n∈ω

Or,

ku ⊂ ωˆ k B =

^

[u(nˆ ) → 1] = 1.

n∈ω

As usual, put Np = { g ∈ 2ω : p ⊂ g}. Then, Definition D.4 p | ` ψ if and only if Np ≤ kψk. Lemma D.5 1. p | ` nˆ ∈ u if and only if for all n0 < n, p(n0 ) = 0 and p(n) = 1. 2. p | ` nˆ 6∈ u if and only if there is n0 < n so that p(n0 ) = 1 or p(n) = 0. Proof : For the first condition: p | ` nˆ ∈ u if and only if Np ≤ knˆ ∈ uk if and only if Np ≤ Bn if and only if p ≤ τn . Now that last condition means that p refines τn , that is to say, p ⊃ τn . For the second condition: p | ` ¬(nˆ ∈ u) if and only if Np ≤ (knˆ ∈ uk B )∗ if and only if Np ≤ Bn∗ if and only if p⊥τn if and only if p 6∈ Bn (seen as a set of finite conditions). The incompatibility is then translated into the condition that either there is an n0 < n so that p(n0 ) = 1 or p(n) = 0. Also, 65

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ˆ n ∈ ω, then for an arbitrary Lemma D.6 If p | ` xˆ = u and for q ≤ p, q | ` nˆ ∈ x, n there is an r ≤ p so that r | ` nˆ ∈ u. Proof : p | ` xˆ = u if and only if Np ≤ k xˆ = uk B . But:

k xˆ = uk B = k∀nˆ ∈ ωˆ (nˆ ∈ xˆ ↔ nˆ ∈ u)k B . Thus, Np ≤

^

k(nˆ ∈ xˆ ) → (nˆ ∈ u)k B .

(4)

n∈ω

Thus, for specific instances of n ∈ ω, Np ≤ knˆ ∈ xˆ → nˆ ∈ uk B .

(5)

Now [3], recall that p | ` α → β implies that for all refinements q ≤ p so that q | ` α, there is a refinement r ≤ q so that r | ` β. We have two possibilities: • n 6∈ x. Then knˆ ∈ xˆ k B = 0, or Np = 0, or finally r = 0, trivially. ˆ and there is a nontrivial r ≤ p so that r | ` nˆ ∈ u. • n ∈ x. Thus V B |= nˆ ∈ x, Now: Proposition D.7 For all x ∈ Pω, V B |= xˆ 6= u. Proof : We have already that V B |= u ⊂ ωˆ (Lemma D.3); the proposition asserts that V B |= u 6∈ ( Pω )ˆ, or, for each x ⊂ ω, V B |= xˆ 6= u. If not, then there is a p 6= 0 such that Np ≤ ku = xˆ k B . Choose an r ≤ p that satisfies the conditions in Lemma D.6. Then r | ` nˆ ∈ u. As usual, put: 1. If n ∈ x, write r 0 = r ∪ {hn, 0i}; 2. If n 6∈ x, write r 0 = r ∪ {hn, 1i}. (But see below.) Case 1: Then r 0 | ` nˆ 6∈ u, from Lemma D.5. Thus, r 0 | ` (nˆ ∈ xˆ ) ∧ (nˆ 6∈ u). A contradiction. Case 2: This situation is a bit more involved, due to Lemma D.5. First notice that as we take equations (4) and (5) in Lemma D.6, it implies that there is an r ≤ p, and r | ` nˆ ∈ u, any n ∈ ω. So, the form of r is restricted by our ˆ Subject to that condition, take r 0 = r ∪ {hn, 1i}. hypothesis that p | ` u = x. 0 Thus r | ` (nˆ 6∈ xˆ ) ∧ (nˆ ∈ u). Again a contradiction. Now let v be the usual generic set in Boolean–valued models (for example, see Bell [3], p. 55, theorem 2.6). Then: Cn = { g ∈ 2ω : g(n) = 1},

(6)

v(nˆ ) = Cn .

(7)

and

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Corollary D.8 V B |= u ⊂ v and inclusion is proper. Proof : From the definition, V B |= ∀nˆ ∈ ωˆ (nˆ ∈ u → nˆ ∈ v). Then, clearly, from equations (6) and (7), as well as from Definition D.1, we have that V B |= u ⊆ v if and only if Bn ≤ Cn , for all n ∈ ω. To show that the inclusion is proper, suppose that V B |= v ⊆ u, or Cn ≤ Bn , for each n. Then kv ⊆ uk B ≤ Cn ⇒ Bn , or kv ⊆ uk B ≤ Cn∗ ∨ Bn , for all n ∈ ω. Thus, ^ kv ⊆ uk B = (Cn∗ ∨ Bn ), n

or

kv ⊆ uk B = (

^

Cn∗ ) ∨ (

n

^

Bn ) = 0.

n

(The second infinite conjunction is zero since the Bn are mutually incompatible; for the first infinite conjunction notice that the complement _

Cn = 1,

n

since we are in a complete algebra out of regular open sets.) So, ku = vk B = 0. Remark D.9 We first used generic–like objects as u above to obtain a ‘faceless’ satisfiability problem within a formalization of their theory.

E Informal arguments about independence The following argument circulates in a folklore–like way; we do not know where it originated, and to our knowledge it has never appeared in print — or has been discussed in depth. Let S be a theory that includes arithmetic, has a recursively enumerable set of theorems, is based on a first–order classical language, and has a model with standard arithmetic. Get some formal version for the P = NP hypothesis as an arithmetic sentence, and note that formal version [ P = NP]. Again [ P < NP] = ¬ [ P = NP]: • Suppose that you’ve proved independence of [ P < NP] and [ P = NP] with respect to S. • Then S + [ P < NP] is consistent. • Better: S + [ P < NP] satisfies the conditions above. Thus we can take it as our new starting theory and may argue for it. • We conclude that S + [ P < NP] + [ P = NP] is consistent. A contradiction. • Thus independence proved in this way actually proves that S ` [ P = NP]. (The argument concludes: as almost nobody believes that P = NP holds true, then any independence proof must contain a flaw somewhere, since (if we follow the preceding argument) it leads to the proof of P = NP.) This is the best–known paradoxical argument. We’ll present some similar arguments in due course. Now: get we get around them?

Difficulties avoided Suppose that independence holds. Consider our arguments involving standard and exotic definitions as in the preceding sections. One would then get, for the paradoxical reasoning sketched at the introduction: • S + [ P < NP] is consistent and arithmetically sound. • Then argue as we do and get that S + [ P < NP] + [ P = NP]F is consistent. • However if independence holds, that theory isn’t arithmetically sound; moreover we cannot add to it as a new axiom [ P < NP] ↔ [ P < NP]F to derive [ P = NP], for we would therefore get a contradiction. The point is, we cannot have [ P = NP] ↔ [ P = NP]F in that theory, or it is trivialized. 68

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Remark E.1 Let’s elaborate on that. If ZFC + [ P < NP] is consistent, then so is ZFC + [ P < NP] + [ P = NP]F . However this latter theory isn’t arithmetically sound, for it only has nonstandard models — and moreover models where [F has a finite domain] holds, so that the equivalence [ P = NP] ↔ [ P = NP]F will never hold. An alternative viewpoint: • We cannot start from a consistent S + [ P < NP] ↔ [ P < NP]F + [ P < NP]. • For S ` [ P < NP]F → [F is total ]. • Thus S + [ P < NP] ↔ [ P < NP]F + [ P < NP] ` [F is total ], and we cannot argue (as in the preceding sections) to show that [ P = NP]F is consistent, for in order to do so we require that [F is total] be unprovable in our theory.

F More about theory S + [ P < NP] This is a rather weird theory. If [ P < NP] is independent of S, if S + [ P < NP] has a model with standard arithmetic, we can consistently add to it a finite set of conditions [ P < NP] ↔ [ P < NP]Fi , i ranging over a finite set. Then we suppose that theory: S + [ P < NP] + ∑ [ P < NP] ↔ [ P < NP]Fi i

has the same provably total recursive functions as S + [ P < NP] (we use here Kreisel’s Lemma [4]). However if we add an infinite set of conditions [ P < NP] ↔ [ P < NP]Fi so that given each intuitively total recursive g there is some Fi in the added set so that Fi dominates g, then the resulting theory becomes a nonrecursive theory, for it proves all arithmetical truths. Another fact: suppose that h is any strictly increasing total recursive function. Form theory S0 = S + [ P < NP]. Then theory S ` FS0 h is consistent. That is to say, FS cannot be precisely placed among all total recursive functions! This isn’t surprising, as the set of all total recursive functions is plagued with undecidabilities of all sorts. Let’s try to understand what has been going here. Define a Busy–Beaver– like function f by the condition “ f dominates all total recursive functions.” As it is well–known, a theory like our S doesn’t “see” that property, as it always has growth–limited total recursive functions. We will require a more sophisticated structure here, a progression of theories in the sense of Kreisel and Feferman [2, 21, 23].

A plausibility argument for one of our conjectures Definition F.1 Any poly algorithm that settles all of S AT is a good algorithm. Our goal here is to argue that: Lemma F.2 S proves the (formal version of the) sentence:

[The set of Gödel numbers of BGS good algorithms is recursively enumerable.]. We note that set K ⊂ ω. Of course, if K = ∅, then the lemma holds. We now consider the nontrivial case, that is, we suppose that S ` [ P = NP] (see below Hypothesis F.5). Preliminary results We need an alternative formulation for [ P = NP]: Lemma F.3

[ P = NP] ↔ ∃e, a ∀ x ∃z ≤ (| x | a + a) [ T (e, x, z) ∧ R( x, U ( T (e, x, z))]. 70

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Define: Remark F.4 For our purposes, we specifically consider in this section Π1 sentences of the form: 1. ∀ x P, P primitive recursive. 2. ∀ x ∃y ≤ g( x ) P, where g, P are primitive recursive. Then:

∀ x ∃z ≤ (| x | a + a) [ T (e, x, z) ∧ R( x, U ( T (e, x, z))] is a Π1 sentence, if a ∈ ω is seen as a constant. Remark F.5 We suppose that S ` [ P = NP]. Corollary F.6 N |= [ P = NP]. Moreover, for integers e, a: Corollary F.7 N |= ∀ x ∃z ≤ (| x | a + a) [ T (e, x, z) ∧ R( x, U ( T (e, x, z))]. Now let PA1 be the (nonrecursive) theory that consists of PA plus all true Π1 sentences. Then clearly: Lemma F.8 If there are constants e, a so that

∀ x ∃z ≤ (| x | a + a) [ T (e, x, z) ∧ R( x, U ( T (e, x, z))] holds true of the standard model for arithmetic, then: PA1 ` [∀ x ∃z ≤ (| x | a + a) [ T (e, x, z) ∧ R( x, U ( T (e, x, z))]]. Corollary F.9 If there are constants e, a so that

∀ x ∃z ≤ (| x | a + a) [ T (e, x, z) ∧ R( x, U ( T (e, x, z))] holds true of the standard model for arithmetic, then: PA1 ` [ P = NP]. For we have:

[∀ x ∃z ≤ (| x | a + a) [ T (e, x, z) ∧ R( x, U ( T (e, x, z))]] → → [∃e, a ∀ x ∃z ≤ (| x | a + a) [ T (e, x, z) ∧ R( x, U ( T (e, x, z))]]. Then by detachment we get [ P = NP] [4].

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Conclusion of the first argument From the arithmetic soundness of S, for a model for S with standard arithmetic N: Corollary F.10 If Remark F.5 holds, then N |= [ P = NP]. We give a waving–hands argument for the assertion: • Suppose S ` [ P = NP]. • Then N |= [ P = NP]. • Therefore it is true that there is at least a (standard) pair e, a ∈ ω so that {e} with bound | x | a + a settles all of S AT. • Let PA1 be Peano Arithmetic plus all true Π1 arithmetic sentences. • Since PA1 ` [ P = NP], then either PA proves it or, at worst, PA0 = PA + ∀ x ∃z ≤ (| x | a + a) [ T (e, x, z) ∧ R( x, U ( T (e, x, z))] trivially proves it. Both are arithmetically sound theories with a recursively enumerable set of theorems. Both PA and PA0 have the same provably total recursive functions, and therefore have (in that sense) the same provability strength. This stems from Kreisel’s Lemma [4, 11]. We now consider BGS poly Turing machines, coded by pairs he, ai (see below). • Then if: K (e, a) ↔Def [∀ x ∃z ≤ (| x | a + a) [ T (e, x, z) ∧ R( x, U ( T (e, x, z))], for infinitely many numerals e, a, the sentences K (e, a) will be theorems of PA0 . • Moreover, for numerals e, a, the set of all PA0 –provable K (e, a) is a recursively enumerable set coded by the pair he, ai. This pair is the BGS code for poly machines. (A BGS pair [1] can be interpreted as a Turing machine {e} coupled to a clock that stops its operation on binary input x before | x | a + a cycles. Every BGS pair then codes a poly machine, and every poly machine can be represented as a BGS pair. Finally, if e runs over all of ω, then the BGS pairs run over ω × ω × ω.) • Now suppose that there is he, ai that settles S AT, that is to say, there is some Turing machine {e} with polynomial bound | x | a + a so that N |= K (e, a), but such that K (e, a) doesn’t come up among the theorems of PA0 .

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• Note that machine P0m . Then there is a machine M so that, for a machine that comes up in the theorem’s enumeration, Pm , we have: P0m = M ◦ P0m . (Of course M must be total and polynomial.) • We then agree that P0m is in the enumeration of theorems: – K (e, a) is a theorem, that is, Pm settles all of S AT. – Then there is a poly Turing machine M so that P0m = M ◦ Pm . – Therefore P0m settles all of S AT, that is, for m0 = he0 , a0 i, K (e0 , a0 ) is a theorem. Proposition F.11 Given Hypothesis F.5, the nonempty set K of all BGS–coded he, ai poly Turing machines so that K (e, a) holds true of the standard model for arithmetic is recursive. For it is recursively enumerable, and its complement (the set of counterexamples over BGS) is also recursively enumerable, as it is the image of a partial recursive function. Proposition F.12 We can effectively compute the Gödel number of an algorithm that recursively enumerates the K (e, a). Begin a listing of all integers, 0, 1, 2, . . .. Input k ∈ ω: ¨ • We first check whether it is the Godel number of a proof in PA0 . If not, go to k + 1. If it is, go to the next step. • Check whether the last code in the proof is a code for K (e, a), for some pair e, a. • If not, go to k + 1. If so, output e, a and go to k + 1. ¨ Let eK be that Godel number. Corollary F.13 Given Hypothesis F.5, for the nonempty set K, we have that S ` [K ⊂ ω is recursive.]. By the representation theorem, given Hypothesis F.5, as we can compute a ¨ Godel number for the algorithmic procedure that recursively enumerates K, we can represent it in PA, and therefore also in S; by definition it is (PA)–provably recursively enumerable, and as the complement of a recursively enumerable set it is recursive in PA and therefore in S. More precisely: • If {eK } is the algorithm that enumerates K as described above, then we can explicitly construct a Diophantine polynomial pK (y, x1 , . . . , xk ) so that it holds of a model N with standard arithmetic that:

[{eK }( a) = b] ↔ ∃ x1 , . . . , xk pK (h a, bi, x1 , . . . , xk ) = 0.

74

da Costa, Doria • We then have that {eK }( a) = b if and only if: S ` ∃ x1 , . . . , xk pK (h a, bi, x1 , . . . , xk ) = 0. • Then (with perhaps some abuse of language): S proves:

[{y ∈ ω : ∃ x1 , . . . , xk pK (y, x1 , . . . , xk ) = 0} is recursively enumerable]. • As its complement is also a recursively enumerable set — it is the set of values of the counterexample function to [ P = NP] — we have that S proves that K is recursive.

G Unbounded theories Recall that S is arithmetically sound if it has a model with a standard arithmetic part. From the arithmetic soundness of S, for a model for it with standard arithmetic (which will always be noted) N: Definition G.1 Let theory S be such that N |= S, and such that we can construct a function FS for it as in Remark 6.107. Then S is unbounded if for any f so that N |= [f is strictly increasing total recursive], then S + [FS dominates f ] is consistent and holds of N. The idea is that we can take the bounding fast–growing function FS as fast– growing as possible. Remark G.2 From here on we will consider a theory S that: • S includes Peano Arithmetic. • S is arithmetically sound, and model N makes it so. • S has a recursively enumerable set of theorems. • S is based on a first–order classical language. • S isn’t unbounded. We notice that both PA and ZFC satisfy the above requirements. Remark G.3 Given independence, then S + [ P < NP] is unbounded. We see this as a kind of consequence of the huge fast–growing behavior of the counterexample function when seen within that theory.

75

H More comments about undecidability and incompleteness in strong theories See the Introduction for more comments on these results. Chaitin’s pioneering results had already convinced us that undecidability and incompleteness are everywhere. The next propositions just add to that. Proposition H.1 There is a real number s whose decimal expansion is proved by ZFC to be generated by an algorithmic procedure and yet such that ZFC cannot compute a single digit of it. Proof : Let r and r 0 be real numbers which are algorithmically generated. Impose also that, for all i, the i–th digits ri 6= ri0 . Then for β as in [8], s = βr + (1 − β)r 0 . A second result: Proposition H.2 There is a real number Ω0 so that ZFC cannot compute a single digit of it. Sketch of proof : List (this isn’t a recursive enumeration!) all instances of the halting problem for some partial recursive algorithm that cannot be proved to diverge within ZFC. Those instances appear as Π1 sentences. Order them by ¨ their respective Godel numbers. This is a noncomputable function, which can be coded as a real number in several ways. ZFC cannot prove by construction a single digit of that number. Another construction: Proposition H.3 There are infinitely many real numbers in ZFC so that none of its decimal places are computable. Proof : Recall that no well–orderings of the real numbers are expressible in ZFC, but there are infinitely many such orderings. Let Γ be one such ordering, and let r be the first real number in Γ so that none of its decimal places are definable — there is one such r, as the set of numbers which have definable decimal places is denumerable. Remark H.4 Still another simple but interesting result about the decidability of Π1 sentences and hierarchies of theories goes as follows. Π1 arithmetic sentences code several interesting facts, e.g. Riemann’s Hypothesis. If we are able to ascertain by some analog procedure [8] that Riemann’s Hypothesis is true of the standard model for arithmetic, then given the Fefermann hierarchy [23] of theories S0 = S, S1 = S + Con S, S2 = S + Con S + Con(S + ConS), . . . , there will be a S j so that S j ` Riemann’s Hypothesis.

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References [1] T. Baker, J. Gill and R. Solovay, “Relativizations of the P =?NP question,” SIAM J. Comp. 4, 431–442 (1975). [2] L. Beklemishev, “Provability and reflection,” Lecture Notes for ESSLLI ’97 (1997). [3] J. L. Bell, Boolean–valued models and independence proofs in set theory, Clarendon Press (1985). [4] S. Ben–David and S. Halevi, “On the independence of P vs. NP, Technical Report # 699, Technion (1991). [5] N. C. A. da Costa and F. A. Doria, “Undecidability and incompleteness in classical mechanics,” Int. J. Theor. Phys 30, 1041–1073 (1991). [6] N. C. A. da Costa and F. A. Doria, “Suppes predicates and the construction of unsolvable Problems in the axiomatized sciences,” in P. Humphreys, ed., Patrick Suppes, Scientific Philosopher, II, 151–191 Kluwer (1994). [7] N. C. A. da Costa and F. A. Doria, “Consequences of an exotic formulation for P = NP,” Applied Mathematics and Computation 145, 655–665 (2003); also “Addendum,” Applied Mathematics and Computation 172, 1364–1367 (2006). [8] N. C. A. da Costa and F. A. Doria, “Computing the future,” in K. Vela Velupillai, ed., Computability, Complexity and Constructivity in Economic Analysis, Blackwell (2005). [9] N. C. A. da Costa and F. A. Doria, “On the provably total recursive functions in Zermelo–Fraenkel set theory plus the axiom [ P < NP]G ↔ [ P < NP],” to appear (2006). [10] N. C. A. da Costa, F. A. Doria and A. F. Furtado–do–Amaral, “A dynamical system where proving chaos is equivalent to proving Fermat’s conjecture,” Int. J. Theoret. Physics, 32, 2187–2206 (1993). [11] N. C. A. da Costa, F. A. Doria and E. Bir, “On the metamathematics of the P vs. NP question,” Applied Mathematics and Computation 189, 1223–1240 (2007). [12] F. Cucker, e–mail messages to the authors (1998). [13] M. Davis, “Hilbert’s Tenth Problem is unsolvable,” in Computability and Unsolvability, Dover (1982). [14] R. A. DeMillo and R. J. Lipton, “Some connections between computational complexity theory and mathematical logic,” in Proc. 12th Ann. ACM Symp. on the Theory of Computing, 153–159 (1979). 77

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[15] R. A. DeMillo and R. J. Lipton, “The consistency of P = NP and related problems with fragments of number theory,” in Proc. 12th Ann. ACM Symp. on the Theory of Computing, 45–57 (1980). [16] F. A. Doria, M. Ribeiro da Silva and A. F. Furtado do Amaral, “A generalization of Einstein’s λ–transformation and gravitational copies,” Lett. Nuovo Cimento 40, 509 (1984). [17] F. A. Doria, S. M. Abrahão and A. F. Furtado do Amaral, “A Dirac–like equation for gauge fields,” Progr. Theoretical Physics 75, 1440 (1986). [18] F. A. Doria, “Is there a simple, pedestrian, arithmetic sentence which is independent of ZFC?” Synthèse 125, # 1/2, 69 (2000). [19] F. A. Doria, “Informal vs. formal mathematics,” Synthèse 154, 401–415 (2007). [20] A. Einstein, The Meaning of Relativity, Methuen (1967). [21] S. Feferman, “Transfinite recursive progressions of axiomatic theories,” J. Symbolic Logic 27, 259–316 (1962). [22] S. Fortune, D. Leivant, M. O’Donnell, “The expressiveness of simple and second–order type structures,” J. ACM 38, 151–185 (1983). [23] T. Franzen, “Transfinite progressions: a second look at completeness,” Bull. Symbolic Logic 10, 367-389 (2004). [24] H. M. Friedman, “Finite functions and the necessary use of large cardinals,” Ann. Math. 148, 803 (1998). ¨ [25] K. Godel, “Undecidable Diophantine propositions [193?],” Collected Works III (1995). ¨ [26] K. Godel, “Some basic theorems on the foundations of mathematics and their implications [1951],” Collected Works III (1995). [27] M. Guillaume, e–mail messages to the authors, 2000–2003. [28] M. Guillaume, “What counts in an exotic formulation. . . ,” unpublished draft (2003). [29] J. Hartmanis and J. Hopcroft, “Independence results in computer science,” SIGACT News, 13, Oct. Dec. (1976). [30] H. Hermes, Enumerability, Decidability, Computability, Springer (1965). [31] J. E. Hopcroft and J. D. Ullman, Formal Languages and their Relation to Automata, Addison–Wesley (1969). [32] D. Joseph and P. Young, “Independence results in computer science?” Proc. 12th Ann. ACM Symp. on the Theory of Computing, 58–69 (1980).

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[33] D. Joseph and P. Young, “Fast programs for initial segments and polynomial time computation in weak models of arithmetic,” STOC Milwaukee 1981, 55–61 (1981). [34] R. Kaye, Models of Peano Arithmetic, Clarendon Press (1991). [35] S. C. Kleene, “General recursive functions of natural numbers,” Math. Ann. 112, 727 (1936). [36] S. C. Kleene, Mathematical Logic, Wiley (1967). [37] W. Kowalczyk, “A sufficient condition for the consistency of P = NP with Peano Arithmetic,” Fund. Inform. 5, 233–245 (1982). [38] G. Kreisel, “On the interpretation of non–finitist proofs,” I, J. Symbol. Logic 16, 241 (1951); II, 17, 43 (1952). [39] G. Kreisel, “On the concepts of completeness and interpretation of formal systems,” Fund. Math. 39, 103–127 (1952). [40] M. Machtey, P. Young, An Introduction to the General Theory of Algorithms, North–Holland (1979). [41] A. Maté, “Nondeterministic polynomial–time computations and models of arithmetic,” J. ACM 37, 175–193 (1990). [42] M. O’Donnell, “A programming language theorem which is independent of Peano Arithmetic,” Proc. 11th Ann. ACM Symp. on the Theory of Computation, 176–188 (1979). [43] J. Paris and L. Harrington, “A mathematical incompleteness in Peano arithmetic,” in J. Barwise, ed., Handbook of Mathematical Logic, North– Holland (1977). [44] T. Rado, “On non–computable functions,” Bell System Tech. J. 91, 877 (1962). [45] H. Rogers Jr., Theory of Recursive Functions and of Effective Computability, reprint, MIT Press (1992). [46] A. S. Sant’Anna, N. C. A. da Costa and F. A. Doria, “The Atiyah–Singer index theorem and the gauge field copy problem,” J. Phys. A 30, 5511 (1997). [47] S. Shapiro, “Incompleteness, mechanism and optimism,” Bull. Symbolic Logic 4, 273–302 (1998). [48] C. Smorynski, Logical Number Theory, I, Springer (1991). ´ [49] J. Spencer, “Large numbers and unprovable theorems,” Amer. Math. Monthly 90, 669 (1983).

Contents Prelude.

9

Introduction

9

The plot.

11

1. Summary A venerable line of research . . . . . . . . . . . . . . . . . . . . . . . . Some notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

12 13 14

2. The exotic version

15

In a nutshell

17

3. Fast–growing functions

18

4. A blueprint

21

5. Σ1 –soundness and F Main theorem, first version . . . . . . . Function F and function G . . . . . . . . Results for G . . . . . . . . . . . . . . . . A possible conclusion of the argument? Kreisel’s lemma . . . . . . . . . . . . . . More stuff . . . . . . . . . . . . . . . . .

. . . . . .

32 35 36 38 39 39 41

6. Good algorithms Preliminary definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . Preliminary results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Conclusion of the argument . . . . . . . . . . . . . . . . . . . . . . . .

42 42 43 44

7. Function f

51

8 Acknowledgments

57

A few half–baked intuitions.

58

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. . . . . .

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A Does a large cardinal imply P < NP ? More on inaccessible cardinals . . . . . . . . . . . . . . . . . . . . . . .

58 59

B Proof strength and Kolmogorov–Chaitin complexity

61

C The halting function and forcing

62

D From ζ to a generic set

65

E Informal arguments about independence Difficulties avoided . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

68 68

F More about theory S + [ P < NP] Preliminary results . . . . . . . . . . . . . . . . . . . . . . . . . . Conclusion of the first argument . . . . . . . . . . . . . . . . . .

70 70 72

G Unbounded theories

75

H More comments about undecidability and incompleteness in strong theories 76

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