We provide some results which emerged from joint research carried out at 
CRM. The theorems are inspired by analogy with situations related to forcin

Friedman, Sy-David

Rathjen, Michael

Weiermann, Andreas

Ghent University Academic Bibliography

Some results on PA-provably recursive functions(Draft)Sy Friedman, Wien, Michael Rathjen, Leeds, Andreas Weiermann, GhentAbstractWe provide some results which emerged from joint research carried outat CRM. The theorems are inspired by analogy with situations related toforcing.1 Vague analogies with forcingIn set theory one has a ”ground model” with a given set of functions from ω toω. Then three things can happen when passing to a larger model, as exemplifiedby:1. Sacks forcing: One adds new functions but any new function is dominatedby an old (ground model) function.2. Cohen forcing: One adds a new function that cannot be dominated by aground model function but no single function which dominates (mod finite) allground model functions. If one adds two such functions f,g using ”Cohen timesCohen” forcing, then in addition any function added by both f and g is in theground model.3. Hechler forcing: One adds a new function that dominates (mod finite)all ground model functions. If one adds two such functions f,g using ”Hechlertimes Hechler” forcing, then again any function added by both f and g is in theground model.The analogy in proof theory is the following: Fix a theory T like PA. Let ustake the ”T-provably recursive” functions to be those with primitive recursivegraph (the honest functions) such that for some choice of primitive recursiverepresentation of that graph, totality of the function is T -provable.Our Theorem 1 says the following: There is a ”natural” total recursive func-tion f with primitive recursive graph which is not PA-provably recursive (viaany primitive recursive representation of its graph) and such that no provablyrecursive function of PA+”f is total ” (expressed using any primitive recursivegraph representation) dominates (mod finite) all provably recursive function ofPA. This is the proof-theoretic analogue of Cohen forcing.1The situation is similar for Theorem 2. It says that there are ”natural”functions f0, f1 with primitive recursive graph which are not provably recursivein PA (via any primitive recursive graph representation), yet any function whichis provably recursive in both PA + Total(f0) and PA + Total(f1) (where thelatter are expressed using primitive recursive graph representations) is in factprovably recursive in PA. This is the analogue of Cohen times Cohen forcing.2 The resultsTheorem 1 There exists an honest number-theoretic function f such that fis not provably recursive in PA and such that any g which is provably total inPA+ Tot(f) does not eventually dominate every PA-recursive function.Proof. (The proof is inspired by [2].) We construct f in stages. d0 := 0. Assumewe are at stage s = 2i and that ds is defined. Assume that f(x) is defined forx < ds. ds+1 := Hε0(ds). We extend f by f(x) := Hε0(x) for ds ≤ x < ds+1.Assume that s = 2i + 1. Let d′s+1 := Hε0(ds) and ds+1 := Hε0(d′s+1). Weextend f by f(x) := d′s+1 + x for ds ≤ x < ds+1. Moreover let fs(x) := f(x)for x < ds and fs(x) := d′s+1 + x for ds ≤ x.Since f(x) = Hε0(x) for infinitely many x we see that f is not provablyrecursive in PA. Assume now that PA + Tot(f) proves Tot(g). Then thereexists an α < ε0 such that for all x we have g(x) < fα(x) wherefα(x) := max({f(x)} ∪ {fβ(fβ(x)) : β < α ∧Nβ ≤ f(Nα+ x)}).Here Nα is defined via N0 := 0 and Nα := Nβ + Nγ + 1 if α has the normalform ωβ + γ.By construction we know that for each odd s we have fs(x) ≤ Hd′s+1(x).Choose an odd stage s with 2 ·Nα+ 12 ≤ d′s+1. Then fαs (d′s+1) ≤ Hε0(d′s+1).For a proof we apply Theorem 4 from [1] (and tacitly Lemma 9 from thesame article). With this we obtain from fs(x) ≤ Hd′s+1(x) (which holds for allx) thatfsα(d′s+1)≤ Hωα+d′s+1+1+8(d′s+1)≤ Hωα+d′s+1+1+8(d′s+1 +Nα+ 10)≤ Hωα+ω (d′s+1 +Nα+ 10)≤ Hωα+ω (Hα+10(d′s+1))≤ Hωα+ω+α+10(d′s+1)≤ Hε0(d′s+1).2Now we prove: (∗) fsα(y) ≤ Hε0(d′s+1)⇒ fα(y) = fsα(y) by induction onα < ε0. Assume that fα(y) = fβ(fβ(y)) for some β < α with Nβ ≤ f(Nα+y).We know that fs(Nα+ y) ≤ Hε0(d′s+1) hence fs(Nα+ y) = f(Nα+ y) ≥ Nβ.Thus fsβ(fsβ(y)) ≤ fsα(y) ≤ Hε0(d′s+1).So fsβ(y) ≤ Hε0(d′s+1) and the i.h. yields fsβ(y) = fβ(y) and hencefsβ(fsβ(y)) = fβ(fsβ)(y)) = fβ(fβ(y)).Since fα(y) = fβ(fβ(y)) = fsβ(fsβ(y)) ≤ fsα(y) we are done with the proofof (∗).Putting things together we obtain for large enough s thatg(d′s+1) < fα(d′s+1) = fsα(d′s+1) ≤ Hωα#ω+α+10(d′s+1).So the function Hωα#ω+α+10 is not eventually dominated by g.Theorem 2 There are two honest recursive functions f0, f1 which are not prov-ably recursive in PA such that if a function g is provably recursive in PA +Tot(f0) and PA+ Tot(f1) then g is provably recursive in PA.Proof. (The proof is inspired by [2].) We construct f0, f1 in stages. d0 := 0.Assume we are at stage s = 2i and that ds is defined. Assume that fi(x) aredefined for x < ds. Put dˆs+1 := Hε0(ds). Define f1,s by f1,s(x) := f1(x) forx < ds and f1,s(x) := f1(ds − 1) + x for x ≥ ds.Putd′s+1 := µn : ∃x < n[x ≥ dˆs+1 ∧ f1,sωi(x) < Hε0(x) ≤ n]and ds+1 := Hε0(d′s+1).Extend f0 by f0(x) := f0(x) for x < ds, f0(x) := dˆs+1 +x for dˆs+1 > x ≥ dsand f0(x) := Hε0(x) for d′s+1 > x ≥ dˆs+1 and f0(x) := ds+1 + x for ds+1 >x ≥ d′s+1. Define f0,s by f0,s(x) := f0(x) for x < d′s+1 and f0,s(x) := ds+1 + xfor x ≥ d′s+1. Extend f1 by f1(x) := f1(x) for x < ds and f1(x) := f1,s(x) fords+1 > x ≥ ds. Assume we are at stage s = 2i + 1 and that ds is defined. Weinterchange the roles of f0 and f1. That means: Assume that fi(x) are definedfor x < ds. Put dˆs+1 := Hε0(ds). Define f0,s by f0,s(x) := f0(x) for x < ds andf0,s(x) := f0(ds − 1) + x for x ≥ ds.Putd′s+1 := µn : ∃x < n[x ≥ dˆs+1 ∧ f0,sωi(x) < Hε0(x) ≤ n]and ds+1 := Hε0(d′s+1). Extend f1 by f1(x) := f1(x) for x < dˆs+1 and f1(x) :=Hε0(x) for d′s+1 > x ≥ dˆs+1 and f1(x) := ds+1 + x for ds+1 > x ≥ d′s+1.Define f1,s by f1(x) := f1(x) for x < d′s+1 and f1,s(x) := ds+1 + x forx ≥ d′s+1. Extend f0 by f0(x) := f0(x) for x < ds and f0(x) := f0,s(x) fords+1 > x ≥ ds.We write f0 ∧ f1 for x 7→ min{f0(x), f1(x)}. Then (f0 ∧ f1)(x) ≤ 2 · x.3Assume that PA + Tot(f1) ` Tot(f0). Then, by [1], there is an α < ε0such that for all x we have f0(x) < fα1 (x). Choose i1 such that α < ωi1 .Then for x ≥ Nα we have (∗∗) f0(x) < fωi11 (x). Assume that s is 2i + 1with i > max{Nα, i1}. Then there is an x < d′s+1 with x ≥ ds such thatf1ωi(x) < Hε0(x) ≤ d′s+1. Since f1 and f1 agree on x < ds+1 we obtainfωi1 (x) = f1ωi(x) < Hε0(x) = f0(x). Contradiction with (∗∗). By a symmetricargument PA+ Tot(f0) does not prove Tot(f1).Now assume that PA+ Tot(f0) ` Tot(g) and PA+ Tot(f1) ` Tot(g). Thenthere exist αi such that for all x we have g(x) < fαii (x). Our idea is now toshow that (fα00 ∧ fα11 )(x) ≤ (f0 ∧ f1)α0#α1(x). Since (f0 ∧ f1)(x) ≤ 2 · x thiswould yield the claim. But the obvious verification doesn’t seem to work.For this purpose let us introduce another iteration hierarchy.fα(x) := max({f(x)} ∪ {fβ(fβ(x)) : β < α ∧Nβ ≤ Nα+ x}).This hierarchy behaves better with respect to the ∧ operator.Indeed for two increasing functions f, h we have(fα ∧ hβ)(x) ≤ (f ∧ h)α#β(x).This is proved by induction on α#β. Assume first that β = 0. Then hβ(x) =h(x). If α = 0 then fα(x) = f(x) and the assertion is clear. If α > 0 thenfα(x) = fγ(fγ(x)) for some γ < α with Nγ ≤ Nα+ x. We have (f ∧ h)α(x) ≥(f ∧h)γ((f ∧h)γ(x)). The induction hypothesis yields (f ∧h)γ(x) ≥ (fγ ∧h)(x).If (fγ ∧h)(x) = h(x) then (f ∧h)α(x) ≥ (f ∧h)γ(h(x)) ≥ h(x) and the assertionfollows. So assume (fγ ∧ h)(x) = fγ(x). We then have (fγ ∧ h)(fγ ∧ h) =(fγ ∧ h)(fγ(x)). If (fγ ∧ h)(fγ(x)) = h(fγ(x)) then the assertion follows fromh(fγ(x)) ≥ h(x). We may assume that (fγ ∧ h)(fγ(x)) = fγ(fγ(x)) = fα(x).Since fα(x) ≥ h(x) the assertion also holds in this case.Assume (by symmetry) for the induction step that (fα ∧ hβ)(x) = hβ(x) =hγ(hγ(x)) for some γ < β with Nγ ≤ Nβ + x. The inequality fα(x) ≥ hβ(x)yields hγ(x) ≤ (fα ∧ hγ)(x) ≤ (f ∧ h)α#γ(x) by the induction hypothesis.The inequality fα(hγ(x)) ≥ fα(x) ≥ hβ(x) yields hγ(hγ(x)) ≤ fα(hγ(x)) ≤(fα ∧ hγ)(hγ(x)) ≤ (f ∧ h)α#γ(hγ(x)) by the induction hypothesis. Puttingthings together we obtain hγ(hγ(x)) ≤ (f ∧h)α#γ((f ∧h)α#γ(x)). Now we haveN(α#γ) ≤ N(α#β) + x. Hence (f ∧ h)α#γ((f ∧ h)α#γ(x)) ≤ (f ∧ h)α#β(x)and we are done.To prove the theorem it suffices to show that fα(x) ≤ fωα+1(x). This isproved by induction on α. Assume that fα(x) = fβ(fβ(x)) with β < α andNβ ≤ f(Nα+x). The induction hypothesis yields fβ(fβ(x)) ≤ fωβ+1(fωβ+1(x)).Thenfβ(fβ(x)) ≤ fωβ+1(fωβ+1(x))≤ fωβ+1(fωβ+1(f(Nα+ x)))4≤ fωα(f(Nα+ x))≤ fωα(fωα(x))≤ fωα+1(x)where we made use of f(Nα + x) ≤ fα(x). The last claim is again proved byinduction on α. For the induction step for proving this claim note that fα(x) =fβ(fβ(x)) for some β < α with Nβ = Nα+x. (It is easily seen that here = hasto hold). Then fβ(fβ(x)) ≥ fβ(x) ≥ f(Nβ+ x) ≥ f(Nα+ x+ x) ≥ f(Nα+ x).References[1] Andreas Weiermann: Classifying the provably total functions of PA. Bul-letin of Symbolic Logic, 12 (2), (2006), 177-90.[2] Lars Kristiansen: Subrecursive Degrees and fragments of Peano arithmetic.Archive for Mathematical Logic 40, (2001), 365-397.5

Some results on PA-provably recursive functions

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Some results on PA-provably recursive functions

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