This paper presents a simplified ordinal analysis of first-order reflection using a new ordinal notation system based on ψ-functions, and demonstrates how provable Σ₁-sentences are bounded via cut-elimination.
Contribution
It introduces a simplified ordinal notation system for analyzing first-order reflection and applies cut-elimination to bound provable Σ₁-sentences.
Findings
01
Ordinal notation system based on ψ-functions
02
Bounded provable Σ₁-sentences on L_{ω₁^{CK}}
03
Simplified approach to ordinal analysis of reflection
Abstract
In this note we give a simplified ordinal analysis of first-order reflection. An ordinal notation system OT is introduced based on ψ-functions. Provable Σ1-sentences on Lω1CK are bounded through cut-elimination on operator controlled derivations.
(b\varepsilon a):\equiv\left\{\begin{array}[]{ll}A(b)&\mbox{{\rm if }}a\equiv[x\in L_{\alpha}:A(x)]\\
b\not\in L_{0}&\mbox{{\rm if }}a\equiv L_{\alpha}\end{array}\right.
(b\varepsilon a):\equiv\left\{\begin{array}[]{ll}A(b)&\mbox{{\rm if }}a\equiv[x\in L_{\alpha}:A(x)]\\
b\not\in L_{0}&\mbox{{\rm if }}a\equiv L_{\alpha}\end{array}\right.
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Full text
A simplified ordinal analysis of first-order reflection
Toshiyasu Arai
Graduate School of Science,
Chiba University
1-33, Yayoi-cho, Inage-ku,
Chiba, 263-8522, JAPAN
[email protected]
Current address:
Graduate School of Mathematical Sciences,
The University of Tokyo,
3-8-1 Komaba, Meguro-ku,
Tokyo 153-8914, JAPAN
[email protected]
Abstract
In this note we give a simplified ordinal analysis of first-order reflection.
An ordinal notation system OT is introduced based on ψ-functions.
Provable Σ1-sentences on Lω1CK are bounded through cut-elimination
on operator controlled derivations.
1 Introduction
Let ORD denote the class of all ordinals, A⊂ORD and α a limit ordinal.
α is said to be Πn-reflecting on A iff
for any Πn-formula ϕ(x) and any b∈Lα, if
⟨Lα,∈⟩⊨ϕ(b),
then there exists a β∈A∩α such that b∈Lβ and
⟨Lβ,∈⟩⊨ϕ(b).
Let us write
α∈rMn(A):⇔α\mboxisΠn\mbox−reflectingonA.
Also α is said to be Πn-reflecting iff
α
is Πn-reflecting on ORD.
It is not hard for us to show that
the assumption that the universe is Πn-reflecting
is proof-theoretically reducible to iterabilities of
the lower operation rMn−1 (and Mostowski collapsings), cf.[3].
In this paper we aim an ordinal analysis of Πn-reflection.
Such an analysis was done by Pohlers and Stegert [7] using reflection configurations introduced in
M. Rathjen[9],
and an alternative analysis in [1, 2, 4]
with the complicated combinatorial arguments of ordinal diagrams and finite proof figures.
Our approach is simpler in view of combinatorial arguments.
In [1], a Πn-reflecting universe is resolved in
ramified hierarchies of lower Mahlo operations, and ultimately in iterations of recursively Mahlo operations.
Our ramification process is akin to a tower, i.e., has an exponential structure.
It is natural that an exponential structure emerges in lowering and eliminating first-order formulas
(in reflections), cf. ordinal analysis for the fragments IΣn−3 of the first-order arithmetic.
Mahlo classes Mhk(ξ) defined in Definition 2.5
to resolve or approximate Πn-reflection are based on similar structure.
As in Rathjen’s analysis for Π3-reflection in [8],
thinning operations are applied on the Mahlo classes Mhk(ξ), and this yields
an exponential structure similar to the one in [1] as follows.
Let us consider the simplest case N=4.
Let
Λ:=εK+1, the next epsilon number above the lease Π4-reflecting ordinal K.
Roughly π∈Mh3(ξ) designates the fact that an ordinal π
is Π3-reflecting on Mh3(ν) for any ν<ξ<Λ.
Suppose a Π3-sentence θ on Lπ is derived from the assumption π∈Mh3(ξ).
We need to find an ordinal κ<π for which Lκ⊨θ holds.
It turns out that κ∈Mh2(Λξa) suffices for an ordinal a<Λ, where
the ordinal κ in the class Mh2(Λξa) is Π2-reflecting
on classes Mh2(Λξb)∩Mh3(ν) for any b<a and any ν<ξ.
Note that the class Mh2(Λξa) is not obtained through iterations of recursively Mahlo operations
since it involves Π4-definable classes Mh3(ν).
The classes Mh3(ν)(ν<ξ) for the assumption π∈Mh3(ξ) are thinned out
with the new classes Mh2(Λξb)(b<Λ), cf. Lemma 5.1.
Our theorem runs as follows.
Let \mboxKPΠN denote the set theory for ΠN-reflecting universes, and
\mboxKPω the Kripke-Platek set theory with the axiom of infinity.
OT is a computable notation system of ordinals defined in section 3,
Ω=ω1CK and ψΩ is a collapsing function such that ψΩ(α)<Ω.
K is an ordinal term denoting the least ΠN-reflecting ordinal in the theorems.
Theorem 1.1
Suppose KPΠN⊢θ
for a Σ1(Ω)-sentence θ.
Then we can find an n<ω such that for α=ψΩ(ωn(K+1)),
Lα⊨θ.
Thus the ordinal ψΩ(εK+1) is seen to be the proof-theoretic ordinal of KPΠN.
Theorem 1.3
[TABLE]
A⊂ORD is Πn1-indescribable in α iff for any Πn1-formula ϕ(X) and any B⊂ORD,
if ⟨Lα,∈;B∩α⟩⊨ϕ(B∩α), then there exists a β∈A∩α such that
⟨Lβ,∈;B∩β⟩⊨ϕ(B∩β).
A regular cardinal π is Πn1-indescribable ifff ORD is Πn1-indescribable in π.
Let us mention the contents of this paper.
In the next section 2 we define simultaneously
iterated Skolem hulls Hα(X) of sets X of ordinals, ordinals
ψκξ(α) for regular cardinals
κ, α<εK+1 and sequences ξ=(ξ2,…,ξN−1)
of ordinals ξi<εK+2,
and classes Mhkα(ξ) under the assumption that a ΠN−21-indescribable cardinal K exists.
It is shown that for 2≤k<N, α<εK+1 and each ξ<εK+2,
(K\mboxisaΠN−21\mbox−indescribablecardinal)→K∈Mhkα(ξ) in
ZF+(V=L).
In section 3 a computable notation system OT of ordinals is extracted.
Following W. Buchholz[6],
operator controlled derivations for \mboxKPΠN is introduced in section 4,
and inference rules for ΠN-reflection are eliminated from
derivations in section 5.
This completes a proof of Theorem 1.1 for an upper bound.
IH denotes the Induction Hypothesis, MIH the Main IH and SIH the Subsidiary IH.
We are assuming tacitly the axiom of constructibility V=L.
Throughout of this paper N≥3 is a fixed integer.
2 Ordinals for ΠN-reflection
In this section
we work in the set theory
ZFLKN obtained from ZFL=ZF+(V=L)
by adding the axiom ∃K(K\mboxisΠN−21\mbox−indescribable)
for a fixed integer N≥3.
For ordinals α, ε(α) denotes the least epsilon number above α.
Let ORD⊂V denote the class of ordinals, K the least ΠN−21-indescribable cardinal, and
Reg the set of regular ordinals below K.
Θ denotes finite sets of ordinals≤K.
u,v,w,x,y,z,… range over sets in the universe,
a,b,c,α,β,γ,… range over ordinals<Λ,
ξ,ζ,ν,μ,ι,… range over ordinals<ε(Λ)=εK+2,
ξ,ζ,ν,μ,ι,… range over finite sequences over ordinals<ε(Λ),
and π,κ,ρ,σ,τ,λ,… range over regular ordinals.
θ denotes formulas.
Let ξ=(ξ0,…,ξm−1) be a sequence of ordinals.
The lengthlh(ξ):=m.
Sequences consisting of a single element (ξ) is identified with the ordinal ξ,
and ∅ denotes the empty sequence.
0 denotes ambiguously a zero-sequence (0,…,0)
with its length
0≤lh(0)≤N−1.
ξ∗μ=(ξ0,…,ξm−1)∗(μ0,…,νn−1)=(ξ0,…,ξm−1,μ0,…,μn−1)
denotes the concatenated sequence of ξ and μ.
Λ=ε(K)=εK+1
denotes the next epsilon number above the least ΠN−2-
indescribable cardinal
K, and
ε(Λ)=εK+2 the next epsilon number above Λ.
Definition 2.1
For a non-zero ordinal ξ<ε(Λ), its Cantor normal form with base Λ
is uniquely determined as
[TABLE]
where
ξm>⋯>ξ0,0<ai<Λ.
K(ξ)={ai:i≤m}∪⋃{K(ξi):i≤m} is the set of components of ξ
with K(0)=∅.
For a sequence ξ=(ξ0,…,ξn−1) of ordinals ξi<ε(Λ),
K(ξ):=⋃{K(ξi):i<n}.
2. 2.
For ξ>1,
te(ξ)=ξ0 in (1) is the tail exponent,
and
he(ξ)=ξm is the head exponent of ξ, resp.
The headHd(ξ):=Λξmam, and the tailTl(ξ):=Λξ0a0 of ξ.
3. 3.
he(i)(ξ) is the i-th head exponent of ξ, defined recursively by
he(0)(ξ)=ξ, he(i+1)(ξ)=he(he(i)(ξ)).
The i-th tail exponentte(i)(ξ) is defined similarly.
4. 4.
ζ is a part of ξ, denoted by
ζ≤ptξ iff
ζ=NF∑i≥nΛξiai=Λξmam+⋯+Λξnan for an n(0≤n≤m+1).
ζ<ptξ:⇔ζ≤ptξ&ζ=ξ.
5. 5.
A sequence μ=(μ0,…,μn) is an iterated tail parts of ξ, denoted by
μ⊂ptξ
iff μ0≤ptξ&∀i<n(μi+1≤ptte(μi)).
6. 6.
ν=(ν0,…,νn)∗0<ξ iff there exists a sequence
μ=(μ0,…,μn) such that
μ⊂ptξ and νi<μi for every i≤n.
7. 7.
Let ν=(ν0,…,νn) and ξ=(ξ0,…,ξn) be
sequences of ordinals in the same length, and 0≤k≤n.
ν<kξ:⇔∀i<k(νi≤ξi)∧(νk,…,νn)<ξk.
8. 8.
ζ is a step-down of ξ, denoted by
ζ<sdξ iff
ζ=Λξmam+⋯+Λξ1a1+Λξ0b+ν for some ordinals
b<a0 and ν<Λξ0.
9. 9.
ν=(ν0,…,νn)∗0<sdξ iff
νi<sdte(i)(ξ) for every i≤n.
10. 10.
ζ≤spξ:⇔∃μ≤ptξ(ζ≤sdμ), and
ζ<spξ:⇔∃μ≤ptξ(ζ<sdμ).
11. 11.
ν<spξ iff
ν<sdμ for a μ≤ptξ.
Let p(ν,ξ) denote the number p(0≤p<m) such that
ξ=NFμ+∑i<pΛξiai
for
μ=Λξmam+⋯+Λξpap and ν<sdμ.
Note that (ν)∗0<ξ⇔ν<ξ, and
(ξ,te(ξ),te(2)(ξ),…)⊂ptξ.
Also ζ<sdξ⇔ζ<ξ if ξ<Λ.
Proposition 2.2
ξ<μ<ε(Λ)⇒te(ξ)≤he(ξ)≤he(μ).
Proposition 2.3
ν<ξ≤ζ⇒ν<ζ.
Proof
by induction on the lengths n=lh(ν).
Let μ=(μ0,…,μn−1) be a sequence for
ν=(ν0,…,νn−1) such that
μ⊂ptξ and ∀i≤n−1(νi<μi),
cf. Definition 2.1.6.
If n=1, then ν0<μ0≤ptξ≤ζ.
ν0<ζ≤ptζ yields ν=(ν0)<ζ.
Let n>1. We have (ν1,…,νn−1)<te(μ0) with (μ1,…,μn−1)⊂ptte(μ0).
We show the existence of a λ such that μ0≤λ≤ptζ and te(μ0)≤te(λ).
Then IH yields (ν1,…,νn−1)<te(λ), and ν<ζ follows.
If μ0≤ptζ, then λ=μ0 works.
Suppose μ0≤ptζ.
On the other hand we have μ0≤ptξ≤ζ.
This means that ξ<ζ and there exists a λ≤ptζ such that μ0<λ and te(μ0)≤te(λ).
□
2.1 Ordinals
Definition 2.4
For i<ω and ξ<ε(Λ),
Λi(ξ) is defined recursively by Λ0(ξ)=ξ and
Λi+1(ξ)=ΛΛi(ξ).
2. 2.
For A⊂ORD, limit ordinals α and i≥0, let
α∈M2+i(A) iff A∩α\mboxisΠi1\mbox−indescribableinα.
3. 3.
κ+ denotes the next regular ordinal above κ.
4. 4.
Ωα:=ωα for α>0, Ω0:=0, and
Ω=Ω1.
Define simultaneously
classes Hα(X), Mhkα(ξ), and
ordinals ψκξ(α) as follows.
We see that these are
Σ1-definable as a fixed point in ZFL,
cf. Proposition 2.7.
Let a<Λ, and φ denote the binary Veblen function.
Let us define a Skolem hull Ha(X) of {0,K}∪X under the functions
+,α↦ωα,(α,β)↦φαβ(α,β<K),α↦Ωα(α<K)
and ψ-functions.
Reg denotes the set of regular ordinals≤K.
Definition 2.5
Ha[Y](X):=Ha(Y∪X) for sets Y⊂K.
(Inductive definition of Ha(X)).
(a)
{0,K}∪X⊂Ha(X).
2. (b)
x,y∈Ha(X)⇒x+y∈Ha(X),
x∈Ha(X)⇒ωx∈Ha(X), and
x,y∈Ha(X)∩K⇒φxy∈Ha(X).
3. (c)
K>α∈Ha(X)⇒Ωα∈Ha(X).
4. (d)
If π∈Ha(X)∩Reg and
b∈Ha(X)∩a, then
ψπ(b)∈Ha(X).
5. (e)
If {b,ξ}⊂Ha(X) with ξ≤b<a, then
κ=ψK0∗(ξ)(b)∈Ha(X), where
lh(0)=N−3.
6. (f)
Let
{π,b,c}⊂Ha(X) with π<K, 2≤k<N−1 an integer,
and ξ=(ξ2,…,ξk,ξk+1)∗0 a sequence of ordinals ξi<ε(Λ) with lh(0)=N−2−k
such that ξk+1=0 and
K(ξ)⊂Ha(X).
Assume
max(K(ξ)∪{c})≤b<a,
and
π∈Mh2b(ξ).
Then
κ=ψπν(b)∈Ha(X) for the sequence
ν=(ξ2,…,ξk+Λξk+1c)∗0 with lh(0)=N−1−k.
7. (g)
Let
{π,b}⊂Ha(X) with π<K, and 0=ξ<ε(Λ) an ordinal
with K(ξ)⊂Ha(X).
Let ν=(ν2,…,νN−1) be a sequence of ordinals<ε(Λ)
such that K(ν)⊂Ha(X).
Assume maxK(ν)≤b<a, K(ν)⊂Hb(π),
π∈Mh2b(ξ), and
ν<ξ, cf. Definition 2.1.6.
Then
κ=ψπν(b)∈Ha(X).
2. 2.
(Definitions of Mhka(ξ) and Mhka(ξ))
First let
K∈MhNa(0):⇔K∈MN⇔K\mboxisΠN−21\mbox−indescribable.
The classes Mhka(ξ) are defined for 2≤k<N,
and ordinals a<Λ, ξ<ε(Λ).
Let π be a regular ordinal≤K. Then for ξ>0
[TABLE]
where ν=(νk,…,νn)(2≤k≤n≤N−1) varies through
non-empty sequences of ordinals<ε(Λ)
and
[TABLE]
By convention, let for 2≤k<N,
π∈Mhka(0):⇔π∈Mh2a(∅):⇔π\mboxisalimitordinal.
Note that by letting ν=(0),
π∈Mhka(ξ)⇒π∈Mk for ξ>0.
Also 0<1, and Mhka(1)=Mk.
3. 3.
(Definition of ψπξ(a))
Let a<Λ be an ordinal, π≤K a regular ordinal and
ξ a sequence of ordinals<ε(Λ) such that
lh(ξ)=N−2.
Then let
[TABLE]
Let
ψπa:=ψπ0a,
where lh(0)=N−2, Mh2a(0)=Lim, and π∈M2, i.e.,
π is a regular ordinal.
Note that
π∈Mhka(ξ)⇒∀ν<ξ(π∈Mk(Mhka(ν))),
since (ν)<ξ holds with (ξ)⊂ptξ for ν<ξ.
Proposition 2.6
b+c∈Ha[Θ](d)⇒c∈Ha[Θ](d), and
ωc∈Ha[Θ](d)⇒c∈Ha[Θ](d).
Each of
x=Ha(y)(a<Λ,y<K),
x=ψκa,
x∈Mhka(ξ) and x=ψκξ(a),
is a Σ1-predicate as fixed points in ZFL.
Proof. This is seen from the facts that there exists a universal Πn1-formula, and by using it,
α∈Mn(x) iff ⟨Lα,∈⟩⊨mn(x∩Lα)
for some Πn+11-formula mn(R) with a unary predicate R.
□
Let A(a) denote the conjunction of
∀u<K∃!x[x=Ha(u)], and
∀ξ∀x(maxK(ξ)≤a&K(ξ)∪{κ,a}⊂x=Ha(κ)→∃!b≤κ(b=ψκξ(a))), where
lh(ξ)=N−2.
Since
the cardinality of the set HεK+1(π) is π for any infinite
cardinal π≤K,
pick an injection f:HΛ(K)→K so that
f"HΛ(π)⊂π for any weakly inaccessibles π≤K.
Lemma 2.8
∀a<ΛA(a).
2. 2.
π∈Mhka(ξ)* is a Πk−11-class on Lπ uniformly for weakly inaccessible cardinals π≤K
and a,ξ.
This means that for each k
there exists a Πk−11-formula mhka(x) such that
π∈Mhka(ξ) iff Lπ⊨mhka(ξ)
for any weakly inaccessible cardinals π≤K with f"({a}∪K(ξ))⊂Lπ.*
3. 3.
K∈MhN−1α(Λ)∩MN−1(MhN−1α(Λ)).
Proof.
2.8.1.
We show that A(a) is progressive, i.e.,
∀a<Λ[∀c<aA(c)→A(a)].
Assume ∀c<aA(c) and a<Λ.
∀b<K∃!x[x=Ha(b)]
follows from IH in ZFL.
∃!b≤κ(b=ψκξa) follows from this.
2.8.2.
Let π be a weakly inaccessible cardinal with f"({a}∪K(ξ))⊂Lπ.
Let f be an injection such that f"HΛ(π)⊂Lπ.
Then for ∀α∈K(ξ)(f(α)∈f"Hα(π)),
π∈Mhka(ξ) iff
for any f(ν)=(f(νk),…,f(νN−1)), each of f(νi)∈Lπ,
if ∀α∈K(ν)(f(α)∈f"Ha(π)) and
ν<ξ, then π∈Mk(Mhka(ν)), where
f"Ha(π)⊂Lπ is a class in Lπ.
2.8.3.
We show the following B(a) is progressive in a<Λ:
[TABLE]
Note that a∈Ha(K) holds for any a<Λ.
Suppose ∀b<aB(b).
We have to show that MhN−1α(a) is ΠN−31-indescribable in K.
It is easy to see that if
π∈MN−1(MhN−1α(a)), then π∈MhN−1α(a)
by induction on π.
Let θ(u) be a ΠN−31-formula such that LK⊨θ(u).
By IH we have ∀b<a[K∈MN−1(MhN−1α(b))].
In other words, K∈MhN−1α(a), i.e., LK⊨mhN−1α(a), where
mhN−1α(a) is a ΠN−21-sentence in Proposition 2.8.2.
Since the universe LK is ΠN−21-indescribable, pick a π<K such that
Lπ enjoys the ΠN−21-sentence θ(u)∧mhN−1α(a),
and {f(α),f(a)}⊂Lπ.
Therefore π∈MhN−1α(a) and Lπ⊨θ(u).
Thus K∈MN−1(MhN−1α(a)).
□
2.2 Normal forms in ordinal notations
In this subsection we introduce an irreducibility of sequences,
which is needed to define a normal form in ordinal notations.
Proposition 2.9
π∈Mhka(ζ)&ξ≤ζ⇒π∈Mhka(ξ).
Proof. (2) for π∈Mhka(ξ) in Definition 2.5.2
follows from π∈Mhka(ζ) and
Proposition 2.3.
□
Lemma 2.10
(Cf. Lemma 3 in [1].)*
Assume K≥π∈Mhka(ξ)∩Mhk+1a(ξ0) with
2≤k≤N−1,
he(μ)≤ξ0 and
{a}∪K(μ)⊂Ha(π).
Then
π∈Mhka(ξ+μ) holds.
Moreover if π∈Mk+1, then
π∈Mk+1(Mhka(ξ+μ))
holds.*
Proof. Suppose π∈Mhka(ξ)∩Mhk+1a(ξ0) and K(μ)⊂Ha(π) with he(μ)≤ξ0.
We show π∈Mhka(ξ+μ)
by induction on ordinals μ.
First note that if b∈Ha(π), then f(b)∈f"HΛ(π)⊂Lπ.
We have K(ξ+μ)⊂Ha(π).
π∈Mk+1(Mhka(ξ+μ)) follows from π∈Mhka(ξ+μ) and π∈Mk+1.
Let
(ζ)∗ν<ξ+μ
and K(ζ)∪K(ν)⊂Ha(π)
for ν=(ν0,…,νn−1).
We need to show that π∈Mk(Mhka((ζ)∗ν)).
By Definition 2.1.6, let (ζ0)∗(μ0,…,μn−1) be a sequence such that
ζ<ζ0≤ptξ+μ, μ0≤ptte(ζ0), ∀i≤n−1(νi<μi),
and ∀i<n−1(μi+1≤ptte(μi)).
If ζ0≤ptξ, then (ζ)∗ν<ξ, and π∈Mk(Mhka((ζ)∗ν))
by π∈Mhka(ξ).
Let ζ0=ξ+ζ1 with 0<ζ1≤ptμ.
If ζ1<ptμ, then by IH with he(ζ1)=he(μ) we have
π∈Mhka(ζ0).
On the other hand we have (ζ)∗ν<ζ0.
Hence π∈Mk(Mhka((ζ)∗ν)).
Finally consider the case when 0<ζ1=μ.
Then we obtain ν<te(ξ+μ)=te(μ)≤he(μ)≤ξ0.
π∈Mhk+1a(ξ0) with Proposition 2.9 yields π∈Mk+1(Mhk+1a(ν)).
On the other side we see π∈Mhka(ζ) as follows.
We have ζ<ξ+μ.
If ζ≤ξ, then this follows from π∈Mhka(ξ) and Proposition 2.9,
and if ζ=ξ+λ<ξ+μ, then IH yields π∈Mhka(ζ).
Since π∈Mhka(ζ) is a Πk−11-sentence holding on Lπ
by Lemma 2.8.2 and {a}∪K(ζ)⊂Ha(π),
we obtain π∈Mk+1(Mhka((ζ)∗ν)), a fortiori π∈Mk(Mhka((ζ)∗ν)).
□
Definition 2.11
For sequences of ordinals ξ=(ξk,…,ξN−1) and ν=(νk,…,νN−1)
and 2≤k,m,n≤N−1,
[TABLE]
Corollary 2.12
Let ν be a sequence defined from a sequence ξ as follows.
∀i<k(νi=ξi), ∀i>k(νi=0), and
νk=ξk+Λξk+1b, where 2≤k<N, b<Λ and ξk+1=0.
Then
Mh2a(ν)≺k+1Mh2a(ξ) holds.
In particular if π∈Mh2a(ξ)
and K(ν)∪{π,a}⊂Ha(π), then
ψπν(a)<π.
Let ν=(ν2,…,νN−1), ξ=(ξ2,…,ξN−1) be sequences of ordinals<ε(Λ) such that ν<kξ for
an integer k with 2≤k≤N−1.
Then Mh2a(ν)≺kMh2a(ξ).
In particular if π∈Mh2a(ξ)
and K(ν)∪{π,a}⊂Ha(π), then
ψπν(a)<π.
Proof. Assume π∈Mh2a(ξ) and K(ν)⊂Ha(π).
We have π∈Mhka(ξk).
By the definition (2) and (νk,…,νN−1)<ξk,
we obtain
π∈Mk(⋂k≤i≤N−1Mhia(νi)).
On the other hand we have π∈⋂i<kMhia(ξi), and hence
π∈⋂i<kMhia(νi)
by ∀i<k(νi≤ξi) and Proposition 2.9.
Since π∈⋂i<kMhia(νi) is a Πk−21-sentence holding in Lπ,
we obtain π∈Mk(⋂i≤N−1Mhia(νi))=Mk(Mh2a(ν)),
a fortiori π∈M2(Mh2a(ν)).
Suppose {π,a}⊂Ha(π).
The set
C={κ<π:Ha(κ)∩π⊂κ,K(ν)∪{π,a}⊂Ha(κ)}
is a club subset of the regular cardinal π.
This shows the existence of a κ∈Mh2a(ν)∩C∩π, and hence
ψπν(a)<π by the definition (3).
□
Proposition 2.14
Let ξ=(ξ2,…,ξN−1) be a sequence of ordinals<ε(Λ) such that
{π,a}∪K(ξ)⊂Ha(π).
Assume Tl(ξi)<Λk(ξi+k+1) for some i<N−1 and k>0.
Then
π∈Mh2a(ξ)⇔π∈Mh2a(μ),
where
μ=(μ2,…,μN−1) with
μi=ξi−Tl(ξi) and μj=ξj for j=i.
Proof. When 0<ξi=Λγmam+⋯+Λγ1a1+Λγ0a0 with
γm>⋯>γ1>γ0,0<ai<Λ,
μi=Λγmam+⋯+Λγ1a1 for Tl(ξi)=Λγ0a0.
If ξi=0, then so is μi=0.
Let π∈Mh2a(μ) and Tl(ξi)<Λk(ξi+k+1).
We obtain ∀j≤k(he(j)(Tl(ξi))<Λk−j(ξi+k+1)), and he(k)(Tl(ξi))≤ξi+k.
On the other hand we have π∈Mhi+ka(ξi+k).
From Lemma 2.10 we see inductively that for any j<k, π∈Mhi+ja(he(j)(Tl(ξi))).
In particular π∈Mhi+1a(he(Tl(ξi))), and once again by Lemma 2.10 and π∈Mhia(μi)
we obtain π∈Mhia(ξi).
Hence π∈Mh2a(ξ).
□
Definition 2.15
A sequence of ordinals ξ=(ξ2,…,ξN−1) is said to be irreducible
iff
∀i<N−1∀k>0(ξi>0⇒Tl(ξi)≥Λk(ξi+k+1)).
**
Proposition 2.16
Let ν=(νk,…,νN−1)=0 be an irreducible sequence,
and k0≥k be the least number such that νk0=0.
Assume νk0<he(k0−k)(ξ). Then
ν<ξ holds in the sense of Definition 2.1.6.
Proof. Let
ℓ<N−k be the largest number such that
νk+ℓ=0.
We show (νk,…,νk+ℓ)<ξ.
Since ν is irreducible, we have Λi(νk0+i+1)≤Tl(νk0).
From νk0<he(k0−k)(ξ) and te(μ)≤he(μ) we obtain
νk0+i<νk0+i+1≤he(i)(νk0)≤he(k0−k+i)(ξ).
Let (μk,…,μN−1)⊂ptξ such that μk=Hd(ξ) and
μi+1=he(μi)=te(Hd(μi)).
Then te(μk+i)=he(μk+i) and μk0+i=he(μk0+i−1)=he(k0−k+i)(ξ) for k0−k+i>0.
Therefore (μk,…,μk+ℓ)⊂ptξ witnesses
(νk,…,νk+ℓ)<ξ.
□
Definition 2.17
Let* ξ=(ξk,…,ξN−1), ν=(νk,…,νN−1)andν=ξ.
Leti≥kbe the minimal number such thatνi=ξi.
Suppose(ξi,…,ξN−1)=0, and letk1≥ibe the minimal number such thatξk1=0. Thenν<lx,kξiff one of the followings holds:*
(νi,…,νN−1)=0.
2. 2.
In what follows assume* (νi,…,νN−1)=0, and letk0≥ibe the minimal number such thatνk0=0(i=min{k0,k1}). Thenν<lx,kξiff one of the followings holds:*
(a)
i=k0<k1* andhe(k1−k0)(νk0)≤ξk1.*
2. (b)
k0≥k1=i* andνk0<he(k0−k1)(ξk1).*
Proposition 2.18
Suppose that both of ν and ξ are irreducible. Then
ν<lx,kξ⇒Mhka(ν)≺kMhka(ξ).
Proof. Let π∈Mhka(ξ), K(ν)⊂Ha(π),
and i≥k be the minimal number such that νi=ξi.
We have π∈⋂k≤j<iMhja(νj), which is a Πi−21-sentence holding on Lπ.
In the case ξi=0, it suffices to show that π∈Mi(⋂j≥iMhja(νj)),
since then we obtain
π∈Mi(Mhka(ν)) by π∈Mhia(ξi)⊂Mi, a fortiori π∈Mk(Mhka(ν)).
If (νi,…,νN−1)=0, then ξi=0 and
⋂j≥iMhja(νj) denotes the class of limit ordinals.
Obviously π∈Mi(⋂j≥iMhja(νj)).
In what follows assume (νi,…,νN−1)=0, and let
k0≥i be the minimal number such that νk0=0,
and k1≥i be the minimal number such that ξk1=0.
Case 1. k0≥k1=i: Then we have νk0<he(k0−k1)(ξk1).
Proposition 2.16 yields
(νk0,…,νN−1)<ξk1=ξi,
which in turn yields
π∈Mi(⋂j≥iMhja(νj)) by the definition (2) of
π∈Mhia(ξi).
Case 2. i=k0<k1:
Then we have he(k1−i)(νi)≤ξk1.
Also νi+p<he(p)(νi) for any p>0 since ν is irreducible and νi=0.
Let j≥k1.
Then νj<he(j−i)(νi)≤he(j−k1)(ξk1).
In particular
νk1<ξk1.
Proposition 2.16 yields
(νk1,…,νN−1)<ξk1.
π∈Mhk1a(ξk1) yields π∈Mk1(⋂j≥k1Mhja(νj)).
Moreover for any p<k1−i, he(k1−i−p)(νi+p)≤ξk1 by Proposition 2.2.
Lemma 2.10 yields π∈⋂k1>j≥iMhja(νj).
Therefore π∈Mk1(Mhka(ν)), a fortiori π∈Mk(Mhka(ν)).
□
Let ν=(ν2,…,νN−1), ξ=(ξ2,…,ξN−1) be
irreducible sequences of ordinals<ε(Λ), and assume that
ψπν(b)<π and ψκξ(a)<κ.
Then β1=ψπν(b)<ψκξ(a)=α1 iff one of the following cases holds:
π≤ψκξ(a).
2. 2.
b<a, ψπν(b)<κ and
K(ν)∪{π,b}⊂Ha(ψκξ(a)).
3. 3.
b>a* and K(ξ)∪{κ,a}⊂Hb(ψπν(b)).*
4. 4.
b=a, κ<π and κ∈Hb(ψπν(b)).
5. 5.
b=a, π=κ, K(ν)⊂Ha(ψκξ(a)), and
ν<lx,2ξ.
6. 6.
b=a, π=κ,
K(ξ)⊂Hb(ψπν(b)).
Proof. If the case (2) holds, then
ψπν(b)∈Ha(ψκξ(a))∩κ⊂ψκξ(a).
If one of the cases (3) and (4) holds, then
K(ξ)∪{κ,a}⊂Ha(ψπν(b)).
On the other hand we have
K(ξ)∪{κ,a}⊂Ha(ψκξ(a)).
Hence ψπν(b)<ψκξ(a).
If the case (5) holds, then Proposition 2.18 yields
Mh2a(ν)≺2Mh2a(ξ)∋ψκξ(a).
Hence ψκξ(a)∈M2(Mh2a(ν)).
Since
K(ν)∪{κ,a}⊂Ha(ψκξ(a)),
the set {ρ<ψκξ(a):Ha(ρ)∩κ⊂ρ,K(ν)∪{κ,a}⊂Ha(ρ)} is club in ψκξ(a).
Therefore
ψπν(b)=ψκν(a)<ψκξ(a)
by (3) in Definition 2.5.3.
Finally assume that the case (6) holds.
Since K(ξ)⊂Ha(ψκξ(a)),
ψπν(b)<ψκξ(a) holds.
Conversely assume that ψπν(b)<ψκξ(a) and
ψκξ(a)<π.
First consider the case b<a.
Then we have K(ν)∪{π,b}⊂Hb(ψπν(b))⊂Ha(ψκξ(a)). Hence (2) holds.
Next consider the case b>a.
K(ξ)∪{κ,a}⊂Hb(ψπν(b)) would yield
ψκξ(a)∈Hb(ψπν(b))∩π⊂ψπν(b), a contradiction ψκξ(a)<ψπν(b).
Hence (3) holds.
Finally assume b=a.
Consider the case κ<π.
κ∈Hb(ψπν(b))∩π would yield
ψκξ(a)<κ<ψπν(b), a contradiction.
Hence κ∈Hb(ψπν(b)), and
(4) holds.
If π<κ, then π∈Hb(ψπν(b))∩κ⊂Ha(ψκξ(a))∩κ, and π<ψκξ(a),
a contradiction, or we should say that (1) holds.
Finally let π=κ.
We can assume that
K(ξ)⊂Hb(ψπν(b)), otherwise
(6) holds.
If ξ<lx,2ν, then by (5)
ψκξ(a)<ψπν(b) would follow.
If K(ν)⊂Ha(ψκξ(a)), then by
(6)
again ψκξ(a)<ψπν(b) would follow.
Hence K(ν)⊂Ha(ψκξ(a))
and ν≤lxξ.
If ν=ξ, then ψκξ(a)=ψπν(b).
Therefore (5) must be the case.
□
Definition 2.20 is utilized to define a computable notation system in the next section 3.
Definition 2.20
A set SD of sequences ξ=(ξ2,…,ξN−1) of ordinals ξi<ε(Λ)
is defined recursively as follows.
0∗(a)∈SD for each a<Λ.
2. 2.
(Cf. Definition 2.1.9.)
Let ξ=(ξ2,…,ξN−1)∈SD, 1≤k<N−1, ζ<ε(Λ) be an ordinal such that
(ξk+1,…,ξN−1)<sdζ, and
(ξ2,…,ξk−1,ξk,ζ)∗0∈SD.
Then for
ζk=ξk+Λζa with an ordinal a<Λ,
(ξ2,…,ξk−1)∗(ζk)∗(ξk+1,…,ξN−1)∈SD
and (ξ2,…,ξk−1)∗(ζk)∗0∈SD.
Proposition 2.21
Let ξ=(ξ2,…,ξN−1)∈SD.
(ξ2,…,ξi)∗0∈SD* for each i with 1≤i<N.*
2. 2.
For 2≤i<j<k<N, if ξi=0 and ξk=0, then ξj=0.
3. 3.
Let ξi=0.
Then
(ξi+1,…,ξN−1)<sdte(ξi).
4. 4.
ξ* is irreducible.*
Proof. Let 1≤k<N−1, ζ<ε(Λ) be an ordinal such that
(ξk+1,…,ξN−1)<sdζ, and
(ξ2,…,ξk−1,ξk,ζ)∗0∈SD.
Also let
ζk=ξk+Λζa with an ordinal a<Λ.
2.21.1 is seen by induction on the recursive definition of ξ∈SD.
2.21.2 is seen by induction on the recursive definition of ξ∈SD.
Suppose ξi=0 for an i<k.
From (ξ2,…,ξk−1,ξk,ζ)∗0∈SD and ζ=0, IH yields ξk=0.
2.21.3 and 2.21.4.
We show these by simultaneous induction on the recursive definition of ξ∈SD.
2.21.3.
We show Proposition 2.21.3 for
the sequence (ξ2,…,ξk−1)∗(ζk)∗(ξk+1,…,ξN−1)∈SD.
The proposition holds for the sequence ξ, and we can assume a=0.
We obtain (ξi+1,…,ξN−1)<sdte(ξi) for i>k if ξi=0, and
(ξk+1,…,ξN−1)<sdte(ζk)=ζ by the assumption.
Let 2≤i<k and ξi=0.
We show
(ξi+1,…,ξk−1)∗(ζk)∗(ξk+1,…,ξN−1)<sdte(ξi).
It suffices to show that ζk<sdte(k−i)(ξi).
By IH we have ξk<sdte(k−i)(ξi).
On the other hand we have ξk=0 by (ξ2,…,ξk−1,ξk,ζ)∗0∈SD,
ζ=0, and Proposition 2.21.2.
Moreover (ξ2,…,ξk−1,ξk,ζ)∗0 is irreducible by Proposition 2.21.4,
and hence Tl(ξk)≥Λζ+1.
Therefore te(ξk)>ζ.
This means that ζk=NFξk+Λζa, and ξk<sdte(k−i)(ξi)
yields ζk<sdte(k−i)(ξi) by Definition 2.1.8.
2.21.4.
If (ξi+1,…,ξN−1)<sdte(ξi) for ξi=0,
then ξi+k<sdte(k)(ξi) for k>0, and
ξi+k+1≤te(k)(ξi).
Hence Λk(ξi+k+1)≤Λte(ξi)≤Tl(ξi), and
ξ is irreducible.
□
3 Computable notation system OT
In this section (except Propositions 3.3)
we work in a weak fragment of arithmetic, e.g., in the fragment IΣ1 or even in the bounded arithmetic S21.
Referring Proposition 2.19
the sets of ordinal terms OT⊂Λ=εK+1
and E⊂ε(Λ)=εK+2
over symbols {0,K,Λ,+,ω,φ,Ω,ψ} are defined recursively.
OT is isomorphic to a subset of HΛ(0).
Simultaneously we define finite sets Kδ(α)⊂OT for δ,α∈OT, and sequences
(mk(α))2≤k≤N−1 for α∈OT∩K, where
in α=ψπν(a), mk(α)=νk, i.e.,
ν=(ν2,…,νN−1)=(m2(α),…,mN−1(α))=(mk(α))k=m(α).
For {α0,…,αm,β}⊂OT,
Kδ(α0,…,αm):=⋃i≤mKδ(αi),
Kδ(α0,…,αm)<β:⇔∀γ∈Kδ(α0,…,αm)(γ<β),
and
β≤Kδ(α0,…,αm):⇔∃γ∈Kδ(α0,…,αm)(β≤γ).
An ordinal term in OT is said to be a regular term if it is one of the form K,
Ωβ+1 or ψπν(a) with the non-zero sequences
ν=0.
K and the latter terms ψπν(a) are Mahlo terms.
α=NFαm+⋯+α0 means that α=αm+⋯+α0
and αm≥⋯≥α0
and each αi is a non-zero additive principal number.
α=NFφβγ means that α=φβγ and β,γ<α.
α=NFωβ means that α=ωβ>β.
α=NFΩβ means that α=Ωβ>β.
Let pd(ψπν(a))=π (even if ν=0).
Moreover for n,
pd(n)(α) is defined recursively by pd(0)(α)=α and
pd(n+1)(α)≃pd(pd(n)(α)).
For terms π,κ∈OT,
π≺κ denotes the transitive closure of the relation
{(π,κ):∃ξ∃b[π=ψκξ(b)]},
and its reflexive closure
π⪯κ:⇔π≺κ∨π=κ⇔∃n(κ=pd(n)(π)).
For each ordinal term α=ψπν(a),
a series (πi)i≤L of ordinal terms is uniquely determined as follows:
πL=α, πi=pd(πi+1) and π0=K.
Let us call the series (πi)i≤L the collapsing series of α=πL.
Then we see that an ordinal term
α=ψπν(a) with ν=0
is constructed by Definition 3.1.2g below
iff L=1.
α is constructed by Definition 3.1.2i
iff L≡1(mod(N−2)).
Otherwise α is constructed by Definition 3.1.2h.
Definition 3.1
ℓα denotes the number of occurrences of symbols
{0,K,Λ,+,ω,φ,Ω,ψ}
in terms α∈OT∪E.
(a)
0∈E.
2. (b)
If 0<a∈OT,
then a∈E.
K(a)={a}.
3. (c)
If {ξi:i≤m}⊂E, ξm>⋯>ξ0>0 and
0<bi∈OT,
then ∑i≤mΛξibi=Λξmbm+⋯+Λξ0b0∈E.
K(∑i≤mΛξibi)={bi:i≤m}∪⋃{K(ξi):i≤m}.
4. (d)
For sequences ν=(ν2,…,νN−1), let
K(ν)=⋃2≤i≤N−1K(νi).
2. 2.
(a)
0,K∈OT.
mk(0)=0 for any k, and Kδ(0)=Kδ(K)=∅.
2. (b)
If α=NFαm+⋯+α0(m>0) with {αi:i≤m}⊂OT,
then
α∈OT, and mk(α)=0 for any k.
Kδ(α)=Kδ(α0,…,αm).
3. (c)
If α=NFφβγ with {β,γ}⊂OT∩K, then
α∈OT, and mk(α)=0 for any k.
Kδ(α)=Kδ(β,γ).
4. (d)
If α=NFωβ with K<β∈OT, then α∈OT,
and mk(α)=0 for any k.
Kδ(α)=Kδ(β).
5. (e)
If α=NFΩβ with β∈OT∩K, then
α∈OT.
m2(α)=1,mk(α)=0 for any k>2 if β is a successor ordinal.
Otherwise
mk(α)=0 for any k.
In each case
Kδ(α)=Kδ(β).
6. (f)
Let α=ψπ(a):=ψπ0(a) where π is a regular term , i.e.,
either π=K or
m(π)=0,
and Kα(π,a)<a.
Then
α=ψπ(a)∈OT. Let
mk(α)=0 for any k.
Kδ(ψπ(a))=∅ if α<δ.
Kδ(ψπ(a))={a}∪Kδ(a,π) otherwise.
7. (g)
Let α=ψKν(a) with
ν=0∗(b)(lh(ν)=N−2) and b,a∈OT
such that 0<b≤a and Kα(b,a)<a.
Then
α=ψKν(a)∈OT.
Let
mN−1(α)=b, mk(α)=0
for k<N−1.
Kδ(ψKν(a))=∅ if α<δ.
Kδ(ψKν(a))={a}∪⋃{Kδ(γ):γ∈K(ν)} otherwise.
8. (h)
Let π∈OT∩K be such that
mk+1(π)=0
and ∀i>k+1(mi(π)=0)
for a k(2≤k≤N−2), and
b,a∈OT such that 0<b≤a.
Let ν=(ν2,…,νN−1) be a sequence
defined by ∀i<k(νi=mi(π)),
νk=mk(π)+Λmk+1(π)b,
and ∀i>k(νi=0).
Then
α=ψπν(a)∈OT if
Kα(π,a,b)∪Kα(K(m(π)))<a.
Let mi(α)=νi for each i.
Kδ(ψπν(a))=∅ if α<δ.
Otherwise
Kδ(ψπν(a))={a}∪Kδ(a,π)∪⋃{Kδ(b):b∈K(ν)}.
9. (i)
Let π∈OT∩K be such that
m2(π)=0 and ∀i>2(mi(π)=0),
and a∈OT.
Let 0=ν=(ν2,…,νN−1)∈SD be a sequence
of ordinal terms νi∈E
such that ν<spm2(π).
Then
α=ψπν(a) if Kα(π,a)<a, and
[TABLE]
Let mi(α)=νi for each i.
Kδ(ψπν(a))=∅ if α<δ.
Otherwise
Kδ(ψπν(a))={a}∪Kδ(a,π)∪⋃{Kδ(b):b∈K(ν)}.
Let {π,a,ξ}⊂Ha(π).
Then ξ=mk(π) is intended to be equivalent to π∈Mhka(ξ).
For Definition 3.1.2h, see Corollary 2.12,
and for
Definition 3.1.2i, see Proposition 2.13.
Proposition 3.2
For each Mahlo term α=ψπν(a)∈OT,
m(α)=ν∈SD for the class SD in
Definition 2.20.
Proposition 3.3
For any α∈OT and any δ such that δ=0,K or δ=ψπν(b) for some π,b,ν,
α∈Hγ(δ)⇔Kδ(α)<γ.
Proof. By induction on ℓα.
□
Lemma 3.4
(OT,<)* is a computable notation system of ordinals.
In particular the order type of the initial segment {α∈OT:α<Ω1}
is less than ω1CK.*
Specifically
each of α<β and α=β is decidable for α,β∈OT,
and α∈OT is decidable for terms α
over symbols {0,K,Λ,+,ω,φ,Ω,ψ}.
Proof. These are shown simultaneously referring Propositions 2.19 and 3.3.
Let us give recursive definitions only for terms Ωα,ψκν(a)∈OT.
First Ωψκν(a)=ψκν(a), i.e.,
Ωα<ψκν(a)⇔α<ψκν(a),
ψκν(a)<Ωα⇔ψκν(a)<α.
Next Ωα<ψΩα+1(a)<Ωα+1.
Finally for ψπν(b),ψκξ(a)∈OT,
ψπν(b)<ψκξ(a) iff one of the following cases holds:
π≤ψκξ(a).
2. 2.
b<a, ψπν(b)<κ, and
Kψκξ(a)({π,b}∪K(ν))<a.
3. 3.
b≥a, and
b≤Kψπν(b)({κ,a}∪K(ξ)).
4. 4.
b=a, π=κ,
Kψκξ(a)(K(ν))<a, and
ν<lx,2ξ.
□
Proposition 3.5
Let β=ψπν(b) with π=ψκξ(a).
Then a<b.
2. 2.
For α=ψπν(a)∈OT,
maxK(ν)≤a holds.
Proof. 3.5.1.
Let β=ψπν(b) with π=ψκξ(a).
Then Kβ({π,b}∪K(ν))<b.
On the other hand we have β<π.
Hence a∈Kβ(π)<b.
3.5.2.
This is seen by induction on ℓα.
Ww have c<a by Proposition 3.5.1 when π=ψσμ(c)
When α is constructed by Definition 3.1.2h,
νk=mk(π)+Λmk+1(π)b holds for b≤a.
By IH we have maxK(m(π))≤c<a when π=ψσμ(c).
Suppose α is constructed by Definition 3.1.2i.
We obtain ν<spm2(π), and hence
maxK(ν)≤maxK(m2(π))≤c<a by IH.
□
4 Operator controlled derivations
In this section,
operator controlled derivations are defined, which are introduced by W. Buchholz[6].
In this and the next sections except otherwise stated
α,β,γ,…,a,b,c,d,… range over ordinal terms in OT⊂HΛ(0),
ξ,ζ,ν,μ,ι,… range over ordinal terms in E,
ξ,ζ,ν,μ,ι,… range over finite sequences over ordinal
terms in E,
and π,κ,ρ,σ,τ,λ,… range over regular ordinal terms
K, Ωβ+1, ψπν(a)
with ν=0.
Reg denotes the set of regular ordinal terms.
We write α∈Ha(β) for Kβ(α)<a.
4.1 Classes of sentences
Following Buchholz[6] let us introduce a language for ramified set theory RS.
Definition 4.1
RS-terms and their levels are inductively defined.
For each α∈OT∩K, Lα is an RS-term of level α.**
2. 2.
If ϕ(x,y1,…,yn) is a set-theoretic formula in the language {∈}, and a1,…,an are
RS-terms of levels<α, then
[x∈Lα:ϕLα(x,a1,…,an)]
is an RS-term of level α.**
Each ordinal term α is denoted by the ordinal term [x∈Lα:x\mboxisanordinal], whose level is α.
Definition 4.2
∣a∣* denotes the level of RS-terms a, and Tm(α) the set of RS-terms of level<α.
Tm=Tm(K) is then the set of RS-terms, which are denoted by a,b,c,d,…*
2. 2.
RS-formulas
are constructed from literalsa∈b,a∈b
by propositional connectives ∨,∧, bounded quantifiers ∃x∈a,∀x∈a and
unbounded quantifiers ∃x,∀x.
Unbounded quantifiers ∃x,∀x are denoted by ∃x∈LK,∀x∈LK, resp.**
3. 3.
For RS-terms and RS-formulas ι,
k(ι) denotes the set of ordinal terms α such that the constant Lα occurs in ι.**
4. 4.
For a set-theoretic Σn-formula ψ(x1,…,xm) in {∈} and a1,…,am∈Tm(κ),
ψLκ(a1,…,am) is a Σn(κ)-formula, where
n=0,1,2,… and κ≤K. Πn(κ)-formulas are defined dually.**
5. 5.
\mboxrk(∃x∈aA(x)):=max{ω\mboxrk(a),\mboxrk(A(L0))+2}* for * a∈Tm(K)∪{LK}.
Proposition 4.5
Let A be a sentence with
A≃⋁(Aι)ι∈J or A≃⋀(Aι)ι∈J.
\mboxrk(A)<K+ω.
2. 2.
∣A∣≤\mboxrk(A)∈{ω∣A∣+i:i∈ω}.
3. 3.
∀ι∈J(\mboxrk(Aι)<\mboxrk(A)).
4. 4.
\mboxrk(A)<λ⇒A∈Σ0(λ)**
4.2 Operator controlled derivations
By an operator we mean a map H,
H:P(OT)→P(OT), such that
∀X⊂OT[X⊂H(X)].
2. 2.
∀X,Y⊂OT[Y⊂H(X)⇒H(Y)⊂H(X)].
For an operator H and Θ,Θ1⊂OT,
H[Θ](X):=H(X∪Θ), and
H[Θ][Θ1]:=(H[Θ])[Θ1], i.e.,
H[Θ][Θ1](X)=H(X∪Θ∪Θ1).
Obviously Hα is an operator for any α, and
if H is an operator, then so is H[Θ].
Sequents are finite sets of sentences, and inference rules are formulated in one-sided sequent calculus.
Let H=Hγ(γ∈OT) be an operator, Θ a finite set of K,
Γ a sequent,
a∈OT and
b∈OT∩(K+ω).
We define a relation (Hγ,Θ)⊢baΓ, which is read ‘there exists an infinitary derivation
of Γ which is Θ-controlled by Hγ, and
whose height is at most a and its cut rank is less than b’.
κ,λ,σ,τ,π ranges over regular ordinal terms.
Definition 4.6
(Hγ,Θ)⊢baΓ holds if
[TABLE]
and one of the following cases holds:
(⋁)
A≃⋁{Aι:ι∈J},
A∈Γ and there exist ι∈J
and
a(ι)<a such that
[TABLE]
and
(Hγ,Θ)⊢ba(ι)Γ,Aι.
(⋀)
A≃⋀{Aι:ι∈J},
A∈Γ and for every
ι∈J there exists an a(ι)<a
such that
(Hγ,Θ∪{k(ι)})⊢ba(ι)Γ,Aι.
(cut)
There exist a0<a and C
such that \mboxrk(C)<b and
(Hγ,Θ)⊢ba0Γ,¬C
and
(Hγ,Θ)⊢ba0C,Γ.
(Ω∈M2)
There exist ordinals aℓ, ar(α) and a sentence C∈Π2(Ω)
such that sup{aℓ+1,ar(α)+1:α<Ω}≤a, b≥Ω,
(Hγ,Θ)⊢baℓΓ,C
and
(Hγ,Θ∪{ωα})⊢bar(α)¬C(α,Ω),Γ
for any α<Ω.
(rfl(π,k,ξ,ν))
There exist a Mahlo ordinal K≥π∈Hγ[Θ]∩(b+1), an integer 2≤k≤N and
sequences ν=(ν2,…,νN−1),ξ=(ξ2,…,ξN−1)∈SD of
ordinals νi,ξi∈E,
ordinals aℓ,ar(ρ),a0,
and a finite set Δ of Σk(π)-sentences enjoying the following conditions:
When π=K,
k=N and
ν=0 with lh(ν)=N−1 hold.
Also let ξ=0 in this case.
When π<K,
ξk=0 with k<N,
0=ξ, and
∀i(ξi≤spmi(π)).
H(ν,π,γ,Θ) denotes the resolvent class for
ν, π, γ and Θ defined as follows:
[TABLE]
for ρ∈Reg∩C(π,γ,Θ).
Then for each ρ∈H(ν,π,γ,Θ),
(Hγ,Θ∪{ρ})⊢bar(ρ)Γ,Δ(ρ,π).
4. 4.
[TABLE]
In the inference rule (rfl(π,k,ξ,ν)) for π=ψσξ(c)<K,
we have π∈Mh2c(ξ).
In particular, π∈⋂i<kMhic(ξi)∩Mhkc(ξk).
Also we are assuming
(νk,…,νN−1)<sdξk, a fortiori (νk,…,νN−1)<ξk.
Since π∈⋂i<kMhic(νi) is a Πk-sentence holding on Lπ,
we obtain π∈Mk(Mh2c(ν)).
Thus the reflection rule (rfl(π,k,ν)) says that
π is Πk-reflecting on the class
H(ν,π,γ,γ0,Θ)
for the club subset
C(π,γ,Θ)
of π,
cf. Proposition 2.13.
On the other side
we see ρ∈Mh2a(ν) from Proposition 2.9
if ∀i(νi≤mi(ρ)) for ρ∈Mh2a(m(ρ)).
We will state some lemmas for the operator controlled derivations.
These can be shown as in [6].
In what follows by an operator H we mean an Hγ for an ordinal γ.
Lemma 4.7
Let (Hγ,Θ)⊢baΓ.
(Hγ′,Θ∪Θ0)⊢b′a′Γ,Δ* for any γ′≥γ,
any Θ0, and any a′≥a, b′≥b such that
k(Δ)∪{a′}⊂Hγ′[Θ∪Θ0].*
2. 2.
Assume Θ1∪{c}=Θ,
c∈Hγ[Θ1].
Then (Hγ,Θ1)⊢baΓ.
Lemma 4.8
(Tautology)*
(H,k(Γ∪{A}))⊢02\mboxrk(A)Γ,¬A,A.*
Lemma 4.9
(Inversion)*
Let A≃⋀(Aι)ι∈J, and
(H,Θ)⊢baΓ with A∈Γ.
Then for any ι∈J,
(H,Θ∪k(ι))⊢baΓ,Aι holds.*
Lemma 4.10
(Boundedness)*
Suppose (H,Θ)⊢caΓ,C for a C∈Σ1(λ),
anda≤b∈H∩λ. Then
(H,Θ)⊢caΓ,C(b,λ).*
Lemma 4.11
(Persistency)*
Suppose (H,Θ)⊢caΓ,C(b,λ) for a C∈Σ1(λ)
and a b<λ∈H[Θ]. Then
(H,Θ)⊢caΓ,C.*
Lemma 4.12
(Predicative Cut-elimination)*
Suppose (H,Θ)⊢c+ωabΓ, a∈H[Θ]
and ]c,c+ωa]∩Reg=∅.
Then (H,Θ)⊢cφabΓ.*
Lemma 4.13
(Embedding of Axioms)
For each axiom A in \mboxKPΠN, there is an m<ω such that
for any operator H=Hγ,
(H,∅)⊢K+mK⋅2A
holds.
Proof. The axiom ¬A,∃zA(z) for ΠN-reflection follows from
A,¬A and
∃zA(z),¬A(ρ) for regular ordinals ρ<K
by an inference (rfl(K,N,0,0)).
□
Lemma 4.14
(Embedding)*
If \mboxKPΠN⊢Γ for sets Γ of sentences,
there are m,k<ω such that for any operator H=Hγ,
(H,∅)⊢K+mK⋅2+kΓ
holds*
5 Lowering and eliminating higher Mahlo operations
In the section inferences (rfl(K,N,0,0))
for ΠN-reflecting ordinals K are eliminated from operator controlled derivations of Σ1-sentences
φLΩ over Ω.
α#β denotes the natural (commutative) sum of ordinal terms α,β.
Lemma 5.1
For a Mahlo term π∈OT, ξ∈SD
denotes a sequence with lh(ξ)=N−2, and 2≤k≤N−1 an integer for which the following hold:
When π=K, let ξ=0 and k=N−1.
Otherwise
ξ=(ξ2,…,ξk+1)∗0 with
ξk+1=0 such that ∀i≤k+1(ξi≤spmi(π)).
For ordinal terms γ,a∈OT
let us define a sequence
ζ(a):=(ζ2(a),…,ζk(a))∗0 with lh(ζ(a))=N−2 as follows.
ζ(a)=0∗(γ+a)
when π=K.
Otherwise ζk(a)=ξk+Λξk+1(γ+a)
and ζi(a)=ξi for i<k.
Let κ∈H(ζ(a),π,γ,Θ) for a finite set Θ⊂OT.
Now suppose
(Hγ,Θ)⊢πaΓ
where {γ,π}∪K(ξ)⊂Hγ[Θ], Θ⊂π,
∀i(K(ξi)⊂HmaxK(ξi)[Θ]),
and Γ⊂Πk+1(π).
Let
γ(a,b)=γ#a#b, β(a,b)=ψπ(γ(a,b)), and c>γ(a,κ).
Then
the following holds:
[TABLE]
Proof by induction on a.
Let κ∈H(ζ(a),π,γ,Θ).
We see ζ(a)∈SD, and from (5) and Θ⊂κ
that
[TABLE]
For any a∈Hγ[Θ], we obtain {γ,π,a,κ}⊂Hγ(π) by
Θ∪{κ}⊂π.
Hence for γ(a,κ)=γ#a#κ, {γ(a,κ),π}⊂Hγ(π), and
{γ(a,κ),π}⊂Hγ(a,κ)(β(a,κ))
by the definition (3).
Therefore κ∈Hγ(a,κ)(β(a,κ))∩π⊂β(a,κ) by Proposition 2.6,
and Θ⊂β(a,κ)<π.
Thus we obtain
[TABLE]
Case 1.
First consider the case when the last inference is a (rfl(π,k+1,ξ,ν)).
We have aℓ∈Hγ[Θ]∩a,
ar(ρ)∈Hγ[Θ∪{ρ}]∩a, and
a finite set Δ of Σk+1(π)-sentences.
We have for each δ∈Δ
[TABLE]
and for each ρ∈H(ν,π,γ,Θ)
[TABLE]
When π<K,
ν=(ν2,…,νN−1)∈SD is a sequence
such that
∀i<k+1(νi=ξi), (νk+1,…,νN−1)<sdξk+1,
K(ν)∪K(ξ)⊂Hγ[Θ],
and
∀i(K(νi)⊂HmaxK(νi)[Θ]),
cf. (7) and (8).
Let Γ0=Γ∩Σk(π) and {∀x∈Lπθi(x):i=1,…,n}(n≥0)=Γ∖Γ0
for Σk(π)-formulas θi(x).
Let us fix d={d1,…,dn}⊂Tm(κ) arbitrarily.
Put k(d)=⋃{k(di):i=1,…,n} and Γ(d)=Γ0∪{θi(di):i=1,…,n}.
By Inversion lemma 4.9 from (13) we obtain for each δ∈Δ
[TABLE]
Let ρ∈C(κ,c,Θ∪{κ}∪k(d)).
We see ρ<κ, and k(d)<ρ
from k(d)<κ.
By Θ∩π⊂Hγ(κ)∩π⊂κ and γ≤c
we obtain C(κ,c,Θ∪{κ}∪k(d))⊂C(π,γ,Θ).
Namely, cf. (9)
[TABLE]
For each ρ∈H(ν,κ,c,Θ∪{κ}∪k(d)),
IH with (14) and (16)
yields for c>γ(ar(ρ),κ)
and κ∈H(ζ(ar(ρ)),π,γ,Θ∪{ρ})
[TABLE]
Let
ρ∈Mℓ:={ρ∈Reg:∀i(ζi(aℓ)≤spmi(ρ))}∩H(ν,κ,c,Θ∪{κ}∪k(d)).
Then
Mℓ⊂H(ζ(aℓ),π,γ,Θ∪k(d)) and
Θ∪k(d)⊂ρ.
For each δ∈Δ,
IH with (15) yields for c>γ(aℓ,ρ)
[TABLE]
From (17) and (18) by several (cut)’s of δ(ρ,π)
with \mboxrk(δ(ρ,π))<κ we obtain for a(ρ)=max{aℓ,ar(ρ)}
and some p<ω
[TABLE]
On the other hand
we have by Tautology lemma 4.8
for each θ(d)(κ,π)∈Γ(d)(κ,π)
[TABLE]
where
2\mboxrk(θ(d)(κ,π))≤κ+p for some p<ω.
Moreover we have
sup{2\mboxrk(θ(d)(κ,π)),β(a(ρ),κ)+p:ρ∈Mℓ}≤β(a0,κ)+p∈Hγ[Θ∪{κ}],
where sup{aℓ,ar(ρ):ρ∈H(ν,π,γ,Θ)}≤a0<a
by (10).
Now let μ=(μ2,…,μN−1)=max{ζ(aℓ),ν} with
μi=max{ζi(aℓ),νi}.
Since νi=ξi≤ptζi(aℓ) for i<k+1, we obtain
\mu_{i}=\left\{\begin{array}[]{ll}\zeta_{i}(a_{\ell})&i\leq k\\
\nu_{i}&i>k\end{array}\right..
We see that
Mℓ=H(μ,κ,c,Θ∪{κ}∪k(d)).
Moreover we have ∀i<k(μi=ξi=ζi(a)) and
(μk,…,μN−1)=(ζk(aℓ))∗(νk+1,…,νN−1)<sdζk(a).
Also
∀i(K(ζi(a))⊂HmaxK(ζi(a))[Θ]) and ∀i(K(μi)⊂HmaxK(μi)[Θ]).
For ¬Γ(d)(κ,π)⊂Πk(κ),
by an inference rule
(rfl(κ,k,ζ(a),μ)) with
its resolvent class
Mℓ,
we conclude
from (20)
and (19) that
(Hc,Θ∪{κ}∪k(d))⊢κβ(a0,κ)+p+1Γ(d)(κ,π),Γ(κ,π).
Since d⊂Tm(κ) is arbitrary,
several (⋀)’s yield (11).
Case 2.
Second consider the case when the last inference is a (rfl(π,j,ξ,ν))
for a j<k+1.
We have
(Hγ,Θ)⊢πaℓΓ,¬δ
for each δ∈Δ⊂Σj(π) with aℓ∈Hγ[Θ]∩a,
and
(Hγ,Θ∪{ρ})⊢πar(ρ)Γ,Δ(ρ,π)
for each
ρ∈H(ν,π,γ,Θ) with ar(ρ)∈Hγ[Θ∪{ρ}]∩a.
ν∈SD is a sequence such that
∀i<j(νi=ξi) and (νj,…,νN−1)<sdξj.
We see that the resolvent class H(ν,κ,c1,Θ∪{κ}) is a subclass of H(ν,π,γ,Θ).
By IH we have
(Hc,Θ∪{κ})⊢κβ(aℓ,κ)Γ(κ,π),¬δ(κ,π)
for each δ∈Δ,
and
(Hc,Θ∪{κ,ρ})⊢κβ(ar(ρ),κ)Γ(κ,π),Δ(ρ,π)
for each ρ∈H(ν,κ,c,Θ∪{κ})
with Δ(ρ,π)=(Δ(κ,π))(ρ,κ).
We claim that ∀i≤j(ξj≤spmi(κ)).
Consider the case when i=j=k.
Then we have ξk≤spmk(π) and ζk(a)≤spmk(κ)
with ξk<ptζk(a).
We obtain ξk≤spmk(κ).
Hence by an inference rule (rfl(κ,j,ξ(j),ν))
for the sequence ξ(j)=(ξ2,…,ξj)∗0∈SD, cf. Proposition 2.21.1,
we obtain (11).
Case 3.
Third consider the case when the last inference is a (rfl(σ,j,μ,ν))
for a σ<π.
We have
(Hγ,Θ)⊢πaℓΓ,¬δ
for each δ∈Δ⊂Σj(σ),
and
(Hγ,Θ∪{ρ})⊢πar(ρ)Γ,Δ(ρ,σ)
for each ρ∈H(ν,σ,γ,Θ).
We obtain σ<κ
by (12) for σ∈Hγ[Θ].
Hence
Δ⊂Σ01(σ)⊂Σ0(κ) and
δ(κ,π)≡δ for any δ∈Δ.
Let H(ν,σ,c,Θ∪{κ}) be the resolvent class for σ, ν, c and Θ∪{κ}.
Then H(ν,σ,c,Θ∪{κ})⊂H(ν,σ,γ,Θ).
From IH
we have
(Hc,Θ∪{κ})⊢κβ(aℓ,κ)Γ(κ,π),¬δ
for each δ∈Δ, and
(Hc,Θ∪{κ,ρ})⊢κβ(ar(ρ),κ)Γ(κ,π),Δ(ρ,σ)
for each ρ∈H(ν,σ,c,Θ∪{κ}).
We obtain
(11)
by an inference rule (rfl(σ,j,μ,ν))
with the resolvent class H(ν,σ,c,Θ∪{κ}).
Case 4.
Fourth consider the case when the last inference (⋀) introduces a Πk+1(π)-sentence
(∀x∈Lπθ(x))∈Γ.
We have
(Hγ,Θ∪k(d))⊢πa(d)Γ,θ(d)
for each d∈Tm(π).
For each d∈Tm(κ),
IH with k(d)<κ yields
(Hc,Θ∪{κ}∪k(d))⊢κβ(a(d),κ)Γ(κ,π),θ(d)(κ,π).
(⋀) yields (11)
for ∀x∈Lκθ(x)(κ,π)≡(∀x∈Lπθ(x))(κ,π)∈Γ(κ,π).
Case 5.
Fifth consider the case when the last inference (⋀) introduces a Σ0(π)-sentence
(∀x∈cθ(x))∈Γ for a c∈Tm(π).
We have
(Hγ,Θ∪k(d))⊢πa(d)Γ,θ(d)
for each d∈Tm(∣c∣).
Then we have ∣d∣<∣c∣<κ by (12).
IH yields (Hc,Θ∪{κ}∪k(d)⊢κβ(a(d),κ)Γ(κ,π),θ(d), and we obtain (11) by an inference (⋀).
Case 6.
Sixth consider the case when the last inference (⋁) introduces a Σk(π)-sentence
(∃x∈Lπθ(x))∈Γ.
We have
(Hγ,Θ)⊢πa0Γ,θ(d)
for a d∈Tm(π).
Without loss of generality we can assume that k(d)⊂k(θ(d)).
Then we see that ∣d∣<κ from (12), and d∈Tm(κ).
Also ∣d∣<κ<β(a,κ) for (6).
IH yields with (∃x∈Lπθ(x))(κ,π)≡(∃x∈Lκθ(x)(κ,π))∈Γ(κ,π),
(Hc,Θ∪{κ})⊢κβ(a0,κ)Γ(κ,π),θ(d)(κ,π),
and we obtain (11) by an inference (⋁).
Case 7.
Seventh consider the case when the last inference is a (cut).
We have (Hγ,Θ)⊢πa0Γ,¬C and
(Hγ,Θ)⊢πa0C,Γ for a0<a with \mboxrk(C)<π.
Then C∈Σ0(π) by Proposition 4.5.4.
On the other side k(C)⊂π holds
by Proposition 4.5.2.
Then k(C)⊂κ
by (12).
Hence C(κ,π)≡C and
\mboxrk(C(κ,π))<κ again by Proposition 4.5.2.
IH yields
(Hc,Θ∪{κ})⊢κβ(a0,κ)Γ(κ,π),¬C(κ,π)
and
(Hc,Θ∪{κ})⊢κβ(a0,κ)C(κ,π),Γ(κ,π).
Hence by a (cut) we obtain (11).
Case 8.
Eighth consider the case when the last inference is an (Ω∈M2).
We have (Hγ,Θ)⊢πaℓΓ,C and
(Hγ,Θ∪{ωα})⊢πar(α)¬C(α,Ω),Γ
for each α<Ω
with sup{aℓ+1,ar(α)+1:α<Ω}≤a and C∈Π2(Ω).
We obtain ωα<κ for α<Ω.
IH with C(κ,π)≡C yields for each α<Ω,
(Hc,Θ∪{κ,ωα})⊢κβ(ar(α),κ)¬C(α,Ω),Γ(κ,π),
and
(Hc,Θ∪{κ})⊢κβ(aℓ,κ)Γ(κ,π),C.
An (Ω∈M2) yields
(11)
All other cases are seen easily from IH.
□
Lemma 5.2
Let λ≤π be a regular ordinal term such that
∀i(K(mi(π))⊂HmaxK(mi(π))[Θ]),
and Γ⊂Σ1(λ).
Suppose for an ordinal term a∈OT
[TABLE]
where {γ,λ,π}⊂Hγ[Θ].
*Assume
*
[TABLE]
Let
a^=γ#ωπ+a+1 and β=ψλ(a^).
Then
the following holds
[TABLE]
Proof by main induction on π with subsidiary induction on a.
We can assume a>0.
Let ξ∈SD be a sequence of ordinals and k a number for which the following hold:
If π=K, then let ξ=0 with lh(ξ)=N−1 and k=N−1.
Let π<K. If m(π)=0, then K(ξ)⊂Hγ[Θ],
ξ≤m(π) and k=max{k≤N−2:ξk+1>0}.
Otherwise let ξ=0 and k=1.
By the assumption (21),
and (5) we obtain
[TABLE]
Case 1.
First consider the case when k≥2.
Let ξ=m(π),
and ζ(a):=(ζ2(a),…,ζk(a))∗0 be the
sequence defined as in Lemma 5.1 from γ,a:
ζ(a)=0∗(γ+a) when π=K, otherwise
ζk(a)=ξk+Λξk+1(γ+a)
and ζi(a)=ξi for i<k.
Also let γ(a,b)=γ#a#b and β(a,b)=ψπγ(a,b).
Let κ:=ψπζ(a)(γ(a,0)).
By the assumption (21)
we have
Θ⊂ψπ(γ#a).
On the other hand we have ψπ(γ#a)=ψπ(γ(a,0))≤κ, and Θ⊂κ.
π∈Hγ[Θ] with Θ⊂π yields
K(ξ)=K(m(π))⊂Hγ[Θ]⊂Hγ(a,0)(κ).
Hence
K(ξ)∪{π,γ(a,0)}⊂Hγ(a,0)(κ), and
κ∈OT by γ(a,0)=γ#a>0 and
Definition 3.1.2h such that
κ<π and Hγ(κ)∩π⊂κ.
Moreover we have
∀i(K(ζi(a))⊂HmaxK(ζi(a))[Θ])
by ∀i(K(mi(π))⊂HmaxK(mi(π))[Θ]) and {γ,a}⊂Hγ[Θ] with Θ⊂κ.
In other words, κ∈H(ζ(a),π,γ,Θ).
By Lemma 5.1 we obtain
(Hγ(a,κ)+1,Θ∪{κ})⊢κβ(a,κ)Γ(κ,π),
and Lemma 4.7.2 with κ∈Hγ(a,0)+1[Θ]
[TABLE]
If λ=π, then Γ(κ,π)⊂Σ1(κ)⊂Σ0(λ).
We have Θ⊂ψπ(a^)=β, and κ∈Ha^(β).
Hence {γ,π,a,κ}⊂Ha^(β), and
γ(a,κ)=γ#a#κ<γ#ωπ+a+1=a^.
Therefore κ<β(a,κ)≤ψπ(a^)=β.
We obtain (22) by Persistency lemma 4.11.
Next consider the case when λ<π.
Then λ<κ and Γ(κ,π)=Γ.
We have for (21),
∀d∀ρ∈[λ,κ](Θ⊂ψρ(γ(a,κ)+1#d)).
By MIH on (24)
we obtain
(Hb0+1,Θ)⊢β0β0Γ
for β0=ψλ(b0) with b0=(γ(a,κ)+1)#ωκ+β(a,κ)+1.
We have b0=γ#a#κ#1#ωβ(a,κ)+1<γ#ωπ+a+1=a^
by β(a,κ)<π.
This yields
ψλ(b0)=β0<β=ψλ(a^) by Θ⊂β and
{γ,κ,π,a}⊂Ha^(β).
Hence (22) follows.
In what follows suppose k=1.
Case 2.
Consider the case when the last inference rule is a (rfl(π,2,ξ,ν)).
We have an ordinal term
aℓ∈Hγ[Θ]∩a, and a finite set Δ of Σ2(π)-sentences
for which
(Hγ,Θ)⊢πaℓΓ,¬δ
holds for each δ∈Δ.
On the other hand we have sequences ν,(ξ2)∗0∈SD such that
ν<sdξ2 and K(ν)∪K(ξ)⊂Hγ[Θ] by (7), and
an ordinal term
ar(ρ)∈Hγ[Θ∪{ρ}]∩a
for which
(Hγ,Θ∪{ρ})⊢πar(ρ)Γ,Δ(ρ,π)
holds for each ρ∈H(ν,π,γ,Θ), where
ξ2≤spm2(π).
Let
ρ:=ψπν(aℓ#π)
for
aℓ=γ#ωπ+aℓ+1.
By the assumption (21) we have
Θ⊂ψπ(aℓ)⊂ρ.
K(ν)∪{π,γ,a}⊂Hγ[Θ]
yields
K(ν)∪{π,aℓ}⊂Haℓ#π(ρ).
Next consider the condition (4).
We have ∀i(K(νi)⊂HmaxK(νi)[Θ]) by (8), and
hence ∀i(K(νi)⊂HmaxK(νi)(ρ)) by Θ⊂ρ.
Therefore ρ∈OT by Definition 3.1.2i.
Moreover ρ∈C(π,γ,Θ), i.e., Hγ(ρ)∩π⊂ρ&Θ∩π⊂ρ.
Hence ρ∈H(ν,π,γ,Θ).
By Inversion lemma 4.9 we obtain for each δ≡(∃x∈Lπδ1(x))∈Δ
and each d∈Tm(ρ) with ∣d∣=max({0}∪k(d)),
(Hγ#∣d∣,Θ∪k(d))⊢πaℓΓ,¬δ1(d).
We have {π,γ,∣d∣}⊂Hγ#∣d∣(π) by ∣d∣<ρ<π, and this yields
∣d∣∈Hγ#∣d∣(ψπ(γ#∣d∣))∩π⊂ψπ(γ#∣d∣).
Hence ∣d∣<ψπ(γ#∣d∣), and
∀e(Θ∪k(d)⊂ψπ(γ#∣d∣#e)), i.e., (21) holds for λ=π
and γ#∣d∣.
Let βd=ψπ(ad) for ad=γ#∣d∣#ωπ+aℓ+1=aℓ#∣d∣.
SIH yields
(Had+1,Θ∪k(d))⊢βdβdΓ,¬δ1(d),
which in turn
Boundedness lemma 4.10 yields
(Haπ+1,Θ∪k(d))⊢βdβdΓ,¬δ1(βd,π)(d)
for
aπ=γ#π#ωπ+aℓ+1=aℓ#π.
By persistency we obtain
(Haπ+1,Θ∪k(d))⊢ρβdΓ,¬δ1(ρ,π)(d)
for βd<ψπ(aπ)=ρ∈Hγ[Θ].
Since d∈Tm(ρ) is arbitrary, (⋀) yields
[TABLE]
Now pick the ρ-th branch from the right upper sequents
Case 2.1.
First consider the case λ=π. Then Δ(ρ,π)⊂Σ0(λ).
Let βρ=ψπ(bρ) with
bρ=aπ#1#ωπ+ar(ρ)+1=γ#ωπ+aℓ+1#ωπ+ar(ρ)+1#π#1.
Then βρ>ρ and ∀d[Θ∪{ρ}⊂ψπ(aπ+1#d)].
SIH yields for (26)
[TABLE]
Several (cut)’s with (27), (25)
yield
(Ha^+1,Θ)⊢βρβρ+pΓ
for βρ≥ρ, aπ<bρ<a^ and some p<ω,
where βρ<β=ψπ(a^)
by bρ<a^.
(22) follows.
Case 2.2.
Next consider the case when λ<π.
Then λ<ρ and Δ(ρ,π)⊂Σ1(ρ+) with ρ+=Ωρ+1.
SIH with (26) yields
(Hbρ+1,Θ∪{ρ})⊢βρ+βρ+Γ,Δ(ρ,π)
for βρ+=ψρ+(bρ)>ρ,
and by Lemma 4.7.2 we obtain
[TABLE]
Several (cut)’s with (25), (28)
yield
(Hb0+1,Θ)⊢βρ+βρ++pΓ
for
βρ+>ρ and
b0=γ#(ωπ+aℓ+1⋅2)#ωπ+ar(ρ)+1#1≥max{bℓ,bρ}.
Predicative cut-elimination lemma 4.12 yields
for β1=φ(βρ+)(βρ++p)<ρ+
[TABLE]
We obtain
λ<ρ∈Hb0+1[Θ]
by γ<aℓ<b0.
MIH with (29)
yields
(Hc+1,Θ)⊢ψλcψλcΓ for c=b0#1#ωρ+β1+1.
We obtain
c=b0#ωρ+β1+1#1=γ#(ωπ+aℓ+1⋅2)#ωπ+ar(ρ)+1#ωρ+β1+1#2<γ#ωπ+a+1=a^
since
aℓ,ar(ρ)<a and ρ,β1<ρ+<π.
Hence
ψλc<ψλ(a^)=β, and (22) follows.
Case 3.
Third consider the case when the last inference introduces a Σ1(λ)-sentence
(∀x∈cθ(x))∈Γ for c∈Tm(λ).
We have
(Hγ,Θ∪k(d))⊢πa(d)Γ,θ(d) for each
d∈Tm(∣c∣).
Then we see from (23) that
∣d∣<∣c∣∈Hγ(ψρ(γ#e))∩ρ⊂ψρ(γ#e) for
any ρ∈[λ,π] and any e.
Hence ∣d∣∈ψρ(γ#e).
(21) is enjoyed for Θ∪k(d).
SIH yields
(Ha^+1,Θ∪k(d))⊢βdβdΓ,θ(d)
for βd=ψλ(a(d)).
(⋀) yields (22) for β=ψλ(a^)>βd.
Case 4.
Fourth consider the case when the last inference introduces a Σ1(λ)-sentence
(∃x∈Lλθ(x))∈Γ.
We have (Hγ,Θ)⊢πa0Γ,θ(d) for a d∈Tm(λ).
SIH yields (Ha^+1,Θ)⊢β0β0Γ,θ(d)
for β=ψλ(a^)>ψλ(a0)=β0.
Without loss of generality we can assume that k(d)⊂k(θ(d)).
Then we see from (23)
that
∣d∣∈Hγ(ψλ(γ+1))∩λ⊂ψλ(γ+1)<β.
Thus is enjoyed in the following inference rule (⋁).
We obtain (Ha^+1,Θ)⊢ββΓ
by a (⋁), which enjoys (6).
Case 5.
Fifth consider the case when the last inference is a (rfl(τ,j,μ,ν))
for a τ∈Hγ[Θ]∩π.
We have an aℓ<a and a finite set Δ of Σj(τ)-sentences such that
(Hγ,Θ)⊢πaℓΓ,¬δ
for each δ∈Δ.
On the other hand we have a sequence ν and
an ordinal term ar(ρ)<a for each ρ∈H(ν,τ,γ,Θ)
such that
(Hγ,Θ∪{ρ})⊢πar(ρ)Γ,Δ(ρ,τ).
By (23),
for any ρ∈H(ν,τ,γ,Θ) we obtain
[TABLE]
Case 5.1.
First consider the case when τ<λ.
Then ρ<ψκ(γ#e) for any κ∈[λ,π] and e.
From SIH with (30)
we obtain
(Ha^+1,Θ)⊢βℓβℓΓ,¬δ
for each δ∈Δ with βℓ=ψλ(aℓ), and
(Ha^+1,Θ∪{ρ})⊢βr(ρ)βr(ρ)Γ,Δ(ρ,τ)
for each ρ∈H(ν,τ,γ,Θ) with βr(ρ)=ψλ(ar(ρ)).
We see max{βℓ,βr(ρ),τ}<β=ψλ(a^), and
an inference rule (rfl(τ,j,μ,ν))
yields (Ha^+1,Θ)⊢ββΓ.
Case 5.2.
Second consider the case when λ≤τ.
Then Δ∪Δ(ρ,τ)⊂Σ1(τ+), and
ρ<ψκ(γ#e) for τ<κ≤π and e by (30).
SIH yields
(Haℓ+1,Θ)⊢β2β2Γ,¬δ
for each δ∈Δ, where
β2=ψτ+(aℓ).
On the other side SIH yields
(Har(ρ)+1,Θ∪{ρ})⊢βρβρΓ,Δ(ρ,τ)
for each ρ∈H(ν,τ,γ,Θ), where βρ=ψτ+(ar(ρ)).
Predicative cut-elimination lemma 4.12 yields
(Haℓ+1,Θ)⊢τδ2Γ,¬δ
and
(Har(ρ)+1,Θ∪{ρ})⊢τδρΓ,Δ(ρ,τ)
for
δ2=φ(β2)(β2) and δρ=φ(βρ)(βρ).
From these with the inference rule (rfl(τ,j,μ,ν)) we obtain
[TABLE]
where sup{δ2,δρ:ρ∈H(ν,τ,a0+1,Θ)}≤δ0:=φ(β0)(β0)∈Ha0+1[Θ]
with
sup{β2,βρ:ρ∈H(ν,τ,γ,Θ)}≤β0:=ψτ+(a0),
and
sup{aℓ,ar(ρ):ρ∈H(ν,τ,γ,Θ)}≤a0∈Hγ[Θ]∩a, cf. (10).
MIH with (31)
yields (Ha^+1,Θ)⊢δδΓ
for δ=ψλ((a0+1)#ωτ+δ0+2)
and (a0+1)#ωτ+δ0+2<a^.
We have δ=ψλ(a0#1#ωτ+δ0+2)<ψλ(a^)=β
by a0<a^ and τ,δ0<τ+<π and τ∈Hγ[Θ].
(22) follows.
Case 6.
Sixth consider the case when the last inference is a (cut).
For an a0<a and a C with \mboxrk(C)<π, we have
(Hγ,Θ)⊢πa0Γ,¬C and
(Hγ,Θ)⊢πa0C,Γ.
Case 6.1. First consider the case when \mboxrk(C)<λ.
Then C∈Σ0(λ).
SIH yields the lemma.
Case 6.2. Second consider the case when λ≤\mboxrk(C)<π.
Let ρ+=(\mboxrk(C))+=min{κ∈Reg:\mboxrk(C)<κ}.
Then C∈Σ0(ρ+) and
λ≤ρ∈Hγ[Θ]∩π.
SIH yields
(Ha0+1,Θ)⊢β0β0Γ,¬C
and
(Ha0+1,Θ)⊢β0β0C,Γ
for β0=ψρ+(a0)∈Ha0+1[Θ].
By a (cut) we obtain
(Ha0+1,Θ)⊢β1β1Γ
for β1=max{β0,\mboxrk(C)}+1 with ρ<β1<ρ+.
Predicative cut-elimination lemma 4.12 yields
(Ha0+1,Θ)⊢ρδ1Γ
for δ1=φ(β1)(β1),
where a0∈Ha0+1[Θ], and
∀e∀τ∈[λ,ρ][Θ⊂ψτ(a0#e)] hold.
Hence MIH with ρ∈Ha0+1[Θ]
yields (Hb+1,Θ)⊢ψλ(b)ψλ(b)Γ
for b=a0#1#ωρ+δ1+1.
We see b<a^ and ψλ(b)<ψλ(a^)=β, and (22) follows.
Case 7.
Seventh consider the case when the last inference is an (Ω∈M2).
We have (Hγ,Θ)⊢πaℓΓ,C for an aℓ<a, and
(Hγ,Θ∪{α})⊢πar(α)¬C(α,Ω),Γ
for an ar(α)<a for each α<Ω,
where C∈Π2(Ω).
The case λ>Ω is seen as in Case 5.1.
The case λ=Ω is seen as in Case 5.2.
□
Proof of Theorem 1.1.
Let KPΠN⊢θ.
By Embedding lemma 4.14 pick an m so that
(H0,∅)⊢K+mK⋅2+mθ.
Predicative cut-elimination lemma 4.12 yields
(H0,∅)⊢Kωm+1(K+1)θ
for ωm(K⋅2+m)<ωm+1(K+1).
Lemma 5.2 yields
(Ha+1,∅)⊢ββθ
for a=ωK+ωm+1(K+1)+1 and β=ψΩ(a).
Predicative cut-elimination lemma 4.12 yields
(Ha+1,∅)⊢0φ(β)(β)θ.
We obtain φ(β)(β)<α:=ψΩ(ωn(K+1)) for n=m+3, and hence
(Hωn(K+1),∅)⊢0αθ.
Boundedness lemma 4.10 yields
(Hωn(K+1),∅)⊢0αθ(α,Ω).
Since each inference rule other than reflection rules (rfl(π,k,ξ,ν)) and (Ω∈M2) is sound,
we see by induction up to α=ψΩ(ωn(K+1)) that
Lα⊨θ.
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