semantics for the intuitionistic propositional …...semantics for the intuitionistic propositional...

Semantics for the intuitionistic propositional

calculus using the Medvedev lattice

By

H.W.A.Duijf

0725889

Supervised by

Dr. S.A.Terwijn, Thesis Advisor

Dr.W.H.M.Veldman, Second Reader

Master Mathematics

Specialization in Algebra & Logic

Radboud University Nijmegen

May 28, 2013

Contents

Introduction 3

Chapter 1. Computability theory 61.1. Turing degrees 61.2. Other notions of degrees by adjusting the notion of T-degrees 71.3. Functional operators 91.4. The Medvedev lattice 131.5. Other reducibilities on mass problems derived from M-reducibility 16

Chapter 2. Algebraic semantics 212.1. Heyting algebras 212.2. Brouwer algebras 222.3. Interpretation of propositional logic on Brouwer algebras 25

Chapter 3. Kripke semantics and Medvedev's Logic 293.1. Linking Kripke frames to Brouwer algebras 293.2. Medvedev's Logic 32

Chapter 4. Semantics for IPC using the Medvedev lattice 39

Bibliography 45

Appendix A. Initial segments of the degrees of unsolvability 47

2

Introduction

The study of intuitionistic logic dates back to Brouwer's objections to the logi-cists and formalists accounts of (mathematical) truth. He pioneered an entirely newbranch of mathematics called Intuitionistic Mathematics. Intuitionistic reasoningcan be studied from the classical point of view. In 1930 A. Heyting [10], a studentof Brouwer who later proved to be a great mathematician, introduced a logical sys-tem that was meant to capture this reasoning. We will refer to this proof-theoreticlogical system as the intuitionistic propositional calculus or just IPC.

Because intuitionistic reasoning di�ers from the classical account, it remainedopen what proper semantics for it would be. Given the proof-theoretic frameworksupplied by Heyting, there have been various approaches to the semantics for intu-itionistic (propositional, but also �rst-order) logic, for instance via algebras (cf. [29,Rasiowa and Sikorski]), realizability theory by Kleene [12], the Logic of Proofs [1,Artemov] and Kripke semantics (cf. [8, Fitting]). But we are going to consideranother attempt introduced by Kolmogorov [14]. Shortly after Heyting publishedhis formal system Kolmogorov informally stated that it had close relations to someunclassi�ed calculus of problems. He stated that a formula was to be understoodas the problem of giving a solution. The logical connectives should then be readas elementary problem-functions. Several attempted to work out this informal ap-proach to connect IPC to some calculus of problems. Ideas of Kolmogorov togetherwith those of Brouwer and Heyting lead to the BHK-interpretation of IPC, whichcan be given as follows (see [14]):

• a solution to φ ∧ ψ consists of both a solution to φ and a solution to ψ,• a solution to φ ∨ ψ consists of a solution of at least one of the disjuncts,• a solution to φ → ψ consists of a method by which you get a solution toψ given any solution to φ.

It seemed that this informal interpretation of the logical connectives coincided withBrouwer's thoughts on (mathematical) truth. From the BHK-interpretation it isobvious that a clear notion of 'method' is needed. One could argue that a solutionto φ → ψ shows that ψ is somehow easier than φ, since in that case from anysolution to φ we can get a solution to ψ. When we put this all together we couldsay that a solution to φ → ψ shows that ψ can be reduced to φ, which meansthat a clear notion of reducibility is essential. This is why we will introduce severalnotions of reducibility to de�ne a calculus of problems.

In the �rst chapter we are going to discuss how computability theory (cf. [27,30]) can be used to make precise the idea of reducibility between a variety ofproblems. As an introduction the chapter starts with elementary computabilitytheory, in particular we introduce the Turing degrees. Subsequently we show thatobvious adjustments to the notion of Turing-reducibility can be used to give rise

3

INTRODUCTION 4

to di�erent notions of reducibility which in turn can be viewed as di�erent calculiof problems in the sense of Kolmogorov. The problems examined up to now aresets of natural numbers, so an evident modi�cation would be to examine sets ofsets of natural numbers. However we jump to sets of (total) functions, calledmass problems. By considering di�erent problems we need to introduce notions ofreducibility. The notions we introduce are closely related to the ones we introducedwhen we were examining reducibilities between sets of natural numbers. We tryto clarify that they show up naturally by altering the de�nition of Turing degrees.The main subject of this thesis is the Medvedev lattice [20], this consists of massproblems ordered by the Medvedev-reducibility. It will be the cause for most of theother treated subjects. We show that the Medvedev lattice constitutes a Brouweralgebra, which will be the source for further investigation of such algebras. Finally,we examine di�erent notions of reducibilities between mass problems. It turnsout that all of them yield the same lattice. So we adjust the problems we areconsidering to mass problems of partial partial functions, which are (as the namealready suggests) sets of partial functions, and show that the previous notions ofreducibility yield di�erent posets. For instance the Dyment lattice emerges fromthese considerations. This last subject is not studied to any further extend, sincewe restrict ourselves to studying the Medvedev lattice.

The second chapter examines an algebraic semantics for IPC (cf. [29]). Westart with an introduction to lattice theory, Heyting algebras and Brouwer algebras.After that we turn to so-called canonical subsets of Brouwer algebras, which is acrucial concept in our investigation of the Medvedev lattice. In order to examinecanonical subsets we introduce the ×-hull of an implicative usl, this will turn outto be a Brouwer algebra in some cases. We continue by showing how Brouweralgebras can be used to interpret propositional formulas, which allows us to speakof the theory of a Brouwer algebra. It is quite remarkable that the �elds of algebra,logic and computability theory meet when one considers the theory of the Medvedevlattice. However, the chapter proceeds with a result that states that if the positivefragment of the theory of a Brouwer algebra equals that of IPC, then the intersectionof the theories of the initial segments of that Brouwer algebra equals IPC (seeLemma 60 on page 27). This result is the cause for a study of the positive fragmentof IPC in the next chapter. Later this result will be shown to be crucial for ourmain result of Chapter 4. This also motivates us to study the initial segments ofthe Medvedev lattice in order to get a semantics for IPC.

Chapter 3 deals with yet another semantic approach to IPC, namely Kripkesemantics (cf. [8]). We �rst discuss the link between Kripke semantics and Brouweralgebras. It turns out that the set of cones or upsets constitutes a Brouwer algebraand the theory of this Brouwer algebra and the underlying Kripke frame coincide.Subsequently, we introduce the Medvedev Logic using Kripke frames and the count-able implicative upper semilattice Cω which are main ingredients for deriving themain result in the �nal chapter. Since Cω is an implicative usl we examine the×-hull of it, because it actually constitutes a Brouwer algebra. Beside that we usethe connection between Kripke semantics and Brouwer algebras to supply algebraicsemantics for the Medvedev Logic. A brief study of these new notions shows thatthe intersection of the theories of initial segments of the Brouwer algebras that givesemantics for the Medvedev Logic equals IPC (see Corollary 77 on page 31). Unfor-tunately, this study relies on a theorem which is referred to as Medvedev's theorem

INTRODUCTION 5

[21]. The unfortunate issue concerning this theorem is that no complete proof couldbe found in the literature. In the second part of this chapter we give a completeproof of this theorem using Kripke frames. This proof starts with a detour in theKripke semantics for IPC, eventually proving that the Jaskowski trees supply goodsemantics. These Jaskowski trees are then used to prove that the positive fragmentof the Medvedev Logic equals that of IPC. This enables us to combine the resultof Chapter 2 with the connection found in this chapter in order to get ever closerto the main result of this thesis.

In Chapter 4 the previous discussions and research will be combined in orderto derive a theorem by Skvortsova (cf. [32]). We succeed in providing a calculus ofproblems that can be used to interpret IPC. However, the resulting calculus seemsquite arti�cial and it is far from intuitively clear that this calculus has this relationwith IPC. However, the outline of the chapter is as follows: we start by recallingthe de�nition of the Medvedev lattice and mass problems. Next, we introducea class of mass problems that are easy to work with, the upper-≤T closed massproblems denoted by M. We examine this class and show that it constitutes animplicative usl and that it is canonical in the Medvedev lattice. After this briefstudy we continue by supplying some embedding results. Most importantly, weshow that any countable implicative usl is isomorphic to a sub-implicative usl ofM. Combined with the ×-hull, this particularly shows that the ×-hull of Cω isisomorphic to a sub-Brouwer algebra of the ×-hull of M. This result unites most ofthe concepts introduced in this thesis, but it relies on an embedding result aboutcountable upper semilattices and the Turing degrees [16, Lachlan and Lebeuf]. Abrief review on the research done on this matter can be found in the appendix.Finally, we prove the main result of this thesis. For this we also use the embeddingresult we had before, but in order to get this beautiful result in addition we needto consider a so-called �xed canonical partition in Cω. The proof eventually givesan initial segment of the Medvedev lattice which theory equals IPC.

Let us brie�y return to the challenge Kolmogorov set us up with and recallwhat our main result implies. Kolmogorov informally stated that some unclassi�edcalculus of problems might provide semantics for IPC. In a way, we succeeded byshowing that there is an initial segment of the Medvedev lattice which can be usedas a calculus of problems to interpret IPC. But we could also argue that the initialsegment we came up with is not that intuitive. So the task which was left open byKolmogorov was meant to have an intuitively clear answer, but we have yet to giveany. The result could also be looked at di�erently, by saying that it shows that itis possible to provide semantics for IPC using the Medvedev lattice and it leavesthe question whether it is possible to provide a more intuitive initial segment to dothe job.

CHAPTER 1

Computability theory

We will assume that the reader is familiar with elementary computability the-ory. We follow mainly [27, Odifreddi] (the reader might also see [30, Rogers]) as weintroduce some notions at the very heart of Medvedev's calculus of mass problemsthat will be given shortly. Let us start by recalling some de�nitions concerningcomputability and then show that Medvedev's calculus shows up pretty naturallyfrom small changes in the de�nitions. Conclusively, we show that there are manyother possible calculi that may be considered.

If one is only interested in the Medvedev lattice and is already familiar withcomputability theory, then it is advisable to only examine Section 1.4 on page 13.In order to get a clearer view on Medvedev-reducibility it could be illuminating toread Section 1.3 on page 9.

1.1. Turing degrees

In this section we will brie�y recall Turing degrees and computable functions.This review will grant us plenty of other possible notions of degrees and reducibil-ities.

We will consider problems as sets of natural numbers. So a solution to aproblem is to get an element of that set. In this context we will construct a calculusof problems using partial computable functions.

Definition 1. [27, 30] The class of partial computable functions is the smallestclass of functions that

(1) contains the initial functions

O(x) = 0

S(x) = x+ 1

Ini (x1, . . . , xn) = xi (1 ≤ i ≤ n);

(2) is closed under composition, primitive recursion and unrestrictedµ-recursion.

From this de�nition we can derive the enumeration theorem:

Theorem 2. [27, 30] For every n ∈ ω there is a computable enumeration{ϕne }e∈ω of the n-ary partial computable functions. We usually write ϕe instead ofϕ1e.

The above de�nition allows us to consider sets of natural numbers and to in-troduce a notion of di�culty. That is to say that problem A is easier than problemB:

6

1.2. OTHER NOTIONS OF DEGREES BY ADJUSTING THE NOTION OF T-DEGREES 7

Definition 3. [27, 30] Let A and B be sets consisting of natural numbers.We say that A Turing reduces to B (notation: A ≤T B) if there is a natural numbere such that ∀x[A(x) = ϕBe (x)], i.e. if cA is computable in cB .

We turn our attention to the powerset of the natural numbers, P(ω), with therelation ≤T . We get the following results:

Proposition 4. [27, 30] The relation ≤T is re�exive and transitive on P(ω).

We can use the relation ≤T to derive an equivalence relation on problems ofthe same di�culty:

Definition 5. [27, 30] We de�ne A ≡T B if A ≤T B and B ≤T A. Then theequivalence classes of ≡T are called Turing degrees and we write degT (A) for theTuring degree of A. Furthermore, this gives a poset DT with the Turing degreesand the obvious ordering.

One of the main result on DT is the following:

Proposition 6. [27, 30] The Turing degrees form an upper semilattice with≤T and not a lattice. Furthermore, there is a least Turing degree which consists ofthe computable functions, but there is no top element.

Remark 7. Let us return to informal concept of calculi of problems given inthe introduction. We could ask whether DT provides us with a calculus of problemswhich gives semantics for IPC. Let φ, ψ and χ denote propositional formulas andfor any φ let Aφ denote the problem corresponding to φ. Just as we de�ned in thischapter (and actually everywhere in this thesis), instead of 'can be reduced to' wewill write ≤ in the following. To answer the question suppose a calculus of problemsgives semantics for IPC. Remark that we have φ → φ ∨ ψ ∈ IPC, which meansAφ∨ψ ≤ Aφ. Similarly we get Aφ∨ψ ≤ Aψ. Furthermore, we have that φ → χ andψ → χ intuitionistically implies φ∨ψ → χ. In other words, if Aχ is a problem suchthat both Aχ ≤ Aφ and Aχ ≤ Aψ holds, then we have Aχ ≤ Aφ∨ψ. This merelymeans that Aφ∨ψ designates the problem which is the greatest lower bound forthe problems Aφ and Aψ. In lattice-theoretic terms, this is to say that Aφ∨ψ is themeet of Aφ and Aψ.

1 In view of this discussion we see that the previous propositionshows that the Turing degrees do not supply us with a calculus of problems thatgives us semantics for IPC in the sense of Kolmogorov.

This concludes our brief survey of the Turing degrees.

1.2. Other notions of degrees by adjusting the notion of T-degrees

We introduce a new notion of reducibility on sets of natural numbers:

Definition 8. [27, 30] Let A and B be sets of natural numbers. Then we saythat A enumeration reduces to B (notation A ≤e B) if there is a c.e. relation Rsuch that A = {x|∃w[Dw ⊆ B ∧R(x,w)]}. Intuitively this amounts to saying thatgiven an enumeration of A we can get an enumeration of B.

We say that A ≤e B via z if A = {x|∃w[Dw ⊂ B ∧ (x,w) ∈Wz}. Remark thatany z de�nes a mapping Φz : 2ω → 2ω given by Φz(A) = {x|∃w[Dw ⊂ A∧ (x,w) ∈Wz}, we call such mappings enumeration operators.

1We refer to Section 2.1 on page 21 for an introduction to lattice-theory.

1.2. OTHER NOTIONS OF DEGREES BY ADJUSTING THE NOTION OF T-DEGREES 8

As in our previous survey of the Turing degrees we continue by stating thefollowing propositions and de�nition:

Proposition 9. [27, 30] The relation ≤e on P(ω) is re�exive and transitive.

Definition 10. [27, 30] Analogous to Turing degrees we de�ne A ≡e B ifA ≤e B and B ≤e A. The equivalence classes of ≡e are called e-degrees and wedenote this poset by De.

Proposition 11. [27, 30] The e-degrees form an upper semilattice with ≤e,they do not form a lattice. Furthermore there is a least partial degree 0e whichcontains exactly the c.e. sets, but it has no top element.

We can even do more. When we considered sets of natural numbers, thatmeans we consider characteristic (total) functions cA. A possible alternative is toconsider partial functions and a notion of reducibility between them. We are goingto examine di�erent 'problems' in the remainder of this section. Instead of setsof natural numbers we are going to consider partial functions. According to theseproblems, we �rst introduce some reducibility notions. These notions can be foundin [27, 30], but the survey [31, Sasso] as well as the work of [3, Casalegno] o�erbetter insights in these matters.

Definition 12. [3, 27, 30, 31] Let φ and ψ be partial functions. We say thatφ Turing reduces to ψ (notation: φ ≤T ψ) if φ(x) can be obtained from ψ and theinitial functions by composition, primitive recursion and unrestricted µ-recursion.Notice the analogy with De�nition 1 of computable functions.

Let τ : P → 2ω map a partial function to its graph. In the following we let udenote a (code for a) �nite function. We say that φ enumeration reduces to ψ (no-tation: φ ≤e ψ) if there is a c.e. set R such that φ ' τ−1({〈x, y〉|∃u[(x, y, u) ∈R ∧ u ⊂ ψ]})2. Recall that this is equivalent to saying τ(φ) ≤e τ(ψ) in thesense of De�nition 8. Moreover, we say φ enumeration reduces to ψ via z ifφ ' τ−1({〈x, y〉|∃u[(x, y, u) ∈Wz ∧ u ⊂ ψ]}), i.e. τ(φ) ≤e τ(ψ) via z.

There is yet another possibility that comes to mind. When there is a natural

number z such that φ ' ϕτ(ψ)z we say φ weakly Turing reduces to ψ via z (notation:

φ ≤wT ψ). This amounts to saying φ ≤T cτ(ψ).3

We continue our survey along the same lines we did when we considered theTuring degrees:

Proposition 13. [3, 27, 30, 31] The relations ≤T , ≤wT and ≤e on P arere�exive and transitive.

Remark 14. Just as before we could de�ne an equivalence relation ≡e on P byφ ≡e ψ i� φ ≤e ψ and ψ ≤e φ. Remark that τ induces a map τ on the equivalenceclasses under ≡e and this in turn induces an embedding τ : P/≡e → De. One couldprove that every e-degree contains the graph of a partial function (cf. [27]), thisimplies that τ is an isomorphism. Hence 〈P/≡e,≤e〉 ∼= De.

Definition 15. [3, 31] We introduce an equivalence relation ≡T on P byφ ≡T ψ if φ ≤T ψ and ψ ≤T φ. We call such an equivalence class a partial

2We identify the function φ with its graph.3This can be found in [27, 31], although the de�nitions there are via recursive operators. In thenext section we will prove that these de�nitions are equivalent.

1.3. FUNCTIONAL OPERATORS 9

T-degree. We also introduce an equivalence relation ≡wT on P by φ ≡wT ψ ifφ ≤wT ψ and ψ ≤wT φ. In literature there is no special name for these equivalenceclasses, but partial wT-degree seems appropriate.

Proposition 16. [31] The partial T-degrees and the partial wT-degrees bothform an upper semilattice and not a lattice. Moreover, there is a least partial degreein both cases and in neither case there is a top element.

From Remark 7 on page 7 it follows that none of the introduced degrees supplya calculus of problems in the sense of Kolmogorov.

1.3. Functional operators

There are various ways to extend this calculus of problems. We will start byintroducing the new kind of problems that will be considered and then discussvarious possibilities for reducibility relations.

We brie�y recall some notation. Let P be the set of partial functions from ωto ω, we denote them by ϕ, φ, etc. Let F be the set of all total function from ωto ω, we denote them by f, g, etc. Finally we let u, u′, . . . denote (codes of) �nitefunctions.

Definition 17. [20] A mass problem A is a collection of total functions. Wedenote mass problems by A, B, etc.

We need to de�ne some kind of reducibility between mass problems.

Definition 18. [27, 30] A mapping Ψ of a subset of P into P is called afunctional operator.

Definition 19. [27, 30] There are various possibilities to give this notion ataste of computability. Probably the �rst that comes to mind is to say that Ψ(φ)(x)can be obtained uniformly from φ and the initial functions (see De�nition 1) bycomposition, primitive recursion and unrestricted µ-recursion. In that case we callΨ a computable functional [27].4 Notice the analogy with De�nition 1 of partialcomputable functions. Also remark that computable functionals are de�ned for ev-ery partial function φ. But this is not all there is to say about functional operators.We may use enumeration reducibility to determine other notions. Let τ be themapping that takes any partial function to its graph, i.e. τ(φ) = {〈x, y〉|φ(x) ' y}.We say that a functional operator Ψ is de�ned by some enumeration operator Φz if

Ψ(φ) '

{τ−1Φzτ(φ) if Φzτ(φ) gives the graph of a partial function,

↑ else.

When a functional operator is de�ned by some enumeration operator Φz we call ita partial recursive operator [30]. Examine that in this case φ ∈ dom(Ψ) impliesthat Ψ(φ) = τ−1({〈x, y〉|∃u[(x, y, u) ∈ Wz ∧ u ⊂ φ]}). We say Ψ is a recursiveoperator [30] if it is a partial recursive operator such that Ψ is total, i.e. Ψ(φ) isde�ned for every partial function φ. When Φz de�nes the recursive operator Ψ weget Ψ(φ) = τ−1({〈x, y〉|∃u[(x, y, u) ∈Wz ∧ u ⊂ φ]}).

The following fundamental operator theorem on partial recursive operators andrecursive operators is crucial.

4In [27] this is called a partial recursive functional.


Theorem 20. [30] There exists a computable function σ such that for every z,if Φz de�nes the partial recursive operator Ψ, then Φσ(z) de�nes a recursive operatorΨ′ such that for every total f we have f ∈ dom(Ψ) implies Ψ(f) = Ψ′(f).

Since mass problems are total objects and because of the previous theorem, wewill mainly consider recursive operators.

We have yet to give any examples of recursive operators. The propositionbelow gives many constructions that provide new recursive operators out of oldones and even shows that the codes of these recursive operators are all computablein the codes of the old ones. First, we recall some basic operations concerning totalfunctions. Let n be a natural number. Then n ∗ f is given by n ∗ f(0) = n andn ∗ f(x+ 1) = f(x). Furthermore, f ⊕ g is given by:

f ⊕ g(x) =

{f(x2 ) if x is even,

g(x−12 ) if x is odd.

We can de�ne these operations analogous for partial functions.

Proposition 21. 5 Let Ψi,Ψj be recursive operators and let k be a naturalnumber. Then for every of the following clauses there is a computable φ (except inthe last clause, where there is a natural number i0) such that:

(1) Ψφ(k,i)(f) ' k ∗Ψi(f);(2) Ψφ(k,i)(f) ' f(k) ∗Ψi(f);(3) Ψφ(k)(f) =' ϕk ◦ f ;(4) Ψφ(k)(f) ' f ◦ ϕk;(5) Ψφ(i,j)(f) ' Ψi(f)⊕Ψj(f);

(6) Ψφ(i,j)(n ∗ f) '

{Ψi(f) if n = 0,

Ψj(f) if n > 0.;

(7) Ψφ(i,j)(f) ' Ψj ◦Ψi(f);(8) Ψi0(n ∗ f) ' Ψn(f).

Proof. We will often identify a function ψ with τ(ψ) ⊂ ω×ω, i.e. we identifya function with its graph. Recall that Ψh(f) = {〈x, y〉|∃u[(x, y, u) ∈Wh ∧u ⊂ f} ifΨh(f) ↓. In order to prove all the clauses we use the S-m-n theorem [27, 30] whichstates that for every m,n ∈ ω there is a computable function smn such that for allx, y1, . . . , ym we have

λz1 · · · zn.ϕ(m+n)x (y1, . . . , ym, z1, . . . , zn) = ϕ

(n)smn (x,y1,...,ym).

(1) Let h be the code of the partial computable function φ1 given by

φ1(k, i, x, y, u) '

0 if x = 0 ∧ y = k,

↑ if x = 0 ∧ y 6= k,

ϕi(x− 1, y, u) else.

Then we see that φ(k, i) ' s23(h, k, i) shows that Ψφ(k,i)(f) = {〈x, y〉|∃u[(x, y, u) ∈

Wφ(k,i) ∧ u ⊂ f} = k ∗Ψi(f).

5Although this proposition looks elementary, it seems impossible to �nd these results in anyliterature. Since these results are so basic, it is doubtful that this is the �rst proof.


(2) Let h be the code of the partial computable function φ2 given by

φ2(k, i, x, y, u) '

0 if x = 0 ∧ y = u(k),

↑ if x = 0 ∧ y 6= u(k),

ϕi(x− 1, y, u) else.

Then we see that φ(k, i) ' s23(h, k, i) shows that Ψφ(k,i)(f) = {〈x, y〉|∃u[(x, y, u) ∈

Wφ(k,i) ∧ u ⊂ f} = f(k) ∗Ψi(f).(3) Let h be the code of the partial computable function φ3 given by

φ3(k, x, y, u) '

{0 if y ' ϕk(u(x))

↑ else.

Then we see that φ(k) ' s13(h, k) shows that Ψφ(k)(f) = {〈x, y〉|∃u[(x, y, u) ∈

Wφ(k) ∧ u ⊂ f} = ϕk ◦ f .(4) Let h be the code of the partial computable function φ4 given by

φ4(k, x, y, u) '

{0 if y ' u(ϕk(x))

↑ else.

Then we see that φ(k) ' s13(h, k) shows that Ψφ(k)(f) = {〈x, y〉|∃u[(x, y, u) ∈

Wφ(k) ∧ u ⊂ f} = f ◦ ϕk.(5) Let h be the code of the partial computable function φ5 given by

φ5(i, j, x, y, u) '

{ϕi(

x2 , y, u) if x is even,

ϕj(x−1

2 , y, u) if x is odd.

Then we see that φ(i, j) ' s23(h, i, j) shows that Ψφ(i,j)(f) = {〈x, y〉|∃u[(x, y, u) ∈

Wφ(i,j) ∧ u ⊂ f} = Ψi(f)⊕Ψj(f).(6) Let h be the code of the partial computable function φ6 given by

φ6(i, j, x, y, u) '

ϕi(x, y, u) if u(0) = 0,

ϕj(x, y, u) if u(0) > 0,

↑ else.

Where u is (a code of the �nite function) given by u(x) = u(x+1). Then we see thatφ(i, j) ' s2

3(h, i, j) shows that Ψφ(i,j)(n ∗ f) = {〈x, y〉|∃u[(x, y, u) ∈ Wφ(i,j) ∧ u ⊂

n ∗ f} =

{Ψi(f) if n = 0,

Ψj(f) if n > 0.

(7) The composition is the most technical clause. Remark that Ψj ◦ Ψi(f) =Ψj({〈x, y〉|∃u[(x, y, u) ∈Wi∧u ⊂ f ]}). So in order to determine the outcome of thecomposition we have to enumerate Ψi(f). The problem when we try to solve thisstraightforward is that in order to determine Ψj({〈x, y〉|∃u[(x, y, u) ∈Wi∧u ⊂ f ]})we have to see if there is any �nite segment u of Ψi(f) such that (x, y, u) ∈ Wj .With the below manipulation we do not need to consider all the �nite segmentsof Ψi(f), but it is enough to approximate Ψi(f) in stages by �nite functions. Weshow that we can make slight changes to the codes i and j such that

(1) (x, y, u) ∈Wφ(j) implies ∀u′[u ⊆ u′ ⇒ (x, y, u′) ∈Wφ(j)];(2) (x, y, u) ∈Wj implies (x, y, u) ∈Wφ(j).


Let h be a code for the partial computable function φ7a given by

φ7a(j, x, y, u) '

{0 if ∃t∃u ⊆ u[t codes a computation of ϕj on (x, y, u)]

↑ else.

Then it follows that φ(j) ' s13(h, j) satis�es (1) and (2). So Ψj(ϕ) ⊆ Ψφ(i)(ϕ) (seen

as graphs), but since Ψj is a recursive operator we have Ψφ(j) = Ψj .To show (7) we also use Ψφ(i) instead of Ψi. Let φ

′ be the computable functionthat does the following given u and i; We start with 〈〉. Then we determine whetherthe computation of ϕφ(i) on (x, y, u) halts in ≤ u steps, for every x, y ≤ u (here uis viewed as a natural number instead of a �nite function), whenever this happensappend 〈x, y〉. Remark that if (x, y, u) ∈ ϕi, then for every u ⊇ u we have (x, y, u) ∈ϕφ(i), in particular the computation of ϕφ(i) on (x, y, u) halts in the same amountof steps that it takes ϕφ(i) to halt on input (x, y, u). So if (x, y, u) ∈ ϕi and if wechoose u such that x, y ≤ u and large enough to assure that the computation ofϕφ(i) on (x, y, u) halts in ≤ u steps, then 〈x, y〉 is appended to φ′(i, u). This meansthat if (x, y, u) ∈ ϕφ(i) then there is a point where 〈x, y〉 is appended by the abovealgorithm.

Let h′′ be a code for the partial computable function φ7c given by

φ7c(i, j, x, y, u) ' ϕφ(j)(x, y, φ′(i, u)).

Finally let φ′′(i, j) ' s23(h′′, i, j). Then we see that this is a code for the composition,

since ∃u[(x, y, u) ∈ Wφ′′(i,j) ∧ u ⊂ f ] i� ∃u[u is a �nite segment of Ψi(f) and(x, y, u) ∈Wj ] i� Ψj ◦Ψi(f)(x) ' y.

(8) Let i0 be the code of the partial computable function φ8 given by

φ8(x, y, u) '

{ϕn(x, y, u) if u(0) = n

↑ else.

Where u is again (a code of the �nite function) given by u(x) = u(x− 1). Then wesee that Ψi0(n ∗ f) = {〈x, y〉|∃u[(x, y, u) ∈Wi0 ∧ u ⊂ n ∗ f} = Ψn(f). �

From the previous proposition it follows that many functional operators arerecursive operators. We can even extend these results to partial recursive operators.

Corollary 22. 6 Let Ψi,Ψj be partial recursive operators and let k be a naturalnumber. Then every clause of the previous proposition de�nes a partial recursiveoperator. Moreover, we can use the same total computable functions for every clauseexcept for composition (7).

Proof. As stated, the only clause that needs revision is the composition. Letφ and φ′ be the same algorithms as in the proof of the composition in the previouspropositiion. Review that ϕφ(j) satis�es:

(1) (x, y, u) ∈Wφ(j) implies ∀u′[u ⊂ u′ ⇒ (x, y, u′) ∈Wφ(j)];(2) (x, y, u) ∈Wj implies (x, y, u) ∈Wφ(j).

Also review that in the case where Ψi is a partial recursive operator φ′(i, u) givesus an approximation to an enumeration of {〈x, y〉|(x, y, u) ∈ Wi}. Remark thatwhen Ψi(f) is unde�ned, then there is a u ⊂ f , an x and there are y, y′ suchthat y 6= y′ and both 〈x, y〉 and 〈x, y′〉 occur in the enumeration φ′(i, u). Wewant to alter the previous proof in such a way that when the enumeration φ′(i, u)

6Similar to Proposition 21, it seems impossible to �nd these results in any literature as well.

1.4. THE MEDVEDEV LATTICE 13

gives an inconsistent function, then we eventually get an inconsistent function aswell. Formally, this means if τ−1(φ′(i, u)) is not the graph of a function then theend result is required to not be the graph of a function as well. Notice that it iscomputable to check whether a �nite segment u is inconsistent. We let φ∗ be thefollowing computable function:

φ∗(x, y, u) '

0 if u is inconsistent,

0 if u is consistent and (x, y, u) ∈Wφ(j),

↑ else.

Now we let h∗∗ be a code for

φ9(i, j, x, y, u) ' φ∗(x, y, φ′(i, u)).

Finally let φ∗∗(i, j) ' s23(h∗∗, i, j). We see that this is a code for the composition.

�

1.4. The Medvedev lattice

Recall De�nition 17 of a mass problem. One says that an element of a massproblem is a solution to that problem. So intuitively a mass problem correspondsto solutions of that problem. This might give us solid grounds for a calculus ofproblems. We already introduced the notion of recursive operator in the previoussection. Now we are ready to introduce new notions of reducibility and de�ne theMedvedev lattice which is the main subject of this thesis.

Definition 23. [25, Muchnik] The �rst notion of reducibility we consider wasintroduced by Muchnik. We say that A Muchnik reduces or weakly reduces to B(notation: A ≤w B) if for every f ∈ A there is a g ∈ B such that f ≤T g. This isto say that from any solution of B we can get a solution of A.

We can de�ne yet another notion of reducibility:

Definition 24. [20, Medvedev] Medvedev proposed another notion of re-ducibility. We say that A Medvedev reduces or M-reduces to B (notation: A ≤M B)if there is a recursive operator Ψ such that Ψ[B] ⊆ A7, i.e. there is a uniform wayto get a solution of A when given a solution of B.

Remark that Muchnik reducibility is possibly weaker than Medvedev reducibil-ity, i.e. A ≤M B implies A ≤w B. It can be shown that it is strictly weaker,i.e. A ≤w B need not imply A ≤M B (see for instance [34, Sorbi]). As with all thenotions of reducibility introduced in Section 1.2 on page 7 we continue similar towhen we studied the Turing-reducibility.

Proposition 25. [20] The relations ≤w and ≤M are re�exive and transitiveon mass problems.

Proof. The relation ≤w is obviously re�exive and transitive, since ≤T is.Re�exivity: Let A be a mass problem. Then obviously A ≤M A with help of

the identity recursive operator.

7Notice that in light of the previous section we also could have put computable functional orpartial recursive operator in this de�nition in order to give the notion of reducibility a computable�avour. In the next section we will investigate if and how these notions can be used to possiblyget di�erent reducibilities.


Transitivity: Suppose A ≤M B and B ≤M C via Ψe and Ψf respectively. Thenwe have that Ψe ◦ Ψf [C] ⊆ Ψe[B] ⊆ A. Since the class of recursive operators isclosed under composition, there is a recursive operator Ψ such that Ψe ◦ Ψf = Ψ.Hence A ≤M C via Ψ. �

Definition 26. [20] We de�ne A ≡M B if A ≤M B and B ≤M A. Theequivalence classes of ≡M are called degrees of di�culty and we write [A] for thedegree of di�culty of A. Furthermore, this gives a poset M = 〈M,≤〉, where M isthe set of degrees of di�culty and [A] ≤ [B] i� A ≤M B. This poset is also referredto as the Medvedev lattice, elements are denoted by A,B, etc.

Definition 27. [25] We can de�ne the Muchnik degrees analogue to the de-grees of di�culty. We write [A]w for the Muchnik degree of A and these degreesobviously form a poset, which we denote by Mw = 〈Mw,≤w〉.

Remark. From this point onward we will be trying to prove several Medvedevreducibilities. The reader should remember (or review if he does not) Proposition21 and see that all the functional operators that are said to be recursive operators,are in fact recursive operators.

There are several basic operations concerning total functions f that we wouldlike to recall. Let f, g be total functions, n be a natural number and A,B be massproblems. Then we de�ne n∗f by n∗f(0) = n and n∗f(x) = f(x−1). Furthermore,

f ⊕ g(x) :=

{f(x2 ) if x is even

g(x−12 ) if x is odd.

n ∗ A := {n ∗ f |f ∈ A}A+ B := {f ⊕ g|f ∈ A and g ∈ B}A × B := 0 ∗ A ∪ 1 ∗ B

[A] + [B] := [A+ B]

[A]× [B] := [A× B].

In order for the last two de�nitions to make sense we need to check that it isindependent of the representative of the degrees:

Suppose [A′] = [A] and [B′] = [B]. Let Ψi,Ψj be recursive operators suchthat Ψi[A] ⊆ A′ and Ψj [B] ⊆ B′. Then there is a recursive operator Ψ given byΨ(f ⊕ g) = Ψi(f) ⊕ Ψj(g) and we see that Ψ[A + B] ⊆ A′ + B′. In other words,[A′ + B′] ≤ [A + B]. Because of the symmetry in our argument we have equality.Furthermore, there is a recursive operator Ψ′ given by

Ψ′(n ∗ f) =

{0 ∗Ψi(f) if n = 0

1 ∗Ψj(f) if n > 0.

It is easy to verify that Ψ′[0 ∗ A ∪ 1 ∗ B] ⊆ 0 ∗ A′ ∪ 1 ∗ B′. Again this means that[A′ × B′] ≤ [A× B] and because of symmetry we actually have an equality. Hencethe de�nitions are independent of the representative of the degrees.

We conclude with two de�nitions: 0 := [A], where the mass problem A containsa computable function, and 1 := [∅]. With help of these de�nitions we are able toprove the following proposition:


Proposition 28. [20] The poset M is a distributive lattice, where the meet is× and the join is +, with 0 and 1.8

Proof. Obviously 0 is the bottom and 1 is the top of M. We continue byshowing that + is the join. Let A,B be mass problems. First notice that both Aand B M-reduce to A+B. Next, suppose that both A and B M-reduce to C via Ψi

and Ψj respectively. Then the recursive operator Ψ given by Ψ(f) = Ψi(f)⊕Ψj(f)satis�es Ψ[C] ⊆ A+ B. Hence A+ B is the join.

We continue by checking that × is the meet. Notice that 0∗A∪1∗B M-reducesto both A and B. Suppose that C ≤M A and C ≤M B via Ψi and Ψj respectively.Then the recursive operator

Ψ(n ∗ f) =

{Ψi(f) if n = 0

Ψj(f) if n > 0

satis�es Ψ[0 ∗ A ∪ 1 ∗ B] ⊆ C. Hence 0 ∗ A ∪ 1 ∗ B is the meet.Finally we need to check distributivity. Let A,B and C be mass problems.

Then Ψ (f ⊕ (n ∗ g)) = n∗(f ⊕ g) is a recursive operator that shows that (A+ B)×(A+ C) ≤M A + (B × C). And of course Ψ (n ∗ (f ⊕ g)) = f ⊕ (n ∗ g) shows theother direction. Furthermore,

Ψ (n ∗ (f ⊕ g)) =

{(n ∗ (f ⊕ g))⊕ (n ∗ (f ⊕ g)) if n = 0

(n ∗ f)⊕ (n ∗ g) if n > 0,

shows that (A× B) + (A× C) ≤M A× (B + C). And for the other direction we canuse

Ψ ((n ∗ f)⊕ (m ∗ g)) =

n ∗ f if n = 0

m ∗ g if n > 0 and m = 0

1 ∗ (f ⊕ g) if n,m > 0.

�

The following theorem will grant us the ability to interpret propositional logicon the Medvedev lattice. This lattice turns out to be a Brouwer algebra, i.e. adistributive lattice such that for every elements A,B the following exists µC.(A+C ≥ B) (this is usually denoted by A→ B). Much more on Brouwer algebras canbe found in the next chapter. In the last chapter we will return to the Medvedevlattice and investigate the propositional logic(s) that can be interpreted on (partsof) it.

Theorem 29. [20] The Medvedev lattice M is a Brouwer-algebra, i.e. for everyA,B there is a C that is minimal with the property that A + C ≥ B.

Proof. For this we de�ne A → B := {n ∗ f |Ψn[A + f ] ⊆ B]} := {n ∗ f |∀g ∈A[Ψn(g ⊕ f) ∈ B]}. Now we show that [A → B] satis�es the required property.

First note that A + (A → B) ≥M B, with help of the recursive operatorΨ(g ⊕ n ∗ f) := Ψn(g ⊕ f). Thus [A] + [A → B] ≥ [B].

Second, we check that A → B is minimal with this property. Suppose C satis�esA + C ≥M B via Ψi. The recursive operator Ψ(f) = i ∗ f satis�es Ψ[C] ⊆ A → B.Hence C ≥M A → B and thus [C] ≥ [A → B].

8For an introduction to lattice-theory we refer to Section 2.1 on page 21. We use the notation ×,+ for meet and join since that gets rid of ambiguity with the logical symbols ∧ and ∨.

1.5. OTHER REDUCIBILITIES ON MASS PROBLEMS DERIVED FROM M-REDUCIBILITY16

Finally, we remark that this shows that [A → B] = [min{C|A + C ≥M B}].In particular this shows that the operator on degrees of di�culty given by [A] →[B] := [A → B] is well de�ned. �

Remark 30. We brie�y return to the informal concept of calculi of problemsgiven in the introduction. We could ask if it is necessary for M to be a Brouweralgebra if we want to use M as a calculus of problems to give semantics for IPC.

Suppose we have a calculus of problems that can be used to interpret IPC.Let φ, ψ and χ be propositional formulas, for any φ let Aφ denote the problemcorresponding to φ and let ≤ be the notion of reducibility. Recall that in Remark 7on page 7 we showed that the problem Aφ + Aψ is the problem that correspondsto φ ∧ ψ.9 We could show in a similar way that the problem Aφ ×Aψ correspondsto φ ∨ ψ. Note that the distributive laws for IPC correspond to the distributivelaw for the operations × and + on our set of problems. Now we turn to theproperty proven in the previous theorem. Since φ ∧ (φ → ψ) intuitionisticallyimplies ψ, we have Aψ ≤ Aφ∧(φ→ψ) = Aφ + Aφ→ψ. Furthermore, let χ be suchthat Aψ ≤ Aφ + Aχ = Aφ∧χ. Because we assume our calculus of problems to bean adequate one for IPC, this means that φ ∧ χ → ψ ∈ IPC. In other words,χ → (φ → ψ) ∈ IPC, which boils down to Aφ→ψ ≤ Aχ. We conclude that thecalculus of problems has to constitute a Brouwer algebra. 10

We would like to remark that the Muchnik lattice which we introduced inDe�nition 27, constitutes both a Brouwer algebra and a Heyting algebra (see [25,Muchnik] and [33, Sorbi]). Furthermore, we would like to go ahead of ourselvesand show that the Medvedev lattice does not provide suitable semantics for IPC:

Let A = [A] be a degree of di�culty and let C = {h}, where h is a computablefunction. Then we see that

A → ∅ = {n ∗ f | Ψn[A+ f ] ⊆ ∅} =

{F if A = ∅,∅ if A 6= ∅.

So we have A → ∅ = F or (A → ∅)→ ∅ = F , i.e. (A → ∅)× ((A → ∅)→ ∅) ≡M C.In other words (A → 1) × ((A → 1) → 1) = 0. In Section 2.3 on page 25 itwill become clear that this means that the theory of the Medvedev lattice does notequal IPC.

This shifts the attention to segments of the Medvedev lattice. We will be tryingto �nd a segment of the Medvedev lattice that supplies us with an adequate calculusof problems in the sense of Kolmogorov.

1.5. Other reducibilities on mass problems derived from M-reducibility

Our di�erent notions of reducibility origine from di�erent notions of functionaloperators. Recall our notions of computable functional, partial recursive operatorand recursive operator from section 1.3, also recall that τ maps a partial functionto its graph. Eventually we are going to introduce partial degrees, for a survey onpartial degrees we refer to [31, Sasso].

9For an introduction to lattice theory we refer to Section 2.1 on page 21. We use ×, + for themeet and the join in order to get rid of ambiguity with the logical connectives ∧, ∨.10To be complete we need to prove that the calculus of problems has both a top and a bottomelement. But it should be clear that A⊥, respectively A> are the right canditates.


We may try to alter the notion of reducibility by using the three notions offunctional operators we introduced in De�nition 19 on page 9.

Proposition 31. 11 We have the following equivalences:

φ ≤T ψ i� there is a computable functional Ψ such that φ ' Ψ(ψ),

φ ≤wT ψ i� there is a recursive operator Ψ such that φ ' Ψ(ψ),

φ ≤e ψ i� there is a partial recursive operator Ψ such that φ ' Ψ(ψ).

Proof. The �rst and third are obvious, as the notions computable functionaland partial recursive operator are formalizations of the notions Turing reducibilityand enumeration reducibility. We prove that φ ≤wT ψ i� there is a recursiveoperator Ψ such that Ψ(ψ) ' φ.

(if) Let Ψz be a recursive operator such that Ψz(ψ) ' φ. Then φ ' τ−1({〈x, y〉|∃u[(x, y, u) ∈ Wz ∧ u ⊂ ψ]}). Let z′ be a natural number such that ϕ

τ(ψ)z′ (x) '

π1(µ〈y, u, t〉.(t codes a computation of ϕz on (x, y, u) and u ⊂ ψ)). Remark thatthis is an appropriate use of unrestricted µ-recursion, since u ⊂ ψ is computablefrom τ(ψ). The fact that Ψz is a recursive operator implies that for all x there is

at most one y such that ∃u[(x, y, u) ∈Wz ∧ u ⊂ ψ]. It follows that ϕτ(ψ)z′ ' φ.

(only if) Let z′ be a natural number such that ϕτ(ψ)z′ ' φ. Consider the

functional operator Ψ de�ned by Ψ(ψ) ' ϕτ(ψ)z′ . First, remark that it is total.

Second, we show that it is de�ned by an enumeration operator. Let z be a codefor the partial computable function that does the following given (x, y, u): check

if u is an initial �nite function. After that compute ϕτ(u)z′ (x). We output ↑ if

the computation queries anything of τ(u) where u was not de�ned and we output

ϕτ(u)z′ (x) else. We see that Ψz(ψ) ' {〈x, y〉|∃u[(x, y, u) ∈ Wz ∧ u ⊂ ψ]} ' ϕ

τ(ψ)z′ '

Ψ(ψ), so the claim follows. �

The previous proposition merely shows that our notions of functionals corre-spond to the notions of reducibility given in De�nition 12 on page 8. In otherswords, the former are generalizations of the latter. When we introduce new notionsof reducibility the following proposition will be used to conclude whether or notsome of these notions coincide.

Proposition 32. [26, Myhill] [31, Sasso]

(1) ϕ ≤T ψ implies ϕ ≤wT ψ, but not conversely,(2) ϕ ≤wT ψ implies ϕ ≤e ψ, but not conversely,(3) ≤T ,≤wT and ≤e all coincide on total functions.

Now let us �rst consider some (possible) di�erent notions of reducibilities onmass problems:

Definition 33. 12In line with Muchnik reducibility we writeA ≤wTw B, A ≤ewB, if for every f ∈ A there is a g ∈ B such that f ≤wT g, f ≤e g respectively.

11This proposition shows that our notion of ≤wT given in De�nition 12 on page 8 coincides withthe one used in the literature. We introduced it di�erently, since our de�nition was more intuitivein our considerations.12These de�nition could not be found in the literature, since both reducibilities turn out to beequal to Muchnik-reducibility.


From Proposition 32 it follows that the notions ≤w,≤wTw and ≤ew coincideon mass problems. So we have not de�ned any new notions yet.

We could try again in line with Medvedev reducibility:

Definition 34. 13We write A ≤TM B, A ≤e B, if there is a computablefunctional, respectively a partial recursive operator, Ψ such that Ψ[B] ⊆ A.

It follows from Theorem 20 on page 10 that ≤M and ≤e coincide on massproblems since these consist of total functions. That leaves the question whether≤M and ≤TM coincide. Given a recursive operator Ψz. We have

Ψz(f) = τ−1({〈x, y〉|∃u[(x, y, u) ∈Wz ∧ u ⊂ f ]}).

We de�ne

Ψ(f)(x) ' π31 (µ〈y, u, t〉.([t codes a computation of ϕz on (x, y, u) and u ⊂ f ])) .

Notice that this is a correct use of µ-recursion, since the predicate between thebrackets [. . .] is computable in f , because f is total. Moreover, Ψ is a computablefunctional and remark that Ψ(f)(x) is de�ned i� there is a y such that Ψz(f)(x) ' y.Also remark that we have no guarantee that this schema will work for partialfunctions.

We conclude that ≤M and ≤wM coincide as well and we have yet to discoverdi�erent notions of reducibility.

The attempts we have done so far did not seem too fruitful, but we could alterthe problems we are considering in order to get more di�erent calculi of problems.For that we will introduce partial degrees.

Definition 35. 14 We say that A is a mass problem of partial functions if itis a collection of partial functions.

We could again de�ne various notions of reducibility.

Definition 36. 15 We write A ≤Tw B, A ≤wTw B, A ≤ew B, if for everyϕ ∈ A there is a ψ ∈ B such that ϕ ≤T ψ, ϕ ≤wT ψ, ϕ ≤e ψ respectively.

Now we are allowed to celebrate, since Proposition 32 shows that these threenotions all di�er! But there is even more. We may alter Medvedev reducibility aswell to get three more di�erent notions of reducibility.

Definition 37. [12, Kleene][30, Rogers]16 We write A ≤TM B, A ≤M B,A ≤e B, if there is a computable functional, recursive operator, partial recursiveoperator respectively, Ψ such that B ⊆ dom(Ψ) and Ψ[B] ⊆ A. The de�nition of≤e is due to [7].

By considering singletons we see that these are three di�erent notions of re-ducibility.

13These de�nitions also could not be found in the literature, but both reducibilities introducedhere turn out to be equal to M-reducibility.14It is unclear to me who was the �rst to introduce mass problems of partial functions.15These notions of reducibility could not be found in literature.16In the introduction of [6, Cooper] it is said that [12] was the �rst to introduce a notion ofreducibility for partial functions. Subsequently the same article mentions that [30] made thismore explicit following [26, Myhill].


Proposition 38. 17All the de�ned reducibilities on mass problems of partialfunctions are re�exive and transitive.

Proof. Re�exivity follows from the fact that the identity functional operatoris a computable functional, recursive operator and a partial recursive operator.Transitivity follows from the fact that a composition of computable functionals,resp. recursive operators, resp. partial recursive operators is again a computablefunctional, resp. recursive operator, resp. partial recursive operator. �

Just as the degrees of di�culty, we may introduce new posets under the equiva-lence classes derived from these reducibilities. We only consider one of them brie�y.

Definition 39. [7, Dyment] [30] We write A ≡e B if both A ≤e B and B ≤e A.The resulting poset is called the Dyment lattice, notated byMe, we denote elementsof it by Ae,Be, . . . and are called degrees of di�culty in Me. Furthermore, we let[A]e be the degree of di�culty in Me of A.

As the de�nition suggests the Dyment lattice is in fact a lattice. The operationsturn out to be similar to those of the Medvedev lattice.

Proposition 40. 18The poset Me constitutes a distributive lattice with re-spect to ×, + (see Section 1.4 on page 13 on the Medvedev lattice ) and 0e =[{φ|φ computable}]e, 1e = [∅]e.

Proof. The proof of this proposition is analogous to the proof of Proposi-tion 28 on page 15. �

We can in fact say more about the structure Me. It turns out to be a Brouweralgebra as well.

Theorem 41. [33] Me constitutes a Brouwer algebra, i.e. for every Ae andBe there is a Ce that is minimal with the property that Ae + Ce ≥e Be.

Proof. We de�ne

A → B := {z ∗ φ|∀ψ ∈ A[ψ ⊕ φ ∈ dom(Ψz) ∧Ψz(ψ ⊕ φ) ∈ B}.

First note that A + (A → B) ≥e B with help of the partial recursive operatorΨ(ψ ⊕ n ∗ φ) ' Ψn(ψ ⊕ φ). Thus [A]e + [A → B]e ≥e [B]e.

Second, we show that this mass problem of partial functions is minimal withthe stated property. Let C be such that A+ C ≥e B. Let Ψz be a partial recursiveoperator such that Ψz[A + C] ⊆ B. That means for every ψ ∈ A and every ϕ ∈ Cwe have Ψz(ψ ⊕ ϕ) ↓ and it is in fact an element of B. Remark that this impliesthat z ∗ϕ ∈ A → B for every ϕ ∈ C. Hence the partial recursive operator Ψ de�nedby Ψ(φ) ' z ∗φ shows that C ≥e A → B. So [A → B]e is minimal with the requiredproperty.

Conclusively, we remark that this shows that [A → B]e = [min{C|A+C ≥e B}]e.In particular this shows that the operator on the degrees of di�culty of Me givenby [A]e → [B]e := [A → B]e is well de�ned. �

17This is such an elementary result that it is hard to point out who was the �rst to derive thisproposition.18No proof of this proposition could be found in the literature, but it is stated that Me is a lattice.


In light of Remark 30 on page 16 the previous proposition shows that theDyment lattice could be studied in order to get a calculus of problems that providessemantics for IPC.

This concludes our study of other reducibilities on mass problems derived fromM-reducibility. We will not study them in more detail here. The main objective ofthis section was to show that there are other possible calculi of problems that couldbe considered in order to succeed in connecting IPC to a calculus of problems inthe sense of Kolmogorov.

CHAPTER 2

Algebraic semantics

In this chapter we will examine algebraic semantics for intermediate logics. The�rst section will brie�y review the well-known notion of Heyting algebras (or pseudo-Boolean algebras, see [8, Fitting] or [29, Rasiowa and Sikorski]). These algebrascan be used very successfully to investigate intermediate logics. Nevertheless, ourmain subject in this chapter will be Brouwer algebras [19, McKinsey and Tarski].We have already stated in section 1.4 on page 13 that the introduced Medvedevlattice (see De�nition 26 on page 14) is a Brouwer algebra. In addition recall that inRemark 30 on page 16 we have shown that any calculus of problems that suppliessemantics for IPC is a Brouwer algebra. We will eventually use the Medvedevlattice to get an interpretation of some intermediate logics. The study of Brouweralgebras done in this chapter is essential to the results we derive concerning theMedvedev lattice.

2.1. Heyting algebras

Heyting algebras are a special kind of lattices. So we start this section byintroducing the notion of lattice:

Definition 42. [29] A lattice L is a set with a partial ordering ≤ and binaryoperations + and ×. For all a, b ∈ L we have a+b = inf{a, b} and a×b = sup{a, b}.In lattice-theoretic terms, we say that + is the join and × is the meet.

A lattice L is called distributive if for all a, b, c ∈ L both a× (b+ c) = (a+ b)×(a+ c) and a+ (b× c) = (a× b) + (a× c) hold.

If a lattice L has a maximal or minimal element, we denote it by 1L or 0Lrespectively and we call them the top element or bottom element respectively.

Remark. In lattice theory it is also common to write ∧ and ∨ instead of ×and + respectively. The reason for us to di�er from this originates from the factthat we will consider the theory of Brouwer algebras. Choosing × and + gets ridof ambiguity with the logical connectives ∧ and ∨.

Definition 43. [8, 29] Let H be a distributive lattice. If for a, b ∈ H thereexists a c ∈ H that is maximal with the property that a×c ≤ b, we denote this c bya→ b (this is also called the pseudo-complement of a relative to b). We say that His a Heyting algebra (or pseudo-Boolean algebra) if H is a distributive lattice with0H , 1H and a→ b exists for all a, b ∈ H. 1

We will brie�y review how we can interpret propositional formulas on a Heytingalgebra and then state the main theorem on the connection between IPC and thepropositional formulas valid on all Heyting algebras.

1In fact, one could prove that if a lattice satis�es the last property then it is also distributive (cf.[29]).

21

2.2. BROUWER ALGEBRAS 22

Definition 44. [8, 29] Let H be a Heyting algebra. Let Form be the setof all propositional formulas and let us denote those formulas by ϕ,ψ, . . .. Wesay that h : Form → H is a H-valuation if it satis�es h(ϕ ∧ ψ) = h(ϕ) × h(ψ),h(ϕ∨ψ) = h(ϕ) + h(ψ), h(ϕ→ ψ) = h(ϕ)→ h(ψ) and h(¬ϕ) = h(ϕ)→ 0H for allϕ,ψ ∈ Form. Furthermore, we say that a propositional formula ϕ is algebraicallyvalid if for every Heyting algebra H and every H-valuation h we have h(ϕ) = 1H .

Theorem 45. [29] For every propositional formula ϕ we have that ϕ ∈ IPC i�ϕ is algebraically valid.

This beautiful and fundamental result concludes this section on Heyting alge-bras.

2.2. Brouwer algebras

We will now turn our attention to Brouwer algebras. Since Heyting algebras arethe number one algebraic tool to examine intermediate logics, little can be foundon Brouwer algebras. We still refer to [29, Rasiowa and Sikorski] since the resultson Heyting algebras can be found there. The fact that the notion of a Brouweralgebra is dual to that of a Heyting algebra, implies that most results carry over.Let us start with the de�nition.

Definition 46. [19, 29] We can brie�y say that B with the ordering ≤ is aBrouwer algebra ifB with the reversed ordering (usually denoted byBop and calledthe dual of B) is a Heyting algebra (see De�nition 43). Normally this is stated asfollows: a lattice B is a Brouwer algebra if the dual lattice is a Heyting algebra.We also spell out this de�nition below.

A Brouwer algebra B is a (distributive) lattice (see De�nition 42) with 0,1,such that for all α, β ∈ B there is an element γ that is minimal with the propertythat α+ γ ≥ β, we denote this element by α→ β.2 We write ¬α for α→ 1.

If L is a lattice such that a → b exists for every a, b ∈ L, then we call Lan implicative lattice. When L is just a set with a partial ordering ≤ such thatsup{a, b} exists for all a, b ∈ L, we call L an upper semilattice (often abbreviatedas usl). If in addition a→ b exists for every a, b ∈ L, then we call L an implicativeusl. Notice that an implicative usl has a bottom element, since a→ a is minimal.

Definition 47. [29] Let ε, ε′ be elements of a Brouwer algebra B such thatε ≤ ε′. Then we write Bε,ε′ for {α ∈ B | ε ≤ α ≤ ε′}, which will also be writtenas [ε, ε′] if the underlying Brouwer algebra is clear from the context. We also writeBε for B0,ε, which can be viewed as an initial segment of B.

Proposition 48. [29] Let ε, ε′ ∈ B and ε ≤ ε′. Then Bε,ε′ is a Brouweralgebra with respect to the operations +, × from B and

α→ε β = (α→ β) + ε,

¬ε,ε′α = α→ε ε′,

0 = ε,

1 = ε′.

2One might wonder whether this is in con�ict with the → operator for Heyting algebras de�nedin De�nition 43. We could show that for any Brouwer algebra B we have α → β is the sameelement as the one in the dual Heyting algebra Bop.


Proof. Let α, β ∈ Bε,ε′ . Then we have ε ≤ α× β ≤ ε′, ε ≤ α+ β ≤ ε′ (sinceε ≤ α ≤ ε′ and ε ≤ β ≤ ε′). We also see that α + ε′ = ε′ ≥ β, so α → β ≤ ε′.This shows that ε ≤ α →ε β ≤ ε′, in particular ¬ε,ε′α ∈ [ε, ε′]. We only needto check that α →ε β = min{γ|ε ≤ γ ≤ ε′ and α + γ ≥ β}. First, remark thatα + α →ε β = α + α → β + ε ≥ β + ε = β. Second, let γ ∈ Bε,ε′ be such thatα + γ ≥ β. Then γ ≥ α → β and γ ≥ ε, so γ ≥ α →ε β. This proves theproposition. �

Corollary 49. [29] Bε is a Brouwer algebra with respect to the operations+, ×, → from B and

¬εα = α→ ε.

We will turn our attention to some other aspects and operations on Brouweralgebras.

Definition 50. [32, Skvortsova] An element α ∈ B is called meet-irreducibleor ×-irreducible if it satis�es for all β, γ ∈ B, α = β × γ implies α = β or α = γ.The set of all ×-irreducible elements is denoted by D(B). In addition we call asubset C ⊆ D(B) canonical (see [32]) if for all α, α′ ∈ C and all β, β′ ∈ B we have

• α+ α′, α→ α′ ∈ C,• α→ (β × β′) = (α→ β)× (α→ β′).

In this case it is clear that C constitutes an implicative usl relative to the operations+,→ originating from B. We show that for any α ∈ C and β, β′ ∈ B we have thatβ × β′ ≤ α implies β ≤ α or β′ ≤ α:

Suppose we have β × β′ ≤ α. Then 0 = α → β × β′ = (α → β) × (α → β′).Since for any α ∈ C we get 0 = α → α ∈ C ⊆ D(B), we have that α → β = 0 orα→ β′ = 0. Hence β ≤ α or β′ ≤ α.

Definition 51. [32] We are going to examine the smallest sub-Brouwer al-gebra of B containing a canonical subset C. For an arbitrary implicative usl A,we de�ne the ×-hull, denoted by ∆(A) as follows. For arbitrary α1, . . . , αn ∈ Awe let bα1, . . . , αnc := {αi|αi is a minimal element of {α1, . . . , αn}}.3 Notice thatthis always gives an antichain and that whenever {α1, . . . , αn} is an antichain thenbα1, . . . , αnc = {α1,. . . , αn}. Let ∆(A) := {bα1, . . . , αnc|n > 0, ∀i[αi ∈ A]}, whichis exactly the set of antichains in A.

Proposition 52. [32]4 Let A be an implicative usl and let {α1, . . . , αn} and{β1, . . . , βm} be antichains in A. The ×-hull ∆(A) constitutes an implicative latticerelative to:

{α1, . . . , αn} ≤∆ {β1, . . . , βm} i� ∀j∃i[αi ≤ βj ]{α1, . . . , αn} ×∆ {β1, . . . , βm} = bα1, . . . , αn, β1, . . . , βmc{α1, . . . , αn}+∆ {β1, . . . , βm} = bαi + βjci,j{α1, . . . , αn} →∆ {β1, . . . , βm} =

∑i∆bαi → βjcj

Furthermore, we can identify A with the singletons in ∆(A). Then A is a sub-implicative usl of ∆(A), in fact A is the set of all ×-irreducibles of ∆(A) and iseven canonical. Finally, if A has a top element, then ∆(A) is a Brouwer algebra.

3This de�nition comes from [32]. However the notation used there is quite confusing, so weintroduced a di�erent notation.4A complete proof of this proposition can not be found in [32]. Here we do give a thorough proof.


Proof. Let {α1, . . . , αn}, {β1, . . . , βm} and {γ1, . . . , γl} be antichains in A. Inthis proof we will refer to them by α, β and γ respectively.

Poset

Obviously α ≤∆ α, so it is re�exive. Suppose α ≤∆ β and β ≤∆ α. The �rstrelation implies that for every 1 ≤ j ≤ m there is a 1 ≤ i ≤ n such that αi ≤ βjand then the second implies that there is an 1 ≤ j′ ≤ m such that βj′ ≤ αi. So wehave βj′ ≤ βj . Since β is an antichain this implies that βj = βj′ , hence αi = βj .Putting this together we get for every 1 ≤ j ≤ m there is a 1 ≤ i ≤ n such thatαi = βj . Of course we can use the same argument to get the converse, so α = βand we conclude that the order is antisymmetric.

Suppose α ≤∆ β and β ≤∆ γ. The �rst relation again implies that for every1 ≤ j ≤ m there is a 1 ≤ i ≤ n such that αi ≤ βj and then the second implies thatthere is an 1 ≤ k ≤ l such that βj ≤ γk. So we have thatαi ≤ γk and thus α ≤∆ γ,hence the order is transitive.

Meet

We review the de�nition to see that α×∆ β = bα1, . . . , αn, β1, . . . , βmc. Remarkthat every αi is either minimal among α1, . . . , αn, β1, . . . , βm or there is a βj thatis minimal such that βj ≤ αi. It follows that α×∆ β ≤∆ α, β.

Now suppose that γ ≤∆ α, β. Then for every 1 ≤ i ≤ n there is a 1 ≤ k ≤ lsuch that γk ≤ αi and for every 1 ≤ j ≤ m there is a 1 ≤ k′ ≤ l satisfying γk′ ≤ βj .Hence γ ≤∆ α×∆ β.

Join

According to the de�nition we have α +∆ β = bαi + βjci,j . Every δ ∈ α +∆ βis of the form αi +βj , so in particular αi, βj ≤ αi +βj = δ. Hence α, β ≤∆ α+∆ β.

Suppose α, β ≤∆ γ. Then for all 1 ≤ k ≤ l there are 1 ≤ i′ ≤ n and 1 ≤ j′ ≤ msuch that αi′ , βj′ ≤ γk. Hence αi′ +βj′ ≤ γk. Since every αi′ +βj′ is either minimalamong the αi+βj 's or there is a minimal αi+βj ≤ αi′ +βj′ , we have α+∆ β ≤∆ γ.

Implication

First we show that α +∆ (α →∆ β) ≥∆ β. Every δ ∈ α →∆ β is of the formα1 → βj1 +. . .+αn → βjn for some 1 ≤ j1, . . . , jn ≤ m. So every ε ∈ α+∆(α→∆ β)is of the form αi +α1 → βj1 + . . .+αn → βjn . Since αi +αi → βji ≥ βji we derivethat βji ≤ ε. This proves the statement.

Next we show that α→∆ β is minimal with respect to that property. Supposeα+∆ γ ≥∆ β. Let 1 ≤ k ≤ l. Then for every 1 ≤ i ≤ n such that αi+γk is minimalamong the αi′ + γk′ 's there is a 1 ≤ j ≤ m such that αi + γk ≥ βj . Hence this alsoholds for every 1 ≤ i ≤ n. So we see that for every 1 ≤ i ≤ n there is a 1 ≤ ji ≤ msuch that γk ≥ αi → βji . So γk ≥ α1 → βj1 + . . . + αn → βjn . Again since everyα1 → βj1 + . . . + αn → βjn is either minimal or there is a δ ∈ α →∆ β of similarform that is less or equal, we derive that α→∆ β ≤∆ γ.

Distributivity

We see that α +∆ (β ×∆ γ) = bαi + βj , αi + γkci,j,k =(α+ β

)× (α+ γ) and(

α× β)

+(α× γ) = bαi+αi, αi+γk, βj +αi, βj +γkci,j,k = bαi, βj +γkci,j,k = α×(β + γ

), where the second equality holds since αi+αi = αi and αi ≤ αi+γk, βj+αi.

Sub-implicative usl

Let the map ϕ : A → ∆(A) be given by α 7→ bαc. Then this map is obviouslyinjective, furthermore bαc ≤∆ bβc implies α ≤ β and vice versa. We also seethat bαc +∆ bβc = bα + βc and bαc →∆ bβc = bα → βc. We say that A is asub-implicative usl., when we actually mean that ϕ[A] is a sub-implicative usl.

2.3. INTERPRETATION OF PROPOSITIONAL LOGIC ON BROUWER ALGEBRAS 25

Canonical

First we check that A is ×-irreducible. Suppose bαc = β × γ. Without loss ofgenerality we can say that α = β1. Thus α is minimal among β1, . . . , βm, γ1, . . . , γl,so in particular this gives us that β = bβ1c = bαc.

Second, remark that bαc +∆ bβc = bα + βc ∈ A and bαc →∆ bβc = bα →βc ∈ A. Finally,

(bαc →∆ β

)× (bαc →∆ γ) = bα→ βjcj ×∆ bα→ γkck = bαc →∆(

β ×∆ γ). So A is canonical.

Brouwer algebra

Suppose A has a top element. Then we only need to check that ∆(A) has a topelement, since we have already proven that it is an implicative lattice. Obviouslyα ≤∆ b1c for all α ∈ ∆(A). So in this case, ∆(A) is indeed a Brouwer algebra. �

The following proposition shows why we are interested in the ×-hull.

Proposition 53. [32] Let C be a canonical set in a Brouwer algebra B. Thenthe smallest sub-implicative lattice in B containing C is Γ(C) = {(α1×. . .×αn)|αi ∈C} and Γ(C) is isomorphic to ∆(C).

Proof. First, notice that any sub-Brouwer algebra containing C has to containΓ(C). Second, we show that Γ(C) is isomorphic to ∆(C): obviously we have Γ(C) ={(α1 × . . . × αn)|αi ∈ C ∀i 6= j[αi � αj ]}. Remark that since C is canonicalwe have that Πn

i=1αi ≤ Πmj=1βj implies that for all βj there is a αi such that

αi ≤ βj (see De�nition 50 on page 23). Conversely, if for all βj there is a αisuch that αi ≤ βj , then Πn

i=1αi ≤ βj for every j, so Πni=1αi ≤ Πm

j=1βj . HenceΠni=1αi ≤ Πm

j=1βj i� ∀j∃i[αi ≤ βj ] i� bα1, . . . , αnc ≤∆ bβ1, . . . , βmc. So themapping (α1 × . . . × αn) 7→ bα1, . . . , αnc preserves the ordering. Furthermore,since C is canonical the operations on Γ(C) have to be de�ned as in the previousproposition. We conclude that the given map is an isomorphism between Γ(C) and∆(C). �

2.3. Interpretation of propositional logic on Brouwer algebras

We will be trying to �nd Brouwer algebras that can be used to interpret in-termediate logics. For that we need to give a thorough account of how to do thisinterpretation:

Definition 54. [29] We start by de�ning a valuation, after that we are able tode�ne validity. Let Var be the set of propositional variables (denoted by p, q, r, . . .),let Form be the set of propositional formulas (denoted by φ, ψ, . . .) and let B be aBrouwer algebra. A B-valuation (or just valuation) is a mapping v : Form → Bsatisfying v(φ∧ψ) = v(φ)+v(ψ), v(φ∨ψ) = v(φ)×v(ψ), v(φ→ ψ) = v(φ)→ v(ψ)and v(¬φ) = v(φ) → 1. We say that a formula φ is valid on B if v(φ) = 0 for allvaluations v. The set of all formulas valid on B is called the logic of (or the theoryof) B and is written as Th(B) = {φ ∈ Form |φ is valid on B}. For any Γ ⊆ Formwe de�ne the positive fragment of Γ by Γπ = {ϕ|ϕ ∈ Γand ϕ does not contain ⊥}.In particular we call Thπ(B) the positive logic of B.

Remark. We could view the set of formulas as a lattice where the logicalsymbols correspond to the lattice-theoretical operations. From that perspective, inthe de�nition of valuations the ordering is reversed. That is exactly the reason whywe have a 1 instead of a 0 in the clause v(¬φ) = v(φ) → 1 and also why we saythat a formula is valid when its valuation equals 0. Also notice the duality with


De�nition 44 on page 22 on valuations on Heyting algebras. This duality could beused to prove that a formula is valid on a Brouwer algebra i� it is algebraicallyvalid on the Heyting algebra of the dual lattice.

Any mapping v : Var → B has a unique extension v : Form → B that isa valuation. Given a valuation v and φ ∈ Form containing only propositionalvariables p1, . . . , pn remark that v(φ) only depends on v � {p1, . . . , pn}, that is ifw is a valuation such that w � {p1, . . . , pn} = v � {p1, . . . , pn}, then w(φ) = v(φ).We will also write φ[β1, . . . , βn] for v(φ), where v is the valuation that extendsthe mapping v : Var → B given by v(pi) = βi for all 1 ≤ i ≤ n and v(q) = 0otherwise. From our discussion above we can conclude that φ[β1, . . . , βn] = w(φ)for any valuation w satisfying w(pi) = βi.

We will mostly study constructions on Brouwer algebras and the e�ect it hason corresponding theories. We state two main results on Brouwer algebras:

Theorem 55. [29] We have IPC ⊆ Th(B).

Proof. Easy calculation using induction on the axioms and rules of inference.�

Theorem 56. [29] We have IPC =⋂Brouwer algebras B Th(B).

These are obviously good results showing that we might be able to use Brouweralgebras in order to interpret intermediate logics and perhaps even the intuitionisticpropositional calculus. We continue by introducing the notion of homomorphismand derive some properties relating homomorphisms to theories of Brouwer alge-bras.

Definition 57. [29] Let A,B be Brouwer algebras. A mapping ξ : A → Bis called a B-homomorphism if it preserves the lattice-theoretic structure and alsopreserves 0 and 1. In other words, if it satis�es ξ(α1 ×A α2) = ξ(α1) ×B ξ(α2),ξ(α1 +A α2) = ξ(α1) +B ξ(α2), ξ(α1 →A α2) = ξ(α1) →B ξ(α2), ξ(0A) = 0B and�nally ξ(1A) = 1B.

We will sometimes say that a mapping is a homomorphism relative to ◦, where◦ ∈ {×,+,→} if it satis�es the corresponding condition above.

Theorem 58. [Folk] Let A, B be Brouwer algebras. Suppose we have a B-homomorphism ξ : A→ B. Then:

(1) if ξ is surjective, then Th(A) ⊆ Th(B);(2) if ξ is injective, then Th(A) ⊇ Th(B);(3) if ξ is an isomorphism, then Th(A) = Th(B).

Proof. We do 1 and 2 by proving the contra position. From 1 and 2 obviously3 follows immediately.

1: Suppose ξ is surjective. Let φ ∈ Form \ Th(B). Then there is a valuationv : Form → B such that v(φ) 6= 0B. Notice that v(φ) only depends on theimages of the propositional variables occurring in φ, let's say p1, . . . , pn. De�neβi := v(pi) for all 1 ≤ i ≤ n. Then there are αi ∈ A such that ξ(αi) = βi. Letw : Var → A be de�ned by w(pi) = αi for 1 ≤ i ≤ n and w(q) = 0A otherwise.This map has a unique A-valuation, say w, that extends it. Then we have thatξ(w(φ)) = v(φ) 6= 0B, so w(φ) 6= 0A. Hence φ /∈ Th(A).


2: Suppose ξ is injective. Let φ ∈ Form \ Th(A). This means that there is avaluation w : Form → A such that w(φ) 6= 0A. Again we can de�ne a valuationv : Form → B satisfying v(φ) = ξ(w(φ)). We also have ξ(w(φ)) 6= 0B, since ξ isinjective. Hence φ /∈ Th(B). �

Corollary 59. [29] Let A and B be Brouwer algebras. Suppose A is a sub-Brouwer algebra of B.5 Then Th(B) ⊆ Th(A).

We have shown in Proposition 48 that Bε is a Brouwer algebra. We get thefollowing result:

Lemma 60. [32, Skvortsova] If Thπ(B) = IPCπ, then⋂ε∈B Th(Bε) = IPC.

Proof. First remark that from Theorem 55 it follows that ∩ε∈BTh(Bε) ⊇IPC. So we need to show the other inclusion. In order to prove this we show thatthe contra position holds. Let φ /∈ IPC and let p1, . . . , pn be the only propositionalvariables occurring in φ. With any formula φ(p1, . . . , pn) we associate the positiveformula φπ(q, p1, . . . , pn) obtained by replacing all occurrences of ⊥ by (q∧p1∧ . . .∧pn), where q is a propositional variable not occurring in φ. Since φ /∈ IPC Theorem56 shows that there is a Brouwer algebra A such that φ /∈ Th(A). That means thereare α1, . . . , αn such that φ[α1, . . . , αn] 6= 0A in A. Remark that φ[α1, . . . , αn] =φπ[1A, α1, . . . , αn] in A, so φπ /∈ Th(A) and thus φπ /∈ IPC. This implies φπ /∈Thπ(B). That means there are β0, . . . , βn ∈ B such that φπ[β0, . . . , βn] 6= 0 in B.De�ne ε := β0 + . . .+ βn. Then we see that the value of φ[β1, . . . , βn] in Bε equalsthe value of φπ[β0, . . . , βn] in B, which is di�erent from 0. Hence φ /∈ Th(Bε). �

Lemma 61. [32] If ε0, ε1, ε2 ∈ B such that ε1 < ε2 and ε2 = ε0 + ε1, thenthere is a surjective B-homomorphism ξ : Bε0 � Bε1,ε2 .

Proof. We de�ne the mapping ξ : Bε0 → Bε1,ε2 by ξ(α) = α + ε1. Now weneed to check that this is a surjective B-homomorphism. That it preserves +, ×follows immediately from distributivity, we also have ξ(α → β) = (α → β) + ε1 =((α+ ε1)→ (β + ε1)) + ε1 = ξ(α)→ε1 ξ(β). The second equality follows from

(α+ ε1) + (α→ β) ≥ β + ε1 and

α+ ((α+ ε1)→ (β + ε1)) + ε1 ≥ β + ε ≥ β,i.e. α → β ≥ ((α+ ε1)→ (β + ε1)) and ((α+ ε1)→ (β + ε1)) + ε1 ≥ α → β. Toconclude it is a B-homomorphism we note that ξ(0) = ε1 = 0Bε1,ε2

and ξ(ε0) =ε0 + ε1 = ε2 = 1Bε1,ε2

. It remains to show that it is surjective. For any α ∈ Bε1,ε2

we have that ξ(α × ε0) = α × ε0 + ε1 = (α + ε1) × (ε0 + ε1) = α × ε2 = α. Weconclude that ξ is a surjective B-homomorphism. �

Corollary 62. [32] If ε0, ε1, ε2 ∈ B such that ε1 < ε2 and ε2 = ε0 +ε1, thenTh(Bε0) ⊆ Th(Bε1,ε2).

Proof. Follows immediately from the previous lemma and Theorem 58. �

Corollary 63. [32] Let ε ∈ B. Then there is a surjective B-homomorphismξ : B� Bε,1.

Proof. Follows from Lemma 61 by setting ε2 = ε0 = 1 and ε1 = ε. �

5We actually did not de�ne this properly, but it should be clear what it means. It could be de�nedas: A ⊆ B and the embedding map is a B-homomorphism.


Lemma 60 is our motivation for studying the positive fragment of IPC in thenext chapter. The latter three statements will be usefull in the following chaptersas well.

CHAPTER 3

Kripke semantics and Medvedev's Logic

In this chapter we will be concerned with another semantic approach to intu-itionistic logic, namely that of Kripke semantics. During this short introduction wefollow mainly [8, Fitting], although some notation and terminology may vary.

Definition 64. [8] A Kripke frame is a poset K = 〈K,≤〉 with a least element0K .

Recall that we have a notion of validity on Kripke frames. For that we needseveral auxiliary de�nitions:

Definition 65. [8] Let K = 〈K,≤〉 be a Kripke frame. Let Var be the setof propositional variables. A K-valuation (or just valuation) V : K → P(Var) isa map such that for all u, v ∈ K u ≤ v implies V (u) ⊆ V (v). We call a triple〈K,≤, V 〉 a Kripke model if 〈K,≤〉 is a Kripke frame and V is a valuation. Letu ∈ K. We de�ne validity of a formula ϕ at u, notation K, V, u ϕ or just u ϕ,by induction on the complexity of the formula as follows:

u p i� p ∈ V (u);u ϕ1 ∧ ϕ2 i� u ϕ1 and u ϕ2;u ϕ1 ∨ ϕ2 i� u ϕ1 or u ϕ2;u ϕ1 → ϕ2 i� for every v ≥ u, if v ϕ1, then u ϕ2.Notice that the de�nition of a valuation together with these clauses for validity

imply upward persistence:

(v ≥ u and u ϕ) implies v ϕ.

We de�ne the theory of a Kripke model by ThKripke(〈K,≤, V 〉) := {ϕ ∈ Form |for every u ∈ K we have K, V, u ϕ}. Furthermore, we de�ne K φ if for everyK-valuation V we have K, V ϕ. Finally we de�ne the theory of a Kripke frameby ThKripke(K) = {ϕ ∈ Form |K ϕ}. We will often write K instead of K in thesede�nitions when the ordering ≤ is clear from the context.

We state one main theorem on the connection between Kripke models and theintuitionistic propositional calculus:

Theorem 66. [8] Let ϕ be any propositional formula. Then ϕ ∈ IPC i� ϕ ∈ThKripke(K) for every �nite Kripke frame K.

This fundamental result will be used in Section 3.2.

3.1. Linking Kripke frames to Brouwer algebras

In this section we show that we can get a Brouwer algebra from a Kripke framein such a way that the theories are equal (see for instance [8]).

29

3.1. LINKING KRIPKE FRAMES TO BROUWER ALGEBRAS 30

Definition 67. [8] Let 〈K,≤〉 be a Kripke frame. We can regard it as atopological space with the order topology: the base consists of cones or upsetsKu = {v ∈ K | v ≥ u}. The open sets are unions of upsets.

Proposition 68. [8] The family of all open sets B(K) ordered by ⊇ constitutesa Brouwer algebra relative to α + β := α ∩ β, α × β := α ∪ β and α → β := {u ∈K | Ku ∩ α ⊆ β}.

Proof. Obviously α+ β = α ∩ β and α× β = α ∪ β. It follows that B(K) isdistributive. For the implication, we have α + (α→ β) = α ∩ (α→ β) = {u ∈ α |Ku ∩ α ⊆ β} ⊆ β. So α+ (α→ β) ≥ β. Now suppose α+ γ ≥ β. Then α ∩ γ ⊆ β.For every u ∈ γ we have Ku ⊆ γ, thus we also have Ku ∩ α ⊆ β. So γ ⊆ α → βand therefore γ ≥ α→ β. It still remains to check that α→ β is an open set. Sinceu ∈ α → β implies Ku ⊆ α → β, we have α → β =

⋃u∈α→βK

u. So α → β is anopen set. This proves the proposition. �

One could ask how the theory of a Kripke model and the Brouwer algebraB(K) correspond to each other. This matter is solved quite straightforward:

Lemma 69. 1 Let 〈K,≤〉 be a Kripke frame and u ∈ K. Then we have thefollowing connections between validity in a Kripke model 〈K,≤, V 〉 at u and validityin the Brouwer algebra B(K):

(1) Let V : K → P(Var) and ϕ = ϕ(p1, . . . , pn). De�ne opens Ki := {v ∈ K |pi ∈ V (v)} for 1 ≤ i ≤ n. Then

〈K,≤, V 〉, u ϕ i� u ∈ ϕ[K1, . . . ,Kn].

(2) Let ϕ = ϕ(p1, . . . , pn) and K1, . . . ,Kn ∈ B(K). De�ne V : K → P(Var)by pi ∈ V (v) i� v ∈ Ki. Then

u ∈ ϕ[K1, . . . ,Kn] i� 〈K,≤, V 〉, u ϕ.

Proof. This can be proven by induction on the complexity of the formula ϕ.All cases follow immediately from the de�nitions. �

Theorem 70. [8] Let 〈K,≤〉 be a Kripke frame. Then the theory of B(K) andthe Kripke frame 〈K,≤〉 coincide, i.e. Th(B(K)) = ThKripke(〈K,≤〉).

Proof. This follows from the previous lemma. �

Corollary 71. [8] The fundamental theorem for Kripke frames can now beused to derive that

IPC =⋂

�nite Kripke frames 〈K,≤〉

Th(B(K)).

With this result we turn our attention to �nite Kripke frames and the derivedBrouwer algebras.

Definition 72. [21, Medvedev (1962)][22, Medvedev (1966)]Let I be a nonempty set. We consider the lattice of subsets 2I = {α | |α ⊆ I}ordered by ⊇. This constitutes a Brouwer algebra with operations α × β = α ∪ β,α + β = α ∩ β, α → β = (I\α) ∪ β and least and greatest elements I and ∅respectively. Consider 2n = 2In , where In = {1, . . . , n}, and the Kripke frame

1The results stated in this lemma could not be found in [8]. However, it is shown that the theoriesof the a Kripke frame K and the Brouwer algebra B(K) coincide.

3.1. LINKING KRIPKE FRAMES TO BROUWER ALGEBRAS 31

2n = 2n\{∅}. We will examine the Medvedev logic of �nite problems LM given by

LM =⋂n>0 ThKripke(2

n) in the next section. We immediately see with help of

Theorem 70 that LM =⋂n>0 Th(B(2n)). In the next section we will prove and

discuss the theorem below which we will refer to as Medvedev's theorem (see [21]):

Theorem. We have LMπ = IPCπ.

Lemma 73. [32, Skvortsova] Let C be a �nite implicative usl with greatestelement 1. Then the Brouwer algebra ∆(C)2 is isomorphic to B(K) for the frameK = C\{1} with the order of C.

Proof. Since K is �nite we have B(K) = {∅} ∪ {Kα1 ∪ . . . ∪ Kαn |n >0 and {α1, . . . , αn} is an antichain in K}. We see that ∪iKαi ⊇ ∪jKβj if andonly if ∀j∃i[Kαi ⊇ Kβj ]. Therefore, the map ξ : ∆(C)→ B(K) given by{

ξ(1) = ∅,ξ(bα1, . . . , αnc) = Kα1 ∪ . . . ∪Kαn for an antichain in C\{1},

is an isomorphism of ∆(C) and B(K) as partially ordered sets, hence it is also aB-isomorphism. �

Definition 74. To prove the main theorem of [32], the countable implicativeusl of all �nite and co�nite sets is essential: Cω := {α ⊆ N | either α or (N\α) is�nite} and ordered by ⊇. The operations are given by α + β = α ∩ β, α → β =(N\α) ∪ β and the least and greatest elements are N and ∅, respectively. Recallfrom Proposition 52 on page 23 that this implies that ∆(Cω) constitutes a Brouweralgebra.

Proposition 75. [32] For every n > 0 there is a surjective B-homomorphismξ : ∆(Cω)→ B(2n).

Proof. Let n > 0. Then εn = {1, . . . , n} ∈ Cω. We see that [εn, ∅]∆(Cω) =∆({α ∈ Cω | εn ⊇ α} and we have a surjective B-homomorphism η : ∆(Cω) →[εn, ∅]∆(Cω) from Corollary 63 on page 27. Furthermore, [εn, ∅]∆(Cω) is obviouslyisomorphic to ∆(2n). By the previous lemma we get an isomorphism between

B(2n) and ∆(2n). If we compose these it gives us a surjective B-homomorphism

ξ : ∆(Cω)→ B(2n). �

Corollary 76. [32] We have Th(∆(Cω)) ⊆⋂n>0 Th(B(2n)) = LM and thus

Thπ(∆(Cω)) = IPCπ.

Proof. This follows from the previous lemma, Theorem 58 on page 26 andMedvedev's theorem (see Theorem 91 on page 38). �

Corollary 77. [32] We have⋂n>0

⋂ε∈B(2n) Th(B(2n)ε) = IPC.

Proof. The proof is similar to that of Lemma 60 on page 27. �

2This is the ×-hull introduced in De�nition 51 on page 23.

3.2. MEDVEDEV'S LOGIC 32

3.2. Medvedev's Logic

This section is devoted to the proof of the following theorem, which will bereferred to as Medvedev's theorem:

Theorem. We have LMπ = IPCπ.

In [21, Medvedev (1962)] Medvedev introduces his notion of �nite problemsand the Medvedev Logic LM. Later, in [22, Medvedev (1966)], he shows that LMis equivalent to the logic of a certain class of Kripke frames. We use this classin order to prove Medvedev's theorem, it will be given in the �rst de�nition ofthis section. In [21] Medvedev asserts without proof the theorem which we callMedvedev's theorem. Nevertheless, some (for instance [4, Chernov]) refer to thisarticle for a proof of it. Our goal is to provide a proof of this theorem using Kripkesemantics.

More can be said about Medvedev's theorem. Medvedev de�nes a class offormulas called critical implications. Again the de�nition can be found in [21]. Hethen claims the following theorem without proof:

Theorem. Let ϕ be a positive formula such that ϕ /∈ IPC. Then there is acritical implication J such that ϕ→ J ∈ IPC.

Let us call this the Critical Implication Theorem for now. One can easily checkthat indeed Medvedev's theorem follows from this. Some (see for example [28,Plisko] and [23, 24, Mezhirov and Vereshchagin]) refer to [5, Chernov et al] for aproof of this. But unfortunately, a complete proof is not given there. It is onlyshown that from Medvedev's theorem the Critical Implication Theorem follows.

A di�erent approach to Medvedev's theorem is made possible by results ofJankov. He introduces the logic of 'the weak of law of excluded middle' Jan =IPC + (¬p ∨ ¬¬p) and proves the following theorems:

Theorem. Let C be the class of all Kripke frames that have a maximal element.Then Jan =

⋂K∈C ThKripke(K).

Theorem. Let L be an intermediate logic. Then we have L ⊆ Jan i� Lπ =IPCπ.

A proof of Medvedev's theorem that relies heavily on these results is given in[18, Maksimova, Skvortsov and Shehtman] and another one in [37, Szatkowski].Hence, a complete proof is not given there either, yet some (for instance [35, Sorbiand Terwijn]) refer to [18] for a proof of Medvedev's theorem.

Let us start our proof with the de�nition of Medvedev's Logic:

Definition 78. [22] For any n ∈ ω we let σn be the frame which consists ofproper subsets of {1, . . . , n}, i.e. E ( {1, . . . n}, and ordered by ⊆. Medvedev'sLogic is given by LM :=

⋂n>0 ThKripke(σn).

In order to prove Medvedev's theorem we make a detour in the semantics forIPC. Most of the results and de�nitions given below are from [2, Blackburn, deRijke, Venema] and/or [9, Gabbay].

Definition 79. [2, 9] A tree is a poset T = 〈T,v〉 that satis�es:(1) There exists a root, i.e. an element of T that is beneath any element. We

denote the root by 0T .


(2) For every 0T 6= t ∈ T there is a unique predecessor, i.e. for every t 6= 0Tthere is a unique s 6= t satisfying ∀t[t @ t implies t v s]

(3) T is acyclic, that means: given t ∈ T and any sequence t1, . . . , tk ∈ T(k > 1) we have that t = t1 v . . . v tk = t implies t = ti for every1 ≤ i ≤ k.

Lemma 80. [2, 9] Let FT =⋂�nite trees T ThKripke(T). Then IPC = FT.

Proof. The method used in this proof is called unfolding, unraveling or un-winding (cf. [2]).

Obviously, ϕ ∈ IPC implies ϕ ∈ FT. For the opposite we let ϕ /∈ IPC and provethat ϕ /∈ FT. From the fundamental Theorem 66 on page 29 on Kripke frames andthe logic IPC it follows that there is a �nite Kripke frame F that contradicts ϕ.So there is a valuation V on F such that F, V,0F 1 ϕ. Now de�ne T to be the setof all �nite increasing sequences of F with 0F as �rst element. Remark that T is�nite. Let the ordering on T be that of 'being equal or an initial segment of', wedenote it by v. For t = (0, . . . , xn) de�ne |t| := xn. Now let W be the valuationon T given by t ∈W (p) i� |t| ∈ V (p).

In the remainder of the proof we write x φ instead of F, V, x φ and t φinstead of T,W, t φ. We show that for every propositional formula φ we haveT,W, t φ i� F, V, |t| φ. This is done by induction on φ. The only interestingcase is that of φ = φ1 → φ2.

(only if) Suppose t φ1 → φ2. Let x ≥F |t| such that x φ1. We de�net = t ∗ x (concatenate t with x) which is in T . By our induction hypothesis we gett φ1. But also t v t. Hence t φ2 and again by our induction hypothesis we getx φ2. We conclude that |t| φ1 → φ2.

(if) Suppose |t| φ1 → φ2. Let t w t such that t φ1. By our inductionhypothesis we get |t| φ1. We also have |t| ≥F |t|. So we get |t| φ2 and againby our induction hypothesis this gives us t φ2. We conclude that t φ1 → φ2.

We conclude that T,W, (0F ) 1 ϕ. So IPC = FT. �

The above lemma lets us use trees instead of arbitrary posets. Trees are con-venient because of their inductive nature. We now introduce maps between Kripkemodels.

Definition 81. A homomorphism (see [9]) or surjective bounded morphism(see [2]) between Kripke models K = 〈K,≤K , V 〉 and L = 〈L,≤L,W 〉 is a mapξ : K → L such that

(1) k ≤K k implies ξ(k) ≤L ξ(k) (homomorphism w.r.t. ≤K);(2) ξ is onto (surjective);(3) ξ(k) ≤L l implies that there is a k such that k ≤K k and ξ(k) = l (the

back condition);(4) V (k) = W (ξ(k)) (k and ξ(k) satisfy the same propositional variables).

Lemma 82. [2, 9] Let K and L be Kripke models as in the above de�nition. Letξ be a homomorphism between K and L. Then we have for every k ∈ K and everyformula φ that K, k φ i� L, ξ(k) φ.

Proof. We prove this by induction on the complexity of the formula. Thecase where φ = p is clear from property (4) of a homomorphism. The cases of ∧and ∨ oppose no serious problems. That leaves the case where φ = φ1 → φ2:


(if) Suppose L, ξ(k) φ1 → φ2. Let k ≥K k be such that K, k φ1. Byour induction hypothesis we have L, ξ(k) φ1. Property (1) of a homomorphismimplies ξ(k) ≥L ξ(k), hence L, ξ(k) φ2. Again by our induction hypothesis weget K, k φ2. We conclude that K, k φ1 → φ2.

(only if) Suppose K, k φ1 → φ2. Let l ≥L ξ(k) be such that L, l φ1.Property (3) of a homomorphism implies that there is a k ≥K k such that ξ(k) = l.So we have L, ξ(k) φ1. By our induction hypothesis this gives us K, k φ1.Since k ≥K k this implies K, k φ2. Again by our induction hypothesis we getL, ξ(k) φ2, but this means L, l φ2. We conclude that L, ξ(k) φ1 → φ2. �

This shows that homomorphisms are very useful tools in order to show thatcertain Kripke models are logically equivalent. We use such homomorphisms toprove that we can narrow the Kripke frames needed for the semantics of IPC downto a special class of trees.

Definition 83. (Originally de�ned by Jaskowski [11], cf. [9]) We de�ne theJaskowski trees J1, J2, . . . by induction:

• J1= 0 .• J2 is given by

0

• Suppose we already have Jn then Jn+1 is de�ned as:

0

Jn(n+ 1 times)

. . .JnJn

The following theorem is proven in Gabbay:

Theorem 84. [9] IPC =⋂n>0 ThKripke(Jn).

Proof. Obviously φ ∈ IPC implies φ ∈ Jn for every n. For the opposite weassume φ /∈ IPC. Lemma 80 shows that there is a �nite tree T such that T 1 φ.We would like to show that for some n there exists a homomorphism from Jn ontoT .

We �rst show that if there exists a homomorphism from Jn onto T then thereexists a homomorphism from Jn+1 onto T . Let ξ : Jn → T be a homomorphism.Recall that Jn+1 is n + 1 copies of Jn connected by a newly added root. De�neι : Jn+1 → T as follows: send 0 to the root of T , for any a > 0 we see that it is ina Jn-component of Jn+1 so set ι(a) = ξ(a). This gives a homomorphism.

Now we continue our proof by induction on the number of points of T . If Thas only one point, then it is equal to J1. Assume T is of the form:

0

Tm. . .T2T1

By our induction hypothesis and what we have just shown, we may assumethat there exists an n ≥ m and homomorphisms ξi : Jn → Ti for every 1 ≤ i ≤ m.


We may de�ne ι : Jn+1 → T by sending 0 to the root of T and then for the i-thJn-component set

ι(a) =

{ξi(a) if i ≤ m,ξm(a) it i ≥ m.

This gives a homomorphism. We conclude that for every �nite tree T there is an nsuch that there exists a homomorphism from Jn onto T .

From Lemma 82 it follows that there is an n such that Jn 1 φ. This proves thetheorem. �

The Jaskowski trees are very useful, since they have a clear simple form and areconstructed inductively. In order to show Medvedev's theorem we use yet anotherresult. We de�ne J+

n to be the Kripke frame Jn with an added maximal elementdenoted by 1. Notice that 1 is a maximal element in the frame-theoretic sense, notlattice-theoretic.

Lemma 85. 3 Let V : Jn → P(Var) be a valuation. Let V + be the valuationon J+

n that is the same as V on Jn and makes every propositional variable true at1. Formally this amounts to V +(a) = V (a) if a ∈ Jn and V +(1) = Var. Then forevery positive formula φ and every a ∈ Jn we have Jn, V, a φ i� J+

n , V+, a φ.

Proof. First remark that for every positive formula φ we have J+n , V

+, 1 φ(this can easily be shown by induction on the complexity of the formula). Now weprove the lemma by induction on the complexity of the formula. The base case andthe cases for ∧ and ∨ present no di�culties. So let φ = φ1 → φ2.

(if) Suppose Jn, V, a 1 φ1 → φ2. Let b ≥ a be such that Jn, V, b φ1 andJn, V, b 1 φ2. By our earlier remark we see that b 6= 1. Hence by induction we haveJ+n , V

+, b φ1 and J+n , V

+, b 1 φ2. So J+n , V

+, a 1 φ.(only if) Suppose J+

n , V+, a 1 φ1 → φ2. Let b ≥ a be such that J+

n , V+, b φ1

and J+n , V

+, b 1 φ2. Then again b 6= 1 and we prove similarly that Jn, V, a 1 φ1 →φ2. �

Corollary 86. Let φ be a positive formula. Then we have φ ∈ IPC i� φ ∈⋂n>0 ThKripke(J

+n ).

Proof. We clearly have that φ ∈ IPC implies φ ∈⋂n>0 ThKripke(J

+n ). For

the converse let φ be a positive formula such that φ /∈ IPC. Theorem 84 impliesthat there is a Jaskowski tree Jn and a valuation V on Jn such that Jn, V 1 φ.From Lemma 85 it follows that J+

n , V+ 1 φ. This proves the corollary. �

This brings us very close to our goal of proving Medvedev's theorem. Ourplan is to make a homomorphism λn : σf(n) → J+

n , where f is some function thatis going to be clear from the de�nition of λn. When we try to construct such ahomomorphism we �rst think of some subset of σf(n) that is isomorphic to Jn andthen try to extend this isomorphism to all of σf(n). That is why we �rst makea map κn : Jn → σf(n) in the other direction, prove that this is an isomorphismbetween Jn and a subset of σf(n) and after that we show that we can extend this

isomorphism (in fact κ−1) to a homomorphism λn : σf(n) → J+n .

Definition 87. We de�ne for every n a mapping κn : Jn → σf(n) by induction:

3The result stated in this lemma seems to be absent in any literature. It is highly doubtful thatthis is the �rst formulation or proof of this result.


• κ1 : J1 → σ1 is de�ned by κ1(0) = ∅. Remark that this is an isomorphism.• κ2 : J2 → σ2 is de�ned by κ2(0) = ∅, κ2(1) = {1} and κ2(2) = {2}.Remark that this is in fact an isomorphism.

• Now suppose we have de�ned κn : Jn → σf(n). We are going to de�neκn+1 shortly, but �rst we introduce some new notions. De�ne J∗n := {E ∈σf(n)+1 | there is an a ∈ Jn such that E = κn(a) ∪ {f(n) + 1}}. Further-more, let i ∈ ω, then i + J∗n = {E ∈ σf(n)+i+1 | there is an F ∈ J∗n suchthat E = {i+ j | j ∈ F}}. Now we will de�ne κn+1 on Jn+1. Recall thatJn+1 has n+ 1 copies of Jn and a root 0. First we set κn+1(0) = ∅. Nextwe will de�ne κn+1 on the Jn-components. First we give a picture of whatwe are planning to do:

0

Jn. . .JnJn

κn+1−−−−−−−→ ∅

n · (f(n) + 1) + J∗n. . .(f(n) + 1) + J∗nJ∗n

For the i-th Jn-component we set κn+1(a) = (κn(a) ∪ {f(n) + 1)}) + (i−1) · (f(n) + 1). It is crucial that the images of the Jn-components aredisjoint, this fact will make sure that we can extend κ−1

n � rng(κn) to ahomomorphism from σf(n) to J

+n .

From this de�nition we see that f : ω → ω is given by f(1) = 1, f(2) = 2 andf(n+ 1) = (n+ 1) · (f(n) + 1).

Example 88. We give some examples of the above de�ned κn : Jn → σf(n).For instance the image of κ3 is

∅

{9}

{8, 9}{7, 9}

{6}

{5, 6}{4, 6}

{3}

{2, 3}{1, 3}

Then J∗3 is given by

{10}

{9, 10}

{8, 9, 10}{7, 9, 10}

{6, 10}

{5, 6, 10}{4, 6, 10}

{3, 10}

{2, 3, 10}{1, 3, 10}

Furthermore f(3) = 9, so J∗3 + (f(n) + 1) is given by

{20}

{19, 20}

{18, 19, 20}{17, 19, 20}

{16, 20}

{15, 16, 20}{14, 16, 20}

{13, 20}

{12, 13, 20}{11, 13, 20}

From this we can derive the image of κ4. Unfortunately, pictures of this becometoo large to handle so we do not provide them.


Lemma 89. The mappings from De�nition 87 have several nice properties. Forevery n ∈ ω and every a, b ∈ Jn we have

(1) a ≤ b i� κn(a) ⊆ κn(b),(2) κn(a) ∩ κn(b) ∈ rng(κn),(3) κn is an isomorphism between Jn and rng(κn).(4) #(κn(a)) < n

Proof. We prove this by induction on n. For n = 1, 2 this is easily checked.Now suppose (1)-(4) hold for κn. We will prove that (1)-(4) hold for κn+1.

(1) If a = 0 then this obviously holds. So suppose a 6= 0. Then a is in one ofthe Jn-components and b is in the same Jn-component, say the i-th. Hence a ≤ b i�κn(a) ⊆ κn(b) i� (κn(a) ∪ {f(n) + 1})+(i−1) ·(f(n)+1) ⊆ (κn(b) ∪ {f(n) + 1})+(i− 1) · (f(n) + 1) i� κn+1(a) ⊆ κn+1(b).

(2) If one of a or b is equal to 0 then the intersection is empty, so the statementis true. Assume that a and b are both di�erent from 0.

(i) Suppose they come from di�erent Jn-components. Let us say that a comesfrom the i-th Jn-component and b form the j-th. From the de�nition of κn+1 itfollows that κn+1(a) = (κn(a) ∪ {f(n) + 1)}) + (i − 1) · (f(n) + 1), in particularm ∈ κn+1(a) implies (i − 1)(f(n) + 1) < m ≤ i · (f(n) + 1). It follows that theintersection is empty when a and b are from di�erent Jn-components.

(ii) So we left the case where a and b are from the same Jn component, saythe i-th. Then we have κn+1(a) ∩ κn+1(b) = (i− 1) · (f(n) + 1)+((κn(a) ∩ κn(b)∪{f(n) + 1})). Now since κn(a)∩κn(b) ∈ rng(κn), say it equals κn(c) for some givenc ∈ Jn, we get κn+1(a)∩κn+1(b) = (i−1) ·(f(n)+1)+κn(c)∪{f(n)+1} = κn+1(c).

(3) From the de�nition of κn it follows that it is injective. It is obviously ontorng(κn). Thus by (1) it follows that it is an isomorphism.

(4) We have #(κn+1(0)) = 0 < n + 1 and for a 6= 0 we get #(κn+1(a)) =#(κn(a) ∪ {f(n) + 1}) = #(κn(a)) + 1 < n+ 1. �

We now use these properties to de�ne a homomorphism λn : σf(n) → J+n . In

particular property (2) will be used in order to de�ne λn(E) as the least elementof rng(κn) that is above E.

Lemma 90. Let n > 2. Let V + be a valuation on J+n . Then there is a valuation

U on σf(n) and a homomorphism λn between the Kripke models 〈σf(n), U〉 and〈J+n , V

+〉.

Proof. In order to de�ne λn : σf(n) → J+n we �rst introduce some notions.

For any E ∈ σf(n) de�ne S(E) = {E′ ∈ σf(n) | E′ ∈ rng(κn) and E ⊆ E′}, whichis the set of successors of E in rng(κn). From the previous lemma we see that⋂E′∈S(E)E

′ ∈ rng(κn). Since κn is an isomorphism between Jn and rng(κn), there

is a unique element aE of Jn that is mapped to⋂E′∈S(E)E

′ by κn. We de�ne

λn(E) =

{1 if S(E) = ∅,aE if S(E) 6= ∅.

That is to say that λn maps a subset E ⊆ {1, . . . , f(n)} to the least F ∈ rng(κn)that contains E if this exists and otherwise maps E to 1. First we show that λnsatis�es (1), (2) and (3) of De�nition 81 of a homomorphism.

(1) We need to show that E ⊆ F implies λn(E) ≤ λn(F ). Suppose E ⊆ F .We see that S(E) ⊇ S(F ). If S(F ) = ∅, then λn(E) ≤ 1 = λn(F ). If S(F ) 6= ∅,


then κn(aE) =⋂E′∈S(E)E

′ ⊆⋂F ′∈S(F ) F

′ = κn(aF ). Lemma 89 on the preceding

page(1) implies that λn(E) = aE ≤ aF = λn(F ).(2) We need to show that λn is onto. Let a ∈ J+

n . If a 6= 1, then κn(a) =⋂E′∈S(κn(a))E

′ so λn(κn(a)) = a. For a = 1 we observe that for every E ∈ rng(κn)

we have #E < n. Since n > 2 implies that f(n) > n we can take any set E ∈ σf(n)

with more than n elements.(3) We need to show that λn(E) ≤ b implies that there is an F ⊇ E such that

λn(F ) = b. Suppose λn(E) ≤ b. First, remark that if b = 1, then again any F ⊇ Ewith more than n elements is mapped to 1. So we suppose that λn(E) ≤ b < 1.Then we have that κn(λn(E)) ⊆ κn(b) and also κn(λn(E)) =

⋂E′∈S(E)E

′ ⊇ E.

Hence E ⊆ κn(b) and λn(κn(b)) = b.All right, now we set U(E) = V +(λn(E)). We only need to prove that this

gives a valuation on σf(n): Let E ⊆ F . Then λn(E) ≤ λn(F ) by (1). Hence

U(E) = V +(λn(E)) ⊆ V +(λn(F ) = U(F ). So U is indeed a valuation. We haveshown

(4) U(E) = V +(λn(E)).(1)-(4) show that λn is the desired homomorphism. �

Now we can put all this together to prove Medvedev's theorem:

Theorem 91. We have LMπ = IPCπ.

Proof. Obviously φ ∈ IPC implies φ ∈ LM. For the converse we prove thecontra position. Let φ be a positive formula such that φ /∈ IPC. Corollary 86 impliesthat there is a Jaskowski tree Jn and a valuation V on Jn such that J

+n , V

+ 1 φ. Wehave seen in Lemma 90 that there is a valuation U on σf(n) and a homomorphism

from 〈σf(n), U〉 to 〈J+n , V

+〉. From Lemma 82 and the fact that J+n , V

+, 0 1 φ itfollows that σf(n), U, ∅ 1 φ. Hence σf(n) 1 φ and thus φ /∈ LM. �

CHAPTER 4

Semantics for IPC using the Medvedev lattice

In this chapter we will prove the main result from [32, Skvortsova]. We willconsider the Medvedev lattice (see De�nition 26 on page 14), which was shown tobe a Brouwer algebra (see De�nition 46 on page 22 and Theorem 29 on page 15).Recall the notion of mass problems given in De�nition 17 on page 9 and the notionof M-reducibility in De�nition 24 on page 13. Moreover, to have a better ideaof M-reducibility one could review De�nition 19 on page 9 and Proposition 21 onpage 10.

We are going to consider a particular section of the Medvedev lattice. For thatwe need to introduce some notions. Arbitrary mass problems will be denoted byA,A′,B, . . ., recursive operators will be denoted by Ψ,Ψe,Ψ

′, . . ..We let the set ofall total functions be F , they will be written as f, g, . . .. Let F be the set of all�nite initial segments, we let f , g, . . . denote (codes for) elements of F . In a way

F ⊆ ω, in fact we choose the coding of �nite initial segments in such a way thatF = ω and such that the graph of a segment is e�ectively constructable given itsnumber.

Definition 92. (cf. [32]) For any mass problem A and any (code of a) �nite

initial segment f , let Af := {f ∈ A | f ⊆ f} and let F (A) = {g ∈ F | Ag 6= ∅}. We

call a mass problem A homogeneous [25, Muchnik] if Af ≤M A for all f ∈ F (A). Inaddition we call A e�ectively homogeneous if there is a partial computable functionϕ such that for every f ∈ F (A) we have that ϕ(f) ↓ and this yields the number ofa recursive operator reducing Af to A.

Definition 93. [32] We call a mass problem C upper-≤T closed if for everyf ∈ C and g ∈ F , f ≤T g implies g ∈ C. The set of all upper-≤T closed degrees,i.e. degrees containing an upper-≤T closed mass problem, will be denoted by M.We denote upper-≤T closed mass problems by C, C′,D, . . . and elements of M byC,D, . . . .

Note that for every nonempty upper-≤T closed C and every f we have f ∗C ⊆ C,so F (C) = F and the recursive operator Ψ(f) = f ∗ f reduces Cf to C. We seethat Proposition 21 on page 10 implies that there is a computable ϕ such thatthe recursive operator Ψ above equals Ψϕ(f). Hence every upper-≤T closed mass

problem is e�ectively homogeneous. We will now examine some properties of M inorder to eventually show that it is a canonical subset of M (recall De�nition 50 onpage 23). For instance, we have

C ≤M A i� A ⊆ C,

since for total functions f, g we have that Ψ(g) = f implies f ≤T g (see Proposi-tion 32 on page 17).

39

4. SEMANTICS FOR IPC USING THE MEDVEDEV LATTICE 40

Proposition 94. [32] Let C, D be upper-≤T closed and let A be a mass prob-lem. Then C + D ≡M C ∩ D and A → C ≡M {f ∈ F | ∀g ∈ A[f ⊕ g ∈ C]}.Furthermore, [C] + [D], [C]→ [D] ∈M.

Proof. First note that C∩D ⊆ C,D, so C,D ≤M C∩D. Hence C+D ≤M C∩D.Moreover, for every f ∈ C and g ∈ D we have that f, g ≤T f ⊕ g. Because C and Dare upper-≤T closed, we have f ⊕ g ∈ C ∩D. This shows that C ∩D ≤M C +D, bymeans of the identity recursive operator. In addition we see that C ∩D is upper-≤Tclosed, so [C] + [D] ∈M.

Second, since {f ∈ F | ∀g ∈ A[f ⊕ g ∈ C]} + A ⊆ C, we see that C ≤M {f ∈F | ∀g ∈ A[f ⊕ g ∈ C]} + A. So A → C ≤M {f ∈ F | ∀g ∈ A[f ⊕ g ∈ C]}.In order to show the other reduction, consider Ψ(n ∗ f) = f . Then we see thatΨ[A → C] = {Ψ(n ∗ f) ∈ F | Ψn[A + f ] ⊆ C} ⊆ {f ∈ F | ∀g ∈ A[f ⊕ g ∈ C]},because if Ψn[A+f ] ⊆ C then for every g ∈ A, Ψn(g⊕f) ∈ C, in particular f⊕g ∈ C,since f ⊕ g ≥T Ψn(g⊕ f). Again A → C is upper-≤T closed, so [A]→ [C] ∈M. �

Proposition 95. [32] Let A,B be arbitrary mass problems and C be an e�ec-

tively homogeneous problem with F (C) = F . Then C → (A× B) ≡M (C → A) ×(C → B).

Proof. First remark that C → (A× B) ≤M (C → A) × (C → B), since C +((C → A)× (C → B)) ≡M (C + (C → A))× (C + (C → B)) ≥M A× B.

For the second reducibility (≥M ) we need to give a recursive operator Ψ. Letϕ be a partial computable function such that Cf ≤M C via Ψϕ(f). Notice that this

implies that ϕ is total, since F can be identi�ed with ω. Now we are going to de�neΨ(n ∗ f):

We begin with searching for a (code of a) �nite initial segment h, using n ∗ f ,such that

(Ψn(f ⊕ h)

)(0) is de�ned and equals either 0 or 1. When we �nd such

a h, then we de�ne Ψ(n ∗ f)(0) =(

Ψn(f ⊕ h))

(0). If this search does not stop

(or does not yield 0 or 1), then let Ψ(n ∗ f) be the partial function that has anempty domain. Remark that if n ∗ f ∈ C → (A× B), then Ψ(n ∗ f)(0) is de�ned

and for every g ∈ C such that h ⊂ g we have λy. (Ψn(f ⊕ g)) (y + 1) ∈ A ∪ B.Moreover, Ψ(n ∗ f)(0) shows whether that function is either in A or in B. Sincethe recursive operator Ψϕ(h) reduces Ch to C, it follows that that for every g ∈ Cwe have h ⊂ Ψϕ(h)(g). This discussion shows that it is su�cient to get a code for

the recursive operator Ψ given by the following composition:

f ⊕ g 7→ f ⊕Ψϕ(h)(g) 7→ Ψn

(f ⊕Ψϕ(h)(g)

)7→ λy.Ψn

(f ⊕Ψϕ(h)(g)

)(y + 1).

From Proposition 21 on page 10 it follows that there is a computable function φthat determines the code of Ψ from 〈n, h〉. Then we let

Ψ(n ∗ f) =

0 ∗ φ(〈n, h〉) ∗ f if

(Ψn(f ⊕ h)

)(0) ' 0)

1 ∗ φ(〈n, h〉) ∗ f if(

Ψn(f ⊕ h))

(0) ' 1)

↑ otherwise.


We see that if n∗f ∈ C → (A× B), then Ψ(n∗f)(0) ↓ and this outcome determines

whether for every g ⊃ h, (Ψn(f ⊕ g)) is either in A or in B. Suppose it is in A.Then Ψ(n ∗ f)(0) ' 0 and we see that for all g ∈ C

Ψφ(〈n,h〉)(f ⊕ g) = λy.Ψn

(f ⊕Ψϕ(〈n,h〉)(g)

)(y + 1).

Since h ⊂ Ψϕ(h)(g), we get Ψn

(f ⊕Ψϕ(h)(g)

)(0) ' 0 and thus Ψφ(〈n,h〉)(f⊕g) ∈ A.

We conclude that Ψ reduces C → (A× B) to (C → A)× (C → B). �

Corollary 96. [32] The set of degrees M is canonical in M and it containsthe top element 1.

Proof. From the discussion following De�nition 93 on page 39 it follows thatevery upper-≤T closed C satis�es the antecedent of the previous proposition, hencewe have C → (A× B) ≡M (C → A)× (C → B). Furthermore, C +D ≡M C ∩D andC → D ≡M {f ∈ F | ∀g ∈ C[f ⊕ g ∈ D]} are both upper-≤T closed. Also remarkthat both 0 and 1 are elements of M. So it remains to show that M ⊆ D(M)1.Let f be a computable function and let A0 := {f}. Notice that [A0] = 0 ∈ D(M),since A0 ≡M A × B (in particular ≥M ) implies that either A or B contains acomputable function. Now suppose we have C ≡M A × B. Then we see from ≥Mthat A0 ≡M C → (A× B) ≡M (C → A)×(C → B). The fact that 0 ∈ D(M) impliesthat either A0 ≡M C → A or A0 ≡M C → B. So either C ≥M A or C ≥M B. Inaddition we have C ≡M A × B ≤M A,B, hence either C ≡M A or C ≡M B. Thisshows that M ⊆ D(M) and is canonical. �

Corollary 97. 2The ×-hull ∆(M)3 constitutes a Brouwer algebra.

Proof. This follows from the previous corollary and Proposition 52 on page 23.�

Lemma 98. [32] Let A be a countable implicative usl. Then it is isomorphic to

a sub-implicative usl of M relative to +, →. Beware that if A has a top element, itmight not be preserved by this isomorphism.

Proof. Let A be a countable implicative usl. From [16, Lachlan and Lebeuf]it follows that A is isomorphic to an initial segment of the Turing degrees DT . Abrief review of the research done on countable order types of initial segments of theTuring degrees is given in Appendix A.4 According to the result of [16] with eachα ∈ A we can associate a function fα ∈ F such that

(1) For all α, β ∈ A, α ≤ β i� fα ≤T fβ ;(2) For all α ∈ A and all f ≤T fα there is a β ∈ A such that f ≡T fβ .

We de�ne D := {f ∈ F | ∃α ∈ A[f ≡T fα]}, D := F\D and for every α ∈ A letDα := D ∪ {f ∈ D | f ≥T fα}. From (2) it follows that D is upper-≤T closedand from that it is clear that Dα is also upper-≤T closed. We will show that A isorder-isomorphic to {Dα | α ∈ A}:

1This is the set of ×-irreducibles (cf. De�nition 50 on page 23).2This corollary is not stated in [32], it is only said that ∆(M) is an implicative lattice.3This is the ×-hull introduced in De�nition 51 on page 23.4We should add that in [32] the result of [15, Lachlan] is used to prove this lemma. It is provenin [15] that every countable distributive lattice is isomorphic to an initial segment of the Turingdegrees.


Since Dα is upper-≤T closed we have for every α ∈ A,

Dα ≤M Dβ i� Dα ⊇ Dβ i� fα ≤T fβ i� α ≤ β.

Hence A is order-isomorphic to {Dα | α ∈ A}.We can extend this order-isomorphism to the order-isomorphism given by α 7→

[Dα]. We want to show that the latter order-isomorphism preserves + and →:For every f ∈ D and any α, β ∈ A we get

fα+β ≤T f i� fα ≤T f and fβ ≤T f :

(if and only if) There is a γ ∈ A such that f ≡T fγ . By (1) we get fα+β ≤T fγi� α + β ≤ γ. We see that α + β ≤ γ i� α, β ≤ γ. Again by (1) we conclude thatfα+β ≤T fγ i� fα ≤T fγ and fβ ≤T fγ .

In particular this implies

fα+β ≡T fα ⊕ fβ :

(≡T ) The reduction≤T follows from the fact that fα, fβ ≤T fα⊕fβ and the previousresult. ≥T follows from the fact that α+ β ≥ α, β and (1).

Furthermore, for every f ∈ D and every α, β ∈ A we have

fα→β ≤T f i� fβ ≤T f ⊕ fα :

(if and only if) There is a γ ∈ A such that f ≡T fγ . From (1), the previous resultand the fact that α → β ≤ γ i� β ≤T α + γ, we get fα→β ≤T f i� fβ ≤T fα+γ i�fβ ≤T f ⊕ fα.

Finally, we remark that for every f ∈ F we have

g ∈ D implies f ⊕ g ∈ D,

since f ⊕ g ∈ D implies f, g ∈ D.The above equivalences imply

Dα+β = D ∪ {f ∈ D | f ≥T fα+β} = D ∪ {f ∈ D | f ≥T fα and f ≥T fβ}= Dα ∩ Dβ ≡M Dα +Dβ

and

Dα → Dβ ≡M {f ∈ F | ∀g ∈ Dα[f ⊕ g ∈ Dβ ]} = D ∪ {f ∈ D | f ⊕ fα ≥T fβ}= D ∪ {f ∈ D | f ≥T fα→β} = Dα→β .

We conclude that A is isomorphic to {[Dα] | α ∈ A} viewed as an implicative uppersemilattice. Finally, remark that when A has a top element 1, then D1 need notequal 1 in M. �

We derive some interesting results.

Corollary 99. [32] Let A be a countable implicative usl. Then its ×-hull∆(A)5 is isomorphic to a sub-implicative lattice of the Brouwer algebra ∆(M).Beware that if A has a top element, it might not be preserved by this isomorphism.

Theorem 100. [32] We have⋂

E∈∆(M) Th(ME) = IPC, in particular the in-

tersection of the logics of the initial segments of the Medvedev lattice is exactly theintuitionistic propositional calculus.

5See De�nition 51 on page 23.


Proof. We consider the countable implicative usl Cω consisting of �nite andco�nite subsets of N and ordered by ⊇ (see De�nition 74 on page 31). Then ∆(Cω)is a countable Brouwer algebra and by the previous lemma and corollary we havea sub-implicative lattice {Eα|α ∈ ∆(Cω)} ⊆ ∆(M) which is isomorphic to ∆(Cω).Hence ∆(Cω) is isomorphic to a sub-Brouwer algebra of

(∆(M)

)E∅

(also preserv-

ing 1). Corollary 59 on page 27 implies that Th((∆(M)

)E

) ⊆ Th(∆(Cω)), using

Corollary 76 on page 31 we get Thπ((∆(M)

)E

) = IPCπ. With help of Lemma 60

on page 27 we may conclude that⋂α∈∆(Cω) Th(MEα) ⊆

⋂α∈∆(Cω) Th(∆(Cω)α) =

IPC. Hence⋂α∈∆(Cω) Th(MEα) = IPC. This proves the theorem. �

Remark 101. [32] Note that we have⋂

E∈M Th(ME) 6= IPC, since for any

E ∈M the following formula KP belongs to Th(ME): KP = (¬p→ q∨r)→ (¬p→q) ∨ (¬p → r). Let E ∈ M and A,B,C ∈ ME. Because M is canonical we have¬A = A→ E ∈M and therefore (¬A→ B ∨C)→ (¬A→ B) ∨ (¬A→ C) = 0.We see that KP ∈ Th(ME).

The equivalences that have been established when proving Lemma 98 will beused to prove the theorem of Skvortsova:

Theorem 102. [32] There exists a degree E0 ∈M such that Th(ME0) = IPC.

Proof. Again we consider Cω. In addition to the result of Lemma 98, onemore ingredient is required in order to obtain this remarkable result.

We would like to make a so called canonical partition in Cω. This is a collectionof segments {[αi, βi]∆(Cω)|i ∈ N} such that for every i:

(1) αi ∈ Cω is �nite,(2) βi =

∏nij=1 βi,j with βi,j ∈ Cω and βi,j ⊂ αi (i.e. βi ∈ [αi,1]∆(Cω)),

(3) i′ 6= i implies αi′ ∩ αi = ∅.

Recall Lemma 73 on page 31 where it is proven that B(2n) is isomorphic to ∆(2n),which in turn is obviously isomorphic to [α,1]∆(Cω) when |α| = n. Let us denote

this last isomorphism by ηα : [α,1]∆(Cω) → B(2n). We �x a canonical partition inCω satisfying:

(A) For any n > 0 and every ε ∈ B(2n) there is a αi and a βi such that|αi| = n and βi =

∏nij=1 βi,j ∈ [αi,1]∆(Cω) is mapped to ε by ηαi .

Remark that this can be done. Then obviously [αi, βi]∆(Cω) is isomorphic toB(2n)εas Brouwer algebras. It follows from Theorem 58 on page 26 that Th([αi, βi]∆(Cω)) =

Th(B(2n)ε). In light of Corollary 77 on page 31, this gives us that IPC equals⋂n>0

⋂ε∈B(2n) Th(B(2n)ε) =

⋂i>0 Th([αi, βi]∆(Cω)).

We also �x an embedding of Cω in M (and ∆(Cω) in ∆(M)) just as in Lemma98. Then we have a collection of upper-≤T closed problems {Dα|α ∈ Cω} satisfyingfor all α, β ∈ Cω

α ⊆ β i� Dα ≥M Dβ i� Dα ⊆ DβDα∩β ≡M Dα +Dβ

Dα +Dβ ⊆ Dα∩β .


Let Dα be the degree of the problem Dα. For any i ∈ N, let Ai := Dαi and Bi :=∏nij=1 Dβi,j . We see that this embedding gives us an embedding of [αi, βi]∆(Cω) into

[Ai,Bi]M, so Th([Ai,Bi]M) ⊆ Th([αi, βi]∆(Cω)), hence⋂i>0 Th([Ai,Bi]M) = IPC.

In order to complete the proof we want to use Corollary 62 and de�ne a massproblem E0 such that for every i we have Bi = Ai +E0. The corollary then showsthat Th(ME0

) =⋂i>0 Th([Ai,Bi]M) = IPC.

Let E0 be the degree of the mass problem E0 = {i∗ j ∗f | i > 0, 1 ≤ j ≤ ni, f ∈Dβi,j}. Fix some i0 > 0. Then obviously E0 ≤M Dβi0,j for every 1 ≤ j ≤ ni0 ,so E0 ≤ Bi0 . It follows that Ai0 + E0 ≤ Bi0 . Now we would like to show theother direction. For that we remark that Bi0 = [Bi0 ], where Bi0 = {j ∗ f | 1 ≤ j ≤ni0 , f ∈ Dβi0,j}. So we are going to show that Dαi0 + E0 ≥M Bi0 . For that we takethe recursive operator de�ned by

Ψ (g ⊕ (i ∗ j ∗ f)) =

{j ∗ f if i = i0

1 ∗ (f ⊕ g) if i 6= i0.

Let h ∈ Dαi0 + E0. Then h = g ⊕ (i ∗ j ∗ f) for some g ∈ Dαi0 , i > 0, 1 ≤ j ≤ niand f ∈ Dβi,j . We see that if i = i0, then Ψ(h) = j ∗ f ∈ j ∗ Dβi,j ⊆ Bi0 . Nowsuppose i 6= i0. Then Ψ(h) = 1 ∗ (f ⊕ g) and αi ∩ αi0 = ∅, since we started ofwith a canonical partition. But also βi,j ⊂ αi, hence f ∈ Dβi,j ⊆ Dαi . Moreover,f ⊕ g ∈ Dβi,j + Dαi0 ⊆ Dαi + Dαi0 ⊆ Dαi∩αi0 = D∅ ⊆ Dβ1,1

. We conclude that

Ψ(h) ∈ 1 ∗ Dβ1,1 ⊆ Bi0 and thus Dαi0 + E0 ≥M Bi0 .We have shown that Bi = Ai + E0 for every i. This implies that Th(ME0

) =IPC and proves the theorem. �

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APPENDIX A

Initial segments of the degrees of unsolvability

In this appendix we will brie�y and super�cially review a part of the researchdone on isomorphic images of lattices in initial segments of the degrees of unsolv-ability. This will accumulate to a result that every countable upper semilatticeswith zero is isomorphic to an initial segment of the degrees of unsolvability. Themain question of this appendix will be what the order types of the initial segmentsof the degrees could be.

It all started with Spector [36] who constructed a minimal degree. From thatpoint on there were many people investigating the main question of this appendix.We pick it up when Thomason [38] remarked that the combinatorial tricks used opto then all relied on a representation for a �nite lattice as the dual of a lattice ofequivalence relations. Probably the most impressive result up to then is by Lachlan[15], stating that every countable distributive lattice is isomorphic to an initialsegment of the degrees. However, we start with Thomason who de�ned a notionof well-representability for a lattice. In order to observe his de�nition we need anauxiliary de�nition.

We �rst de�ne the dual lattice of equivalence relations.

Definition A.1. Let S be any set. The dual equivalence lattice E(S) of the setS consists of all equivalence relations on S ordered by reversed inclusion , i.e. θ ≤ ϑif for all a, b ∈ S we have that aϑb implies aθb. Furthermore, the join (+) and meet(×) are given by

a(θ + ϑ)b i� aθb and aϑb

a(θ × ϑ)b i� there are a1, . . . , an ∈ S satisfying

aθa1ϑa2θ . . . θanϑb.

Finally, 0 is the universal relation and 1 is the identity relation.

Definition A.2. [38, Thomason] A �nite lattice L is well-representable if thereis a �nite set S = {s0, . . . , sk−1} and an isomorphism α 7→ θα of L onto a sublatticeof E(S) satisfying the following property

(1.1) for every u, v ∈ S and i1, i2 ∈ {0, . . . , k − 1} if∀α ∈ L[si1θαsi2 implies uθαv],

then there are s′0, . . . , s′k−1 ∈ S such that s′i1 = u and s′i2 = v and for every

i, j ∈ {0, . . . , k − 1},∀α ∈ L[siθαsj implies s′iθαs

′j ].

Thomason then eventually proves the following theorems:

Theorem A.3. [38] Let L be a �nite lattice. Either of the following conditionsis su�cient for L to be well-representable:

47

A. INITIAL SEGMENTS OF THE DEGREES OF UNSOLVABILITY 48

(1) L is distributive,(2) L = PG(F,m − 1) = (the subspace lattice of an m-dimensional vector

space over the �nite �eld F )

Theorem A.4. [38] If L is any �nite lattice, then there is a sublattice of degreesisomorphic to L.

Theorem A.5. [38] If L is a well-representable lattice, then L is isomorphicto an initial segment of the degrees.

So Thomason successfully used his notion of well-representability in order toexamine the order types of the degrees. But he did not manage to derive theultimate goal to characterize the countable initial segments of the degrees. However,his attempt inspired Lerman [17] to introduce yet another notion of representability.

Definition A.6. [17, Lerman] Let L be a lattice. A representation of thelattice L is an ordered pair 〈F,U〉 where U is a set and F is a lattice isomorphismfrom L onto a subset of E(U). A weak representation of the lattice L is an orderedpair 〈F,U〉 where U is a non-empty set and F is an order isomorphism from L ontoa subset of E(U), such that F (0) = 0, F (1) = 1 and for all x, y ∈ L we have

F (x+ y) = F (x) + F (y).

Whitman [39] proved that every lattice L has a representation 〈F,U〉 and ifL is �nite, then U can be chosen to be �nite or countable. Lerman introducesa nested sequence of approximations to a representation of a lattice L, via weakrepresentations having certain homomorphism properties. Such a sequence withouthomomorphism properties can be used to prove Whitman's theorem. We proceedwith the de�nition of this nested sequence of Lerman, this de�nition can be foundin [16, Lachlan and Lebeuf].

Definition A.7. [17] Let L be a �nite lattice. Consider two weak represen-tations 〈G,V 〉 and 〈F,U〉. We say that 〈G,V 〉 extends 〈F,U〉 if U ⊆ V and forevery c ∈ L, G(c) � U = F (c). Next we call L sequentially representable if thereis a sequence {〈Fi, Ui〉|i ∈ ω} of weak representations of L such that for every i,〈Fi+1, Ui+1〉 extends 〈Fi, Ui〉 and

(2.1) If a, b, c ∈ L and a × b = c, u0, u1 ∈ Ui and u0F (c)u1 then there is asequence {v0, v1, v2} of elements of Ui+1 such that u0Fi+1(a)v0Fi+1(b)v1Fi+1(a)v2

Fi+1(b)u1;(2.2) If u0, u1, v0, v1 ∈ Ui and for all b ∈ L we have that u0Fi(b)u1 implies

v0Fi(b)v1, then there are w0 and embeddings h0, h1 of 〈Fi, Ui〉 in 〈Fi+1, Ui+1〉 suchthat h0(u0) = v0, h0(u1) = h1(u0) = v1 and h1(u1) = v1.

The two extra homomorphism properties are used to prove certain technicalissues when Lerman is proving that every �nite lattice can be embedded in aninitial segment of the degrees. After this de�nition Lerman turns to proving thefollowing theorem. A better readable proof can be found in [16].

Theorem A.8. [17] Every �nite lattice L is sequentially representable.

Theorem A.9. [17] Every sequentially representable lattice is isomorphic toan initial segment of the degrees of unsolvability.

We will now continue by following Lachlan and Lebeuf [16]. They de�ne yetanother notion of representability. In the following we let L be a countable upper

A. INITIAL SEGMENTS OF THE DEGREES OF UNSOLVABILITY 49

semilattice with zero. We may suppose it has a top element, or else we can addone. We can approximate L by an ascending chain of �nite lattices {Li|i ∈ ω} eachwith the same 0 and 1, such that the join in Li+1 extends the join in Li.

Definition A.10. [16, Lachlan and Lebeuf] It is shown that we can constructan array

〈〈〈Fi,0, Ui,0〉|φ(0) ≤ i < ω〉, . . . , 〈〈Fi,j , Ui,j〉|φ(j) ≤ i < ω〉, . . .〉,where φ(0) = 0 and φ(j) ≤ φ(j+1) for all j, and satisfying the following conditions:

(3.1) For each j the sequence 〈〈Fi,j , Ui,j〉|φ(j) ≤ i < ω〉 is almost a sequentialrepresentation of Lj . That means there is a strictly increasing sequence φ(j) =ψ(0) < ψ(1) < . . . such that 〈〈Fψ(i),j , Uψ(i),j〉|i < ω〉 is a sequential representationof Lj and for all i,

ψ(i) ≤ l < ψ(i+ 1) implies 〈Fl,j , Ul,j〉 = 〈Fψ(i),j , Uψ(i),j〉.(3.2) For each pair 〈i, j + 1〉 such that i ≥ φ(j + 1) and i, j < ω, for all

u, v ∈ Ui,j+1 and for each c ∈ Lj ,Ui,j ⊃ Ui,j+1 and uFi,j+1(c)v ⇔ uFi,j(c)v.

(3.3) Each column is recursive, i.e. for each j there is a bijection g from⋃{Ui,j |φ(j) ≤ i < ω} onto ω such that 〈g(Ui+φ(j),j)|i < ω〉 is a strongly c.e. se-

quence of �nite sets, and g−1(x)Fy+φ(j),j(c)g−1(z) is a recursive predicate of x, y, z

for each c ∈ Lj .Lachlan and Lebeuf do not provide a name for this, but let us say that a lattice

for which such an array exists is called array representable.

Theorem A.11. [16] Every countable upper semilattice is array representable.

Theorem A.12. [16] Every array representable upper semilattice is isomorphicto an initial segment of the degrees.

We should remark that this completely characterizes all the countable initialsegments of the degrees, since every countable initial segment of the degrees willin fact constitute a countable upper semilattice. This result concludes our briefreview of the research done on the order types of countable initial segments of thedegrees of unsolvability.

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