At the beginning of the course we built (recursively in *0′*) two incomparable sets *A,B*. It follows that *A* and *B* have degrees intermediate between *0* and *0′*. The construction required that we fixed finite initial segments of *A* and *B* during an inductive construction and so it is not clear whether they are c.e. or not. (A c.e. construction would add elements to a set and we would not have complete control on what is kept out of the set). Post asked whether there is an intermediate c.e. degree and this was solved by Friedberg and Muchnik using what is now called a finite injury priority construction. We show this construction; again, 2 incomparable sets *A* and *B* are built and the construction explicitly shows they are c.e. This implies they are not recursive and have degree strictly below *0′*.

## Archive for the ‘117b’ Category

### 117b – Unsolvable problems – Lecture 3

March 12, 2007### 117b – Unsolvable problems – Lecture 2

March 2, 2007We briefly discuss Church’s and Trakhtenbrot’s theorems: The set of logically valid sentences and the set of sentences valid in all finite models are undecidable. More precisely: If *L* is a language that contains either a binary symbol or the language of arithmetic, then the set of theorems of *Q* reduces to the set of *validities* of *L*, so this set is (r.e. and) undecidable. For the same *L*, we can axiomatize the behavior of a given Turing machine on a given input (by a similar argument to the association of a semi-Thue system to each Turing machine) in such a way that this axiomatization has a finite model iff the machine converges on this input, thus the halting problem reduces to the complement of the set of *finite validities*, so this last set is (co-r.e. and) undecidable.

We define *Post correspondence systems* and show that the problem of determining whether such a system admits a solution is unsolvable. Two applications to the theory of *formal languages* follow: There is no algorithm to test whether 2 given *context-free grammars* have disjoint languages, and there is no algorithm to test whether a given context-free grammar is ambiguous.

### 117b – Unsolvable problems – Lecture 1

March 1, 2007We define *semi-Thue processes* and show (Post, 1947) that for each *e* there is a semi-Thue process such that the word problem for is Turing-above the halting problem for .

We then proceed to define *Thue processes* and show that they give rise to semigroups. It follows (Post; Markov) that there is a presentation of a semigroup with an unsolvable word problem.

It is easy to modify a Thue process so the semigroup it generates is indeed a group. A modification of the arguments for semigroups shows (Novikov; Boone) that the word problem for groups is unsolvable. In fact (Adjan,Rabin) if ** G** is a class of groups closed under isomorphisms and subgroups which neither contains nor is disjoint from the class of finitely presented groups, then there is no algorithm that determines from a given group presentation whether it is in

*.*

**G**Additional references:

*Modular machines, the word problem for finitely presented groups, Collins’ theorem,*by Aanderaa, Cohen. In**Word problems II. The Oxford book**, Adjan, Boone, Higman, eds. North-Holland*(1980)*, 1-16*.**Modular machines and the Higman-Clapham-Valiev embedding theorem,*by Aanderaa, Cohen. In**Word problems II. The Oxford book**, Adjan, Boone, Higman, eds. North-Holland*(1980)*, 17-28*.*

### 117b – Homework 8

March 1, 2007### 117b – Some additional remarks on provably recursive functions

February 28, 2007There are quite a few examples of natural combinatorial statements not provable in other than the ones discussed in lecture. The references listed below present several of them and provide a guide to the existent literature; I especially recommend the beautiful expository paper by Stephen Simpson and in general the whole collection where that paper appeared. I will ask you to abstain from looking at the Kanamori-McAloon paper for the time being, since its main result is the content of the last homework set.

There are at least two `combinatorial’ methods for showing that a function is not provably recursive. One is to use *indicators, *this technique, due to Paris, requires a bit of model theory, and is the method of the proof you are asked to provide in the homework.

The other method comes from a careful analysis of which functions are provably total in . For this, one defines a *fast-growing* hierarchy of recursive functions. This requires some understanding of the concept of an *ordinal*. The *fast-growing* (or *Grzegorczyk*) hierarchy is defined as follows:

- .
- , where the superindex means
*iterated times*. - , if is limit.

Here, is some (natural) sequence converging to (if you are not aware of ordinals, ignore for the moment what the last clause means).

For example, , , grows like a stack of powers of and grows like an iterated stack of powers of , there being such iterations. It is easy to see that each is primitive recursive and strictly increasing, that < implies that is eventually dominated by (i.e., that < for all but finitely many values of ), and that every primitive recursive function is eventually dominated by some . Thus, if a function eventually dominates each , it cannot be primitive recursive.

The ordinals provide us with a method for continuing this sequence while preserving that its members are (eventually) increasing and the sequence itself is increasing under eventual domination. Say that a linearly ordered set is* well-ordered* if it has no infinite descending chains. For our purposes, the ordinals are simply a way to talk about well-ordered sets (classically, ordinals denoted the *order types* of the well-ordered sets, i.e., the equivalence classes of these sets under the relation of being order-isomorphic. The modern viewpoint is different and we don’t need it here). We use to denote the order type of any linearly ordered set of elements, and use to represent the order type of the natural numbers. We `add’ two order types and by forming the disjoint union of a set of order type followed by a set of order type . We also say that is smaller than if any set of order type has a (proper) initial segment of order type . We can then continue the sequence of the natural numbers, as follows:

where we use to represent , etc, to represent and so on, and we can continue even beyond:

Call (epsilon-0) the first ordinal obtained this way such that

(so ).

One can show that any ordinal below admits a unique `pure base ‘ representation, from which there is a natural way of defining the sequence converging to whenever is a limit, which means that it is not of the form for any . For example,

, etc,

are all limits. So, for example, the sequence corresponding to is just and the corresponding function is easily seen to grow as the (diagonal) Ackermann function ; the sequence corresponding to is , etc. As before, each is (eventually) strictly increasing and if < then eventually dominates . Each , for < (and in fact, for many more ) is recursive.

The relation of this sequence with lies in that *any* function provably recursive in is eventually dominated by one of the with <. So, one purely combinatorial way of showing that a function is not provably recursive is to show that it eventually dominates each such . This kind of analysis was developed by Solovay and Ketonen.

There are also other ways of relating to this number , and *proof theorists *have a lot more to say about this relation.

Additional references:

*Lecture notes on enormous integers,*by H. Friedman*. (2001)**Unprovable theorems,*by H. Friedman*. (2005)*See www.math.ohio-state.edu/~friedman/manuscripts.html for additional manuscripts and details.*Unprovable theorems and fast-growing functions,*by S. Simpson. In**Logic and combinatorics**, S. Simpson, ed. AMS*(1987).**On Gödel incompleteness and finite combinatorics,*by A. Kanamori and K. McAloon*. Annals of Pure and Applied Logic***33***(1987),*23-41*.*

### 117b – Undecidability and incompleteness – Lecture 11

February 27, 2007Let be r.e. and suppose there is a sentence such that does not prove . Then is more powerful than in two ways:

- It proves
*more*theorems than ; for example, . - It provides
*much shorter*proofs of theorems of ; for example: For any (total) recursive function there is a sentence provable in but such that in there is a proof of such that no proof of in is shorter than .

This leads to considering those recursive functions of which can prove that they are total. One calls such functions *provably recursive* in . It is easy to show that for any -sound , there are recursive functions that are not provably recursive in . The statement that is total,

,

is easily seen to be and for one can provide natural examples of such recursive but not provably recursive functions , thus providing natural (combinatorial) examples of independent -sentences. Actually, this can also be done for much stronger systems than , like , and Harvey Friedman has provided several such examples.

For , the most famous examples are perhaps the following:

- Goodstein (1944). The
*pure base- representation of a number*is obtained by writting in base , writing the exponents of this representation also in base , writing the exponents of these representations in base , and so on. For let and for let be the result of writing the pure base- representation of , replacing each by , and subtracting . The*Goodstein sequence*of is the sequence This sequence eventually converges to*0*. The function that to assigns the number of steps necessary for this to occur is recursive but not provably recursive in . In fact, given any provably recursive in , eventually dominates . - Kirby, Paris (1982). Define a
*hydra*to be a finite rooted tree. A*head*of the hydra is any node (other than the head) with only one edge connected to it,*together*with this edge. Hercules fights the hydra by chopping off its heads one by one by stages. At stage , the hydra grows new heads as follows: from the node that used to be attached to the head just removed, traverse one edge towards the root and from this node just reached, sprout*copies*of the part of the hydra above the edge just traversed. For any hydra, no matter how Hercules battles it, he eventually wins (i.e., the hydra is reduced to a single node). Moreover, given a hydra there is an absolute bound on the number of stages that it takes Hercules to win. Coding (in any reasonable fashion) hydras by numbers, the function that assigns this number to a given hydra is recursive but eventually dominates all functions provably recursive in . - Paris, Harrington (1977). Given a set , let denote the collection of its subsets of size . Given a function , say that a subset of is homogeneous for iff restricted to is constant, and say that it is
*large*if . Ramsey’s theorem, provable in , states that for any there is an such that given any of size , any of size and any , there is a homogeneous for and of size at least . The function that to assigns the least such is well known to be primitive recursive (just about any standard proof of Ramsey theorem in any combinatorics book shows that it roughly grows exponentially). Paris and Harrington showed that if we also require that the homogeneous set be large, then the new statement is also true but the resulting function (clearly recursive) eventually dominates all functions provably recursive in .

### 117b – Homework 7

February 22, 2007### 117b – Undecidability and incompleteness – Lecture 10

February 22, 2007A theory *T* (r.e., extending *Q*) is *reflexive* iff it proves the consistency of all its finite subtheories. It is *essentially reflexive* iff each r.e. extension is reflexive.

Then *PA* is essentially reflexive and therefore no consistent extension of *PA* is finitely axiomatizable. This is obtained by showing that, in spite of Tarski’s undefinability of truth theorem, there are (provably in *PA*) *S ^{0}_{n}*–

*truth predicates*for all

*n*.

We define Rosser sentences and show their undecidability. We also show Löb’s theorem that if *T|-PA* is an r.e. theory and* T*|-Pr_{T}(j)®j, then *T*|- j. This gives another proof of the second incompleteness theorem.

Finally, we show that the length of proofs of P* ^{0}_{1}*-sentences is not bounded by any recursive function: For any

*T|-Q*r.e. and consistent, and any recursive function

*f*, there is a P

*-sentence j provable in*

^{0}_{1}*T*but such that any proof of j in

*T*has length >

*f(*j

*).*

### 117b – Undecidability and incompleteness – Lecture 9

February 20, 2007Let *S* and *T* be r.e. theories in the language of arithmetic. Assume that *S* is strong enough to binumerate all primitive recursive relations. This suffices to prove the *fixed point lemma*:

For any formula g(x) there is a sentence j such that *S* |- (j ↔g(j)),

moreover, j can be chosen of the same complexity as g.

The *(Löb)* *Derivability conditions* for *S,T* are the following three statements:

- For any j, if
*T*|-j, then*S*|- Pr_{T}(j). *S*|- For any j, y ( Pr_{T}(j®y) Ù Pr_{T}(j) ® Pr_{T}(y) ).*S*|-For any j ( Pr_{T}(j)®Pr_{T}(Pr_{T}(j)) ).

For example,* S=Q* suffices for the proof of the fixed point lemma and of the first derivability condition and *S=PA *suffices for the other two.

**Gödel incompleteness theorems: **

- Let
*S*be r.e. and strong enough to satisfy the fixed point lemma and the first derivability condition. Let*T*|-*S*be r.e. and consistent. Then there is a*true*P^{0}_{1}sentence j such that*T*does not prove j. If*T*is S^{0}_{1}-sound, then*T*does not prove ¬j either. - If in addition
*S*satisfies the other two derivability conditions, then*T*does not prove Con(*T*), the statement asserting the consistency of*T*.

As a corollary, *Q* is essentially undecidable: Not only it is undecidable, but any r.e. extension is undecidable as well. Gödel’s original statement replaced S^{0}_{1}-soundness with the stronger assumption of w–*consistency*.