117b – Some additional remarks on provably recursive functions

February 28, 2007

There are quite a few examples of natural combinatorial statements not provable in mbox{sf PA} 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 mbox{sf PA}. 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:

  • F_0(n)=n+1.
  • F_{alpha+1}(n)=F_alpha^n(n), where the superindex means iterated n times.
  • F_lambda(n)=F_{lambda(n)}(n), if lambda is limit.

Here, {lambda(n): n =0,1,dots} is some (natural) sequence converging to lambda (if you are not aware of ordinals, ignore for the moment what the last clause means).

For example, F_1(n)=2n, F_2(n)=2^n n, F_3(n) grows like a stack of n powers of 2 and F_4(n) grows like an iterated stack of powers of 2, there being n such iterations. It is easy to see that each F_n is primitive recursive and strictly increasing, that n<m implies that F_n is eventually dominated by F_m (i.e., that F_n(k)<F_m(k) for all but finitely many values of k), and that every primitive recursive function is eventually dominated by some F_n. Thus, if a function eventually dominates each F_n, 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 X 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 n to denote the order type of any linearly ordered set of n elements, and use omega to represent the order type of the natural numbers. We `add’ two order types alpha and beta by forming the disjoint union of a set of order type alpha followed by a set of order type beta. We also say that alpha is smaller than beta if any set of order type beta has a (proper) initial segment of order type alpha. We can then continue the sequence 0,1,2,dots of the natural numbers, as follows: 

omega,omega+1,omega+2,dots,omega 2,omega 2+1,dots,omega 3,dots,omega^2,dots,omega^3,dots

where we use omega2 to represent omega+omega, etc, omega^2 to represent omegaomega and so on, and we can continue even beyond:

omega^omega,dots,omega^{omega^omega},dots

Call epsilon_0 (epsilon-0) the first ordinal alpha obtained this way such that

alpha=omega^alpha (so epsilon_0=omega^{omega^{omega^{...}}}).

One can show that any ordinal below epsilon_0 admits a unique `pure base omega‘ representation, from which there is a natural way of defining the sequence {lambda(n):n=0,1,dots} converging to lambda whenever lambda is a limit, which means that it is not of the form alpha+1 for any alpha. For example,

omega,omega 3,omega^{omega^2 3+omega 17+8} 5+omega^3 2, etc,

are all limits. So, for example, the sequence {lambda(n): n =0,1,dots} corresponding to lambda=omega is just lambda(n)=n and the corresponding function F_omega is easily seen to grow as the (diagonal) Ackermann function A(n,n); the sequence corresponding to lambda=omega^3 4 is lambda(n)=omega^3 3+omega^2 n, etc. As before, each F_alpha is (eventually) strictly increasing and if alpha<beta then F_beta eventually dominates F_alpha. Each F_alpha, for alpha<epsilon_0 (and in fact, for many more alpha) is recursive.

The relation of this sequence with mbox{sf PA} lies in that any function provably recursive in mbox{sf PA} is eventually dominated by one of the F_alpha with alpha<epsilon_0. So, one purely combinatorial way of showing that a function is not provably recursive is to show that it eventually dominates each such F_alpha. This kind of analysis was developed by Solovay and Ketonen.

There are also other ways of relating mbox{sf PA} to this number epsilon_0, 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, 2007

Let Tvdash Q be r.e. and suppose there is a sentence varphi such that T does not prove varphi. Then T+varphi is more powerful than T in two ways:  

  1. It proves more theorems than T; for example,  varphi.
  2. It provides much shorter proofs of theorems of T; for example: For any (total) recursive function f there is a sentence theta provable in T but such that in T+varphi there is a proof p of theta such that no proof of theta in T is shorter than f(p).

This leads to considering those recursive functions f of which T can prove that they are total. One calls such functions provably recursive in T. It is easy to show that for any Sigma^0_1-sound Tvdashmbox{sf PA}, there are recursive functions that are not provably recursive in T. The statement that f is total,

forall xexists y,f(x)=y,

is easily seen to be Pi^0_2 and for T=mbox{sf PA} one can provide natural examples of such recursive but not provably recursive functions f, thus providing natural (combinatorial) examples of independent Pi^0_2-sentences. Actually, this can also be done for much stronger systems than mbox{sf PA}, like mbox{sf ZFC}, and Harvey Friedman has provided several such examples.

For mbox{sf PA}, the most famous examples are perhaps the following:

  • Goodstein (1944). The pure base-n representation of a number m is obtained by writting m in base n, writing the exponents of this representation also in base n, writing the exponents of these representations in base n, and so on. For nge2 let G_n(0)=0 and for mge1 let G_n(m) be the result of writing the pure base-n representation of m, replacing each n by n+1, and subtracting 1. The Goodstein sequence of m is the sequence m, G_2(m), G_3(G_2(m)), G_4(G_3(G_2(m))),dots This sequence eventually converges to 0. The function F that to m assigns the number of steps necessary for this to occur is recursive but not provably recursive in mbox{sf PA}. In fact, given any f provably recursive in mbox{sf PA}, F eventually dominates f.
  • 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 n, 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 n 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 M 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 M=F(n) to a given hydra n is recursive but eventually dominates all functions provably recursive in mbox{sf PA}.
  • Paris, Harrington (1977). Given a set A, let A^{[n]} denote the collection of its subsets of size n. Given a function f:A^{[n]}to B, say that a subset C of A is homogeneous for f iff f restricted to C^{[n]} is constant, and say that it is large if |C|gemin(C). Ramsey’s theorem, provable in mbox{sf PA}, states that for any c,n,b there is an M such that given any A of size M, any B of size b and any f:A^{[n]}to B, there is a C homogeneous for f and of size at least c. The function that to c,n,b assigns the least such M 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 C be large, then the new statement is also true but the resulting function (clearly recursive) eventually dominates all functions provably recursive in mbox{sf PA}.

117b – Homework 7

February 22, 2007

Due Thursday, March 1 at 1pm.

Homework 7


117b – Undecidability and incompleteness – Lecture 10

February 22, 2007

A theory T (r.e., extending {sf 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 {sf PA} is essentially reflexive and therefore no consistent extension of {sf PA} is finitely axiomatizable. This is obtained by showing that, in spite of Tarski’s undefinability of truth theorem, there are (provably in {sf PA}Sigma^0_ntruth predicates for all n.

We defined Rosser sentences and showed their undecidability. We also showed Löb’s theorem that if Tvdash{sf PA} is an r.e. theory and Tvdash{rm Pr}_T(varphi)tovarphi, then Tvdashvarphi. This gives another proof of the second incompleteness theorem.

Finally, we showed that the length of proofs of Pi^0_1-sentences is not bounded by any recursive function: For any Tvdash {sf Q} r.e. and consistent, and any recursive function f, there is a Pi^0_1-sentence varphi provable in T but such that any proof of varphi in T has length > f(varphi).


117b – Undecidability and incompleteness – Lecture 9

February 20, 2007

Let 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 gamma(x) there is a sentence varphi  such that Svdash(varphileftrightarrowgamma(varphi)),

moreover, varphi can be chosen of the same complexity as gamma.

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

  • For any varphi, if Tvdashvarphi, then Svdash{rm Pr}_T(varphi).
  • Svdashforallvarphi,psi,({rm Pr}_T(varphitopsi)land {rm Pr}_T(varphi)longrightarrow{rm Pr}_T(psi)). 
  • Svdashforallvarphi,({rm Pr}_T(varphi)to{rm Pr}_T({rm Pr}_T(varphi))).

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

Gödel incompleteness theorems:

  1. Let S be r.e. and strong enough to satisfy the fixed point lemma and the first derivability condition. Let Tvdash S be r.e. and consistent. Then there is a true Pi^0_1 sentence varphi such that T does not prove varphi. If T is Sigma^0_1-sound, then T does not prove lnotvarphi either.
  2. If in addition S satisfies the other two derivability conditions, then T does not prove {rm Con}(T), the statement asserting the consistency of T.

As a corollary, {sf Q} is essentially undecidable: Not only it is undecidable, but any r.e. extension is undecidable as well. Gödel’s original statement replaced Sigma^0_1-soundness with the stronger assumption of omegaconsistency.


117b- Homework 6

February 15, 2007

Due Thursday, February 22 at 1pm.

Homework 6


117b – Undecidability and incompleteness – Lecture 8

February 15, 2007

We introduced Robinson’s arithmetic {sf Q}. It is a weak (and sound) theory that proves all true Sigma^0_1 sentences. We also introduced Peano Arithmetic {sf PA}. {sf PA} is sound and designed to “capture” all true first order statements about {mathbb N}. (The incompleteness theorems will show that it does not suceed: There are true statements that {sf PA} does not prove.)

We showed that {sf Q} represents all c.e. sets and binumerates all recursive sets, in particular all primitive recursive functions. {sf PA} actually proves that the representations of primitive recursive functions define functions, so in any (maybe non-standard) model of {sf PA} it makes sense to talk, for example, of the exponential function, or the function enumerating the primes.

We proved the fixed point lemma and deduced from it Tarski’s theorem on the undefinability of truth, thus formally solving the liar’s paradox.