Tom Bogue, tbogue2 CS 579 Final Project
Recursive Functions and the Arithmetical Hierarchy
In this presentation, I will give a (very) brief introduction to the theory of recursive functions, describe the arithmetical hierarchy, and prove that it does not collapse. I will draw connections to decision theory and the polynomial hierarchy in the process. Many of the definitions, examples, and proofs I will use are thanks to Anush Tserunyan’s course notes1. Recursive Functions
We will begin by defining the search operator µ:
Definition 1: (Search Operator) For a relation R Nk+1, and for each a Nk,definethesearch operator ✓ 2 µx(R(a, x)) = min x N : R(a, x) holds . If, for every a Nk,thereisax N such that R(a, x) holds, we say the search is{ well-defined2 . } 2 2
This is enough to allow us now to define recursive functions.
Definition 2: (Recursive Functions) Consider a function f : Nk N.Wesayf is recursive if it can be obtained inductively from the following rules: !
(1. Primitives) The following functions are recursive:
Addition and multiplication, that is, + : N2 N, and : N2 N, • ! · ! 1ifx y : N2 N,where (x, y)= • ! 0 otherwise ⇢ k k k Projections P : N N defined as P (x1,...,xk)=xi, for each i 1,...,k . • i ! i 2{ }
m k (2. Composition) If g : N N and h1,...,hm : N N are all recursive functions, then so is their ! k ! composition f = g(h1,...,hm): N N,definedbyf(a)=g(h1(a),...,hm(a)). ! (3. Well-defined search) If g : Nk+1 N is recursive, and for every a Nk, there is an x N such that ! 2 2 g(a, x) = 0, then f : Nk N defined by f(a)=µx(g(a, x) = 0) is recursive. ! k k A relation R N is called recursive if its characteristic function R : N 0, 1 is recursive. Let Rec be the class containing✓ all recursive relations. !{ }
To get a feel for the power of recursive functions, we will show the following:
Lemma 1.a: The relations N2 and = N2 are recursive. ✓ ✓ 2 2 Proof: (x, y)= (P2 (x, y),P1 (x, y)). =(x, y)= (x, y) (x, y). ⌅ · Lemma 1.b: The constant functions Ck : Nk N,definedbyCk(a)=n for all a Nk, are recursive. n ! n 2 k k+1 k k+1 k+1 Proof: By induction. C0 (a)=µx(P (a, x) = 0). Cn+1(a)=µx( (Cn (a, x),P (a, x)) = 0). ⌅ k+1 k+1 1http://www.math.uiuc.edu/~anush/2015F:Math570--Logic/logic_lectures.pdf particularly sections 6.C, 7.A, and 7.B
1 Lemma 1.c: If P, Q Nk are recursive relations, then so are P = Nk P and P Q = P Q. ✓ ¬ \ ^ \ k Proof: P (a)= =( P (a),C0 (a)). P Q(a)= P (a) Q(a). ⌅ ¬ ^ · Lemma 1.c shows us that Rec is closed under complementation and intersections (and thus unions.) Notice that this allows us to put any relation in the search operator as long as the search is well-defined.
k+1 Lemma 1.d: (Bounded quantification) If R N is recursive, then so is Q(a, y)= x