Double Recursion Theorems

32 Chapter I. Recursive Enumerability and Recursivityra, if (x, y) is the nth term, then the (n + l)th term isThe first 13 terms of the enumeration are (0,0), (0,1), (1,0), (0,2),(1,1), (2,0), (0,3), (1,2), (2,1), (3,0), (0,4), (1,3), (2,2), (3,1),(4.0). We then define J(x,y) to be that number n such that (x,y)is the nth element of the enumeration.One can show that(see Exercise 4 below), and so the function J(x,y) is recursive. NotethatSince each number x is J(xi,xz) for exactly one pair (a;i,a;2), wedefine K(x) — x\ and L(x) — x%. Then J(Kx,Lx) — J(xi^x^) = a;,and so J(Kx,Lx} = x.Also, for any numbers x and y, if we let z = J(x, y), then Kz — xand Lz — y—in other words, KJ(x,y) = x and LJ(x,y) = y.Since J(x, y) is recursive, so are the functions Kx and Lx becauseandWe thus have:Proposition 8. There is a 1-1 recursive function J(x,y) and recursivefunctions Kx and Lx such that for all numbers x and y:(1) J(Kx,Lx) = x,(2) KJ(x,y) = x and LJ(x,y) = y.The functions Kx and Lx are called the inverse functions ofJ(x,y).Corollary. For any recursive function f(x,y), there is a recursivefunction 4>(x) such that for all x,y : f(x,y) =