30 Computability Theory function given by the pair of equations h(0) = m = 1 h(y + 1) = g(h(y), y) = h(y) · (y + 1). Using this pair of equations, we can proceed to calculate the values of the function h: h(0) = m = 1 h(1) = g(h(0), 0) = g(1, 0) = 1 h(2) = g(h(1), 1) = g(1, 1) = 2 h(3) = g(h(2), 2) = g(2, 2) = 6 h(4) = g(h(3), 3) = g(6, 3) = 24 And so forth. In order to calculate h(4), we first need to know h(3), and to find that we need h(2), and so on. The function h in this example is, of course, better known as the factorial function, h(x) = x!. It should be pretty clear that given any number m and any two-place function g, there exists a unique function h obtained by primitive recursion from g by using m.