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**Extra resources for Computability and incompleteness. Lecture notes**

**Sample text**

What is the lambda calculus doing in a chapter on models of computation? The point is that it does provide us with a model of the computable functions, although, at first, it is not even clear how to make sense of this statement. To talk about computability on the natural numbers, we need to find a suitable representation for such numbers. Here is one that works surprisingly well. 3 For each natural number n, define the numeral n to be the lambda term λxy (x(x(x(. . x(y))))), where there are n x’s in all.

Mn−1 has no normal form otherwise. 52 CHAPTER 2. 5 A function f is a partial computable function if and only if it is represented by a lambda term. This theorem is somewhat striking. As a model of computation, the lambda calculus is a rather simple calculus; the only operations are lambda abstraction and application! From these meager resources, however, it is possible to implement any computational procedure. The “if” part of the theorem is easier to see: suppose a function, f , is represented by a lambda term X.

G k−1 (x0 , . . , xl−1 )). In other words, the language of the lambda calculus is well suited to represent composition as well. When it comes to primitive recursion, we finally need to do some work. We will have to proceed in stages. As before, on the assumption that we already have terms g and h representing functions g and h, respectively, we want a term f representing the function f defined by f (0, z) = g(z) f (x + 1, z) = h(z, f (x, z), z). So, in general, given lambda terms G and H , it suffices to find a term F such that F (0, z) ≡ G (z) F (n + 1, z) ≡ H (n, F (n, z), z) for every natural number n; the fact that G and H represent g and h means that whenever we plug in numerals m for z, F (n + 1, m) will normalize to the right answer.