Computability and Incompleteness by Avigad J.

By Avigad J.

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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.

Then for every x, we have f (x) = output(µs CompSeq(M, x, s)). This shows that f is partial recursive. Let me remind you that we have intentionally set issues of efficiency aside. In practice, it would be ludicrous to compute the function f above by checking each natural number, 0, 1, 2, . . to see if it codes a halting computation sequence of Turing machine M on the given input. The proof that every Turing computable function is partial recursive goes a long way towards explaining why we believe that every computable partial function is partial recursive.

Suppose h is the kth function in the enumeration; what can we say about h(k)? The following theorem hones in on the difference between the last two theorems. 7 Let f (k, x) = 1 if Un(k, x) is defined 0 otherwise. Then f is not computable. Since, in our construction, Un(k, x) is defined if and only if the Turing machine coded by k halts on input x, the theorem asserts that the question as to whether a given Turing machine halts on a given input is computationally undecidable. I will provide two proofs below.

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