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can anyone explain this concept to me? and help me out in figuring out the answer? as I don't quite get it yet but any help will do, thank you

There exists two really cool primitive recursive functions: T(z, x1, x2, ..., xn, y) returns 0 if z is an encoding of a Turing Machine, and its computation with inputs x1, ... xn encodes to y. The function returns 1 otherwise. U(y) returns the result of the computation y.

Goal:Use these two functions and the µ(mu) operator to get a definition of a function f(x1, x2, ...xn) which returns the output of the TM encoded by z

what I have so far:

using these two functions and the µ operator we can then generate a definition for the function f(x1,x2, ...xn) that returns the output of the TM encoded by z such that we have a TM that we can define as T(n) where this computes the number of primes which are less than or equal n. z = 0 which is the number of primes found, y = 1 which is for each y we test whether it is prime, while k <= n:, j = 1 are the possible divisors of y, d = 0 the number of divisors of k found, while j <= n: if y % j == 0:, d = d + 1, j = j + 1, if d== 2:, z= z+1,y=y+1, then return z which is the output of the TM encoded by z

DeCastroAj
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  • This question is missing a lot of notational context, but it sure looks like mu y(T(z,x1,...,xn, y)=0) is the encoding of a TM that simulates machine z running on inputs x1,...,xn. Therefore U( mu y(T(z,x1,...,xn)=0) ) is the desired function. Or am my missing something? – Gene May 02 '17 at 03:12
  • This question is for https://cs.stackexchange.com – HEKTO May 02 '17 at 19:56
  • Please, check the text of your exercise. Usually, T is for Kleene's function, while U is the universal machine. If that is the case, search the web for kleene's normal form. – Andrea Asperti May 04 '17 at 05:58

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