Does the normal form theorem imply that every partially computabe function is primitive recursive?

Question

This is Normal Form Theorem (Second Edition of Computability, Complexity, and Languages written by Martin Davis page 75):

Let $f(x_1,...,x_n)$ be a partially computable function. Then there is a primitive recursive predicate $R(x_1,...,x_n,y)$ such that:

$f(x_1,...,x_n) = L(min R(x_1,...,x_n,z)_z)$ (minimization is on z)

So I think an immediate result of this theorem is that every partially computable function is primitive recursive and every primitive recursive function is partially computable. is it true?

Answer 1

No it not true. Primitive recursive functions are a subset of computable functions. The minimization operator from $\mu$-calculus is not primitive recursive and not guaranteed to terminate. The constantly false function for instance would cause minimization to loop infinitely. As you likely know primitive recursion is guaranteed to terminate.

What does seem to follow from this is that every computable function is $\mu$-recursive as primitive recursion is a subset of $\mu$-recursion. In fact I happen to know this to be true because $\mu$-recursion is turing complete.

You can check out primitive recursion and $\mu$-recursion on Wikipedia.

I also found a wikipedia page with some related matters. In particular they give another form of the Normal Form Theorem that seems slightly stronger; see here.

Answer 2

You read the theorem wrong; the right-hand side contains this $L(\min \dots)$ -- this is not primitive recursive! Note that, in particular, such $z$ does not necessarily exist (the "algorithm" will loop.)

$L(\min \_)$ in that theorem is also called the $\mu$-operator. It can not be expressed with primitive recursion.

This theorem states that one application of the $\mu$-operator is enough for Turing-completeness. In terms of WHILE programs, this means that you can program any function with using at most one WHILE loop.

Answer 3

The classical example of a computable function which is not primitive recursive is the Ackermann function. You can also construct such a function using diagonalization: given some effective enumeration $f_n$ of the primitive recursive functions, the function $n \mapsto f_n(n) + 1$ is computable but not primitive recursive.