I am a little bit confused with the unbounded minimization ($\mu$ operator of the $\mu$ recursive functions).

The $\mu$ operator is $\mu(f)(x) = \min(n | f(x, n) > 0)$ and the operator is said safe if we only apply it to functions such that for every $x$, $f(x, n) > 0$ for some $n$. In that case, every recursive function is total and every such function is computable with a Turing machine. The proof simulates every initial function and every operator with a Turing machine. Particularly, for $\mu$, the machine applies $f(x, n)$ for each $n \in \mathbb{N}$ until $f(x, n)$ is not 0.

This can be done because the operator is safe. The machine halts on every input.

When the operator is not safe, we can create partial functions as $\mu(f)(x)$ may be infinite (and thus undefined). Considering Turing machine, partial functions is not a problem as we can state that the machine does not stop if the input is not defined for the fonction. However, I cannot prove that the unbounded and unsafe minimization operator can be simulated by a Turing machine.

Indeed, if $f$ is partial, we can define $\mu$ as $\mu(f)(x) = \min(n | f(x, n)$ is defined and $f(x, n) > 0)$. In that case, there may exist $n = \mu(f)(x)$ and $m < n$ such that $f(x, m)$ is not defined. And, in that case, the machine that was used on the safe case does not halt anymore. Due to the halting problem indecidability, it is not possible to determine if $f(x, n)$ is defined or not.

So, my question is : Are the partial recursive functions (with unsafe operator $\mu$) equivalent to partial functions computable by Turing machines?

  • $\begingroup$ (Indeed. I made the change.) $\endgroup$
    – mouton5000
    Commented May 9, 2018 at 8:27

2 Answers 2


Unsafe $\mu$ is strictly stronger than partial TMs.

Indeed, define $f$ such that $f(x,0)=1$ if the $x$-th Turing machine halts on empty input, let $f(x,0)$ be undefined otherwise, and let $f(x,n)=1$ for all $n>0$. Then $f$ is a computable partial function. Let $g:=\mu(f)$. Then $g(x)=0$ if the $x$-th Turing machine halts and $g(x)=1$ otherwise. Hence $\mu(f)$ is not computable.


The closure of partial recursive functions under unsafe $\mu$ seems to be the arithmetical hierarchy. (In order to make that statement precise, we have to translate between partial functions and subsets of $\mathbb N$.)


No, your definition of $\mu$ makes $\mu f$ not partial recursive, in general, even when we assume that $f$ is partial recursive, as @kne shows above.

If you want a minimization operator such that $\mu f$ is always partial recursive when $f$ is, you can take

$$ \mu f(x) = \min\{n | f(x, n) > 0 \mbox{ and $\forall m<n.\ f(x, m)$ is defined} \} $$


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