The definition of partial recursive function says that given a partial recursive function $G(x,y)$, a new partial recursive function can be generated as$$F(x)\simeq\mu y[G(x,y)=0]$$
$F(x)$ is equal to the least $y$ such that $G(x,y)=0$ and for all $y'$ less than $y$, $G(x,y')$ is defined and is not zero, otherwise $F(x)$ is undefined. I don't think the $\forall y'
Why μ of partial recursive functions requires values of least zero's predecessors be defined
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computability
recursion
1 Answers
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The problem is that when you find the first $y$ such that $P(x, y) = 0$ using dovetailing technique, it doesn't mean that there is no $z < y$ with $P(x, z) = 0$. There are two cases:
- such $z < y$ exists, but the computation for $P(x, z)$ just takes more time than for $P(x, y)$ (and the only way to "check" this is to wait for the convergence of $P(x, t)$ for all $t < y$);
- there is no such $z$, and $y$ is indeed the least such number.
There is the halting problem inside. Let's show that there is a partial recursive $G(x, y)$ such that $F(x) = \min(y)[G(x, y) = 0]$ defined as "the least $y$ such that $G(x, y) = 0$" is not recursive.
Consider the following function $$ G(x, y) = \begin{cases} 0,& \text{if } y = 1 \text{ or } (y = 0 \text{ and } \varphi_x(x)\!\downarrow),\\ \uparrow,& \text{otherwise.} \end{cases} $$
Clearly, $$F(x) = \begin{cases} 0,& \text{if } \varphi_x(x)\!\downarrow,\\ 1,& \text{otherwise}, \end{cases}$$ which is the characteristic function of the halting set $K = \{x \mid \varphi_x(x)\!\downarrow\}$. Given that $K$ is not recursive, we have that $F(x)$ is not recursive too.