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For all $f\colon \mathbb{N}^2 \rightarrow \mathbb{N}$ partial recursive there exists partial recursive $g\colon \mathbb{N} \rightarrow \mathbb{N}$ such that

a) $x \in \operatorname{Dom}(g) \Leftrightarrow \exists y[(x,y) \in \operatorname{Dom}(f)]$

b) $\exists y(x,y) \in \operatorname{Dom}(f) \Rightarrow (x, g(x)) \in \operatorname{Dom}(f)$

My solution is as follows: Let $g(x)= \mu_{y}((x, y ) \in \operatorname{Dom}(f))$, I claim this function is partial recursive because $\operatorname{Dom}(f)$ is recursive. Is this correct?

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  • $\begingroup$ Does the function $g$ need to be recursive or partially recursive? $\endgroup$ – Yuval Filmus Dec 4 '18 at 4:58
  • $\begingroup$ Partially is enough! $\endgroup$ – Tom Ryddle Dec 4 '18 at 5:06
  • $\begingroup$ Why do you think that the domain of $f$ is recursive? $\endgroup$ – Andrej Bauer Dec 4 '18 at 22:54
  • $\begingroup$ We know $f$ is a function recursive iff $Gra(f)$ is a set recursive. Then $Dom(f)=(\pi_1 Gra,\pi_2 Gra)$, isn't? $\endgroup$ – Tom Ryddle Dec 4 '18 at 23:07

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