I pretty sure this is true but I couldn't find it stated in my text, so I just wanted to verify it.
Is the following true?
If $\phi$ is general recursive in $\Psi$ and $\Psi$ is general recursive, then $\phi$ is general recursive.
Asked
Active
Viewed 27 times
1 Answers
1
Yes, this is true (if a bit garbled - I think "each are" should be replaced with "is"). Informally, suppose I have a computable procedure $\pi$ for computing $A$ from $B$, and I separately have a computable procedure $\rho$ for building $B$. Then I'll compute $A$ as follows: build more and more of $B$ via $\rho$, and keep checking whether the amount of $B$ we've built so far is enough to determine - via $\pi$ - the desired "bit" of $A$. Eventually we'll see $\pi$ "make up its mind" since it can only use finitely many "bits" of $B$ in a single computation, and the resulting answer is the desired "bit" of $A$.
It's a good exercise to give a fully-detailed proof of this fact (basically, you'll "compose" a Turing reduction and a Turing computation).
Noah Schweber
- 245,398
-
I also assume that this is true for partial recursive functions (being the same class of functions as computable functions). So if $\phi$ is partial recursive in $\Psi$ and $\Psi$ is completely defined partial recursive, then $\phi$ is partial recursive. – BENG Jun 29 '19 at 03:45
-
@BENG Yes, but we have to be a bit careful here: recursion relative to a partial object is complicated. So the right thing to do at least as first is to always restrict attention to total objects, at least as far as the oracle goes. – Noah Schweber Jun 29 '19 at 03:46