Proving an universal function for which there is computable total composition is principal
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Let $U(n, x)$ be an universal function for the class of computable functions. Let $c(p, q)$ be a total computable function such that $forall p, q, x : U(c(p, q), x) = U(p, U(q, x))$ — that is, given $U$-indices of any two functions, $c$ produces an $U$-index of their composition. It's easy to show that if $U$ is principal, then $c$ exists. But how can we prove the other direction: if $c$ exists, then $U$ defines a principal numbering of computable functions? That is, we need to prove that for every other computable $V(n, x)$ (not necessarily universal) there exists a total computable $s$ such that $forall n, x: V(n, x) = U(s(n), x)$.
logic computability
asked Dec 1 '18 at 17:26
0xd34df00d0xd34df00d
412212
$endgroup$
add a comment |
$begingroup$
Let $U(n, x)$ be an universal function for the class of computable functions. Let $c(p, q)$ be a total computable function such that $forall p, q, x : U(c(p, q), x) = U(p, U(q, x))$ — that is, given $U$-indices of any two functions, $c$ produces an $U$-index of their composition. It's easy to show that if $U$ is principal, then $c$ exists. But how can we prove the other direction: if $c$ exists, then $U$ defines a principal numbering of computable functions? That is, we need to prove that for every other computable $V(n, x)$ (not necessarily universal) there exists a total computable $s$ such that $forall n, x: V(n, x) = U(s(n), x)$.
logic computability
asked Dec 1 '18 at 17:26
0xd34df00d0xd34df00d
412212
$endgroup$
add a comment |
$begingroup$
Let $U(n, x)$ be an universal function for the class of computable functions. Let $c(p, q)$ be a total computable function such that $forall p, q, x : U(c(p, q), x) = U(p, U(q, x))$ — that is, given $U$-indices of any two functions, $c$ produces an $U$-index of their composition. It's easy to show that if $U$ is principal, then $c$ exists. But how can we prove the other direction: if $c$ exists, then $U$ defines a principal numbering of computable functions? That is, we need to prove that for every other computable $V(n, x)$ (not necessarily universal) there exists a total computable $s$ such that $forall n, x: V(n, x) = U(s(n), x)$.
logic computability
asked Dec 1 '18 at 17:26
0xd34df00d0xd34df00d
412212
$endgroup$
Let $U(n, x)$ be an universal function for the class of computable functions. Let $c(p, q)$ be a total computable function such that $forall p, q, x : U(c(p, q), x) = U(p, U(q, x))$ — that is, given $U$-indices of any two functions, $c$ produces an $U$-index of their composition. It's easy to show that if $U$ is principal, then $c$ exists. But how can we prove the other direction: if $c$ exists, then $U$ defines a principal numbering of computable functions? That is, we need to prove that for every other computable $V(n, x)$ (not necessarily universal) there exists a total computable $s$ such that $forall n, x: V(n, x) = U(s(n), x)$.
logic computability
logic computability
asked Dec 1 '18 at 17:26
0xd34df00d0xd34df00d
412212
asked Dec 1 '18 at 17:26
0xd34df00d0xd34df00d
412212
edited Dec 1 '18 at 17:50
0xd34df00d
asked Dec 1 '18 at 17:26
0xd34df00d0xd34df00d
412212
asked Dec 1 '18 at 17:26
0xd34df00d0xd34df00d
412212
asked Dec 1 '18 at 17:26
0xd34df00d0xd34df00d
412212
412212
add a comment |
add a comment |
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