Is the set {i | $phi_i$ is total } recursively enumerable?
$begingroup$
Is the set {i | $phi_i$ is total } recursively enumerable?
and can you please tell me why ?
Bests
Norman
computability
asked Jan 5 at 11:56
NormanNorman
257
$endgroup$
add a comment |
$begingroup$
Is the set {i | $phi_i$ is total } recursively enumerable?
and can you please tell me why ?
Bests
Norman
computability
asked Jan 5 at 11:56
NormanNorman
257
$endgroup$
add a comment |
$begingroup$
Is the set {i | $phi_i$ is total } recursively enumerable?
and can you please tell me why ?
Bests
Norman
computability
asked Jan 5 at 11:56
NormanNorman
257
$endgroup$
Is the set {i | $phi_i$ is total } recursively enumerable?
and can you please tell me why ?
Bests
Norman
computability
computability
asked Jan 5 at 11:56
NormanNorman
257
asked Jan 5 at 11:56
NormanNorman
257
asked Jan 5 at 11:56
NormanNorman
257
asked Jan 5 at 11:56
NormanNorman
257
asked Jan 5 at 11:56
NormanNorman
257
257
add a comment |
add a comment |
1 Answer
1
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$begingroup$
No, its not by the theorem of Rice-Shapiro:
Let $A$ be a class of monadic partial recursive functions whose corresponding
index set ${rm prog}(A) = {xin{Bbb N}_0mid phi_xin A}$ is recursively enumerable. Then a monadic partial recursive function $f$ lies in $A$ iff there is a finite function $g in A$ such that $g subseteq f$ (i.e., $f$ is an extension of $g$).
If the set $A={xmid phi_xmbox{ total}}$ given is r.e., it will contain finite functions (functions with finite domain), which is not possible.
answered Jan 5 at 12:16
WuestenfuxWuestenfux
4,7921513
$endgroup$
$begingroup$
Thank you so much
$endgroup$
– Norman
Jan 5 at 12:42
add a comment |
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1 Answer
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1 Answer
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$begingroup$
No, its not by the theorem of Rice-Shapiro:
Let $A$ be a class of monadic partial recursive functions whose corresponding
index set ${rm prog}(A) = {xin{Bbb N}_0mid phi_xin A}$ is recursively enumerable. Then a monadic partial recursive function $f$ lies in $A$ iff there is a finite function $g in A$ such that $g subseteq f$ (i.e., $f$ is an extension of $g$).
If the set $A={xmid phi_xmbox{ total}}$ given is r.e., it will contain finite functions (functions with finite domain), which is not possible.
answered Jan 5 at 12:16
WuestenfuxWuestenfux
4,7921513
$endgroup$
$begingroup$
Thank you so much
$endgroup$
– Norman
Jan 5 at 12:42
add a comment |
$begingroup$
No, its not by the theorem of Rice-Shapiro:
Let $A$ be a class of monadic partial recursive functions whose corresponding
index set ${rm prog}(A) = {xin{Bbb N}_0mid phi_xin A}$ is recursively enumerable. Then a monadic partial recursive function $f$ lies in $A$ iff there is a finite function $g in A$ such that $g subseteq f$ (i.e., $f$ is an extension of $g$).
If the set $A={xmid phi_xmbox{ total}}$ given is r.e., it will contain finite functions (functions with finite domain), which is not possible.
answered Jan 5 at 12:16
WuestenfuxWuestenfux
4,7921513
$endgroup$
$begingroup$
Thank you so much
$endgroup$
– Norman
Jan 5 at 12:42
add a comment |
$begingroup$
No, its not by the theorem of Rice-Shapiro:
Let $A$ be a class of monadic partial recursive functions whose corresponding
index set ${rm prog}(A) = {xin{Bbb N}_0mid phi_xin A}$ is recursively enumerable. Then a monadic partial recursive function $f$ lies in $A$ iff there is a finite function $g in A$ such that $g subseteq f$ (i.e., $f$ is an extension of $g$).
If the set $A={xmid phi_xmbox{ total}}$ given is r.e., it will contain finite functions (functions with finite domain), which is not possible.
answered Jan 5 at 12:16
WuestenfuxWuestenfux
4,7921513
$endgroup$
No, its not by the theorem of Rice-Shapiro:
Let $A$ be a class of monadic partial recursive functions whose corresponding
index set ${rm prog}(A) = {xin{Bbb N}_0mid phi_xin A}$ is recursively enumerable. Then a monadic partial recursive function $f$ lies in $A$ iff there is a finite function $g in A$ such that $g subseteq f$ (i.e., $f$ is an extension of $g$).
If the set $A={xmid phi_xmbox{ total}}$ given is r.e., it will contain finite functions (functions with finite domain), which is not possible.
answered Jan 5 at 12:16
WuestenfuxWuestenfux
4,7921513
answered Jan 5 at 12:16
WuestenfuxWuestenfux
4,7921513
answered Jan 5 at 12:16
WuestenfuxWuestenfux
4,7921513
answered Jan 5 at 12:16
WuestenfuxWuestenfux
4,7921513
4,7921513
$begingroup$
Thank you so much
$endgroup$
– Norman
Jan 5 at 12:42
add a comment |
$begingroup$
Thank you so much
$endgroup$
– Norman
Jan 5 at 12:42
$begingroup$
Thank you so much
$endgroup$
– Norman
Jan 5 at 12:42
$begingroup$
Thank you so much
$endgroup$
– Norman
Jan 5 at 12:42
add a comment |
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