Recursive and recursively enumerable sets.












0












$begingroup$


I am trying to solve the following problem :



Let $A = {n in mathbb{N} : mbox{ for } mbox{ all } xin mathbb{N}, mbox{ if } {phi}_{n}^{1}(x) mbox{ is } mbox{ defined } mbox{ then } {phi}_{n}^{1}(x) leq 10 }$.



1) Prove that $A$ and $mathbb{N} - A$ are not recursive.



2) Prove that $mathbb{N} - A$ is recursively enumerable and use it to explain why $A$ is not r.e.



I am not very familiar with this area. So any help would be really appreciated.



Thank you in advance!






logic computability







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What does $phi$ mean here ?
    $endgroup$
    – Peter
    May 30 '17 at 17:19






  • 1




    $begingroup$
    It is a fuction from $mathbb{N}$ to $mathbb{N}$
    $endgroup$
    – Angela
    May 30 '17 at 17:51






  • 2




    $begingroup$
    Surely $phi^1_n$ is an enumeration of the recursive functions $mathbb{N}tomathbb{N}$...
    $endgroup$
    – Alex Kruckman
    May 30 '17 at 17:53










  • $begingroup$
    Yes exactly! Do you have any hints for this problem?
    $endgroup$
    – Angela
    May 30 '17 at 23:33










  • $begingroup$
    What's your definition of a recursive function? There are a lot of equivalent definition out there and depending on which one you know, a given proof might be unrecognizable to you unless we work with similar notions.
    $endgroup$
    – Stefan Mesken
    May 31 '17 at 16:47


















0












$begingroup$


I am trying to solve the following problem :



Let $A = {n in mathbb{N} : mbox{ for } mbox{ all } xin mathbb{N}, mbox{ if } {phi}_{n}^{1}(x) mbox{ is } mbox{ defined } mbox{ then } {phi}_{n}^{1}(x) leq 10 }$.



1) Prove that $A$ and $mathbb{N} - A$ are not recursive.



2) Prove that $mathbb{N} - A$ is recursively enumerable and use it to explain why $A$ is not r.e.



I am not very familiar with this area. So any help would be really appreciated.



Thank you in advance!






logic computability







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What does $phi$ mean here ?
    $endgroup$
    – Peter
    May 30 '17 at 17:19






  • 1




    $begingroup$
    It is a fuction from $mathbb{N}$ to $mathbb{N}$
    $endgroup$
    – Angela
    May 30 '17 at 17:51






  • 2




    $begingroup$
    Surely $phi^1_n$ is an enumeration of the recursive functions $mathbb{N}tomathbb{N}$...
    $endgroup$
    – Alex Kruckman
    May 30 '17 at 17:53










  • $begingroup$
    Yes exactly! Do you have any hints for this problem?
    $endgroup$
    – Angela
    May 30 '17 at 23:33










  • $begingroup$
    What's your definition of a recursive function? There are a lot of equivalent definition out there and depending on which one you know, a given proof might be unrecognizable to you unless we work with similar notions.
    $endgroup$
    – Stefan Mesken
    May 31 '17 at 16:47
















0












0








0





$begingroup$


I am trying to solve the following problem :



Let $A = {n in mathbb{N} : mbox{ for } mbox{ all } xin mathbb{N}, mbox{ if } {phi}_{n}^{1}(x) mbox{ is } mbox{ defined } mbox{ then } {phi}_{n}^{1}(x) leq 10 }$.



1) Prove that $A$ and $mathbb{N} - A$ are not recursive.



2) Prove that $mathbb{N} - A$ is recursively enumerable and use it to explain why $A$ is not r.e.



I am not very familiar with this area. So any help would be really appreciated.



Thank you in advance!






logic computability







share|cite|improve this question











$endgroup$




I am trying to solve the following problem :



Let $A = {n in mathbb{N} : mbox{ for } mbox{ all } xin mathbb{N}, mbox{ if } {phi}_{n}^{1}(x) mbox{ is } mbox{ defined } mbox{ then } {phi}_{n}^{1}(x) leq 10 }$.



1) Prove that $A$ and $mathbb{N} - A$ are not recursive.



2) Prove that $mathbb{N} - A$ is recursively enumerable and use it to explain why $A$ is not r.e.



I am not very familiar with this area. So any help would be really appreciated.



Thank you in advance!






logic computability




logic computability






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited May 31 '17 at 16:41









Carl Mummert

67.1k7133251




67.1k7133251










asked May 30 '17 at 16:51









AngelaAngela

605




605








  • 1




    $begingroup$
    What does $phi$ mean here ?
    $endgroup$
    – Peter
    May 30 '17 at 17:19






  • 1




    $begingroup$
    It is a fuction from $mathbb{N}$ to $mathbb{N}$
    $endgroup$
    – Angela
    May 30 '17 at 17:51






  • 2




    $begingroup$
    Surely $phi^1_n$ is an enumeration of the recursive functions $mathbb{N}tomathbb{N}$...
    $endgroup$
    – Alex Kruckman
    May 30 '17 at 17:53










  • $begingroup$
    Yes exactly! Do you have any hints for this problem?
    $endgroup$
    – Angela
    May 30 '17 at 23:33










  • $begingroup$
    What's your definition of a recursive function? There are a lot of equivalent definition out there and depending on which one you know, a given proof might be unrecognizable to you unless we work with similar notions.
    $endgroup$
    – Stefan Mesken
    May 31 '17 at 16:47
















  • 1




    $begingroup$
    What does $phi$ mean here ?
    $endgroup$
    – Peter
    May 30 '17 at 17:19






  • 1




    $begingroup$
    It is a fuction from $mathbb{N}$ to $mathbb{N}$
    $endgroup$
    – Angela
    May 30 '17 at 17:51






  • 2




    $begingroup$
    Surely $phi^1_n$ is an enumeration of the recursive functions $mathbb{N}tomathbb{N}$...
    $endgroup$
    – Alex Kruckman
    May 30 '17 at 17:53










  • $begingroup$
    Yes exactly! Do you have any hints for this problem?
    $endgroup$
    – Angela
    May 30 '17 at 23:33










  • $begingroup$
    What's your definition of a recursive function? There are a lot of equivalent definition out there and depending on which one you know, a given proof might be unrecognizable to you unless we work with similar notions.
    $endgroup$
    – Stefan Mesken
    May 31 '17 at 16:47










1




1




$begingroup$
What does $phi$ mean here ?
$endgroup$
– Peter
May 30 '17 at 17:19




$begingroup$
What does $phi$ mean here ?
$endgroup$
– Peter
May 30 '17 at 17:19




1




1




$begingroup$
It is a fuction from $mathbb{N}$ to $mathbb{N}$
$endgroup$
– Angela
May 30 '17 at 17:51




$begingroup$
It is a fuction from $mathbb{N}$ to $mathbb{N}$
$endgroup$
– Angela
May 30 '17 at 17:51




2




2




$begingroup$
Surely $phi^1_n$ is an enumeration of the recursive functions $mathbb{N}tomathbb{N}$...
$endgroup$
– Alex Kruckman
May 30 '17 at 17:53




$begingroup$
Surely $phi^1_n$ is an enumeration of the recursive functions $mathbb{N}tomathbb{N}$...
$endgroup$
– Alex Kruckman
May 30 '17 at 17:53












$begingroup$
Yes exactly! Do you have any hints for this problem?
$endgroup$
– Angela
May 30 '17 at 23:33




$begingroup$
Yes exactly! Do you have any hints for this problem?
$endgroup$
– Angela
May 30 '17 at 23:33












$begingroup$
What's your definition of a recursive function? There are a lot of equivalent definition out there and depending on which one you know, a given proof might be unrecognizable to you unless we work with similar notions.
$endgroup$
– Stefan Mesken
May 31 '17 at 16:47






$begingroup$
What's your definition of a recursive function? There are a lot of equivalent definition out there and depending on which one you know, a given proof might be unrecognizable to you unless we work with similar notions.
$endgroup$
– Stefan Mesken
May 31 '17 at 16:47












1 Answer
1






active

oldest

votes


















1












$begingroup$

For 1) you can either use Rice's theorem or prove it directly by showing that some non-recursive set, for example, $K = {n in mathbb{N} mid varphi_x(x)!downarrow}$ is $m$-reducible to $mathbb{N} setminus A$, i.e., there exists a recursive function $f(x)$ such that $n in K Leftrightarrow f(n) in mathbb{N} setminus A$. Consider the following function
$$
g(x, y) = begin{cases}
11,& text{if } varphi_x(x)!downarrow,\
uparrow,& text{otherwise.}
end{cases}
$$
By s-m-n theorem there exists a recursive function $f(x)$ such that $varphi_{f(x)}(y) = g(x, y)$. We also have
$$
n in K Leftrightarrow exists y, (varphi_{f(n)}(y) = g(n, y) = 11) Leftrightarrow f(n) in mathbb{N} setminus A.
$$
This shows that $K leqslant_m mathbb{N}setminus A$, and since $K$ is not recursive, so is $mathbb{N}setminus A$. The complement of a recursive set is recursive, so $A$ also can't be recursive.



For 2), Post's theorem says that a set is recursive if both itself and its complement are recursively enumerable. So if we show that $mathbb{N}setminus A$ is r.e. we are done. Showing that $mathbb{N} setminus A$ is r.e. depends on your definition of an r.e. set. We have
$$
n in mathbb{N}setminus A Leftrightarrow exists x,
(varphi_n(x)!downarrow wedge varphi_n(x) > 10)
Leftrightarrow f(n)!downarrow,
$$
where $f(n) = mu z(z = (z_1, z_2) wedge varphi_n(z_1)!downarrow text{in $z_2$ steps} wedge varphi_n(z_1) > 10)$. Intuitively, the program for $f(n)$ at step $m$ runs $varphi_n(0), dots, varphi_n(m)$ each for $m$ steps and checks the condition $varphi_n(x) > 10$ for convergent computations. If it finds one it halts, otherwise it keeps searching. This shows that $mathbb{N} setminus A = mathrm{dom},f$ for some recursive $f$ and hence this set is r.e.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Your answer is very understanding and in detail. It helped me a lot to understand the whole idea of this problem. Thank you so much!
    $endgroup$
    – Angela
    Jun 8 '17 at 19:16













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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

For 1) you can either use Rice's theorem or prove it directly by showing that some non-recursive set, for example, $K = {n in mathbb{N} mid varphi_x(x)!downarrow}$ is $m$-reducible to $mathbb{N} setminus A$, i.e., there exists a recursive function $f(x)$ such that $n in K Leftrightarrow f(n) in mathbb{N} setminus A$. Consider the following function
$$
g(x, y) = begin{cases}
11,& text{if } varphi_x(x)!downarrow,\
uparrow,& text{otherwise.}
end{cases}
$$
By s-m-n theorem there exists a recursive function $f(x)$ such that $varphi_{f(x)}(y) = g(x, y)$. We also have
$$
n in K Leftrightarrow exists y, (varphi_{f(n)}(y) = g(n, y) = 11) Leftrightarrow f(n) in mathbb{N} setminus A.
$$
This shows that $K leqslant_m mathbb{N}setminus A$, and since $K$ is not recursive, so is $mathbb{N}setminus A$. The complement of a recursive set is recursive, so $A$ also can't be recursive.



For 2), Post's theorem says that a set is recursive if both itself and its complement are recursively enumerable. So if we show that $mathbb{N}setminus A$ is r.e. we are done. Showing that $mathbb{N} setminus A$ is r.e. depends on your definition of an r.e. set. We have
$$
n in mathbb{N}setminus A Leftrightarrow exists x,
(varphi_n(x)!downarrow wedge varphi_n(x) > 10)
Leftrightarrow f(n)!downarrow,
$$
where $f(n) = mu z(z = (z_1, z_2) wedge varphi_n(z_1)!downarrow text{in $z_2$ steps} wedge varphi_n(z_1) > 10)$. Intuitively, the program for $f(n)$ at step $m$ runs $varphi_n(0), dots, varphi_n(m)$ each for $m$ steps and checks the condition $varphi_n(x) > 10$ for convergent computations. If it finds one it halts, otherwise it keeps searching. This shows that $mathbb{N} setminus A = mathrm{dom},f$ for some recursive $f$ and hence this set is r.e.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Your answer is very understanding and in detail. It helped me a lot to understand the whole idea of this problem. Thank you so much!
    $endgroup$
    – Angela
    Jun 8 '17 at 19:16


















1












$begingroup$

For 1) you can either use Rice's theorem or prove it directly by showing that some non-recursive set, for example, $K = {n in mathbb{N} mid varphi_x(x)!downarrow}$ is $m$-reducible to $mathbb{N} setminus A$, i.e., there exists a recursive function $f(x)$ such that $n in K Leftrightarrow f(n) in mathbb{N} setminus A$. Consider the following function
$$
g(x, y) = begin{cases}
11,& text{if } varphi_x(x)!downarrow,\
uparrow,& text{otherwise.}
end{cases}
$$
By s-m-n theorem there exists a recursive function $f(x)$ such that $varphi_{f(x)}(y) = g(x, y)$. We also have
$$
n in K Leftrightarrow exists y, (varphi_{f(n)}(y) = g(n, y) = 11) Leftrightarrow f(n) in mathbb{N} setminus A.
$$
This shows that $K leqslant_m mathbb{N}setminus A$, and since $K$ is not recursive, so is $mathbb{N}setminus A$. The complement of a recursive set is recursive, so $A$ also can't be recursive.



For 2), Post's theorem says that a set is recursive if both itself and its complement are recursively enumerable. So if we show that $mathbb{N}setminus A$ is r.e. we are done. Showing that $mathbb{N} setminus A$ is r.e. depends on your definition of an r.e. set. We have
$$
n in mathbb{N}setminus A Leftrightarrow exists x,
(varphi_n(x)!downarrow wedge varphi_n(x) > 10)
Leftrightarrow f(n)!downarrow,
$$
where $f(n) = mu z(z = (z_1, z_2) wedge varphi_n(z_1)!downarrow text{in $z_2$ steps} wedge varphi_n(z_1) > 10)$. Intuitively, the program for $f(n)$ at step $m$ runs $varphi_n(0), dots, varphi_n(m)$ each for $m$ steps and checks the condition $varphi_n(x) > 10$ for convergent computations. If it finds one it halts, otherwise it keeps searching. This shows that $mathbb{N} setminus A = mathrm{dom},f$ for some recursive $f$ and hence this set is r.e.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Your answer is very understanding and in detail. It helped me a lot to understand the whole idea of this problem. Thank you so much!
    $endgroup$
    – Angela
    Jun 8 '17 at 19:16
















1












1








1





$begingroup$

For 1) you can either use Rice's theorem or prove it directly by showing that some non-recursive set, for example, $K = {n in mathbb{N} mid varphi_x(x)!downarrow}$ is $m$-reducible to $mathbb{N} setminus A$, i.e., there exists a recursive function $f(x)$ such that $n in K Leftrightarrow f(n) in mathbb{N} setminus A$. Consider the following function
$$
g(x, y) = begin{cases}
11,& text{if } varphi_x(x)!downarrow,\
uparrow,& text{otherwise.}
end{cases}
$$
By s-m-n theorem there exists a recursive function $f(x)$ such that $varphi_{f(x)}(y) = g(x, y)$. We also have
$$
n in K Leftrightarrow exists y, (varphi_{f(n)}(y) = g(n, y) = 11) Leftrightarrow f(n) in mathbb{N} setminus A.
$$
This shows that $K leqslant_m mathbb{N}setminus A$, and since $K$ is not recursive, so is $mathbb{N}setminus A$. The complement of a recursive set is recursive, so $A$ also can't be recursive.



For 2), Post's theorem says that a set is recursive if both itself and its complement are recursively enumerable. So if we show that $mathbb{N}setminus A$ is r.e. we are done. Showing that $mathbb{N} setminus A$ is r.e. depends on your definition of an r.e. set. We have
$$
n in mathbb{N}setminus A Leftrightarrow exists x,
(varphi_n(x)!downarrow wedge varphi_n(x) > 10)
Leftrightarrow f(n)!downarrow,
$$
where $f(n) = mu z(z = (z_1, z_2) wedge varphi_n(z_1)!downarrow text{in $z_2$ steps} wedge varphi_n(z_1) > 10)$. Intuitively, the program for $f(n)$ at step $m$ runs $varphi_n(0), dots, varphi_n(m)$ each for $m$ steps and checks the condition $varphi_n(x) > 10$ for convergent computations. If it finds one it halts, otherwise it keeps searching. This shows that $mathbb{N} setminus A = mathrm{dom},f$ for some recursive $f$ and hence this set is r.e.






share|cite|improve this answer









$endgroup$



For 1) you can either use Rice's theorem or prove it directly by showing that some non-recursive set, for example, $K = {n in mathbb{N} mid varphi_x(x)!downarrow}$ is $m$-reducible to $mathbb{N} setminus A$, i.e., there exists a recursive function $f(x)$ such that $n in K Leftrightarrow f(n) in mathbb{N} setminus A$. Consider the following function
$$
g(x, y) = begin{cases}
11,& text{if } varphi_x(x)!downarrow,\
uparrow,& text{otherwise.}
end{cases}
$$
By s-m-n theorem there exists a recursive function $f(x)$ such that $varphi_{f(x)}(y) = g(x, y)$. We also have
$$
n in K Leftrightarrow exists y, (varphi_{f(n)}(y) = g(n, y) = 11) Leftrightarrow f(n) in mathbb{N} setminus A.
$$
This shows that $K leqslant_m mathbb{N}setminus A$, and since $K$ is not recursive, so is $mathbb{N}setminus A$. The complement of a recursive set is recursive, so $A$ also can't be recursive.



For 2), Post's theorem says that a set is recursive if both itself and its complement are recursively enumerable. So if we show that $mathbb{N}setminus A$ is r.e. we are done. Showing that $mathbb{N} setminus A$ is r.e. depends on your definition of an r.e. set. We have
$$
n in mathbb{N}setminus A Leftrightarrow exists x,
(varphi_n(x)!downarrow wedge varphi_n(x) > 10)
Leftrightarrow f(n)!downarrow,
$$
where $f(n) = mu z(z = (z_1, z_2) wedge varphi_n(z_1)!downarrow text{in $z_2$ steps} wedge varphi_n(z_1) > 10)$. Intuitively, the program for $f(n)$ at step $m$ runs $varphi_n(0), dots, varphi_n(m)$ each for $m$ steps and checks the condition $varphi_n(x) > 10$ for convergent computations. If it finds one it halts, otherwise it keeps searching. This shows that $mathbb{N} setminus A = mathrm{dom},f$ for some recursive $f$ and hence this set is r.e.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jun 5 '17 at 12:00









Random JackRandom Jack

2,1161519




2,1161519












  • $begingroup$
    Your answer is very understanding and in detail. It helped me a lot to understand the whole idea of this problem. Thank you so much!
    $endgroup$
    – Angela
    Jun 8 '17 at 19:16




















  • $begingroup$
    Your answer is very understanding and in detail. It helped me a lot to understand the whole idea of this problem. Thank you so much!
    $endgroup$
    – Angela
    Jun 8 '17 at 19:16


















$begingroup$
Your answer is very understanding and in detail. It helped me a lot to understand the whole idea of this problem. Thank you so much!
$endgroup$
– Angela
Jun 8 '17 at 19:16






$begingroup$
Your answer is very understanding and in detail. It helped me a lot to understand the whole idea of this problem. Thank you so much!
$endgroup$
– Angela
Jun 8 '17 at 19:16




















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.mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid
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