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













Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2303041%2frecursive-and-recursively-enumerable-sets%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






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




















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2303041%2frecursive-and-recursively-enumerable-sets%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Bressuire

Image
.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 ...
Read more

Cabo Verde

Image
  Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
Read more

Gyllenstierna

Image
.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 ...
Read more