Is the set {i | $phi_i$ is total } recursively enumerable?












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Is the set {i | $phi_i$ is total } recursively enumerable?



and can you please tell me why ?



Bests
Norman










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    0












    $begingroup$


    Is the set {i | $phi_i$ is total } recursively enumerable?



    and can you please tell me why ?



    Bests
    Norman










    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Is the set {i | $phi_i$ is total } recursively enumerable?



      and can you please tell me why ?



      Bests
      Norman










      share|cite|improve this question









      $endgroup$




      Is the set {i | $phi_i$ is total } recursively enumerable?



      and can you please tell me why ?



      Bests
      Norman







      computability






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 5 at 11:56









      NormanNorman

      257




      257






















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          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.






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          • $begingroup$
            Thank you so much
            $endgroup$
            – Norman
            Jan 5 at 12:42











          Your Answer





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          1 Answer
          1






          active

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          1 Answer
          1






          active

          oldest

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          active

          oldest

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          active

          oldest

          votes









          0












          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you so much
            $endgroup$
            – Norman
            Jan 5 at 12:42
















          0












          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you so much
            $endgroup$
            – Norman
            Jan 5 at 12:42














          0












          0








          0





          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$



          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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 5 at 12:16









          WuestenfuxWuestenfux

          4,7921513




          4,7921513












          • $begingroup$
            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




          $begingroup$
          Thank you so much
          $endgroup$
          – Norman
          Jan 5 at 12:42


















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