A property of representable functions?












2












$begingroup$


Recall that a function $f:mathbb{N}^krightarrowmathbb{N}$ is representable in Peano arithmetic if there exists a formula $varphi(x_1,dots ,x_k,y)$ such that for every $n_1,dots,n_k,minmathbb{N}$,




  • If $m=f(n_1,dots,n_k)$ then $mathcal{A}vdash varphi(underline{n_1},dots ,underline{n_k},underline{m})$

  • If $mneq f(n_1,dots,n_k)$ then $mathcal{A}vdash negvarphi(underline{n_1},dots ,underline{n_k},underline{m})$


Here, $mathcal{A}$ is the theory (set of axioms) of Peano arithmetic, and $underline{n}$ is the term $S(S(dots S(0)dots))$ (with $n$ operations of $S$).



I have two related questions regarding this definition.
Suppose that $f$ is representable by a formula $varphi(x_1,dots ,x_k,y)$ (as defined above);




  1. Is it necessarily true that for every $n_1,dots,n_k$, $$mathcal{A}vdashexists !y(varphi(underline{n_1},dots ,underline{n_k},y))$$?

  2. Is it necessarily true that $$mathcal{A}vdashforall x_1cdotsforall x_kexists !y(varphi(x_1,dots ,x_n,y))$$?


I wanted to prove these by contradiction, i.e assume that $mathcal{A}vdashpsi$ doesn't hold, but the problem is that I can't deduce that $mathcal{A}vdashnegpsi$ since Peano arithmetic isn't complete.



Other than that I'm quite clueless so I'd appriciate your help.






first-order-logic peano-axioms







share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    Recall that a function $f:mathbb{N}^krightarrowmathbb{N}$ is representable in Peano arithmetic if there exists a formula $varphi(x_1,dots ,x_k,y)$ such that for every $n_1,dots,n_k,minmathbb{N}$,




    • If $m=f(n_1,dots,n_k)$ then $mathcal{A}vdash varphi(underline{n_1},dots ,underline{n_k},underline{m})$

    • If $mneq f(n_1,dots,n_k)$ then $mathcal{A}vdash negvarphi(underline{n_1},dots ,underline{n_k},underline{m})$


    Here, $mathcal{A}$ is the theory (set of axioms) of Peano arithmetic, and $underline{n}$ is the term $S(S(dots S(0)dots))$ (with $n$ operations of $S$).



    I have two related questions regarding this definition.
    Suppose that $f$ is representable by a formula $varphi(x_1,dots ,x_k,y)$ (as defined above);




    1. Is it necessarily true that for every $n_1,dots,n_k$, $$mathcal{A}vdashexists !y(varphi(underline{n_1},dots ,underline{n_k},y))$$?

    2. Is it necessarily true that $$mathcal{A}vdashforall x_1cdotsforall x_kexists !y(varphi(x_1,dots ,x_n,y))$$?


    I wanted to prove these by contradiction, i.e assume that $mathcal{A}vdashpsi$ doesn't hold, but the problem is that I can't deduce that $mathcal{A}vdashnegpsi$ since Peano arithmetic isn't complete.



    Other than that I'm quite clueless so I'd appriciate your help.






    first-order-logic peano-axioms







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      Recall that a function $f:mathbb{N}^krightarrowmathbb{N}$ is representable in Peano arithmetic if there exists a formula $varphi(x_1,dots ,x_k,y)$ such that for every $n_1,dots,n_k,minmathbb{N}$,




      • If $m=f(n_1,dots,n_k)$ then $mathcal{A}vdash varphi(underline{n_1},dots ,underline{n_k},underline{m})$

      • If $mneq f(n_1,dots,n_k)$ then $mathcal{A}vdash negvarphi(underline{n_1},dots ,underline{n_k},underline{m})$


      Here, $mathcal{A}$ is the theory (set of axioms) of Peano arithmetic, and $underline{n}$ is the term $S(S(dots S(0)dots))$ (with $n$ operations of $S$).



      I have two related questions regarding this definition.
      Suppose that $f$ is representable by a formula $varphi(x_1,dots ,x_k,y)$ (as defined above);




      1. Is it necessarily true that for every $n_1,dots,n_k$, $$mathcal{A}vdashexists !y(varphi(underline{n_1},dots ,underline{n_k},y))$$?

      2. Is it necessarily true that $$mathcal{A}vdashforall x_1cdotsforall x_kexists !y(varphi(x_1,dots ,x_n,y))$$?


      I wanted to prove these by contradiction, i.e assume that $mathcal{A}vdashpsi$ doesn't hold, but the problem is that I can't deduce that $mathcal{A}vdashnegpsi$ since Peano arithmetic isn't complete.



      Other than that I'm quite clueless so I'd appriciate your help.






      first-order-logic peano-axioms







      share|cite|improve this question









      $endgroup$




      Recall that a function $f:mathbb{N}^krightarrowmathbb{N}$ is representable in Peano arithmetic if there exists a formula $varphi(x_1,dots ,x_k,y)$ such that for every $n_1,dots,n_k,minmathbb{N}$,




      • If $m=f(n_1,dots,n_k)$ then $mathcal{A}vdash varphi(underline{n_1},dots ,underline{n_k},underline{m})$

      • If $mneq f(n_1,dots,n_k)$ then $mathcal{A}vdash negvarphi(underline{n_1},dots ,underline{n_k},underline{m})$


      Here, $mathcal{A}$ is the theory (set of axioms) of Peano arithmetic, and $underline{n}$ is the term $S(S(dots S(0)dots))$ (with $n$ operations of $S$).



      I have two related questions regarding this definition.
      Suppose that $f$ is representable by a formula $varphi(x_1,dots ,x_k,y)$ (as defined above);




      1. Is it necessarily true that for every $n_1,dots,n_k$, $$mathcal{A}vdashexists !y(varphi(underline{n_1},dots ,underline{n_k},y))$$?

      2. Is it necessarily true that $$mathcal{A}vdashforall x_1cdotsforall x_kexists !y(varphi(x_1,dots ,x_n,y))$$?


      I wanted to prove these by contradiction, i.e assume that $mathcal{A}vdashpsi$ doesn't hold, but the problem is that I can't deduce that $mathcal{A}vdashnegpsi$ since Peano arithmetic isn't complete.



      Other than that I'm quite clueless so I'd appriciate your help.






      first-order-logic peano-axioms




      first-order-logic peano-axioms






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      asked Jan 2 at 12:49









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