First Order predicate logic exponents

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I want to "translate" an exponent function in predicate logic, but Im having some trouble, because the structure I have is $(mathbb R, +, times)$.



And if for example I want to represent $x=5^y$ ,
I dont have $5$ (which however I managed to get), but how can I represent the multiplication of $5 y$ times? ($y$ can be $0$ or larger)



Thank you.







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  • I think that it is possible, but quite compelx... See Richard Kaye, Models of Peano arithmetic, page 67.
    – Mauro ALLEGRANZA
    Jan 11 '17 at 13:05










  • May you give us some more context ? Book, page,...
    – Mauro ALLEGRANZA
    Jan 11 '17 at 13:06














up vote
0
down vote

favorite












I want to "translate" an exponent function in predicate logic, but Im having some trouble, because the structure I have is $(mathbb R, +, times)$.



And if for example I want to represent $x=5^y$ ,
I dont have $5$ (which however I managed to get), but how can I represent the multiplication of $5 y$ times? ($y$ can be $0$ or larger)



Thank you.







share|cite|improve this question






















  • I think that it is possible, but quite compelx... See Richard Kaye, Models of Peano arithmetic, page 67.
    – Mauro ALLEGRANZA
    Jan 11 '17 at 13:05










  • May you give us some more context ? Book, page,...
    – Mauro ALLEGRANZA
    Jan 11 '17 at 13:06












up vote
0
down vote

favorite









up vote
0
down vote

favorite











I want to "translate" an exponent function in predicate logic, but Im having some trouble, because the structure I have is $(mathbb R, +, times)$.



And if for example I want to represent $x=5^y$ ,
I dont have $5$ (which however I managed to get), but how can I represent the multiplication of $5 y$ times? ($y$ can be $0$ or larger)



Thank you.







share|cite|improve this question














I want to "translate" an exponent function in predicate logic, but Im having some trouble, because the structure I have is $(mathbb R, +, times)$.



And if for example I want to represent $x=5^y$ ,
I dont have $5$ (which however I managed to get), but how can I represent the multiplication of $5 y$ times? ($y$ can be $0$ or larger)



Thank you.









share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 9 at 18:09









Bram28

55.2k33982




55.2k33982










asked Jan 11 '17 at 10:46









vedsil

566




566











  • I think that it is possible, but quite compelx... See Richard Kaye, Models of Peano arithmetic, page 67.
    – Mauro ALLEGRANZA
    Jan 11 '17 at 13:05










  • May you give us some more context ? Book, page,...
    – Mauro ALLEGRANZA
    Jan 11 '17 at 13:06
















  • I think that it is possible, but quite compelx... See Richard Kaye, Models of Peano arithmetic, page 67.
    – Mauro ALLEGRANZA
    Jan 11 '17 at 13:05










  • May you give us some more context ? Book, page,...
    – Mauro ALLEGRANZA
    Jan 11 '17 at 13:06















I think that it is possible, but quite compelx... See Richard Kaye, Models of Peano arithmetic, page 67.
– Mauro ALLEGRANZA
Jan 11 '17 at 13:05




I think that it is possible, but quite compelx... See Richard Kaye, Models of Peano arithmetic, page 67.
– Mauro ALLEGRANZA
Jan 11 '17 at 13:05












May you give us some more context ? Book, page,...
– Mauro ALLEGRANZA
Jan 11 '17 at 13:06




May you give us some more context ? Book, page,...
– Mauro ALLEGRANZA
Jan 11 '17 at 13:06










1 Answer
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Kurt Godel showed how you could do this using $beta$-functions. If you really want to know about this, look for this in the literature, but the rough idea is that for your function $x = 5^y$ we can consider the sequence $5^0,5^1,5^2,...,5^y,...$, and if you have some kind of upper bound to this, i.e. the sequence is $5^0,5^1,5^2,...,5^y,...,5^n$ then it turns out that that finite sequence can be encoded with only two numbers, and the entries of the sequence (i.e. the $5^y$ values that you want) can be recovered using those two numbers and basic addition and multiplication operations (actually, you also do need either a successor function or a constant $1$).



But as Mauro says it is rather complicated, and probably not something you want to use for any kind of practical use. If you want to use exponentiation in practice, I would bite the bullet and just introduce a new symbol for exponentiation ... but at least you can define it recursively in terms of $+$ and $times$ ... and maybe that's acceptable to you:



$forall x : x^0 = 1$ (or $forall x : x^0 = s(0)$ if you have $s$ instead of $1$)



$forall x forall y : x^y+1 = x^y*x$






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    1 Answer
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    active

    oldest

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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    1
    down vote



    accepted










    Kurt Godel showed how you could do this using $beta$-functions. If you really want to know about this, look for this in the literature, but the rough idea is that for your function $x = 5^y$ we can consider the sequence $5^0,5^1,5^2,...,5^y,...$, and if you have some kind of upper bound to this, i.e. the sequence is $5^0,5^1,5^2,...,5^y,...,5^n$ then it turns out that that finite sequence can be encoded with only two numbers, and the entries of the sequence (i.e. the $5^y$ values that you want) can be recovered using those two numbers and basic addition and multiplication operations (actually, you also do need either a successor function or a constant $1$).



    But as Mauro says it is rather complicated, and probably not something you want to use for any kind of practical use. If you want to use exponentiation in practice, I would bite the bullet and just introduce a new symbol for exponentiation ... but at least you can define it recursively in terms of $+$ and $times$ ... and maybe that's acceptable to you:



    $forall x : x^0 = 1$ (or $forall x : x^0 = s(0)$ if you have $s$ instead of $1$)



    $forall x forall y : x^y+1 = x^y*x$






    share|cite|improve this answer
























      up vote
      1
      down vote



      accepted










      Kurt Godel showed how you could do this using $beta$-functions. If you really want to know about this, look for this in the literature, but the rough idea is that for your function $x = 5^y$ we can consider the sequence $5^0,5^1,5^2,...,5^y,...$, and if you have some kind of upper bound to this, i.e. the sequence is $5^0,5^1,5^2,...,5^y,...,5^n$ then it turns out that that finite sequence can be encoded with only two numbers, and the entries of the sequence (i.e. the $5^y$ values that you want) can be recovered using those two numbers and basic addition and multiplication operations (actually, you also do need either a successor function or a constant $1$).



      But as Mauro says it is rather complicated, and probably not something you want to use for any kind of practical use. If you want to use exponentiation in practice, I would bite the bullet and just introduce a new symbol for exponentiation ... but at least you can define it recursively in terms of $+$ and $times$ ... and maybe that's acceptable to you:



      $forall x : x^0 = 1$ (or $forall x : x^0 = s(0)$ if you have $s$ instead of $1$)



      $forall x forall y : x^y+1 = x^y*x$






      share|cite|improve this answer






















        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        Kurt Godel showed how you could do this using $beta$-functions. If you really want to know about this, look for this in the literature, but the rough idea is that for your function $x = 5^y$ we can consider the sequence $5^0,5^1,5^2,...,5^y,...$, and if you have some kind of upper bound to this, i.e. the sequence is $5^0,5^1,5^2,...,5^y,...,5^n$ then it turns out that that finite sequence can be encoded with only two numbers, and the entries of the sequence (i.e. the $5^y$ values that you want) can be recovered using those two numbers and basic addition and multiplication operations (actually, you also do need either a successor function or a constant $1$).



        But as Mauro says it is rather complicated, and probably not something you want to use for any kind of practical use. If you want to use exponentiation in practice, I would bite the bullet and just introduce a new symbol for exponentiation ... but at least you can define it recursively in terms of $+$ and $times$ ... and maybe that's acceptable to you:



        $forall x : x^0 = 1$ (or $forall x : x^0 = s(0)$ if you have $s$ instead of $1$)



        $forall x forall y : x^y+1 = x^y*x$






        share|cite|improve this answer












        Kurt Godel showed how you could do this using $beta$-functions. If you really want to know about this, look for this in the literature, but the rough idea is that for your function $x = 5^y$ we can consider the sequence $5^0,5^1,5^2,...,5^y,...$, and if you have some kind of upper bound to this, i.e. the sequence is $5^0,5^1,5^2,...,5^y,...,5^n$ then it turns out that that finite sequence can be encoded with only two numbers, and the entries of the sequence (i.e. the $5^y$ values that you want) can be recovered using those two numbers and basic addition and multiplication operations (actually, you also do need either a successor function or a constant $1$).



        But as Mauro says it is rather complicated, and probably not something you want to use for any kind of practical use. If you want to use exponentiation in practice, I would bite the bullet and just introduce a new symbol for exponentiation ... but at least you can define it recursively in terms of $+$ and $times$ ... and maybe that's acceptable to you:



        $forall x : x^0 = 1$ (or $forall x : x^0 = s(0)$ if you have $s$ instead of $1$)



        $forall x forall y : x^y+1 = x^y*x$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 11 '17 at 14:20









        Bram28

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