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.
logic predicate-logic logic-translation
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up vote
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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.
logic predicate-logic logic-translation
I think that it is possible, but quite compelx... See Richard Kaye, Models of Peano arithmetic, page 67.
â Mauro ALLEGRANZA
Jan 11 '17 at 13:05
May you give us some more context ? Book, page,...
â Mauro ALLEGRANZA
Jan 11 '17 at 13:06
add a comment |Â
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.
logic predicate-logic logic-translation
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.
logic predicate-logic logic-translation
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.
â Mauro ALLEGRANZA
Jan 11 '17 at 13:05
May you give us some more context ? Book, page,...
â Mauro ALLEGRANZA
Jan 11 '17 at 13:06
add a comment |Â
I think that it is possible, but quite compelx... See Richard Kaye, Models of Peano arithmetic, page 67.
â Mauro ALLEGRANZA
Jan 11 '17 at 13:05
May you give us some more context ? Book, page,...
â 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.
â 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.
â Mauro ALLEGRANZA
Jan 11 '17 at 13:05
May you give us some more context ? Book, page,...
â Mauro ALLEGRANZA
Jan 11 '17 at 13:06
May you give us some more context ? Book, page,...
â Mauro ALLEGRANZA
Jan 11 '17 at 13:06
add a comment |Â
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 $1$)
$forall x forall y : x^y+1 = x^y*x$
add a comment |Â
1 Answer
1
active
oldest
votes
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 $1$)
$forall x forall y : x^y+1 = x^y*x$
add a comment |Â
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 $1$)
$forall x forall y : x^y+1 = x^y*x$
add a comment |Â
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 $1$)
$forall x forall y : x^y+1 = x^y*x$
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 $1$)
$forall x forall y : x^y+1 = x^y*x$
answered Jan 11 '17 at 14:20
Bram28
55.2k33982
55.2k33982
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I think that it is possible, but quite compelx... See Richard Kaye, Models of Peano arithmetic, page 67.
â Mauro ALLEGRANZA
Jan 11 '17 at 13:05
May you give us some more context ? Book, page,...
â Mauro ALLEGRANZA
Jan 11 '17 at 13:06