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I'm trying to express this idea using FO+TC (First-order Logic with Transitive closure) with the signature $\Sigma = \{+, s\}$ (+ addition function, s is a successor function) over the standard structure of the natural numbers:

$y = x^N$

($x$ is a power on $N$, for some natural $N$)

N.B: I expressed GCD using FO+TC, so i tried to play with something like "the GCD of $x$ and $y$ is $x$" but actually its just a starting direction i have.

Any ideas how can i express it ?

Yan
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  • This relation is more properly described as "$y$ is an integral power of $x$". What's your notation for "$S$ is the transitive closure of $R$"? It's guaranteed to exist for any relation-defining formula $\varphi(x,y)$? – BrianO Nov 21 '15 at 07:41
  • I'm using transitive closure as follows: $y(x,y) := TC(a,b)(some formula)(x,y)$ , where $a := x$ and $b := y$ (a and b are used inside the TC formula). – Yan Nov 21 '15 at 10:04

1 Answers1

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Consider the formula $$ \varphi(x, y,z) := z = x*y. $$ Then $$ \varphi(1,x,x), \varphi(x,x,x^2), \dotsc, \varphi(x^n, x, x^{n+1}), \text{ and }(\forall y)\,\varphi(y, x, x*y). $$ Let $T$ abbreviate the transitive closure of $\varphi(a,x,b)$, which uses $x$ as a parameter, so that $$ T(x,y) := \operatorname{TC}(a,b)\varphi(a,x,b)(x,y). $$ Then $$ y = x^N \text{ for some $N\ge 0$}\iff (y = 1 \lor T(1,y)). $$

BrianO
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  • Thanks, I got it. lets say i could only use addition in my signature. Can it be also expresed with an addition instead of multiplication? – Yan Nov 21 '15 at 09:53
  • That is true, a result of Arnon Avron, but the proof is not so obvious. See the footnote [*] to this answer: http://math.stackexchange.com/questions/619131/can-equinumerosity-by-defined-in-monadic-second-order-logic/620714#620714. In FOL+TC, it goes like this: (1) using TC, you can define $x|y$, i.e. "$x$ divides $y$:,as $(TC_{a,b} b=a+x)(0,y)$. (2) $y=x^2$ is definable: $y=x^2\iff (x+y) = lcm(y,y+1)$. Note that $lcm$ is definable in terms of "divides" and $<$, which is definable in terms of successor (+1) and TC. Finally, (3) $x = y*z \iff (y+z)^2 = y^2 + z^2 + x+x$ is definable. – BrianO Nov 21 '15 at 13:14
  • I garbled (2): that should be "$x = y^2 \iff ...$". Clearly, $12\ne lcm(9,10$, but $12=lcm(3,4)$. – BrianO Nov 21 '15 at 22:01