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Intuitively, one can say that $S(n) > n$. But how do we prove it using the Peano Axioms. It seems like I need a formal statement as to what $>$ means.

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    @Shahab: That phrasing doesn't work if we're working in Peano Artithmetic, since PA is$a$separate first-order theory that doesn't let you speak about "sets", "ordered pairs" and so forth.2012-10-16

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Usually, it's $\leq$ which gets defined first, not $>$. In the case of PA, you can define $\leq$ as $ a \leq b \leftrightarrow \exists c\: (b = a+ c) $

But of cource, once you've defined one of the relations $\leq$, $<$, $>$, $\geq$, definitions for the others follow immediately. You e.g. have $ a > b \leftrightarrow (a \neq b) \land (b \leq a) $

Or you can define $>$ directly as $ a > b \leftrightarrow \exists c\: (c \neq 0) \land (a = b + c) $

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    @timvermeulen: we're working in PA. Everything's nonnegative.2013-07-20
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Here is a definition that uses the order axiom:

For an ordered field $\mathbb{F}$, there is a unique subset $P$ satisfying the following conditions.

  1. if $a,b\in{P}$ then $a+b,ab\in{P}$.

  2. for all $a$ in $\mathbb{F}$, one and only one of the following is true: $a\in{P}$, $-a\in{P}$, or $a=0$.

We say that $p \in \mathbb{F}$ is positive iff $p \in P$.

The subset $P$ is the positive numbers of the field. From here, the relation of 'strictly greater than' can be defined as follows.

$a>b\iff{a-b}\in{P}$

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    You're welcome!2013-07-20