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.
Is there a formal definition of "Greater Than"
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1@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
2 Answers
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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1@timvermeulen: we're working in PA. Everything's nonnegative. – 2013-07-20
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.
if $a,b\in{P}$ then $a+b,ab\in{P}$.
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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0You're welcome! – 2013-07-20