Quite the opposite; in *no* non-standard model of Peano arithmetic is there a minimal non-standard number.
Consider the formula \\(\phi(x) = \left(x = 0\right) \vee \exists y \left(x = S(y)\right)\\). The first-order induction axiom for \(\phi\) is
\\[
\phi(0) \wedge \forall x \bigl(\phi(x) \Rightarrow \phi(S(x))\bigr) \Rightarrow \forall x \phi(x)
\\]
\\(\phi(0)\\) is trivially true in any model. If we have \\(\phi(x)\\), and there exists a \\(y\\) such that \\(x = S(y)\\), then \\(S(x) = S(S(y))\\), giving us \\(\phi(S(x))\\). From the induction axiom, \\(\phi\\) holds for all numbers—every number is 0 or has a predecessor, in any model with first-order Peano induction.
The non-existence of a minimal non-standard number follows by contradiction: any number with a standard predecessor is itself standard, any number with a non-standard predecessor is not minimal, and any number with no predecessor is zero.
r~~
·
2024-02-12T23:12:48Z (3 months ago)