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Thanks, that makes sense.

If you have time for another question, I'm still confused about how it applies to second-order arithmetic, though, since the Peano axioms are well-known and easily listed on a single piece of paper. What difficulty is there be in determining whether a statement is a second-order Peano axiom?

It seems particularly strange since there are apparently fewer axioms than in first-order Peano arithmetic (by replacing an axiom schema with a single induction axiom).

[1] https://math.stackexchange.com/questions/106635/why-does-the...



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