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I am really confused if the Peano axioms are supposed to be strictly a set theory / first order thing or how are we supposed to state them. In English? Can we use purely logical expressions? What about induction? Does this correctly state the axioms?

  1. $0 \in \mathbb{N}$

  2. $\forall a \in \mathbb{N}, S(a) \in \mathbb{N}$

  3. $\forall a \in \mathbb{N}, S(a) \neq 0$

  4. $\forall a, b \in \mathbb{N}, S(a) = S(b) \to a = b$

  5. $\forall P(P(0) \land \forall k(P(k) \to P(k+1)) \to \forall n (P(n)))$

That fifth one I am pretty sure is second order logic, which I don't know if that's the right way to do it or not. Not sure if the correct one is some first-order thing instead.

Are these the correct way to state axioms? I see so many different formulations and a lot of them are just in English, is that the way it's normally stated, informally like that, or is there a formal and standard way to say "these are the Peano axioms for defining how natural numbers work"?

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  • $\begingroup$ Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. $\endgroup$ – dantopa Jun 22 '19 at 19:37
  • $\begingroup$ The fifth is not a single axiom, but an "axiom schema." That is, for any statement $P(x)$ with a variable $x,$ you have an instance of $5.$ $\endgroup$ – Thomas Andrews Jun 22 '19 at 19:41
  • $\begingroup$ I know it is an axiom schema, it's a little unrelated to my question though $\endgroup$ – user684073 Jun 22 '19 at 19:44
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    $\begingroup$ @user684073 The difference between an axiom and axiom schema is exactly how you resolve the question of whether it's second-order or not. In secord-order logic, the formula you wrote is a single axiom. In first-order logic, it is not a well-formed formula, but we could instead interpret it as an axiom schema, in which case we get an axiom for each formula $P$ and the axiom schema then stands for an infinite number of individual axioms. $\endgroup$ – Derek Elkins left SE Jun 22 '19 at 21:20
  • $\begingroup$ Is an axiom schema still technically first order? Or even axiom schemas in propositional logic? Or does it have no bearing on the order? A little confused by that comment @DerekElkins $\endgroup$ – user684073 Jun 22 '19 at 21:21
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The axioms you've written are essentially those of Peano. He worked in an informal set theory, and began with an axiom saying that $\mathbb N$ is a set (or a class --- the distinction didn't yet exist). Also, he started his $\mathbb N$ with $1$, not $0$. So, from a historical point of view, your axioms have a reasonable right to be called the Peano axioms.

Instead of set theory, one can base these axioms on second-order logic. In particular, second-order logic allows the quantification over predicates, $\forall P$, in the induction axiom.

More recently, though, the phrase "Peano arithmetic" (with the abbreviation PA) has come to be used almost universally for a first-order approximation to these second-order axioms --- replacing the second-order induction axiom, which refers to all predicates, with a schema of first-order axioms (as in J.G.'s answer), referring just to the (parametrically) definable predicates.

There are major differences between the original second-order axioms and the first-order theory PA. In particular, the second-order axioms admit only one model up to isomorphism, namely the standard natural numbers. In contrast, PA, like any first-order theory with an infinite model, admits many non-isomorphic models.

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  • $\begingroup$ AS J.G. pointed out, first-order PA needs additional axioms to describe addition and multiplication. In second-order arithmetic, these operations would be definable, using quantifiers over predicates. $\endgroup$ – Andreas Blass Jun 22 '19 at 20:19
  • $\begingroup$ So we have Peano's original formulation of the "Peano axioms" which are more set theory, a second-order formulation of "Peano axioms" that are in terms of second order logic, and a first-order approximation "Peano arithmetic" that's in terms of first order logic? $\endgroup$ – user684073 Jun 22 '19 at 20:26
  • $\begingroup$ @user684073 That's correct, but somewhat anachronistic, since Peano's work came before any formal systems of set theory. So I tend to blur the distinction between set-theory and second-order logic here. $\endgroup$ – Andreas Blass Jun 22 '19 at 20:30
  • $\begingroup$ Is there an idiot-friendly outline of all three of these systems and why they needed to be made in the first place? I feel like as I Google around for this stuff I am seeing a mix of systems and it is mind-bogglingly confusing to keep things properly separated. I can't tell what's what, hence the issues in my answer apparently. $\endgroup$ – user684073 Jun 22 '19 at 20:33
  • $\begingroup$ @user684073 I"m away from most of my books at the moment, but if I weren't then the first place I'd look for this sort of material is William Hatcher's book "Foundations of Mathematics". As for "why they needed to be made in the first place", remember that people worked successfully with the natural numbers for millennia before any axiomatizations were proposed.The value of the axiomatizations is in showing just what assumptions underlie what facts of arithmetic. $\endgroup$ – Andreas Blass Jun 22 '19 at 20:38
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The Peano axioms are a first-order theory in which the only objects are natural numbers (in this context defined to start at $0$, not $1$), so you can't have $\Bbb N$ suddenly show up. A better statement of your axioms would be:

  1. $\exists 0\forall a (a+0=a)$
  2. $\forall a\exists b(Sa=b)$
  3. This one's fine as-is for now
  4. This one too
  5. Add one axiom per choice of $P$, rather than quantifying over $P$ (that's as forbidden as having $\Bbb N$ appear); for better or worse, we need infinitely many axioms.

But if you want to really go to town, the axioms here add a lot more. They break down the rules for $=$ and $+$, and even $\times$ (because we're not just doing Presburger arithmetic here, so e.g. you need $aSb=ab+a$). In particular, the axioms for $+,\,\times$ provide more information about $S$.

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  • $\begingroup$ I am really confused then because my other axioms do use $\mathbb{N}$, why can't I mention it if I am then allowed to bring in something like $0$? I also don't know what you mean about 5. $\endgroup$ – user684073 Jun 22 '19 at 20:10
  • $\begingroup$ And are the rules for equality, addition, multiplication, etc really part of peano axioms? $\endgroup$ – user684073 Jun 22 '19 at 20:12
  • $\begingroup$ @user684073 To comment 1, I could just have easily written $\exists z\forall a(a+z=a)$, but if I can use the symbol $z$ for it I can use $0$, whereas in your attempt $\Bbb N$'s appearance violates the theory's first-order status, as does your final $\forall P$ usage, because neither sets of natural numbers nor unary predicates over natural numbers are legal objects herein.$$\cdots$$To comment 2, we need them because the shortlist of axioms you had in mind has models that violate the other axioms you've proposed doing without. But these models don't fit our intuitions about $\Bbb N$. $\endgroup$ – J.G. Jun 22 '19 at 20:15
  • $\begingroup$ Why does $\mathbb{N}$ violate first order status? $\endgroup$ – user684073 Jun 22 '19 at 20:17
  • $\begingroup$ @user684073 What do you think first-order means? $\endgroup$ – J.G. Jun 22 '19 at 20:18

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