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Sep 19, 2022 at 9:17 answer added memorial timeline score: 2
Sep 17, 2022 at 22:46 comment added kaya3 Is it not true that every $\mathbb{Z}/n\mathbb{Z}$ satisfies all of these axioms?
Sep 17, 2022 at 13:08 vote accept CommunityBot
Sep 17, 2022 at 13:08 comment added user155516 @JoelDavidHamkins I am referring to the first order theory that includes S (see Formulario Mathematico)
Sep 17, 2022 at 11:06 comment added Joel David Hamkins Some people refer to Dedekind arithmetic as "Peano arithmetic" or as "second-order Peano arithmetic", in light of Peano's brilliant presentation of the theory, but Peano himself credits Dedekind.
Sep 17, 2022 at 11:04 comment added Joel David Hamkins I think there is some confusion in the comments and the question between Peano arithmetic and Dedekind arithmetic. Dedekind defined his (second-order) arithmetic in the signature with only 0 and successor S, and then proves that + and * are second-order definable by a recursive definition. What is called Peano arithmetic PA, in contrast, is a first-order theory in the signature with +,*,0,1,< (and not S), but one needs the defining axioms for + and *, and the induction axiom stated as a scheme over first-order formulas.
Sep 17, 2022 at 11:00 answer added Joel David Hamkins timeline score: 15
Sep 17, 2022 at 10:41 answer added Andrej Bauer timeline score: 2
Sep 17, 2022 at 6:20 answer added Emil Jeřábek timeline score: 13
Sep 17, 2022 at 6:08 comment added Emil Jeřábek Axiom 3 follows from axiom 4. The extra axiom $Px+y=P(x+y)$ follows from $Sx+y=S(x+y)$ and Axiom 4. You need a primitive ordering predicate to state the induction axiom.
Sep 17, 2022 at 5:57 history became hot network question
Sep 17, 2022 at 2:22 answer added user44143 timeline score: 22
Sep 17, 2022 at 0:56 comment added Gerald Edgar I do not know of a "standard" axiom system of this kind. I would guess that you need to consider two sets, $\mathbf Z$ and a subset $\mathbf N$. This is closer to the Peano axioms than including operations like $+$ among the axioms.
Sep 16, 2022 at 23:11 review Close votes
Sep 22, 2022 at 4:01
Sep 16, 2022 at 22:54 comment added user155516 @LSpice Well it's precisely what I've just done
Sep 16, 2022 at 22:53 history edited user155516 CC BY-SA 4.0
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Sep 16, 2022 at 22:53 comment added LSpice I think you need to state the entire axiom system you have in mind, rather than modifying the question each time I comment. Peano's axioms as usually stated do not, and do not need to, include addition, since it can be defined in terms of $S$ using induction.
Sep 16, 2022 at 22:49 history edited user155516 CC BY-SA 4.0
added 104 characters in body
Sep 16, 2022 at 22:47 comment added user155516 @LSpice Actually $0$ would be singled out by the additive axiom $\forall x,x+0=0$
Sep 16, 2022 at 22:45 comment added LSpice Unless you single out $0$ as a distinguished constant, any structure of the form you define is at best a torsor under $\mathbb Z$. (The Peano axioms do not need to single out $0$, since it is the unique natural number that is not the successor of any natural number.) Anyway, I think currently the question is probably not sufficiently well formed for MO, although I can imagine that it is possible to make it so.
Sep 16, 2022 at 22:42 history edited user155516 CC BY-SA 4.0
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Sep 16, 2022 at 22:36 comment added user155516 I edited the question to require $SPx=x$. I have no other goal than extending Peano to $\mathbf Z$. I am not sure about the best way to formalize the induction axiom, maybe I have to include extra axioms
Sep 16, 2022 at 22:35 history edited user155516 CC BY-SA 4.0
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Sep 16, 2022 at 22:03 comment added LSpice What would your exact proposed induction axiom be? I imagine that, among other things, you would run into trouble by not requiring that $S P x = x$. Regardless, what is your goal?
Sep 16, 2022 at 21:54 history asked user155516 CC BY-SA 4.0