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I've checked formulation of Peano arithmetic second order induction axiom in several books and everywhere it is +- the same:

Let $P (n)$ be any property pertaining to a natural number n. Suppose that P (0) is true, and suppose that whenever P (n) is true, $P (n++)$ is also true. Then $P (n)$ is true for every natural number $n$.

I have checked this discussion A question on Terence Tao's representation of Peano Axioms but I still can't accept arguments from top rated answer because as for me it is actually still possible that there exist such natural numbers $a,b$ and $c$: a++ = b, b++ = c, c++ = a. After all this axiom only states that if some predicate $P$ satisfies the requirements from the axiom then I can say that $P(n)$ is true for every natural number and I understand intuition behind this axiom that $P(0)$ $\implies$ P(0++) $\implies$ P((0++)++) $\ldots$ so that it is true for all increments of 0, but I don't see in the formulation of axiom any explicit restrictions for what natural number can be. However, this formulation is used everywhere, so most likely I just do not understand something... moreover I am quite new to logic. Could you please explain to me where I'm wrong?

Addition to question: I will outline the big picture as I see it. I assume that "true" natural numbers exist, but I would like to have a strict formal description of them in the form of a system of axioms. Such a system is Peano's system of axioms. I know that Peano's system of axioms is categorical. I also know that these axioms have been proven consistent within the ZFC axiom system. The last piece of the puzzle for me is the answer to the question: do these axioms really describe "true" natural numbers and only them? You propose to consider the predicate "not to be rogue", but this is a rather vague formulation of the predicate, I do not understand what is hidden behind it, so I do not quite trust the proof using it. An acceptable proof for me would be 1) to prove that all properties of "true" natural numbers follow from Peano's axioms 2) to prove that properties that "true" natural numbers do not have do not follow from these axioms (it is the violation of this point that should lead to the appearance of "rogue" natural numbers). I accept the 1st point, but not the 2nd point. In order to consider the 2nd point satisfied, I must be sure that whatever predicate that gives false for "true" natural numbers I take, I must be able to prove its falsity using Peano's axioms. The catch is that there are infinitely many such predicates, so you need to use some more "low-level" methods to prove it. And at this point I'm stuck, I don't know what results to rely on to prove to myself that this 2nd point is true.

user341
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    claim: no natural number is equal to a, b, or c. Proof: by induction. Base: a, b, c, can't be 0 since they are all successors. Step: suppose x is not a, b, c. S(x) can't be a because then S(x) = S(c), so x=c, contradicting the inductive hypothesis. Similarly S(x) is not b or c. – Matthew Towers Mar 25 '23 at 20:35
  • @MatthewTowers thank you for your comment. Actually the point is not in this particular case with a, b, c. My question is about such all possible cases with "rogue" elements. Do I understand correctly that you mean that the purpose of induction axiom is to give a way to prove all possible predicates that don't contradict other Peano axioms for all natural numbers? – user341 Mar 25 '23 at 21:19
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    Consider applying induction to the property "is not a rogue element." – Noah Schweber Mar 25 '23 at 22:06
  • @NoahSchweber as far as I understand Peano axioms define what natural number is, so it seems to me that the whole point of induction axiom is just to give opportunity to prove or disprove any proposition for all natural numbers based on other Peano axioms. Otherwise property "is not a rogue element" is quite vague. – user341 Mar 25 '23 at 22:20
  • @RossMillikan No, it's second order - the quantification over arbitrary properties ("Let $P$ be any property") is just quantification over sets rephrased. To the OP, the (second-order) Peano axioms do not define the natural numbers, they describe the natural numbers (quite well as a matter of fact). But let's say you don't want to adopt a Platonist attitude w/r/t the naturals; ok, then the point of my comment above is that any "roguish" behavior you can articulate is ruled out by the PA axioms. – Noah Schweber Mar 25 '23 at 22:49
  • @NoahSchweber is it really proved that any "roguish" behavior is ruled out by the PA axioms? To be honest I can't even suppose how it can be proved. – user341 Mar 25 '23 at 23:24
  • @NoahSchweber (addition to previous comment) just as far as I understand second order PA is categorical. Then if PA really describes natural numbers and only them everything becomes +- clear – user341 Mar 25 '23 at 23:33
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    @user341 I don't really understand your question, to be honest. If "rogue-ness" is vague, then it makes no sense to ask your question in the first place; if "rogue-ness" is not vague, then we can apply induction to the property "is not rogue." And we can leave the term "rogue" behind: if you are comfortable with the idea of "true" $\mathbb{N}$ existing at the outset it should be easy to see that second-order PA is totally categorical. For a model $M$ of second-order PA consider the property "is in the image of the unique embedding of $\mathbb{N}\rightarrow M$" (this is just "non-roguishness"). – Noah Schweber Mar 25 '23 at 23:37
  • Suppose we have a set $X$ and $x_0 \in X$ and $f: X \to X$. It can be formally proven (in 228 lines) that induction holds on $X$ with successor function $f$ and "first element" $x_0$ if and only if every element of $X$ but $x_0$ itself can be reached by a process repeated succession starting at $x_0$. There are no isolated, inaccessible "rogue" terms. – Dan Christensen Mar 26 '23 at 00:32
  • $X$ need not be infinite, e.g. $X={0,~1}, f(0)=1,$ and $f(1)=0$. Then induction holds on $X$ with successor function $f$ and "first element" $0$. That is, we have: $\forall P\subset X : [0\in P \land \forall x\in P:[f(x)\in P]\implies P=X$. Hint: There are only 4 subsets of $X$ to consider. – Dan Christensen Mar 26 '23 at 01:38
  • @NoahSchweber I've edited the question, the text doesn't fit in the comment. Check it please. – user341 Mar 26 '23 at 11:39
  • @DanChristensen could you please share this proof somehow (here or in PM) – user341 Mar 26 '23 at 11:52
  • As requested by @user341 : http://www.dcproof.com/InductionIffAccessible.htm – Dan Christensen Mar 26 '23 at 15:52
  • Whether it is the set of all natural numbers $N$ or the set ${ 0,~1 }$, saying that induction holds on a set effectively rules out the existence any "rogue" (or isolated) elements of that set, i.e. any elements that cannot be reached by repeated succession starting from the designated "first element" (assuming the existence a successor function on that set). – Dan Christensen Mar 26 '23 at 22:43
  • Put another way, if we have $f: X \to X$ and $x_0 \in X$, then induction is said to hold on $(X, f, x_0)$ if and only if $X = {x_0,~f(x_0),~f(f(x_0)), \cdots ~}$. – Dan Christensen Mar 27 '23 at 00:40
  • You may also find it reassuring to know that given any pair of structures $(N,S,0)$ and $(N',S',0')$, that satisfy Peano's Axioms will be essentially the same structures, differing only in the names used. A formal proof (720 lines) is available. – Dan Christensen Mar 27 '23 at 01:42
  • Example: Here are the usual Peano Axioms: https://mathworld.wolfram.com/PeanosAxioms.html The Principle of Mathematical Induction (axiom 5) forbids the existence of a "rogue" number that is its own successor. – Dan Christensen Mar 27 '23 at 15:57

2 Answers2

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Axiom 2.3 ensures $0$ is not the successor of another number. Define $s^0=0$ and $s^{n+1}=s^n++$ and assume $s^{n}$ is the first element to be repeated. That is to say there exists some chain $$s^{n-1} \rightarrow s^n \rightarrow \dots \rightarrow s^{n+k} \rightarrow s^n$$. This means $s^{n-1}$ and $s^{n+k}$ have the same successor contradicting axiom 2.4. From that we can conclude no elements are repeated.

Note that I've tacitly used induction here. I check that $0$ isn't a rogue element and then use the fact $s^{n}$ is the successor of some other number to finish the proof. Also, because you're building understanding I decided to use counting notation even though counting hasn't technically been defined yet. Some familiarity can improve understanding when working this close to the axioms.

CyclotomicField
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You wrote:

I don't see in the formulation of axiom any explicit restrictions for what natural number can be.

Here are some well known, supposedly self-evident properties of the natural numbers:

  1. $0$ is a natural number
  2. Every natural number has a unique successor that is a natural number
  3. Different natural numbers have different successors
  4. $0$ is not the successor of any natural number
  5. Every natural number but $0$ itself can be reached by a process of repeated succession starting at $0$

These properties of the natural numbers can be formalized respectively in the language of set theory as follows (Peano's Axioms for $(N,S,0))$:

  1. $0\in N$
  2. $\forall x\in N: S(x) \in N$
  3. $\forall x, y \in N: [S(x)=S(y) \implies x=y]$
  4. $\forall x \in N: S(x)\neq 0$
  5. $\forall P\subset N:[0\in P \land \forall x\in P:[S(x)\in P] \implies P=N]$

Where

$N$ = the set of natural numbers

$S$ = the successor function on $N$

(1) through (4) here are straightforward translations. (5) may be not be so obvious. For what it is worth, here is formal proof (228 lines) of this equivalence for the skeptics.

Are we missing anything? Maybe some other essential properties of the natural numbers? It seems unlikely that additional axioms will be required since every algebraic system that satisfies Peano's Axioms as listed above will essentially be the be the same system, only the names will differ.

Informally, we can match up the elements of $N$ and $N'$ quite naturally as follows:

$$0 \longleftrightarrow 0'$$

$$S(0) \longleftrightarrow S'(0)$$

$$S(S(0)) \longleftrightarrow S'(S'(0'))$$

$$\vdots$$

This matching up would be uniquely given by the function f mapping $N$ to $N'$ such that:

$$f: N\to N'$$

$$f(0)=0'$$

$$\forall a\in N: f(S(a))=S'(f(a))$$

Not for the faint of heart, you can find formal proof of this here (723 lines).

Dan Christensen
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  • Thanks for the detailed answer, Dan! I would like to clarify one last thing before closing the question. Do I understand correctly that this proof is built on the basis of ZFC axioms? – user341 Mar 28 '23 at 01:11
  • Not exactly. It is based on what might be called a simplified, more user-friendly version of ZFC. It is robust enough to avoid the contradictions that arise, for example, from Russell's Paradox. – Dan Christensen Mar 28 '23 at 02:13