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The principle of induction can be stated, in first-order logic, as follows. Let $S\subseteq\mathbb N$, and suppose that

  • $0\in S$.
  • $\forall n:n\in S\to n+1\in S$.

Then, $S=\mathbb N$. Now, suppose instead we adopt the following axiom for the natural numbers:

Every natural number other than $0$ can be written as $n+1$ for some natural number $n$.

Doesn't this axiom make the principle of induction "self-evident"? We know that $0\in S$, and that $0\in S\to1\in S$ (universal instantiation). Therefore, by modus ponens, $1\in S$. Then, $1\in S$, and $1\in S\to2\in S$, so $2\in S$, and so forth, meaning that $S=\mathbb N$. I don't quite understand the error in this "proof"—perhaps it is that, if it were written down formally, then it would be infinite in length, which is not allowed. Or is there another, more fundamental, error in this line of reasoning?

Joe
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  • Is your axiom adopted in addition to the other one? –  Oct 03 '21 at 13:37
  • @SaucyO'Path: No, it is meant as an alternative to the axiom of induction. I have edited my post for clarity. – Joe Oct 03 '21 at 13:38
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    Your axiom is satisfied by an ordering that looks like a copy of N followed by copy of Z, but induction doesn't necessarily work for it. – Ned Oct 03 '21 at 13:39
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    Another technical point (which isn't important for your question, but that I thought I'd mention) is that the principle of induction can't be stated like that in first order logic because you're quantifying over subsets of $ \mathbb N $ in your statement of it. – Ege Erdil Oct 03 '21 at 13:57
  • @EgeErdil: I don't understand, sorry. How does the statement $\forall n:n\in S\to n+1\in S$ involve quantifying over subsets of $\mathbb N$, rather than individual elements of $\mathbb N$? – Joe Oct 03 '21 at 14:00
  • @Joe That statement doesn't, but for it to be "the principle of induction" you need to add another universal quantification over all $ S $ to that statement. Otherwise, you're just saying that the statement holds for a particular set $ S $ which is a constant symbol in your theory - not very interesting. – Ege Erdil Oct 03 '21 at 14:08
  • @EgeErdil: Oh, I see. I suppose if we turn the principle of induction into an axiom schema (one axiom for each subset $S$ of $\mathbb N$), then that also fixes the problem. – Joe Oct 03 '21 at 14:09
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    @Joe You can't actually do that because there's no way to make those kinds of statements in first order logic. What you have to do is to "encode" the set $ S $ as the collection of natural numbers for which a particular first order formula $ \varphi $ holds, and then you can have an axiom schema, one axiom per $ \varphi $. The obvious problem that there are uncountably many subsets of $ \mathbb N $ but only countably many first order formulae is what leads to the inability of first order theories of arithmetic to control the size of their models. – Ege Erdil Oct 03 '21 at 14:12
  • Perhaps if you used Peano's 'successor of $n$' and replaced $n+1$ in your axiom with $n^\prime$ it would avoid any confusion with addition. – John Wayland Bales Oct 03 '21 at 15:20
  • @Eric: I think it is close enough for you to cast a close vote. – Joe Oct 03 '21 at 15:57
  • So you think induction should hold for, e.g. the model $S=\mathbb{N}\cup(\mathbb{Z}+\frac12)$? – user10354138 Oct 03 '21 at 15:59
  • @Joe: Alright, I refrained from casting a vote immediately since I have a gold badge so I would unilaterally close it. – Eric Wofsey Oct 03 '21 at 16:00
  • You could tweak it a bit. Something like: Every natural number but zero can be reached by a process of repeated succession starting at zero. $N={ 0, S(0), S(S(0)), \cdots }$ – Dan Christensen Oct 04 '21 at 16:13

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