Can anyone explain the difference between induction as it's stated in first order logic and that from second order logic? I don't understand the difference as it pertains to things like Peano axioms.

5Secondorder induction is an axiom which applies to any set $X\subseteq \mathbb{N}$: if $0\in X$ and $\forall n(n \in X \rightarrow n+1\in X)$, then $X=\mathbb{N}$. Firstorder induction is an axiom (or, to be rigorous, a scheme of axioms) which works the same way but only applies to those set defined by some firstorder formula $\phi$, i.e., sets of the form $X=\{ n\in \mathbb{N}: \phi(n)\ \mathrm{is\ true}\}$. – realdonaldtrump Nov 04 '18 at 16:34

@realdonaldtrump: of course, when we work in secondorder arithmetic, the "first order induction scheme" is typically defined to include the formula "$n \in X$", and so the secondorder induction axiom is just one particular axiom in the induction scheme. – Carl Mummert Nov 04 '18 at 19:41

@user525966  it is not completely clear to me what you're asking. Which induction statement in second order logic are you looking at? What kind of second order theories are you interested in? The induction axioms aren't really part of logic, they are specific to theories of arithmetic. – Carl Mummert Nov 04 '18 at 19:43

1I suggest you don't touch secondorder arithmetic (including secondorder induction) before you can handle firstorder logic. However, my answer to your other question (which incidentally is on the hotnetworkquestions list) does in fact answer this one as well; firstorder induction (and the only one that ever matters) always involves properties that **can be written down**. In any practical foundational system you can write down only countably many properties, and hence firstorder induction says nothing about the subsets of naturals that do not correspond to a property you can write down. – user21820 Nov 05 '18 at 07:52
1 Answers
The informal statement of induction is:
For any property $P$ of natural numbers, if $P(0)$ holds, and $P(n)$ implies $P(n+1)$ for all $n$, then $P(n)$ holds for all $n$.
Of course, this raises the question: What exactly do we mean by a "property of natural numbers"?
One natural interpretation is to identify properties of natural numbers with sets of natural numbers. That is, for any property $P$, we can form the set of all natural numbers satisfying that property. And for any set of natural numbers $X$, we can consider the property of being in $X$. For example, the property of being a prime number corresponds to the set $\{n\in \mathbb{N}\mid n\text{ is prime}\}$
Another natural interpretation is to identify properties of natural numbers with formulas in one free variable in some logic (in this discussion, let's just talk about firstorder logic in the language of arithemetic). Here the syntax of the logic gives us a language for writing down properties of natural numbers. For example, the property of being a prime number corresponds to the formula $\lnot (x= 1)\land \forall y\, (\exists z\, (y\cdot z = x) \rightarrow (y = 1 \lor y = x))$.
Induction under the interpretation "properties are sets" can be formalized as follows:
$\forall P\subseteq \mathbb{N}: ((0\in P\land \forall n\in \mathbb{N}: (n\in P \rightarrow (n+1)\in P))\rightarrow \forall n\in \mathbb{N}: n\in P)$
This is a sentence of secondorder logic, since it involves a quantification $\forall P\subseteq \mathbb{N}$ over subsets of $\mathbb{N}$.
The interpretation "properties are formulas" leads to the following formalization of induction:
$(\varphi(0)\land \forall n\, (\varphi(n)\rightarrow \varphi(n+1)) \rightarrow \forall n\,\varphi(n)$
Here we have an infinite schema of sentences of firstorder logic, one for each firstorder formula $\varphi(x)$. It's firstorder because the quantifiers only range over elements of $\mathbb{N}$, not subsets, and the formulas $\varphi(x)$ are themselves firstorder.
It's worth noting that secondorder induction is much stronger than firstorder induction. Secondorder induction applies to all subsets, while firstorder induction only applies to those which can be defined by some firstorder formula (and since there are are $2^{\aleph_0}$many subsets of $\mathbb{N}$ and only $\aleph_0$many firstorder formulas, there are many subsets which are not definable).
The secondorder Peano axioms (which consist of some basic rules of arithmetic, together with the secondorder induction axiom above) suffice to pin down $\mathbb{N}$ up to isomorphism.
The firstorder Peano axioms (which consist of some basic rules of arithmetic, together with the firstorder induction axiom schema above) cannot hope to pin down $\mathbb{N}$ up to isomorphism (thanks to the LöwenheimSkolem theorems). That is, there are "nonstandard models" of the firstorder Peano axioms, which satisfy induction for all firstorder definable properties, but not for arbitrary subsets.
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1It is true that secondorder induction is stronger than firstorder, but only when we are working in a context where we have strong comprehension axioms as well. For example, if we are working syntactically, and we want to prove some formula using secondorder induction, we will have to construct the set $P$ in our formal proof before we can apply induction to it, and so the strength of the induction axiom will depend on which sets we can construct in our proofs  essentially reducing things to the question of which formulas have a comprehension axiom. – Carl Mummert Nov 04 '18 at 19:27

Of course, when we work semantically with full second order semantics, in essence we have comprehension for *all* subsets of N, and so secondorder induction is much stronger than the firstorder induction scheme in that setting. But when we begin to look at particular formal theories, the relationship is more complex. – Carl Mummert Nov 04 '18 at 19:31

1@CarlMummert Yes, of course when I write "secondorder" in this answer, I mean with the full semantics. – Alex Kruckman Nov 04 '18 at 19:49