TITLE: Deducing PA's axioms in ZFC
QUESTION [14 upvotes]: Recently I've stumbled across this claim:

Peano axioms can be deduced in ZFC

I found a lot of info regarding this claim (e.g. what would (one version of) the natural numbers look like within the universe of sets: $0 = \emptyset$, $n + 1 = n \cup \{n\}$), but not what the deductions (ZFC $\vdash$ PA) actually look like. What do they look like?

REPLY [7 votes]: The most direct, model-theoretic method to prove the existence of a model of PA within ZFC is as follows:

First, we formalize the syntax of PA within ZFC. The method is similar to the one used for formalize syntax for the second incompleteness theorem. The main result is a formula $S(n,m)$ which says that $n,m\in \omega$, and $n$ codes a formula $\phi_n(x_1)$ in the language of PA with one free variable and  $$(\omega, +,\times,0,1, =) \vDash \phi_n(\ulcorner m\urcorner).$$  Developing this formula $S$ takes some time but the methods are completely standard
Then, using the axiom of separation, one uses ZFC to form the set $$I = \{(n,m) \in \omega^2 : \text{$m$ is minimal such that } S(n,m) \}$$
Now one must verify in a single ZFC proof that $(\omega, +,\times,0,1,=) \vDash PA$. This goes as follows. Using the formalization from part 1, ZFC proves, for some $k$, that there are coded axioms $n_1, \ldots, n_k$ of PA such that for any coded axiom $n$ of PA, either $n=n_i$ for some $i < k$ or $n$ is a coded induction axiom. In the first case, we can do a proof by cases to verify that $(\omega, +,\times,0,1,=)$ satisfies the axiom coded by $n_i$. In the latter case, where $n$ is an induction axiom, we use the single set $I$ from part 2 as a parameter to verify that the axiom coded by $n$ holds in $(\omega, +,\times,0,1,=)$. Namely: if the induction axiom failed for some coded formula $\phi(x)$ then if we let $\phi_r(x)$ be the negation of $\phi(x)$ the set $\{ m \in \omega : (r,m) \in I\}$ would have no least element.