Do induction axioms of Peano Arithmetic have other simple equivalent forms?

Question

For the example of Presburger Arithmetic, which consists of basic axioms for successor, addition, as well as an infinite set of induction axioms. However, it is well known that Presburger Arithmetic can also be formalized with the following axioms:

1. Axioms for ordered Abelian groups;
2. $0 < 1$;
3. $\forall x\ (x \leq 0\vee x \geq 1)$;
4. $\forall x\ (P_n(x) \leftrightarrow \exists y\ x = ny)$, for $n=2,3,...$;
5. $\forall x\ \bigvee_{i=0}^{n-1}\big[ P_n(x+i) \wedge \bigwedge_{j\neq i} \neg P_n(x+j) \big]$.

These axioms are exactly what Cooper's algorithm needs to decide any sentence of Presburger Arithmetic: by converting the sentence into divisibility-equality normal form and then eliminate existential quantifiers.

On the other hand, the set of axioms of divisibility seems “more natural" than induction axioms, since it's very hard to come up with the correct induction axioms to prove a theorem. My question is that, is there any set of "more natural" axioms that is equivalent to induction axioms in Peano Arithmetic? Hopefully, it would seem like this (I borrow the idea from Presburger Arithmetic):

1. Axioms for discretely ordered ring;
2. $\forall n\forall m>0\ \big[\exists! p\exists! q\ (n = pm+q \wedge 0\leq q < m)\big]$.
3. etc...

New edit: I am not seeking for a finite axiomatization since the divisibility axioms of Presburger Arithmetic are infinite schemes. Noah Schweber mentioned that full PA would need arbitrarily large nested quantifier axioms. So would the situation be more possible for $I\Delta_0$, the case that only bounded formulas are used to be induction axioms?

Answer 1

PA is not finitely axiomatizable, so there's no way to replace the induction scheme with a single simpler axiom. Moreover, restricting the induction scheme to $\Sigma_n$ formulas for fixed $n\in\mathbb{N}$ also yields a strictly weaker theory (called "I$\Sigma_n$"), so any equivalent formulation of the induction scheme will have to involve arbitrary-quantifier-complexity formulas. Indeed, any axiomatization of PA has to involve sentences of arbitrarily high quantifier complexity.

While your question is too subjective to really admit a definitely negative answer, I think the above facts give good evidence that there won't be a truly satisfying way to rephrase induction along the lines you're looking for.