# Why Does Induction Prove Multiplication is Commutative?

Andrew Boucher's General Arithmetic (GA2) is a weak sub-theory of second order Peano Axioms (PA2).

GA has second order induction and a single successor axiom:

$$\forall x \forall y \forall z\bigr((Sx=y \land Sx=z)\to(y=z)\bigl)$$

Boucher proves multiplication is commutative in GA2. Why does induction prove multiplication is commutative? GA2 has many finite models. The rings $\mathbb Z/n\mathbb Z$ are models. If we remove induction from GA2 it is easy to see GA-Ind is sub-theory of Ring Theory (RT). RT has finite non-commutative models. Why aren't these finite non-commutative rings models of GA? Would a first order version of GA also prove multiplication is commutative?

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