Can An Axiom Schema be Independent?

Question

Consider the following theory: Ring Theory (RT) + $\forall x(Sx=x+1)$ + first order induction (Ind). The finite rings $Z/nZ$ are models of this theory. Now consider RT + $\forall x(Sx=x+1)$ + Not(Ind). The finite non-commutative rings $M_2(Z/nZ)$ are models of this theory. Induction is false in non-commutative rings because we can prove $\forall x \forall y(xy = yx)$ using induction. Does this prove induction is independent of the axioms RT + $\forall x(Sx=x+1)$? What does it mean to say an axiom schema is independent of other axioms? Saying a schema is independent is almost like saying the schema is useless. We can't know the schema is always correct.

Answer 1

Your examples do show that first-order induction is independent of RT + $( \forall x ) ( Sx = x + 1 )$.

Saying that an axiom $\varphi$ (or an axiom schema) is independent of some collection $\Sigma$ of axioms just means that the axioms of $\Sigma$ can neither prove nor refute $\varphi$. Far from making it useless, what it means is that the addition of $\varphi$ to $\Sigma$ results in a strictly stronger theory (i.e., you can prove more theorems), speaking about a strictly smaller class of models. This in turn means that sometimes there are theorems in $\Sigma + \varphi$ which cannot be proved without using $\varphi$: that's the antithesis of useless!

As a more basic example, the statement of commutativity, $( \forall x ) ( \forall y ) ( x \cdot y = y \cdot x )$, is independent of the usual axioms of group theory (as evidenced by the existence of both abelian and non-abelian groups). As such we know more about the structure of abelian groups than groups in general, but the study of abelian groups is important in its own right.

As a more advanced example, (assuming the consistency of ZF) the Axiom of Choice is independent of the other axioms of set theory. But there are very few mathematicians who think that the Axiom of Choice is useless.