# Axiom schemas vs second order axioms: which first-order predicates “exist”?

Axiom schemas, such as the PA induction schema, differ from second-order axioms in that they only hold for "definable" predicates, rather for "all" predicates. As a result, you can have non-standard models of first-order PA for which induction holds for all definable predicates, but does not hold true in general.

This leads to questions about which predicates "exist." In general, for every possible infinite combination of natural numbers (or more generally, objects in the domain), is there considered to "exist" a unary predicate that is "true" for only those numbers, even if the predicate is undefinable? And likewise with n-ary predicates?

This would seem to be a semantic thing. Is there an "axiom of first-order logic" that determines which of these predicates we consider to "exist?" (Or would this perhaps be more pertinent to second-order logic, being related to things like Henkin semantics?)

• (Unary) Predicates define subsets of the domain (or, more generally, subobjects but I won't go there...) What's happening is not every subset is necessarily characterized by a predicate. I recommend terminologically separating predicates from the subsets (relations, in the general $n$-ary case) that are their semantics. You get: "the interpretation of the induction schema means induction holds for the interpretation of any predicate but not for all subsets." – Derek Elkins left SE Mar 10 '19 at 7:12
• One small but important point is that the second-order induction axiom is a specific instance of the induction scheme for second-order arithmetic. It is the induction instance for the formula $n \in A$. This axiom is only strong in circumstances where other choices require many sets to exist. – Carl Mummert Mar 10 '19 at 13:52
• @CarlMummert - well, I was thinking more of things like $A(n)$ than $n \in A$, but I guess they're the same thing. But then what other formulas would be in the "scheme"? Any formula which you could write using predicates and logical primitives (and/or/not) basically just defines another predicate, so wouldn't the single axiom get everything? – Mike Battaglia Mar 10 '19 at 14:34
• The axiom scheme for induction in second-order arithmetic has (the universal closure of) every formula of the form $\phi(0) \land (\forall n) [ \phi(n) \to \phi(n+1)] \to (\forall n)\phi(n)$ where $\phi(x)$ is a formula in the language of second-order arithmetic. – Carl Mummert Mar 10 '19 at 15:13

Semantically, if we work with "full models" then, in each model, we require the second-order part to include all subsets of the first order part. If the model also satisfies the induction axiom $$\forall A( 0 \in A \land (\forall n)[n \in A \to n+1 \in A])$$ then the model will satisfy induction for every subset of the first-order part (which, in conjunction with a few other first-order axioms, will force the first-order part to be the standard natural numbers $$\omega$$).