> That seems implausible. All natural numbers are finite. I can easily express that in F.O.L. Just for the sake of clear terminology, I'll take "finite natural

> That seems implausible. All natural numbers are finite. I can easily express that in F.O.L. Just for the sake of clear terminology, I'll take "finite natural number" to mean that it's either 0 or formed by applying the successor function to zero finitely many times. Here's an easy proof for why you can't prove that every number is finite in PA: Take the theory PA' = PA + the infinitely many sentences "c > 0", "c > S 0", "c > S (S 0)", etc. (with a new constant c) We can probably agree that the standard natural numbers aren't a model of PA' because there is no natural number greater than every other natural number. However, the natural numbers are clearly a model of any finite subset of that theory. And here's where we use the compactness theorem (which is easily proven from completeness): https://en.m.wikipedia.org/wiki/Compactness_theorem https://en.m.wikipedia.org/wiki/Compactness_theorem By compactness, PA' has a model. But that model can't be the standard model of arithmetic. Yet it is also a model of PA. Thus we have a model of PA in which infinite numbers exist - so we can't prove that all natural numbers are finite from PA. To rule out such models - and therefore prove that every natural number is finite - you need the second-order induction axiom.