Is it possible to prove the absolute consistency of a theory by giving a finite model?

Question

Most consistency results are relative, they say that if theory $T$ is consistent, then theory $T'$ is consistent. However, is it possible to prove the "absolute" consistency of a theory, by giving a finite (preferably a small finite) model? For example, the theories of groups and Boolean algebras are absolutely consistent, because there are small finite models for both of them. Has any book or paper or text talked about this notion of absolute consistency?

Answer 1

I disagree with the premise that most consistency results are relative. I think there's a theory/meta-theory conflation going on here. Here are two standard examples of consistency results:

1. It's a theorem that if $\mathsf{ZFC}$ is consistent, then $\mathsf{NBG}$ (see here) is consistent. This consistency result is relative, since we're assuming consistency of a theory whose consistency we can't prove (assuming our meta-theory is like $\mathsf{ZFC}$ or weaker, we can't prove $\mathrm{Con}(\mathsf{ZFC})$, by Gödel). The usual way to prove this would be to take a model of $\mathsf{ZFC}$ and produce from it a model of $\mathsf{NBG}$.

2. It's a theorem that $\mathsf{PA}$ is consistent. This consistency result is not relative. The usual way to prove this would be to point to a model of $\mathsf{PA}$ (like $\mathbb{N}$), which we can do without assuming consistency of any other theory.

Now, you might still say that the consistency of $\mathsf{PA}$ is not absolute, because it relies on believing in the soundness of the meta-theory (i.e., that the meta-theory doesn't prove anything false), or because not every meta-theory can carry out the proof (e.g., if $\mathsf{PA}$ is our meta-theory, we cannot carry out this proof - or any proof - of $\mathrm{Con}(\mathsf{PA})$, by Gödel). But if this is your view, then there are no absolute theorems whatsoever in mathematics! Proving a theorem always relies on meta-theoretic reasoning, and different meta-theories can always disagree about what reasoning is allowable.

Now, you might argue that certain "very concrete" facts are more "absolute" than $\mathrm{Con}(\mathsf{PA})$, like $1+1 = 2$ or "there is a prime number greater than $10$". Sure, I agree, but saying that every meta-theory should prove these facts just amounts to putting some reasonable minimum requirements on what should count as a meta-theory, like "the meta-theory should be sound and complete for such and such class of very concrete statements". That is, you want the meta-theory to at least contain some base very weak system, and then the "absolute" theorems will be the ones provable in that system.

So maybe a better framing of your question is: How weak can a meta-theory be and still carry out model-theoretic consistency proofs for theories with explicit finite models?

To prove, for example, that the theory of groups is consistent by exhibiting the trivial group as a model, our meta-theory needs to be able to (a) talk about strings of symbols and recognize proofs from finite theories, so as to even formulate $\mathrm{Con}(T)$, (b) define "model of $T$" (at least for models with a single element!) and recognize that the trivial group is a model of the group axioms, and (c) prove the soundness theorem (at least for models with a single element!) - that is, prove by induction on the length of proofs that nothing provable from the group axioms is false in the trivial group. I'm not an expert in weak theories of arithmetic, but all that can certainly be carried out in a very weak meta-theory, with just a little bit of induction.