Finiteness can’t be captured in a sound, complete, finitary proof system

Consider the sentence “This blog has finitely many posts.” Do you understand what this sentence means? Then no set of rules (even infinite, even uncomputable!) can model your reasoning. This claim may sound shocking, but it can be justified on solid metamathematical grounds.

Another example: the sentence “There are finitely many planets in the universe.” You don’t have to think it’s true, you just have to think you understand what it means. What’s the common theme? It’s the notion of there being ‘finitely many’ of some class of objects. Let’s imagine building a language that has the expressive resources of first-order logic (which are quite modest), plus an additional quantifier F, whose semantics are given by the following rule: Fx φ(x) is satisfied by a model iff there are only finitely many objects in that model that satisfy φ(x).

It turns out that any language consisting of first order logic plus the quantifier F can’t be axiomatized in any sound, complete, and finitary proof system. Notice that effective enumerability of the rules of the proof system is not a requirement here! So long as the language is strong enough to express the semantics of {∧, ¬, ∀, F, variables xn, and relations Rn}, no set of sentences and sentence-manipulation rules in that language will suffice to capture these semantics.

Here’s a proof: consider the first-order theory of Peano arithmetic. This theory has nonstandard models (as any theory of arithmetic must have in a logic that is compact). All of these nonstandard models have the following feature: that there are numbers that are larger than infinitely many numbers. Think about what this means: this is a common feature of all nonstandards, so if we could write a sentence to express this feature then we could rule them all out! This is where the quantifier F steps in. With F, we can write the following simple sentence and add it to PA as an axiom:

∀x Fy (y < x)

In English: every number has finitely many numbers less than it. And with that, we’ve ruled out all nonstandard models! So now we have a theory that is categorical for ℕ. And that’s a big deal, metamathematically speaking!

Why? Well, as I’ve talked about in a previous post, you can prove some pretty strong limitative results about any logic that can produce a theory that’s categorical for ℕ. In particular, if we can produce such a theory then its logic cannot be compact. Quick proof: suppose a set of sentences Σ models ℕ. Add to Σ a constant c and the axioms “c ≠ 0”, “c ≠ 1”, “c ≠ 2”, and so on, and call this new set Σ’. Every finite subset of Σ’ models ℕ. So by compactness, Σ’ has a model. But this model is nonstandard – it contains numbers that aren’t natural numbers. And since Σ is a subset of Σ’, any model of Σ’ is also a model of Σ.

So compactness implies that no theory is categorical for ℕ. But compactness follows from the following three properties: a sound and complete proof system (Σ ⊢ α if and only if Σ ⊨ α), and that all proofs are only finitely long (try expressing this property without F!). Quick proof: If a set of sentences is finitely satisfied, then every finite subset of it has a model (by definition), so no finite subset of it can be refuted (by soundness), so the entire set can’t be refuted (by finite proofs), so the entire set is satisfied (by completeness).

So soundness + completeness + finiteness ⇒ compactness ⇒ the existence of nonstandard models of arithmetic in any theory that models ℕ. Which means that the semantics of F cannot be captured in any sound, complete, and finite proof system!

Take your pick: either you don’t really understand the semantics of the “finitely many” quantifier F, or no set of rules (not even requiring this set to be finite or computable) can fully capture your reasoning in finite-length proofs.

More information about related extensions of first-order logic and their limitations can be found here. The result I describe here is a rephrasing of results discussed there.

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