Inferentialism and the completeness of geometry

The Quinean criterion for existential commitment is that we incur existential commitment precisely by affirming existentially quantified sentences. But what’s an existential quantifier?

The inferentialist answer is that an existential quantifier is anything that behaves logically like an existential quantifier by obeying the rules of inference associated with quantifiers in classical logic.

Here is a fun little problem with the pairing of the above views. Tarski proved that, with an appropriate axiomatization, Euclidean geometry is complete and consistent, i.e., for every geometric sentence ϕ, exactly one of ϕ and its negation is provable from the axioms. Now let us stipulate a philosophically curious language L^*. Syntactically, the symbols of L^* are the symbols of L but with asterisks added after every logical connective, and the sentences are of L^* are the sentences of L with an asterisk added after every connective and predicate. The semantics of L^* are as follows: the sentence ϕ of L^* means that the sentence of L formed by dropping the asterisks from ϕ is provable from the axioms of Euclidean geometry.

Inferentially, the asterisked connectives of L^* behave exactly like the corresponding non-asterisked connectives of L.

Consider the sentence ϕ of L^* that is written ∃^*x(x=^*x). This sentence, by stipulation, means that ∃x(x=x) is provable from the axioms of Euclidean geometry. According to the Quinean criterion plus inferentialism, it incurs existential commitment, because ∃^*x, since it behaves inferentially just like an existential quantifier, is an existential quantifier. Now, it is intuitively correct that ∃^*x(x=^*x) does incur existential commitment: it claims that there is a proof of ∃x(x=x), so it incurs existential commitment to the existence of a proof. So in this case, the inferentialist Quinean gets right that there is existential commitment. But rather clearly only coincidentally so! For now consider the sentence ψ that is written ∀^*x(x=^*x). Since ∀^*x behaves inferentially just like ∀x, by inferentialist Quineanism it incurs no existential commitment. But ψ means that there is a proof of ∀x(x=x), and hence incurs exactly the same kind of existential commitment as ϕ did, which said that there was a proof of ∃x(x=x).

What can the inferentialist Quinean respond? Perhaps this: The language L^* is syntactically and inferentially compositional, but not semantically so. The meaning of p∨^*q, namely that the unasterisked version of p∨^*q has a proof, is not composed from the meanings of p and of q, which respectively mean that p has a proof and that q has a proof. But that’s not quite right. For meaning-composition is just a function from meanings to meanings, and there is a function from the meanings of p and of q to the meaning of p∨^*q—it’s just a messy function, rather than the nice function we normally associate with disjunction.

Perhaps what the inferentialist Quinean should do is to insist on the intuitive non-inferentialist semantic compositional meanings for the truth-functional connectives, but not for the quantifiers. This feels ad hoc.

Even apart from Quineanism, I think the above constitutes an argument against inferentialism about logical connectives. For the asterisked connectives of L^* do not mean the same thing as their unasterisked variants in L.