Meta-inferences and Supervaluationism

Abstract

Many classically valid meta-inferences fail in a standard supervaluationist framework. This allegedly prevents supervaluationism from offering an account of good deductive reasoning. We provide a proof system for supervaluationist logic which includes supervaluationistically acceptable versions of the classical meta-inferences. The proof system emerges naturally by thinking of truth as licensing assertion, falsity as licensing negative assertion and lack of truth-value as licensing rejection and weak assertion. Moreover, the proof system respects well-known criteria for the admissibility of inference rules. Thus, supervaluationists can provide an account of good deductive reasoning. Our proof system moreover brings to light how one can revise the standard supervaluationist framework to make room for higher-order vagueness. We prove that the resulting logic is sound and complete with respect to the consequence relation that preserves truth in a model of the non-normal modal logic NT. Finally, we extend our approach to a first-order setting and show that supervaluationism can treat vagueness in the same way at every order. The failure of conditional proof and other meta-inferences is a crucial ingredient in this treatment and hence should be embraced, not lamented.

Appendix

We prove that the instances of the Δ-Tolerance Principle (i.e. ⊖ΔFa_n ∧Δ¬Fa_n+ 1 for any n) are derivable from the Borderline, Trichotomy, Successor and Monotonicity Principles in QSML⁻ augmented with the Rigidity Rule. The Rigidity Rule says that identity is rigid across precisifications.

For the proof, we require the following derived rules; note that (+ Δ I.) may be used in the subderivations of (⊕∃E.) and (⊕∨E.).

Proof of (⊕∃E.):

Proof of (⊕∨E.):

The derivations of the (⊕∧)-rules are straightforward applications of (Weak Inference) and the (+ ∧)-rules.

Now recall that from Zardini’s proof we get the following.

$$(\text{Z})\quad + \forall m \forall n (\neg{{\varDelta}} Fa_{m} \land \neg{{\varDelta}}\neg Fa_{m} \wedge {{\varDelta}} Fa_{n}\land{{\varDelta}}\neg Fa_{n+1}) \rightarrow a_{m} = a_{n} \vee a_{m} = a_{n+1}.$$

Then note that (∗) ⊕ (a_j = a_i), +ΔFa_i ∧Δ¬Fa_i+ 1,⊕¬ΔFa_j ∧¬Δ¬F_aj ⊩⊥.

By an analogous proof, (∗′) ⊕ (a_j = a_i+ 1), +ΔFa_i ∧Δ¬Fa_i+ 1,⊕¬ΔFa_j ∧¬Δ¬Fa_j ⊩⊥. Then derive any instance of the Δ-Tolerance Principle as follows.

The argument requires the use of (⊕∃E.) and (⊕∨E.) since the proofs of ⊥ from ⊕ (a_j = a_i) and ⊕ (a_j = a_i+ 1) require the Δ-Intersubstitutivity Rule, which is not licit under their counterparts with + . For the same reason, one cannot apply reductio in the final step of the proof to derive the F-Gap Principle.