Field’s logic of truth

Appendix on validity

We want to show that, for arguments with countably many premises, ℵ₀-validity coincides with validity. Then we’ll use the proof to determine the complexity of the set of valid sentences.

To say that the argument from a countable premiss set Γ to φ is invalid is to say that there is a fixed-point model with standard integers in which all the members of Γ have value 1 but φ does not. This can be formalized as a Σ₁ sentence of the language of set theory. If it’s true, then it’s true in some model of the form <V_λ,∈>, where λ is a strong limit cardinal. Taking the transitive collapse of a countable elementary submodel, we’ll still get a fixed-point model with standard integers in which all the members of Γ have value 1 but φ does not, only now the fixed-point model will be countable. The existence of such a model shows that the inference is not ℵ₀-valid.

The same argument shows that, for any α, an argument with ℵ_α or fewer premisses is valid iff it’s ℵ_α-valid.

Turning to examine the set of valid sentences, it is clear that a complete axiomatization is out of the question. For a theory of truth to be any use at all, it needs to be able to identify sentences syntactically. If we are encoding the syntax by Gödel numbers, having a syntactic theory is tantamount to having an arithmetical theory. To avoid fretting over irrelevant pathologies that arise in models that misrepresent the syntax, Field lays it down that the only models to be considered are ones that have standard integers. This ensures the set of set of valid sentences will be at least as complex as the set of arithmetical truths. It turns out that it’s significantly more complex, specifically, complete $\Uppi_{2}^{1}$.

φ is invalid iff there is a well-founded model of Zermelo set theory⁹ in which there is a fixed-point model that doesn’t assign φ the value 1. The Löwenheim–Skolem argument shows that this happens iff there is a well-founded model of Zermelo set theory with domain the set of natural numbers in which there is a fixed-point model in which φ is assigned a value different from 1. This can be formalized as a $\Upsigma_{2}^{1}$ arithmetical statement.¹⁰

We need to show that every $\Uppi_{2}^{1}$ set of natural numbers is 1-reducible to the set of valid sentences. For i a natural number and A a set of natural numbers, define the ith tree in A to be the set of all finite sequences x with property that x and all its initial segments are in the set whose characteristic function is calculated by the ith oracle Turing machine with an oracle for A, or to be the empty set if the machine doesn’t calculate a characteristic function. The set of numbers i such that there is an infinite path through the ith tree in A is a complete $\Upsigma_{1}^{1}$-in-A set,¹¹ so {i: for each set A, there is an infinite path through the ith tree in A} is a complete $\Uppi_{2}^{1}$ set. We want to show that this set is 1-reducible to the set of valid sentences.

In the language obtained from the language of arithmetic by adding, in addition to the truth predicate “T” and the new conditional “→,” a new monadic predicate “R,” form, using Gödel’s self-referential technique, a formula σ(i , x) equivalent to:

x is in the ith tree in the set of natural numbers that satisfy R ∧ (T(^┌σ(i,x)^┐) → (∃y) (y is a member of the ith tree in the set of natural numbers that satisfy R ∧ y extends x ∧ T(^┌σ(i,y)^┐))).

Where < > is the empty sequence, we want to see that σ(i , < >) is valid iff, for each set A, there in an infinite path through the ith tree in A.

Suppose that (∀A)(there is an infinite path through the ith tree for A). Take a fixed-point model, and let A be the set of natural numbers the model places into the extension of “R.” We know that there is an infinite path through the ith tree in A. We want to see that, for that every node x along the path, σ(i , x) is assigned the value 1 by the model; this will show that σ(i, < >) is assigned the value 1 by the model. We know that σ(i , x) can’t be assigned the value 0, because that would give us a conditional with value 0 whose antecedent had value 0, which is impossible. The possibility we have to worry about is that σ(i , x) has value ½. If σ(i , x) has value ½, then we have a ½-valued conditional with a ½-valued antecedent, so the consequent must have value 0 or ½. If we take z to be a node on the path further along than x, σ(i , z) can’t have the value 1, since if it did “(∃y) (y is a member of the ith tree in the set of natural numbers that satisfy R ∧ y extends x ∧ T(^┌σ(i,y)^┐))” would have value 1. It can’t have value 0, for the same reason σ(i , x) can’t have value 0. So σ(i , z) must have value ½, which implies that “(∃y) (y is a member of the ith tree in the set of natural numbers that satisfy R ∧ y extends x ∧ T(^┌σ(i , y)^┐))” has value ½. Where Δ is an acceptable ordinal, both the antecedent and the consequent of the conditional “(T(^┌σ(x)^┐) → (∃y) (y is a member of the ith tree in the set of natural numbers that satisfy R ∧ y extends x ∧ T(^┌σ(i , y)^┐)))” have the value ½ at level Δ, which means that the conditional, and hence its antecedent and consequent, will assume the value 1 at level Δ+1. Once these sentences achieve the value 1, they’ll maintain the value 1 at every subsequent level, with z serving as a witness to the consequent’s existential claim. So the ultimate value of σ(i , x) is 1, contrary to hypothesis.

Now for the other direction, suppose that there is a set A for which there is no infinite path through the ith tree for A. This means that the partial order on the nodes of the tree that we get by stipulating that y is less than x iff y extends x is well-founded, so that we can do inductions. We want to show that, for every node x of the tree, σ(i , x) is assigned a value different from 1 in the fixed-point model obtained by taking the ground model to be the natural number system and taking A as the extension of “R.” We may assume the tree is nonempty, since otherwise σ(i , x) has value 0, for every x. Suppose that x is a node of the tree and that, for every y below x, σ(i , y) is assigned a value different from 1. Then the consequent of the conditional “T(^┌σ(i,x)^┐) → (∃y) (y is a member of the ith tree in the set of natural numbers that satisfy R ∧ y extends x ∧ T(^┌σ(i,y)^┐))” has a value different from 1. If σ(i , x) had value 1, we’d be assigning the value 1 to a conditional whose antecedent had value 1 and whose consequent had a value different from 1, which is absurd. It follows by induction that the value of σ(i,< >) is different from 1.¹²