Notes on Stratified Semantics

Abstract

In 1988, Kit Fine published a semantic theory for quantified relevant logics. He referred to this theory as stratified semantics. While it has received some attention in the literature (see, e.g. Mares, Studia Logica 51(1), 1–20, 1992; Mares & Goldblatt, Journal of Symbolic Logic 71(1), 163–187, 2006), stratified semantics has overall received much less attention than it deserves. There are two plausible reasons for this. First, the only two dedicated treatments of stratified semantics available are (Fine, Journal of Philosophical Logic 17(1), 27–59, 1988; Mares, Studia Logica 51(1), 1–20, 1992), both of which are quite dense and technically challenging. Second, there are a number of prima facie reasons to be worried about stratified semantics. The purpose of this paper is to revitalize research on stratified semantics. I will do so by giving a ‘user friendly’ presentation of the semantics, and by giving reasons to think that the prima facie reasons to be worried about it are too simplistic.

Appendices

Appendix A: Soundness For R W Z

Let $M=\langle D,S,N,R,\delta ,\mathcal {E}^{+},\mathcal {E}^{-}\rangle$ be an RWZ-model.

Lemma 1 (Horizontal Heredity)

Ifa ≤ bthen if 1 ∈ M^a(ϕ) then 1 ∈ M^b(ϕ) and if 0 ∈ M^a(ϕ) then 0 ∈ M^b(ϕ).

Proof

The proof is by induction on the complexity of ϕ. If ϕ is atomic, then the result follows from atomic horizontal heredity. Most of the induction cases are trivial. We present only the conditional cases here, and leave the rest to the reader.

1 ∈ M^a(ϕ → ψ) iff for all c and d, if Racd and 1 ∈ M^c(ϕ), then 1 ∈ M^d(ψ), and if Racd and 0 ∈ M^c(ψ), then 0 ∈ M^d(ϕ). But if Rbcd then since a ≤ b, Racd as well. The desired result follows from this observation almost immediately.

On the other hand, 0 ∈ M^a(ϕ → ψ) iff for some c and d with Rcda, 1 ∈ M^c(ϕ) and 0 ∈ M^d(ψ). Since a ≤ b, there is an n ∈ N such that Rnab. Thus, by Permutation, Ranb as well. But from Rcda and Ranb it follows by B’ that there is a z so that Rcnz and Rdzb. From Rdzb and Rcnz it follows by permutation that Rzdb and Rncz. From the latter, we see that c ≤ z. Thus, by the inductive hypothesis, 1 ∈ M^z(ϕ). But since we already know that 0 ∈ M^d(ψ) and that Rzdb, this tells us that 0 ∈ M^b(ϕ → ψ). □

Theorem 1

Ifα ∈RWZ,thenαisRWZ-valid.

Proof

The proof is straightforward: we show each axiom is valid, then that the rules preserve validity. For the axioms, we present here only the argument for A8 and leave the rest to the reader. For rules, we only examine R1.

A8::

$M\vDash (\alpha \to \beta )\to ((\beta \to \gamma )\to (\alpha \to \gamma ))$ iff 1 ∈ Mⁿ[(α → β) → ((β → γ) → (α → γ))] for all n ∈ N. This happens, in turn, iff if Rnab, then (a) if 1 ∈ M^a(α → β) then 1 ∈ M^b[(β → γ) → (α → γ)], and (b) if 0 ∈ M^a[(β → γ) → (α → γ)] then 0 ∈ M^b(α → β).

For (a) suppose 1 ∈ M^a(α → β). Then by heredity 1 ∈ M^b(α → β). Note that 1 ∈ M^b[(β → γ) → (α → γ)] iff for all c and d if Rbcd then (i) if 1 ∈ M^c(β → γ) then 1 ∈ M^d(α → γ), and (ii) if 0 ∈ M^c(α → γ) then 0 ∈ M^d(β → γ).

Suppose Rbcd. For (i) let 1 ∈ M^c(β → γ). To show that 1 ∈ M^d(α → γ) we need to show that for all e and f if Rdef then (i^‡) if 1 ∈ M^e(α) then 1 ∈ M^f(γ) and (i^‡‡) if 0 ∈ M^e(γ) then 0 ∈ M^f(α). For (i^‡), suppose Rdef and 1 ∈ M^e(α). Since Rbcd and Rdef, there is some z so that Rczf and Rbez. Since Rbez, 1 ∈ M^b(α → β) and 1 ∈ M^e(α), we get that 1 ∈ M^z(β). Then since Rczf, 1 ∈ M^c(β → γ) and 1 ∈ M^z(β), we get that 1 ∈ M^f(γ) as required. For (i^‡‡), suppose Rdef and 0 ∈ M^e(γ). Since Rbcd, Rcbd as well. With this and Rdef we get that there is some z so that Rcez and Rbzf. From Rcez, 1 ∈ M^c(β → γ) and 0 ∈ M^e(γ), we get that 0 ∈ M^z(β). Finally, from Rbzf, 1 ∈ M^b(α → β) and 0 ∈ M^z(β), we get that 0 ∈ M^f(α) as required.

For (ii), suppose Rbcd (so by permutation Rcbd as well) and 0 ∈ M^c(α → γ). Then there are e and f so that Refc, 1 ∈ M^e(α) and 0 ∈ M^f(γ). By B’, since Refc and Rcbd there is a z so that Rebz and Rfzd. From Rebz, permutation gives Rbez. This together with 1 ∈ M^b(α → β) and 1 ∈ M^e(α) gives us that 1 ∈ M^z(β). From Rfzd permutation gives Rzfd. Since 1 ∈ M^z(β) and 0 ∈ M^f(γ), we get that 0 ∈ M^d(β → γ) as required.

For (b), suppose 0 ∈ M^a[(β → γ) → (α → γ)]. It follows by heredity that 0 ∈ M^b[(β → γ) → (α → γ)]. So for some c and d, Rcdb and 1 ∈ M^c(β → γ) and 0 ∈ M^d(α → γ). Thus, for some e and f , Refd, 1 ∈ M^e(α) and 0 ∈ M^f(γ). From Rcdb we get Rdcb while from Refd we get Rfed. Together Rfed and Rdcb give that there is a z so that Rfcz and Rezb. But from Rfcz we get Rcfz, and since 1 ∈ M^c(β → γ) and 0 ∈ M^f(γ), we get that 0 ∈ M^z(β). Then from Rezb, 1 ∈ M^e(α) and 0 ∈ M^z(β) we get 0 ∈ M^b(α → β) as required.

R1::

Suppose $M\vDash \alpha$ and $M\vDash \alpha \to \beta$. Let n ∈ N. Since n ≤ n, for some m ∈ N, Rmnn. Since m ∈ N and $M\vDash \alpha \to \beta$, 1 ∈ M^m(α → β). Since n ∈ N, and $M\vDash \alpha$, 1 ∈ Mⁿ(α). Thus, since Rmnn, 1 ∈ Mⁿ(β). Since n was an arbitrary element of N, $M\vDash \beta$ as required.

□

Appendix B: Completeness For R W Z

Our completeness proof is via the Post-Lindenbaum route. Explicitly, this means we will use the proof theory itself to construct a canonical model. Before we get going, we’ll need a variety of fairly standard definitions. Let Σ ∪Π∪Δ∪{ϕ}⊆LZ.

• Write Σ ⊩ ϕ when there is a sequence ϕ₁,…,ϕ_n = ϕ such that each ϕ_i is either in RWZ ∪Σ or follows from previous members of the sequence by R1 or by R2.

• Write Π_→ for the set of sentences in Π that are of the form α → β.

• Write Σ ⊩_Πϕ when Σ ∪Π_→⊩ ϕ.

• Say Σ is a Π-theory when (i) α ∈Σ and ⊩_Πα → β only if β ∈Σ and (ii) α ∈Σ and β ∈Σ only if α ∧ β ∈Σ.

• Say Σ is prime when α ∨ β ∈Σ only if α ∈Σ or β ∈Σ.

• Say that Σ is closed under disjunction when α ∈Σ and β ∈Σ only if α ∨ β ∈Σ.

• Write Σ ⊩_ΠΔ when there are δ₁,…,δ_n all in Δ such that Σ ⊩_Πδ₁ ∨⋯ ∨ δ_n.

• Write ⊩_ΠΣ→Δ when there are σ₁,…,σ_n all in Σ and δ₁,…,δ_m all in Δ such that ⊩_Π(σ₁ ∧⋯ ∧ σ_n) → (δ₁ ∨⋯ ∨ δ_m).

• Say that 〈Σ,Δ〉 is a Π-partition when Σ ∪Δ = LZ and ⊯_ΠΣ →Δ.

• Say that Π is normal when RWZ ⊆Π.

2.1 B.1 Facts About Provability

The first thing we need are some facts about the provability relation. Let Π ∪{α,β,γ,ρ}⊆LZ. Then, as can be verified by tediously constructing derivations, all of the following hold:

Fact 1::

⊩_Πα →¬¬α

Fact 2::

If ⊩_Πα → β and ⊩_Πβ → γ, then ⊩_Πα → γ

Fact 3::

⊩_Π(α → (β → γ)) → (β → (α → γ))

Fact 4::

⊩_Π(α → β) → (¬β →¬α)

Fact 5::

⊩_Π¬α →¬(α ∧ β)

Fact 6::

⊩_Π¬β →¬(α ∧ β)

Fact 7::

⊩_Π¬(α ∧ β) → (¬α ∨¬β)

Fact 8::

⊩_Π(α → (β → γ)) → (α → (¬γ →¬β))

Fact 9::

⊩_Π(¬α → β) → (¬β → α)

Fact 10::

⊩_Π(γ → ρ) → ((α → (β → γ)) → (α → (β → ρ)))

Fact 11::

⊩_Π((α → β) ∧ (δ → ρ)) → ((α ∧ δ) → (β ∧ ρ))

Fact 12::

⊩_Π((α → δ) ∧ (β → δ)) → ((α ∨ β) → δ)

2.2 B.2 Partition and Primality Lemmas

The proofs of these results are fairly standard (see for example Section 11.3 of [24]). But the techniques they introduce are ubiquitous, so they are worth quickly going through.

Lemma 2

If 〈Σ,Δ〉 is a Π-partition,then Σ is a prime Π-theory.

Proof

Suppose α and β are in Σ. If α ∧ β∉Σ, then since Σ ∪Δ = LZ, α ∧ β ∈Δ. But then since (α ∧ β) → (α ∧ β) ∈RWZ, ⊩_Π(α ∧ β) → (α ∧ β), so ⊩_ΠΣ→Δ. This contradicts 〈Σ,Δ〉 being a Π-partition, so α ∧ β ∈Σ. Now suppose α ∈Σ and ⊩_Πα → β. Then if β ∈Δ, then ⊩_ΠΣ→Δ. So since ⊯_ΠΣ →Δ, we must have β∉Δ. But then β ∈Σ. So Σ is a theory.

Now suppose α ∨ β ∈Σ but that α∉Σ and β∉Σ. Then α and β are in Δ. Since (α ∨ β) → (α ∨ β) ∈RWZ, ⊩_Π(α ∨ β) → (α ∨ β). But then ⊩_ΠΣ→Δ. So since ⊯_ΠΣ →Δ, either α ∈Σ or β ∈Σ. So Σ is prime. □

Lemma 3

If ⊯_ΠΣ →Δ then there is a Π-partition〈Σ^′,Δ^′〉 with Σ^′⊇Σ and Δ^′⊇Δ.

Proof

Let ϕ₁, ϕ₂,… be an enumeration of LZ, Σ₀ = Σ, Δ₀ = Δ, and for i > 0 define Σ_i and Δ_i as follows:

• If ⊯_ΠΣ_i− 1 ∪{ϕ_i− 1}→Δ_i− 1, then Σ_i = Σ_i− 1 ∪{ϕ_i− 1} and Δ_i = Δ_i− 1.

• If ⊩_ΠΣ_i− 1 ∪{ϕ_i− 1}→Δ_i− 1, then Σ_i = Σ_i− 1 and Δ_i = Δ_i− 1 ∪{ϕ_i− 1}.

Let ${\Sigma }^{\prime }=\cup _{i = 0}^{\infty }{\Sigma }_{i}$ and ${\Delta }^{\prime }=\cup _{i = 0}^{\infty }{\Delta }_{i}$. It’s clear that Σ^′∪Δ^′ = LZ. It claim that in fact 〈Σ^′,Δ^′〉 is a Π-partition. To see this, note that if it isn’t, then ⊩_ΠΣ^′→Δ^′. So for some $\phi _{i_{1}},\dots ,\phi _{i_{n}}$ all in Σ and $\phi _{j_{1}},\dots ,\phi _{j_{m}}$ all in Δ, $\vdash _{\Pi }(\phi _{i_{1}}\land \dots \land \phi _{i_{n}})\to (\phi _{j_{1}}\lor \dots \lor \phi _{j_{m}})$. Letting M = max{i₁,…,i_n, j₁,…,j_m}, we then have that ⊩_ΠΣ_M →Δ_M. Thus there is a least k for which ⊩_ΠΣ_k →Δ_k. Call this number k₀.

By assumption, k₀≠ 0. So ${\Sigma }_{k_{0}-1}$ and ${\Delta }_{k_{0}-1}$ are defined. By construction either ${\Sigma }_{k_{0}}={\Sigma }_{k_{0}-1}$ or ${\Delta }_{k_{0}}={\Delta }_{k_{0}-1}$. The latter case only occurs if we also have that ${\Sigma }_{k_{0}}={\Sigma }_{k_{0}-1}\cup \{\phi _{k_{0}-1}\}$ and $\not \vdash _{\Pi }{\Sigma }_{k_{0}-1}\cup \{\phi _{k_{0}-1}\}\to {\Delta }_{k_{0}-1}$. But then clearly $\not \vdash _{\Pi }{\Sigma }_{k_{0}}\to {\Delta }_{k_{0}}$, which is a contradiction.

On the other hand, ${\Sigma }_{k_{0}}={\Sigma }_{k_{0}-1}$ only if both

• (a) $\vdash _{\Pi }{\Sigma }_{k_{0}-1}\cup \{\phi _{k_{0}-1}\}\to {\Delta }_{k_{0}-1}$, and

• (b) ${\Delta }_{k_{0}}={\Delta }_{k_{0}-1}\cup \{\phi _{k_{0}-1}\}$.

It then follows from $\vdash _{\Pi }{\Sigma }_{k_{0}}\to {\Delta }_{k_{0}}$, (b), and the fact that ${\Sigma }_{k_{0}}={\Sigma }_{k_{0}-1}$ that

• (c) $\vdash _{\Pi }{\Sigma }_{k_{0}-1}\to {\Delta }_{k_{0}-1}\cup \{\phi _{k_{0}-1}\}$.

Also, since k₀ is minimal, we have

• (d) $\not \vdash _{\Pi }{\Sigma }_{k_{0}-1}\to {\Delta }_{k_{0}-1}$.

It follows from (a) and (d) that for some $\sigma \in {\Sigma }_{k_{0}-1}$ and $\delta \in {\Delta }_{k_{0}-1}$ we have

• (e) $\vdash _{\Pi }(\sigma \land \phi _{k_{0}-1})\to \delta$.

And it follows from (c) and (d) that for some $\sigma ^{\prime }\in {\Sigma }_{k_{0}-1}$ and $\delta ^{\prime }\in {\Delta }_{k_{0}-1}$ we have

• (f) $\vdash _{\Pi }\sigma ^{\prime }\to (\delta ^{\prime }\lor \phi _{k_{0}-1})$.

From (e) and (f) it follows (by a mildly annoying derivation) that $\vdash _{\Pi }{\Sigma }_{k_{0}-1}\to {\Delta }_{k_{0}-1}$, which contradicts k₀’s minimality. So 〈Σ^′,Δ^′〉 is a Π-partition. □

Lemma 4

If Σ is a Π-theory,Δ is closed under disjunction, and Σ ∩Δ = ∅then there is a prime Π-theoryΣ^′⊃Σ with Σ^′∩Δ = ∅.

Proof

If ⊩_ΠΣ→Δ, then ⊩_Π(σ₁ ∧⋯ ∧ σ_n) → (δ₁ ∨⋯ ∨ δ_m) for some σ₁,…,σ_n ∈Σ and δ₁,…,δ_m ∈Δ. Since Σ is a Π-theory and each σ_i ∈Σ, clearly σ₁ ∧⋯ ∧ σ_n ∈Σ. It follows that δ₁ ∨⋯ ∨ δ_m ∈Σ. But Δ is closed under disjunction, so δ₁ ∨⋯ ∨ δ_m ∈Δ. This contradicts Σ ∩Δ = ∅, so ⊯_ΠΣ →Δ. Lemmas 2 and 3 then give the result. □

Lemma 5

Suppose Σ⊯_Πα. Then there is a prime Π-theoryΣ^′⊇Σ withα∉Σ^′.

Proof

If Σ is empty, then ⊯_ΠΣ →{α}. On the other hand, if Σ is nonempty and ⊩_ΠΣ→{α}, then ⊩_Π(σ₁ ∧⋯ ∧ σ_n) → α for some σ₁,…,σ_n in Σ. Since each σ_i ∈Σ, clearly Σ ⊩_Πσ₁ ∧⋯ ∧ σ_n. It then follows that Σ ⊩_Πα. So since Σ⊯α, it is also the case that ⊯_ΠΣ →{α}. Thus by Lemma 3, there is a Σ^′⊇Σ and Γ ⊇{α} so that 〈Σ^′,Γ〉 is a Π-partition. But then by Lemma 2, Σ^′ is a prime Π-theory and since ⊩_Πα → α, it follows from the definition of a Π-partition that α∉Σ^′. □

2.3 B.3 Extension Lemmas

First, for Σ ∪Δ∪Γ⊆LZ, define the relation $\mathcal {R}$ by saying $\mathcal {R}{\Sigma }{\Delta }{\Gamma }$ iff if δ → γ ∈Σ and δ ∈Δ, then γ ∈Γ. Each of the following lemmas proves that under certain conditions, if $\mathcal {R}{\Sigma }{\Delta }{\Gamma }$, then we can extend one or more of Σ, Γ, or Δ to a prime theory without breaking the relation. Each is also proved in generally the same way: first, we construct a set Θ of sentences that we want to avoid. We then complete the required theory(ies) away from Θ using Lemma 4.

Lemma 6

If Σ,Γ,and Δ are Π-theories,$\mathcal {R}{\Sigma }{\Gamma }{\Delta }$andΔ is a prime Π-theory,then there is a prime Π-theoryΓ^′⊇Γ such that$\mathcal {R}{\Sigma }{\Gamma }^{\prime }{\Delta }$

Proof

We want to extend Γ^′ to a prime theory without adding to it the antecedent of any conditional in Σ whose consequent isn’t in Δ. So we let Θ, the set of sentences we want to avoid, be {α : α → β ∈Σ for some β∉Δ}. I claim Θ is closed under disjunctions. To see this, notice that if α₁ and α₂ are in Θ, then there are β₁ and β₂ not in Δ such that α₁ → β₁ and α₂ → β₂ are in Σ and since Δ is prime and β₁ and β₂ are not in Δ, β₁ ∨ β₂∉Δ. By an instance of A8 we have that ⊩_Π(α₁ → β₁) → ((β₁ → (β₁ ∨ β₂)) → (α₁ → (β₁ ∨ β₂))). So by Fact 3, ⊩_Π(β₁ → (β₁ ∨ β₂)) → ((α₁ → β₁) → (α₁ → (β₁ ∨ β₂))). But using Fact 2, the definition of ‘∨’ and instances of Facts 1 and 5, we get that ⊩_Πβ₁ → (β₁ ∨ β₂). So ⊩_Π(α₁ → β₁) → (α₁ → (β₁ ∨ β₂)). Since α₁ → β₁ ∈Σ and Σ is a Π-theory, it then follows that α₁ → (β₁ ∨ β₂) ∈Σ. So β₁ ∨ β₂ ∈Θ.

Suppose α ∈Γ∩Θ. Then since α ∈Θ, for some β∉Δ, α → β ∈Σ. But from $\mathcal {R}{\Sigma }{\Gamma }{\Delta }$, α → β ∈Σ and α ∈Γ it follows that β ∈Δ, a contradiction. So Γ ∩Θ must be empty. So by Lemma 4, there is a prime Π-theory Γ^′⊇Γ with Γ^′∩Θ = ∅. Now suppose α → β ∈Σ and α ∈Γ^′. If β∉Δ, then α ∈Θ. But then α ∈Γ^′∩Θ, which is a contradiction. So β ∈Δ. Thus $\mathcal {R}{\Sigma }{\Gamma }^{\prime }{\Delta }$. □

Lemma 7

If Σ,Γ,and Δ are Π-theories,$\mathcal {R}{\Sigma }{\Gamma }{\Delta }$,and Δ is a prime Π-theory,then there is a prime Π-theoryΣ^′⊇Σ such that$\mathcal {R}{\Sigma }^{\prime }{\Gamma }{\Delta }$

Proof

The hard part here is identifying the sentences we want to avoid. The idea to have is this: we don’t want to add to Σ anything that forces us to also add to it some conditional whose antecedent is in Γ, but whose consequent isn’t in Δ. Thus, this time we let Θ be {α :⊩_Πα → (γ → β) for some γ ∈Γ and β∉Δ}. By almost the same argument as in the previous lemma, we see that Θ is closed under disjunction.

Suppose that α ∈Σ∩Θ. Then ⊩_Πα → (γ → β) for some γ ∈Γ and β∉Δ. Since Σ is a Π-theory, it follows that γ → β ∈Σ. But then β ∈Δ because $\mathcal {R}{\Sigma }{\Gamma }{\Delta }$, γ → β ∈Σ and γ ∈Γ. Since this is a contradiction, Σ ∩Θ = ∅. So by Lemma 4, there is a prime Π-theory Σ^′ with Σ^′⊇Σ and Σ^′∩Θ = ∅.

Finally, suppose α → β ∈Σ^′. Since Σ^′∩Θ = ∅, α → β∉Θ. Since ⊩_Π(α → β) → (α → β), it follows from the definition of Θ that if α ∈Γ and β∉Δ, then α → β ∈Θ. Since this is impossible, if α ∈Γ, then β ∈Δ. So $\mathcal {R}{\Sigma }^{\prime }{\Gamma }{\Delta }$. □

Lemma 8

Let Σ be a prime Π-theoryandγ → δ∉Σ. Then there are prime Π-theoriesΓ and Δ such that$\mathcal {R}{\Sigma }{\Gamma }{\Delta }$,γ ∈Γ,andδ∉Δ.

Proof

To begin, let Γ^′ = {α :⊩_Πγ → α} and Δ^′ = {β : α → β ∈Σ for some α ∈Γ^′}. We leave it to the reader to verify that Γ^′ is a Π-theory. It’s clear from these definitions that $\mathcal {R}{\Sigma }{\Gamma }^{\prime }{\Delta }^{\prime }$ and γ ∈Γ^′. To see that δ∉Δ^′, notice that if δ were in Δ^′, then there would be an α ∈Γ^′ with α → δ ∈Σ. But if α ∈Γ^′, then ⊩_Πγ → α. But also by A8, ⊩_Π(γ → α) → ((α → δ) → (γ → δ)). So if ⊩_Πγ → α, then ⊩_Π(α → δ) → (γ → δ). Since Σ is a Π-theory, it would then follow that γ → δ ∈Σ, which is a contradiction. So δ∉Δ^′.

To see that Δ^′ is a Π-theory, first let β₁ ∈Δ^′ and β₂ ∈Δ^′. Then for some α₁ ∈Γ^′ and α₂ ∈Γ^′, α₁ → β₁ ∈Σ and α₂ → β₂ ∈Σ. Since Γ^′ is a theory, α₁ ∧ α₂ ∈Γ^′. Since Σ is a theory, (α₁ → β₁) ∧ (α₂ → β₂) ∈Σ. By Fact 12, ⊩_Π((α₁ → β₁) ∧ (α₂ → β₂)) → ((α₁ ∧ α₂) → (β₁ ∧ β₂)). Thus (α₁ ∧ α₂) → (β₁ ∧ β₂) ∈Σ. So β₁ ∧ β₂ ∈Δ^′. Now suppose β ∈Δ^′ and ⊩_Πβ → ρ. Since β ∈Δ^′, α → β ∈Σ for some α ∈Γ^′. Applying Fact 3 to an instance of A8, we see that ⊩_Π(β → ρ) → ((α → β) → (α → ρ)). Thus, since ⊩_Πβ → ρ, it follows that ⊩_Π(α → β) → (α → ρ) as well. So since Σ is a theory, α → ρ ∈Σ. Thus ρ ∈Δ^′.

Let Θ be the closure of {δ} under disjunction. A trivial induction using instances of Fact 12 shows that if 𝜃 ∈Θ, then ⊩_Π𝜃 → δ. Thus, if 𝜃 ∈Δ^′∩Θ, then since Δ^′ is a theory, δ ∈Δ^′. So since δ∉Δ^′, Δ^′∩Θ = ∅. It follows by Lemma 4 that there is a prime theory Δ ⊇Δ^′ with Δ ∩Θ = ∅. In particular δ∉Δ, and since Δ ⊇Δ^′, $\mathcal {R}{\Sigma }{\Gamma }^{\prime }{\Delta }$. Thus by Lemma 6, there is a prime Π-theory Γ ⊃Γ^′ so that $\mathcal {R}{\Sigma }{\Gamma }{\Delta }$. Since γ ∈Γ^′, γ ∈Γ as well, finishing the proof. □

2.4 B.4 Completeness

If Π is a prime theory, then the canonical premodel $\mathfrak {C}_{\Pi }$ is the septuple $\langle D,S,N,\mathcal {R},\delta ,\mathcal {E}^{+}_{\Pi },\mathcal {E}^{-}_{\Pi }\rangle$ such that

• D is the set of names,

• S is the set of prime Π-theories,

• N is the set of normal prime Π-theories,

• δ is the identity function,

• If P is an i-ary predicate, then $\mathcal {E}^{+}_{\Pi }(P,a)$ is the set of all i-tuples 〈n₁,…,n_i〉 such that Pn₁…n_i ∈ a, and

• If P is an i-ary predicate, then $\mathcal {E}^{-}_{\Pi }(P,a)$ is the set of all i-tuples 〈n₁,…,n_i〉 such that ¬Pn₁…n_i ∈ a.

Obviously this is a premodel. To see it’s a model, we must verify that it meets the ordering, monotonicity, closure, rearranging, and horizontal atomic heredity conditions. Of these, ordering, closure, and heredity follow immediately from the following lemma:

Lemma 9

In$\mathfrak {C}_{\Pi }$,a ≤ biffa ⊆ b

Proof

Let n be a normal prime Π-theory and suppose $\mathcal {R} nab$. Let α ∈ a. Then since n is normal, α → α ∈ n. So α ∈ b. Thus if a ≤ b, then a ⊆ b.

Now suppose a ⊆ b. If α → β ∈RWZ and α ∈ a, then since a is a Π-theory, and ⊩_Πα → β, β ∈ a. Thus β ∈ b. So $\mathcal {R}\mathbf {RWZ} ab$, and thus a ≤ b. So if a ⊆ b, then a ≤ b. □

For the remaining conditions, we argue as follows:

Monotonicity::

If $\mathcal {R} abc$, a^′≤ a, b^′≤ b, and c ≤ c^′, then by Lemma 9 a^′⊆ a, b^′⊆ b, and c ⊆ c^′. So if ϕ → ψ ∈ a^′ and ϕ ∈ b^′, then ϕ → ψ ∈ a and ϕ ∈ b, and since $\mathcal {R} abc$ it follows from these that ψ ∈ c. So since c ⊆ c^′, ψ ∈ c^′. So $\mathcal {R} a'b'c'$.

B::

Suppose $\mathcal {R} abcd$ – that is, that for some Π-theory x, $\mathcal {R} abx$ and $\mathcal {R} xcd$. Let y^′ = {α : γ → α ∈ b for some γ ∈ c}. The reader will easily verify that y^′ is a theory, and it is clear by definition that $\mathcal {R}{}bcy'$. I claim that $\mathcal {R}{}ay'd$ as well.

To see this, suppose ϕ → ψ ∈ a and ϕ ∈ y^′. Since ϕ ∈ y^′, for some γ ∈ c we have that γ → ϕ ∈ b. Using A8 and Fact 3, we see that ⊩_Π(ϕ → ψ) → ((γ → ϕ) → (γ → ψ)). Thus since ϕ → ψ ∈ a and a is a Π-theory, (γ → ϕ) → (γ → ψ) ∈ a as well. Thus since $\mathcal {R}{}abx$ and γ → ϕ ∈ b, γ → ψ ∈ x. And since γ → ψ ∈ x and γ ∈ c, we can conclude that ψ ∈ d as required.

Lemma 6 then gives that there is a Π-theory y ⊇ y^′ so that $\mathcal {R}{}ayd$. Clearly since y ⊇ y^′ and $\mathcal {R}{}bcy'$ it is also the case that $\mathcal {R}{}bcy$, completing the proof.

B’::

Again, suppose $\mathcal {R} abcd$. This time let y^′ = {α : γ → α ∈ a for some γ ∈ c}. Mutatis mutandis, the same argument as before works here.

C::

With the same assumptions as in the prior case, all that is required is that we show that $\mathcal {R}{}y'bd$. So let ϕ → ψ ∈ y^′ and ϕ ∈ b. Since ϕ → ψ ∈ y^′, γ → (ϕ → ψ) ∈ a for some γ ∈ c. By Fact 3, since γ → (ϕ → ψ) ∈ a and a is a Π-theory, ϕ → (γ → ψ) ∈ a as well. Thus, since ϕ ∈ b and $\mathcal {R}{}abx$, γ → ψ ∈ x. And thus since γ ∈ c, ψ ∈ d. So $\mathcal {R}{}y'bd$. The proof then finishes as before, replacing Lemma 6 with Lemma 7.

Lemma 10

$1\in \mathfrak {C}_{\Pi }^{a}(\phi )$iffϕ ∈ aand$0\in \mathfrak {C}_{\Pi }^{a}(\phi )$iff¬ϕ ∈ a.

Proof

By simultaneous induction on the complexity of ϕ in both (a) and (b). The base case and the cases involving conjunctions and negations are entirely straightforward so omitted.

For the ‘1’ part of the conditional case, first suppose ϕ → ψ ∈ a, $\mathcal {R}{}abc$, and $1\in \mathfrak {C}_{\Pi }^{b}(\phi )$. Then by the inductive hypothesis, ϕ ∈ b. So since $\mathcal {R}{}abc$, ψ ∈ c. It follows that $1\in \mathfrak {C}_{\Pi }^{c}(\psi )$. Now suppose instead that $0\in \mathfrak {C}_{\Pi }^{b}(\psi )$. Then by the inductive hypothesis, ¬ψ ∈ b. By Fact 4, ⊩_Π(ϕ → ψ) → (¬ψ →¬ϕ). So since ϕ → ψ ∈ a, ¬ψ →¬ϕ ∈ a as well. Thus since $\mathcal {R} abc$, ¬ϕ ∈ c. Thus, again by the inductive hypothesis, $0\in \mathfrak {C}_{\Pi }^{c}(\phi )$. Together these two pieces give that $1\in \mathfrak {C}_{\Pi }^{a}(\phi \to \psi )$ as required. For the other direction, note that if ϕ → ψ∉a, then by Lemma 8 there are b and c so that $\mathcal {R} abc$, ϕ ∈ b but ψ∉c. By the inductive hypothesis, then, $1\in \mathfrak {C}^{b}_{\Pi }(\phi )$ and $1\not \in \mathfrak {C}^{c}_{\Pi }(\psi )$. So $1\not \in \mathfrak {C}_{\Pi }^{a}(\phi \to \psi )$.

For the ‘0’ part of the conditional case, suppose $0\in \mathfrak {C}_{\Pi }^{a}(\phi \to \psi )$. Then for some b and c with $\mathcal {R} bca$, $1\in \mathfrak {C}^{b}_{\Pi }(\phi )$ and $0\in \mathfrak {C}^{c}_{\Pi }(\psi )$. By the inductive hypothesis, then, ϕ ∈ b and ¬ψ ∈ c. Since ⊩_Πϕ → ((ϕ → ψ) → ψ) and b is a Π-theory, (ϕ → ψ) → ψ ∈ b. By Fact 4, ⊩_Π((ϕ → ψ) → ψ) → (¬ψ →¬(ϕ → ψ)), so ¬ψ →¬(ϕ → ψ) ∈ b too. Thus, since $\mathcal {R} bca$ and ¬ψ ∈ c, we get that ¬(ϕ → ψ) ∈ a.

On the other hand, suppose ¬(ϕ → ψ) ∈ a. Let b = {β :⊩_Πϕ → β} and let c = {γ :⊩_Π¬ψ → γ}. Clearly ϕ ∈ b and ¬ψ ∈ c. For familiar reasons (see, e.g. the proof of Lemma 8), both of these are Π-theories. It is also true that $\mathcal {R}{}bca$. To see this, let β₁ → β₂ ∈ b and let β₁ ∈ c. Then by definition,

• (a) ⊩_Πϕ → (β₁ → β₂) and

• (b) ⊩_Π¬ψ → β₁.

From (a), an instance of Fact 8 gives

• (c) ⊩_Πϕ → (¬β₂ →¬β₁).

From (b), an instance of Fact 9 gives

• (d) ⊩_Π¬β₁ → ψ.

(c) and (d) together with an instance of, first, Fact 10 and, second, Fact 3 then give

• (e) ⊩_Π¬β₁ → (ϕ → ψ)

Thus, by Fact 10 again, ⊩_Π¬(ϕ → ψ) → β₁. So since a is a Π-theory and ¬(ϕ → ψ) ∈ a, β₁ ∈ a as required.

Since a is a prime Π-theory, Lemmas 6 and 7 then allow us to apply the inductive hypothesis and finish the proof. □

Theorem 2

IfαisRWZ-valid,thenα ∈RWZ.

Proof

Suppose α∉RWZ. It follows from this that RWZ⊯_∅α. But then by Lemma 5, there is a prime ∅-theory Π ⊇RWZ such that α∉Π. Since Π ⊇RWZ, Π is normal. By Lemma 10, since α∉Π, it follows that $1\not \in \mathfrak {C}^{\Pi }_{\Pi }(\alpha )$. So α is not true in $\mathfrak {C}_{\Pi }$ and thus not RWZ-valid. □

Appendix C: Soundness For R W Q

Let $M=\langle D, {\Omega }, \delta ,\mathcal {M},\Downarrow \rangle$ be a varying-domain RWQ-model and ϕ ∈LQ. Each of the first three lemmas we need can be proved with straightforward inductions which we leave to the reader.

Lemma 11 (Stability)

Ifd ∈ D_Xandνdoes not occur freely inϕ,then$1\in {M^{a}_{X}}(\mathtt {va},\phi )$iff$1\in {M^{a}_{X}}(\mathtt {va}^{\nu }_{d},\phi )$and$0\in {M^{a}_{X}}(\mathtt {va},\phi )$iff$0\in {M^{a}_{X}}(\mathtt {va}^{\nu }_{d},\phi )$.

Lemma 12 (Symmetry)

Ifa ∈ S_Xis symmetric in m and n, andis X-coherent then$1\in {M^{a}_{X}}(\mathtt {va}^{\nu }_{m},\phi )$iff$1\in {M^{a}_{X}}(\mathtt {va}^{\nu }_{n},\phi )$and$0\in {M^{a}_{X}}(\mathtt {va}^{\nu }_{m},\phi )$iff$0\in {M^{a}_{X}}(\mathtt {va}^{\nu }_{n},\phi )$.

Lemma 13 (Evaluation)

Ifa ∈ S_X,is X-coherent, and$d=\varepsilon ^{\mathtt {va}}_{X}(\tau )$,then$1\in {M^{a}_{X}}(\mathtt {va}^{\nu }_{d},\phi )$iff$1\in {M^{a}_{X}}(\mathtt {va},\phi (\tau /\nu ))$and$0\in {M^{a}_{X}}(\mathtt {va}^{\nu }_{d},\phi )$iff$0\in {M^{a}_{X}}(\mathtt {va},\phi (\tau /\nu ))$.

Lemma 14 (Horizontal Heredity)

If a and b are inS_X,is X-coherent, anda ≤ bthen if$1\in {M^{a}_{X}}(\mathtt {va},\phi )$then$1\in {M^{b}_{X}}(\mathtt {va},\phi )$,and if$0\in {M^{a}_{X}}(\mathtt {va},\phi )$then$0\in {M^{b}_{X}}(\mathtt {va},\phi )$.

Proof

Generally as in Lemma 1. The only novelty concerns the quantifiers. For these, note that $1\in {M^{a}_{X}}(\mathtt {va},\forall \nu \phi )$ iff for some $Y\supsetneq X$ and i ∈ Y − X, for all c ∈ S_Y if $c{\downarrow ^{Y}_{X}}=a$ then $1\in {M^{c}_{Y}}(\mathtt {va}^{\nu }_{\omega _{i}},\phi )$. Since a ≤ b, Rnab for some n ∈ N_X. Suppose $d{\downarrow ^{Y}_{X}}=b$. Then by the Extension Condition, there are e and f in S_Y such that Refd, $e{\downarrow ^{Y}_{X}}=n$ and $f{\downarrow ^{Y}_{X}}=a$. By the Normality Condition, $e{\downarrow ^{Y}_{X}}=n$ gives that e ∈ N_Y. So f ≤ d. Also, since $f{\downarrow ^{Y}_{X}}=a$, it follows that $1\in {M^{f}_{Y}}(\mathtt {va}^{\nu }_{\omega _{i}},\phi )$. It then follows by the inductive hypothesis that $1\in {M^{d}_{Y}}(\mathtt {va}^{\nu }_{\omega _{i}},\phi )$. Since d was an arbitrary setup in S_Y such that $d{\downarrow ^{Y}_{X}}=b$, it follows that $1\in {M^{b}_{X}}(\mathtt {va},\forall \phi )$. The zero case is essentially the same. □

Lemma 15 (Vertical Heredity)

Ifis X-coherent and${b{\downarrow ^{Y}_{X}}}=a$,then$1\in {M^{b}_{Y}}(\mathtt {va},\phi )$iff$1\in {M^{a}_{X}}(\mathtt {va},\phi )$and$0\in {M^{b}_{Y}}(\mathtt {va},\phi )$iff$0\in {M^{a}_{X}}(\mathtt {va},\phi )$.

Proof

By induction on the complexity of ϕ. The base case and the cases involving the extensional connectives are straightforward and omitted. The conditional case is simply a matter of keeping track of all the details, so is also omitted. This leaves the quantifier cases.

To begin, suppose $1\in {M^{b}_{Y}}(\mathtt {va},\forall \nu \phi )$. Then for some $Z\supsetneq Y$ and i ∈ Z − Y, for all c ∈ S_Z, if $c{\downarrow ^{Z}_{Y}}=b$, then $1\in {M^{c}_{Y}}(\mathtt {va}^{\nu }_{\omega _{i}},\phi )$. Following the proof of Lemma 4 in [13], let W = X ∪ (Z − Y ).¹ Notice that i ∈ W, W ∩ Y = X, W ∪ Y = Z, and that since is X-coherent, $\mathtt {va}^{\nu }_{\omega _{i}}$ is W-coherent and Z-coherent.

Let d ∈ S_W be such that $d\!{\downarrow ^{W}_{X}}=a$. Then notice $d\!\!\!\downarrow ^{W}_{W\cap Y}=d{\downarrow ^{W}_{X}}=a=b{\downarrow ^{Y}_{X}}=b\!\!\downarrow ^{Y}_{W\cap Y}$. Thus, by the Lifting condition, there is an e ∈ S_W∪Y = S_Z such that $e{\downarrow ^{Z}_{W}}\leq d$ and $b\leq e{\downarrow ^{Z}_{Y}}$. Since $b\leq e{\downarrow ^{Z}_{Y}}$, for some n ∈ N_Y, $Rnbe{\downarrow ^{Z}_{Y}}$. By the extension and normality conditions, then, for some m ∈ N_Z and f ∈ S_Z, Rmfe, $m{\downarrow ^{Z}_{Y}}=n$, and $f{\downarrow ^{Z}_{Y}}=b$. Since $f{\downarrow ^{Z}_{Y}}=b$, $1\in {M^{f}_{Z}}(\mathtt {va}^{\nu }_{\omega _{i}},\phi )$. So by Horizontal Heredity, $1\in {M^{e}_{Z}}(\mathtt {va}^{\nu }_{\omega _{i}},\phi )$. The inductive hypothesis then gives that $1\in M^{e{\downarrow ^{Z}_{W}}}_{W}(\mathtt {va}^{\nu }_{\omega _{i}},\phi )$. Since $e{\downarrow ^{Z}_{W}}\leq d$, Horizontal Heredity then gives that $1\in {M^{d}_{W}}(\mathtt {va}^{\nu }_{\omega _{i}},\phi )$. But d was an arbitrary element of S_w such that $d{\downarrow ^{W}_{X}}=a$, so all such elements have this feature. So $1\in {M^{a}_{X}}(\mathtt {va},\forall \nu \phi )$.

Now suppose $1\in {M^{a}_{X}}(\mathtt {va},\forall \nu \phi )$. Then for some $Z\supsetneq X$ and i ∈ Z − X, for all c ∈ S_Z, if $c{\downarrow ^{Z}_{X}}=a$, then $1\in {M^{c}_{Z}}(\mathtt {va}^{\nu }_{\omega _{i}},\phi )$. Let W = Y ∪ Z and let d ∈ S_W be such that $d{\downarrow ^{W}_{Y}}=b$. Notice that $d{\downarrow ^{W}_{Z}}{\downarrow ^{Z}_{X}}=d{\downarrow ^{W}_{X}}= d{\downarrow ^{W}_{Y}}{\downarrow ^{Y}_{X}}=b{\downarrow ^{Y}_{X}}=a$. Thus since for all c ∈ S_Z if $c{\downarrow ^{Z}_{X}}=a$, then $1\in {M^{c}_{Z}}(\mathtt {va}^{\nu }_{\omega _{i}},\phi )$, it follows that $1\in M^{d{\downarrow ^{W}_{Z}}}_{Z}(\mathtt {va}^{\nu }_{\omega _{i}},\phi )$. But then by the inductive hypothesis, $1\in {M^{d}_{W}}(\mathtt {va}^{\nu }_{\omega _{i}},\phi )$, so $1\in {M^{b}_{Y}}(\mathtt {va},\forall \nu \phi )$.

The zero case is analogous, replacing each instance of ‘1 ∈’ with ‘0∉’. □

Theorem 3

Ifα ∈RWQ,thenαisRWQ-valid.

Proof

As expected, we show that every axiom is valid and that the rules preserve validity. We will only cover QA1 and QA2, leaving the remaining axioms and all the rules to the reader.

QA1::

$M\vDash \forall \nu \phi \to \phi (\tau /\nu )$ iff if is X-coherent and n ∈ N_X, then $1\in {M_{X}^{n}}(\mathtt {va},\forall \nu \phi \to \phi (\tau /\nu ))$, iff for all a and b, if Rnab then (i) if $1\in {M_{X}^{a}}(\mathtt {va},\forall \nu \phi )$ then $1\in {M_{X}^{b}}(\mathtt {va},\phi (\tau /\nu ))$, and (ii) if $0\in {M_{X}^{a}}(\mathtt {va},\phi (\tau /\nu ))$ then $0\in {M_{X}^{b}}(\mathtt {va},\forall \nu \phi )$.

For (i), suppose $1\in {M_{X}^{a}}(\mathtt {va},\forall \nu \phi )$. Then for some $Y\supsetneq X$ and y ∈ Y − X, for all c ∈ S_Y, if ${c{\downarrow ^{Y}_{X}}}=a$, then $1\in {M^{c}_{Y}}(\mathtt {va}^{\nu }_{\omega _{y}},\phi )$. Since is X-coherent, $\varepsilon ^{\mathtt {va}}_{X}(\tau )\in D_{X}$ and ω_y ∈ D_Y − D_X, the symmetry condition guarantees that there is a c ∈ S_Y that is symmetric in ω_y and $\varepsilon ^{\mathtt {va}}_{X}(\tau )$ such that $c{\downarrow ^{Y}_{X}}=a$. It follows that $1\in {M^{c}_{Y}}(\mathtt {va}^{\nu }_{\omega _{y}},\phi )$. Since c is symmetric in ω_y and $\varepsilon ^{\mathtt {va}}_{X}(\tau )$, the Symmetry Lemma gives that $1\in {M^{c}_{Y}}(\mathtt {va}^{\nu }_{\varepsilon ^{\mathtt {va}}_{X}(\tau )},\phi )$. So by the Evaluation Lemma, $1\in {M^{c}_{Y}}(\mathtt {va},\phi (\tau /\nu ))$. Thus, since $c{\downarrow ^{Y}_{X}}=a$, the Heredity Lemma gives that $1\in {M_{X}^{a}}(\mathtt {va},\phi (\tau /\nu ))$. Since a ≤ b, Horizontal Heredity then gives $1\in {M^{b}_{X}}(\mathtt {va},\phi (\tau /\nu ))$.

For (ii), suppose $0\in {M_{X}^{a}}(\mathtt {va},\phi (\tau /\nu ))$. Let $Y\supsetneq X$ and y ∈ Y − X. By the symmetry condition, there is a c ∈ S_Y that is symmetric in ω_y and $\varepsilon ^{\mathtt {va}}_{X}(\tau )$ such that $c{\downarrow ^{Y}_{X}}=a$. By the Heredity Lemma, $0\in {M^{c}_{Y}}(\mathtt {va},\phi (\tau /\nu ))$. Thus, by the Evaluation Lemma, $0\in {M^{c}_{Y}}(\mathtt {va}^{\nu }_{\varepsilon ^{\mathtt {va}}_{X}(\tau )},\phi )$. So, by the Symmetry Lemma, $0\in {M^{c}_{Y}}(\mathtt {va}^{\nu }_{\omega _{y}},\phi )$. Since Y and y were arbitrary, this suffices to demonstrate that $0\in {M_{X}^{a}}(\mathtt {va},\forall \nu \phi )$. So $0\in {M_{X}^{b}}(\mathtt {va},\forall \nu \phi )$ follows by Horizontal Heredity.

QA2::

$M\vDash \forall \nu (\phi \to \psi )\to (\phi \to \forall \nu \psi )$ iff if is X-coherent and n ∈ N_X, then $1\in {M_{X}^{n}}[\mathtt {va},\forall \nu (\phi \to \psi )\to (\phi \to \forall \nu \psi )]$. This happens iff for all a and b, if Rnab then (i) if $1\in {M^{a}_{X}}[\mathtt {va},\forall \nu (\phi \to \psi )]$, then $1\in {M^{b}_{X}}(\mathtt {va},\phi \to \forall \nu \psi )$, and (ii) if $0\in {M^{a}_{X}}[\mathtt {va},\phi \to \forall \nu \psi ]$, then $0\in {M^{b}_{X}}[\mathtt {va},\forall \nu (\phi \to \psi )]$

For (i), suppose $1\in {M^{a}_{X}}[\mathtt {va},\forall \nu (\phi \to \psi )]$. Then by Horizontal Heredity, $1\in {M^{b}_{X}}(\mathtt {va},\forall \nu (\phi \to \psi ))$. We need to show that $1\in {M^{b}_{X}}(\mathtt {va},\phi \to \forall \nu \psi )$. To do so we must show that if Rbcd, then (a) if $1\in {M^{c}_{X}}(\mathtt {va},\phi )$, then $1\in {M^{d}_{X}}(\mathtt {va},\forall \nu \psi )$, and (b) if $0\in {M^{c}_{X}}(\mathtt {va},\forall \nu \psi )$, then $0\in {M^{c}_{X}}(\mathtt {va},\phi )$. In either case, we also have that $1\in {M^{b}_{X}}(\mathtt {va},\forall \nu (\phi \to \psi ))$. Thus for some $Y\supsetneq X$ and y ∈ Y − X, for all e ∈ S_Y, if $e{\downarrow ^{Y}_{X}}=b$, then $1\in {M^{e}_{Y}}(\mathtt {va}^{\nu }_{\omega _{y}},\phi \to \psi )$.

For (a), suppose Rbcd and $1\in {M^{c}_{X}}(\mathtt {va},\phi )$. Let f ∈ S_Y be such that $f{\downarrow ^{Y}_{X}}=d$. By the extension condition, there are g and h in S_Y such that Rghf, $g{\downarrow ^{Y}_{X}}=b$, and $h{\downarrow ^{Y}_{X}}=c$. Since $g{\downarrow ^{Y}_{X}}=b$, $1\in {M^{g}_{Y}}(\mathtt {va}^{\nu }_{\omega _{y}},\phi \to \psi )$. Since $h{\downarrow ^{Y}_{X}}=c$ and $1\in {M^{c}_{X}}(\mathtt {va},\phi )$, Vertical Heredity gives that $1\in {M^{h}_{Y}}(\mathtt {va},\phi )$. Since ν does not occur freely in ϕ, the Stability Lemma then gives that $1\in {M^{h}_{Y}}(\mathtt {va}^{\nu }_{\omega _{y}},\phi )$. So $1\in {M^{f}_{Y}}(\mathtt {va}^{\nu }_{\omega _{y}},\psi )$. Since f was an arbitrary element of S_Y such that $f{\downarrow ^{Y}_{X}}=d$, it follows from this that $1\in {M^{d}_{X}}(\mathtt {va},\forall \nu \psi )$.

For (b), suppose Rbcd and $0\in {M^{c}_{X}}(\mathtt {va},\forall \nu \psi )$. It follows that for some f ∈ S_Y, $f{\downarrow ^{Y}_{X}}=c$ and $0\in {M^{f}_{Y}}(\mathtt {va}^{\nu }_{\omega _{y}},\psi )$. Since $f{\downarrow ^{Y}_{X}}=c$, the extension condition gives that for some g and h in S_Y, Rgfh, $g{\downarrow ^{Y}_{X}}=b$ and $h{\downarrow ^{Y}_{X}}=d$. Since $g{\downarrow ^{Y}_{X}}=b$, $1\in {M^{g}_{Y}}(\mathtt {va}^{\nu }_{\omega _{y}},\phi \to \psi )$. Then, since $0\in {M^{f}_{Y}}(\mathtt {va}^{\nu }_{\omega _{y}},\psi )$, $0\in {M^{h}_{Y}}(\mathtt {va}^{\nu }_{\omega _{y}},\phi )$. Since ν does not occur free in ϕ, the Stability Lemma then gives that $0\in {M^{h}_{Y}}(\mathtt {va},\phi )$. Thus, by the Heredity Lemma, $0\in {M^{c}_{X}}(\mathtt {va},\phi )$.

For (ii) Suppose $0\in {M^{a}_{X}}[\mathtt {va},\phi \to \forall \nu \psi ]$. Then for some c and d, Rcda, $1\in {M^{c}_{X}}(\mathtt {va},\phi )$ and $0\in {M^{d}_{X}}(\mathtt {va},\forall \nu \psi )$. Let $Y\supsetneq X$ be and y ∈ Y − X. Since $0\in {M^{d}_{X}}(\mathtt {va},\forall \nu \psi )$, there is an e ∈ S_Y with $e{\downarrow ^{Y}_{X}}=d$ and $0\in {M^{e}_{Y}}(\mathtt {va}^{\nu }_{\omega _{y}},\psi )$. Since $e{\downarrow ^{Y}_{X}}=d$ and Rcda, there are f and g in S_Y such that $f{\downarrow ^{Y}_{X}}=c$, $g{\downarrow ^{Y}_{X}}=a$, and Rfeg.

By the Heredity Lemma, $1\in {M^{f}_{Y}}(\mathtt {va},\phi )$. Since ν does not occur freely in ϕ and $\mathtt {va}^{\nu }_{\omega _{y}}$ is Y -coherent, it follows that $1\in {M^{f}_{Y}}(\mathtt {va}^{\nu }_{\omega _{y}},\phi )$. Thus, since Rfeg and $0\in {M^{e}_{Y}}(\mathtt {va}^{\nu }_{\omega _{y}},\psi )$, $0\in {M^{g}_{Y}}(\mathtt {va}^{\nu }_{\omega _{y}},\phi \to \psi )$. Since Y and y were arbitrary, it follows that $0\in {M^{a}_{Y}}[\mathtt {va},\forall \nu (\phi \to \psi )]$. Thus, by Horizontal Heredity, $0\in {M^{b}_{Y}}[\mathtt {va},\forall \nu (\phi \to \psi )]$.

□

Appendix D: Completeness For R W Q

In this appendix we construct our canonical model in a way that differs mildly from the construction found in Appendix B. There are two reasons for doing this: first, it makes the proof of Symmetry in Lemma 27 a bit easier. But second, and more importantly, it’s nice to have a demonstration of the different methods available.

To begin we settle some notation. Let {v₁, v₂,… } be the set of variables in LQ. For ϕ ∈LQ, let fv(ϕ) be the set of variables that occur freely in ϕ. For each set of numbers V, let LQ_V = {ϕ ∈LQ : fv(ϕ) ⊆{v_i}_i∈V}. Let Σ ∪{ϕ}⊆LQ.

4.1 D.1 Definitions

• Write Σ_V for Σ ∩LQ_V.

• Write ${\Sigma }\Vdash \phi$ when there is a sequence ϕ₁, ϕ₂,…,ϕ_n such that ϕ_n = ϕ and for all 1 ≤ i ≤ n, either ϕ_i ∈Σ; or ϕ_i follows by one of these two rules:

  DR1:

  For some j < i and k < i, ϕ_i = ϕ_j ∧ ϕ_k;

  DR2:

  For some j < i, ϕ_j → ϕ_i ∈RWQ.

  We say that any such sequence witnesses${\Sigma }\Vdash \phi$.²

• Write c(Σ) for $\{\phi :{\Sigma }\Vdash \phi \}$. Write c_V(Σ) for (c(Σ))_V.

• Say Σ is a theory when ${\Sigma }\Vdash \phi$ only if ϕ ∈Σ.

• Say Σ is a V-theory when Σ ⊆LQ_V and for all ϕ ∈LQ_V, ${\Sigma }\Vdash \phi$ only if ϕ ∈Σ.

• Say a V-theory Σ is normal when RWQ_V ⊆Σ.

• Say Σ is prime when α ∨ β ∈Σ only if α ∈Σ or β ∈Σ.

We adopt the usual abbreviations. For example, we write $\alpha \Vdash \beta$ instead of $\{\alpha \}\Vdash \beta$ and ${\Delta },\alpha \Vdash \beta$ instead of ${\Delta }\cup \{\alpha \}\Vdash \beta$.

4.2 D.2 Basic Lemmas

Lemma 16 (Deduction Lemma)

If$\alpha \Vdash \beta$,thenα → β ∈RWQ.

Proof

By induction on the length of the shortest witness of $\alpha \Vdash \beta$. If there is a witness of length 1, then β is α. Since α → α ∈RWQ, the base case is established. Now suppose ϕ₁,…,β = ϕ_n+ 1 witnesses $\alpha \Vdash \beta$. There are two cases to consider.

• Case 1: There are j < n + 1 and k < n + 1 such that β = ϕ_j ∧ ϕ_k. Then ϕ₁,…,ϕ_j witnesses $\alpha \Vdash \phi _{j}$ and ϕ₁,…,ϕ_k witnesses $\alpha \Vdash \phi _{k}$. By the inductive hypothesis, α → ϕ_j ∈RWQ and α → ϕ_k ∈RWQ. From here a short derivation gives that α → (ϕ_j ∧ ϕ_k) = α → β ∈RWQ as required.

• Case 2: There is a j < n + 1 such that ϕ_j → β ∈RWQ. Then ϕ₁,…,ϕ_j witnesses $\alpha \Vdash \phi _{j}$. By the inductive hypothesis, α → ϕ_j ∈RWQ. From here a short derivation gives α → β ∈RWQ as well.

□

Lemma 17

If${\Sigma }\Vdash \phi (\nu )$,νnot free in Σ but free for x inϕ(ν),then${\Sigma }\Vdash \forall x\phi (x)$.

Proof

If ${\Sigma }\Vdash \phi (\nu )$, then for some σ₁,…,σ_n all in Σ, $\sigma _{1}\land \dots \land \sigma _{n}\Vdash \phi (\nu )$. So by the Deduction Lemma, (σ₁ ∧⋯ ∧ σ_n) → ϕ(ν) ∈RWQ. So by RQ1, ∀ν[(σ₁ ∧⋯ ∧ σ_n) → ϕ(ν)] ∈RWQ. Since ν is not free in Σ, ∀ν[(σ₁ ∧⋯ ∧ σ_n) → ϕ(ν)] → [(σ₁ ∧⋯ ∧ σ_n) →∀νϕ(ν)] ∈RWQ. Thus (σ₁ ∧⋯ ∧ σ_n) →∀νϕ(ν) ∈RWQ, and hence ${\Sigma }\Vdash \forall \nu \phi (\nu )$. But ∀νϕ(ν) →∀xϕ(x) ∈RWQ, so ${\Sigma }\Vdash \forall x\phi (x)$ as required. □

Lemma 18 (Lindenbaum)

Suppose Δ is closed under disjunction, Γ is a theory, and Γ ∩Δ = ∅. Then there is a prime theory Γ^′⊃Γ such that Γ^′∩Δ = ∅.

Proof

To begin, choose an enumeration of all the formulas of the form α ∨ β. Let α_i ∨ β_i be the ith member of this list. Define ${{\Gamma }^{i}_{j}}$ as follows:

• ${{\Gamma }^{0}_{0}}={\Gamma }$.

• ${\Gamma }^{i}_{j + 1}= \left \{ \begin {array}{rl} {{\Gamma }^{i}_{j}} & \text {if} \ \alpha _{j}\lor \beta _{j}\not \in c({{\Gamma }^{i}_{j}})\\[1ex] {{\Gamma }^{i}_{j}}\cup \{\alpha _{j}\} & \text {if} \ \alpha _{j}\lor \beta _{j}\in c({{\Gamma }^{i}_{j}})\ \text { and} \ c({{\Gamma }^{i}_{j}}\cup \{\alpha _{j}\})\cap {\Delta }=\emptyset \\[1ex] {{\Gamma }^{i}_{j}}\cup \{\beta _{j}\} & \text {otherwise} \end {array} \right .$

• ${\Gamma }^{i + 1}_{0}=\cup _{j = 0}^{\infty }{{\Gamma }^{i}_{j}}$

Let ${\Gamma }^{\prime }=\cup _{i = 0}^{\infty }{{\Gamma }^{i}_{0}}$. Clearly ${\Gamma }={{\Gamma }^{0}_{0}}\subseteq {\Gamma }^{\prime }$. It remains to show that Γ^′ is prime, is a theory, and that Γ^′∩Δ = ∅.

To see Γ^′ is prime, let α ∨ β ∈Γ. Then for some i, ${{\Gamma }^{i}_{0}}\Vdash \alpha _{j}\lor \beta _{j}$. But then by the definition of ${\Gamma }^{i}_{j + 1}$, either $\alpha _{j}\in {\Gamma }^{i}_{j + 1}$ or $\beta _{j}\in {\Gamma }^{i}_{j + 1}$. Either way, α_j ∈Γ^′ or β_j ∈Γ^′.

To see that Γ^′ is a theory, notice that if ${\Gamma }^{\prime }\Vdash \alpha$, then ${\Gamma }^{\prime }\Vdash \alpha \lor \alpha$. Thus for some i, ${{\Gamma }^{i}_{0}}\Vdash \alpha \lor \alpha$. By construction, we then see that α ∈Γⁱ ⊆Γ^′.

To see that Γ^′∩Δ = ∅, suppose otherwise, and let $i_{0}=\inf \{i:c({{\Gamma }^{i}_{j}})\cap {\Delta }\neq \emptyset \ \text {for some}\ j\}$ and let $j_{0}=\inf \{j:c({\Gamma }^{i_{0}}_{j})\cap {\Delta }\neq \emptyset \}$. Clearly either i₀ = 0 or i₀≠ 0. In the former case, since ${{\Gamma }^{0}_{0}}={\Gamma }$ is a theory and Γ ∩Δ = ∅, clearly j₀≠ 0. In the latter case, since ${\Gamma }^{i + 1}_{0}=\cup ^{\infty }_{j = 0}{{\Gamma }^{i}_{j}}$, if $c({\Gamma }^{i_{0}}_{0})\cap {\Delta }\neq \emptyset$, then since proofs are finite, for some j, $c({\Gamma }^{i_{0}-1}_{j})\cap {\Delta }\neq \emptyset$, contradicting the minimality of i₀. So again we can conclude that j₀≠ 0. Thus, in either case, j₀≠ 0. So ${\Gamma }^{i_{0}}_{j_{0}-1}$ is defined. Also if $\alpha _{j_{0}-1}\lor \beta _{j_{0}-1}\not \in c({\Gamma }^{i_{0}}_{j_{0}-1})$, then ${\Gamma }^{i_{0}}_{j_{0}}={\Gamma }^{i_{0}}_{j_{0}-1}$. But then $c({\Gamma }^{i_{0}}_{j_{0}-1})\cap {\Delta }\neq \emptyset$, contradicting j₀’s minimality. So $\alpha _{j_{0}-1}\lor \beta _{j_{0}-1}\in c({\Gamma }^{i_{0}}_{j_{0}-1})$.

Now suppose $c({\Gamma }^{i_{0}}_{j_{0}-1}\cup \{\alpha _{j_{0}-1}\})\cap {\Delta }=\emptyset$. Then ${\Gamma }^{i_{0}}_{j_{0}}={\Gamma }^{i_{0}}_{j_{0}-1}\cup \{\alpha _{j_{0}-1}\}$. But then $c({\Gamma }^{i_{0}}_{j_{0}})\cap {\Delta }=\emptyset$, contradicting our assumption. Thus some sentence $\alpha \in c({\Gamma }^{i_{0}}_{j_{0}-1}\cup \{\alpha _{j_{0}-1}\})\cap {\Delta }$ It follows that ${\Gamma }^{i_{0}}_{j_{0}}={\Gamma }^{i_{0}}_{j_{0}-1}\cup \{\beta _{j_{0}-1}\}$. Since $c({\Gamma }^{i_{0}}_{j_{0}-1})\cap {\Delta }=\emptyset$ (by j₀’s minimality), but $c({\Gamma }^{i_{0}}_{j_{0}})\cap {\Delta }\neq \emptyset$, it follows that for some β ∈Δ, ${\Gamma }^{i_{0}}_{j_{0}-1},\beta _{j_{0}-1}\Vdash \beta$.

By short derivations, we can establish that α → (α ∨ β) and β → (α ∨ β) are in RWQ. Thus, since ${\Gamma }^{i_{0}}_{j_{0}-1},\alpha _{j_{0}-1}\Vdash \alpha$ and ${\Gamma }^{i_{0}}_{j_{0}-1},\beta _{j_{0}-1}\Vdash \beta$, it is also the case that ${\Gamma }^{i_{0}}_{j_{0}-1},\alpha _{j_{0}-1}\Vdash \alpha \lor \beta$ and ${\Gamma }^{i_{0}}_{j_{0}-1},\beta _{j_{0}-1}\Vdash \alpha \lor \beta$. By a series of tedious manipulations, one sees that it follows from these that ${\Gamma }^{i_{0}}_{j_{0}-1},\alpha _{j_{0}-1}\lor \beta _{j_{0}-1}\Vdash \alpha \lor \beta$.

But ${\Gamma }^{i_{0}}_{j_{0}-1}\Vdash \alpha _{j_{0}-1}\lor \beta _{j_{0}-1}$ by assumption. So in fact ${\Gamma }^{i_{0}}_{j_{0}-1}\Vdash \alpha \lor \beta$. But α ∨ β ∈Δ because Δ is closed under disjunctions. This contradicts the minimality of i₀ and j₀, completing the proof. □

The following lemma is important corollary of the Lindenbaum Lemma:

Lemma 19

If Γ is a theory and {Γ_i}_i∈Iis the set of all prime theories that contain Γ,then Γ = ∩_i∈IΓ_i

Proof

Clearly Γ ⊆∩_i∈IΓ_i. Now suppose δ∉Γ. Let Δ be the closure of {δ} under disjunction. Since Γ is a theory and δ∉Γ, Γ does not intersect Δ. So by the Lindenbaum Lemma, there is a prime theory containing Γ that does not intersect Δ. So δ∉ ∩_i∈IΓ_i. □

4.3 D.3 A Preview

The plan is for the canonical model to be the quintuple $\langle D,{\Omega },\delta ,\mathcal {M},\Downarrow \rangle$ such that

• D is the set of names.

• Ω is the set of variables.

• δ is the identity function.

• $\mathcal {M}:X\mapsto M_{X}=\langle D_{X},S_{X},N_{X},R_{X},\delta ,\mathcal {E}^{+}_{X},\mathcal {E}^{-}_{X}\rangle$ with S_X the set of all prime X-theories; N_X the set of all normal prime X-theories; R_XΣΓΔ iff β ∈Δ whenever α → β ∈Σ and α ∈Γ; and $\mathcal {E}^{+}_{X}$ and $\mathcal {E}^{-}_{X}$ exactly as in Appendix B.

• $a{\downarrow ^{X}_{Y}}=a_{Y}=a\cap \text {LQ}_{Y}$.

For this to do any work for us, we need to locate RWQ_X among the N_X. We will do this by showing RWQ to be prime. To show this, one standardly uses what are called metavaluations. This technique was pioneered in [21] and extended in [32]. A recent survey of applications of this technique can be found in [7].

4.4 D.4 Metavaluations

We take a metavaluation to be a pair of functions M and M^∗ mapping LQ to {T,F}. The intuitive picture to have is that M(ϕ) = T just if ϕ ∈RWQ and M^∗(ϕ) = T just if ¬ϕ∉RWQ. Explicitly, we require the functions to obey the following conditions:

MV1::

If ϕ is atomic, then M(ϕ) = F and M^∗(ϕ) = T

MV2::

M(ϕ ∧ ψ) = T iff M(ϕ) = T and M(ψ) = T.

MV3::

M^∗(ϕ ∧ ψ) = T iff M^∗(ϕ) = T and M^∗(ψ) = T.

MV4::

M(¬ϕ) = T iff ¬ϕ ∈RWQ and M^∗(ϕ) = F.

MV5::

M^∗(¬ϕ) = T iff M(ϕ) = F.

MV6::

M(ϕ → ψ) = T iff (i) ϕ → ψ ∈RWQ, (ii) M(ϕ) = T materially implies M(ψ) = T, and (iii) M^∗(ϕ) = T materially implies M^∗(ψ) = T.

MV7::

M^∗(ϕ → ψ) = T iff M(ϕ) = T materially implies M^∗(ψ) = T.

MV8::

M(∀xϕ(x)) = T iff M(ϕ(τ)) = T for all terms τ that are free for x in ϕ(x).

MV9::

M^∗(∀xϕ(x)) = T iff M^∗(ϕ(τ)) = T for all terms τ that are free for x in ϕ(x).

We now prove two lemmas that show these conditions force M and M^∗ to agree with our intuitive picture. For all of these, it helps to first notice that if ⊩_∅ϕ, then by the definition of RWQ, ϕ ∈RWQ, so each of Facts 1-12 has an analogue in terms of elements of RWQ. In what follows we will often cite the fact when, technically speaking, we should be citing its analogue in this sense.

Lemma 20

M(ϕ) = Tmaterially impliesϕ ∈RWQ.

Proof

By induction on the complexity of ϕ. The base case holds vacuously; the cases for the connectives are straightforward. For the quantifier, suppose M(∀xϕ(x)) = T. Then M(ϕ(t)) = T for all terms t that are free for x in ϕ. Letting y be a variable that is free for x in ϕ, ϕ(y) is in RWQ. But then by QR1, so is ∀yϕ(y). Thus, since ∀yϕ(y) →∀xϕ(x) ∈RWQ, ∀xϕ(x) ∈RWQ as required. □

Lemma 21

M^∗(ϕ) = Fmaterially implies ¬ϕ ∈RWQ.

Proof

By a slightly more subtle induction than in the previous lemma. The base case is still vacuously true. Now suppose M^∗(ϕ ∧ ψ) = F. Then either M^∗(ϕ) = F or M^∗(ψ) = F. So, by the inductive hypothesis, either ¬ϕ ∈RWQ or ¬ψ ∈RWQ. Thus, by either Fact 5 or Fact 6, ¬(ϕ ∧ ψ) ∈RWQ.

Suppose M^∗(¬ϕ) = F. Then M(ϕ) = T. So by Lemma 20, ϕ ∈RWQ. Thus, since ϕ →¬¬ϕ ∈RWQ by Fact 1, ¬¬ϕ ∈RWQ.

Suppose M^∗(ϕ → ψ) = F. Then M(ϕ) = T and M^∗(ψ) = F. By Lemma 20, from the first conjunct we can conclude that ϕ ∈RWQ. Since ϕ → ((ϕ → ψ) → ψ) is an instance of A9, it’s in RWQ. Thus so is (ϕ → ψ) → ψ. By Fact 1 and Fact 3, it then follows that (ϕ → ψ) →¬¬ψ ∈RWQ, whence by an instance of A7, so is ¬ψ →¬(ϕ → ψ). But since M^∗(ψ) = F, by the inductive hypothesis, ¬ψ ∈RWQ, and thus ¬(ϕ → ψ) is as well.

Finally, suppose M^∗(∀xϕ(x)) = F. Then for some term t, M^∗(ϕ(t)) = F. Thus by the inductive hypothesis, ¬ϕ(t) ∈RWQ. But by QA1, ∀xϕ(x) → ϕ(t) ∈RWQ. So by a similar argument to the one in the previous case, ¬ϕ(t) →¬∀xϕ(x) ∈RWQ as well. Thus so is ¬∀xϕ(x). □

Before proving the next lemma, we need a tool:

Lemma 22

Suppose x occurs freely inϕ(x). Then (a) ifM(ϕ(x)) = T,thenM(ϕ(τ)) = Tfor all termsτthat are free for x inϕ(x) and (b) ifM^∗(ϕ(x)) = F,thenM^∗(ϕ(τ)) = Ffor all termsτthat are free for x inϕ(x).

Proof

By a simultaneous induction (on the complexity of ϕ) in both (a) and (b). The atomic cases are vacuously true and the conjunction case is straightforward. If ϕ has the form ¬ψ(x), then if M(ϕ(x)) = T, then ¬ψ(x) ∈RWQ and M^∗(ψ(x)) = F. Since ¬ψ(x) ∈RWQ, by an instance of QR1, ∀x¬ψ(x) ∈RWQ, whence by QA1, for all terms τ that are free for x in ψ(x), ¬ψ(τ) ∈RWQ. And since M^∗(ψ(x)) = F, the inductive hypothesis gives that for all such terms τ, M^∗(ψ(τ)) = F. Together these show that for all terms τ that are free for x in ψ(x), M(ϕ(τ)) = T. Aside from the need to keep track of more details, the conditional and quantifier cases are essentially the same. □

Lemma 23

ϕ ∈RWQmaterially impliesM(ϕ) = T.

Proof

We will first show that if ϕ is an axiom of RWQ, then M(ϕ) = T. We only examine A4 and QA2, leaving the remaining cases to the reader.

For this case, notice that M(((α → β) ∧ (α → γ)) → (α → (β ∧ γ))) = T iff (a) ((α → β) ∧ (α → γ)) → (α → (β ∧ γ)) ∈RWQ; (b) M((α → β) ∧ (α → γ)) = T materially implies M(α → (β ∧ γ)) = T; and (c) M^∗((α → β) ∧ (α → γ)) = T materially implies M^∗(α → (β ∧ γ)) = T. (a) is obviously true. For (b), we reason as follows (using ‘$\sqsupset$’ for our metalinguistic material conditional); (c) then follows by a similarly tedious ‘metaderivation’:

1	M((α → β) ∧ (α → γ)) = T	Assumption
2	(a) M(α → β) = T and (b) M(α → γ) = T	From 1 by MV2
3	(i) α → β ∈RWQ, (ii) $M(\alpha )=\mathsf {T} \sqsupset M(\beta )=\mathsf {T}$, and	From 2(a) by MV6
	(iii) $M^{*}(\alpha )=\mathsf {T} \sqsupset M^{*}(\beta )=\mathsf {T}$
4	(i) α → γ ∈RWQ, (ii) $M(\alpha )=\mathsf {T} \sqsupset M(\gamma )=\mathsf {T}$, and	From 2(b) by MV6
	(iii) $M^{*}(\alpha )=\mathsf {T} \sqsupset M^{*}(\gamma )=\mathsf {T}$
5	(α → β) ∧ (α → γ) ∈RWQ	R2 applied to 3(i) and 4(i)
6	α → (β ∧ γ) ∈RWQ	R1 applied to A4, 5
7	M(α) = T	Assumption
8	M(β) = T and M(γ) = T	Via 3(ii) and 7; 4(ii) and 7
9	M(β ∧ γ) = T	MV2 applied to 8
10	$M(\alpha )=\mathsf {T} \sqsupset M(\beta \land \gamma )=\mathsf {T}$	7-9, discharging 7
11	$M^{*}(\alpha )=\mathsf {T} \sqsupset M^{*}(\beta \land \gamma )=\mathsf {T}$	7-10, mutatis mutandis
12	M(α → (β ∧ γ)) = T	MV6 applied to 6, 10, 11
13	$M((\alpha \to \beta )\land (\alpha \to \gamma ))=\mathsf {T}\sqsupset M(\alpha \to (\beta \land \gamma ))=\mathsf {T}$	1-12, discharging 1.

For QA2, suppose ν is not free in ϕ. M(∀ν(ϕ → ψ(ν)) → (ϕ →∀νψ(ν))) = T iff (a) ∀ν(ϕ → ψ(ν)) → (ϕ →∀νψ(ν)) ∈RWQ; (b) M(∀ν(ϕ → ψ(ν))) = T materially implies M(ϕ →∀νψ(ν)) = T; and (c) M^∗(∀ν(ϕ → ψ(ν))) = T materially implies M^∗(ϕ →∀νψ(ν)) = T. (a) is obviously true. For (b), we reason as follows; (c) then follows by a similarly tedious ‘metaderivation’:

1	M(∀ν(ϕ → ψ(ν))) = T	Assumption
2	M(ϕ → ψ(t)) = T	From 1 by MV8
3	(i) ϕ → ψ(t) ∈RWQ, (ii) $M(\phi )=\mathsf {T}\sqsupset M(\psi (t))=\mathsf {T}$, and	From 2 by MV6
	(iii) $M^{*}(\phi )=\mathsf {T}\sqsupset M^{*}(\psi (t))=\mathsf {T}$
4	M(ϕ) = T	Assumption
5	M(ψ(t)) = T	By 4 and 3(ii)
6	M(ψ(τ)) = T for all appropriate τ	From 5
7	M(∀νψ(ν)) = T	From 6 by MV8
8	$M(\phi )=\mathsf {T}\sqsupset M(\forall \nu \psi (\nu ))=\mathsf {T}$	4-7, discharging 4.
9	$M^{*}(\phi )=\mathsf {T}\sqsupset M^{*}(\forall \nu \psi (\nu ))=\mathsf {T}$	4-8, mutatis mutandis
10	M(ϕ →∀νϕ(ν)) = T	From 3(i), 8, and 9 by MV6

In lines 2, 3, and 5 we take t to be an arbitrary term that is free for the appropriate variable. We complete the proof by showing that each of the rules preserves T. This is immediately obvious for R1 and R2. For QR1, notice that if ν occurs free in ψ(ν) and M(ψ(ν)) = T, then by Lemma 22, M(ψ(τ)) = T for all terms τ. So by MV8, M(∀νψ(ν)) = T as well. □

With our metavaluation in hand, the following result is nearly immediate:

Lemma 24

For each X,RWQ_Xis prime.

Proof

We prove that RWQ itself is prime, since the result immediately follows from this. To that end, let α ∨ β ∈RWQ. Then by Lemma 23, M(α ∨ β) = T. So M(¬(¬α ∧¬β)) = T. Thus M^∗(¬α ∧¬β) = F. So M^∗(¬α) = F or M^∗(¬β) = F. Thus by Lemma 21, ¬¬α ∈RWQ or ¬¬β ∈RWQ. So by A6, either α ∈RWQ or β ∈RWQ. □

4.5 D.5 Modelhood

With that out of the way, we can now prove that the canonical model is a premodel. This requires showing the models M_X satisfy the ordering, monotonicity, closure, rearranging, and heredity constraints. But these are proved exactly as in Appendix B; we note here the following two Lemmas:

Lemma 25

a ≤ biffa ⊆ b

Lemma 26

The canonical model is an R W Q -premodel

Now we turn to modelhood proper.

Lemma 27

The canonical model is an R W Q -model.

We’ve already seen that the canonical model is an RWQ-premodel. It remains to show that it satisfies the vertical atomic heredity, normality, lifting, homomorphism, extension, and symmetry conditions. Atomic heredity and homomorphism are straightforward; for the remainder we argue as follows:

Normality::

It’s clear that if a ∈ N_X, then $a{\downarrow ^{X}_{Y}}\in N_{Y}$. Now suppose a∉N_X. Let ϕ ∈RWQ_X − a. Let x₁,…,x_n be the variables that occur in ϕ that are not in {v_i}_i∈Y. Since ϕ ∈RWQ, repeated application of QR1 gives that ∀x₁…∀x_nϕ ∈RWQ as well. Notice that ∀x₁…∀x_nϕ ∈LQ_Y. Thus ϕ ∈RWQ_Y.

Now suppose ∀x₁…∀x_nϕ ∈ a. Then clearly by repeated application of QA1, ϕ ∈ a as well. But ϕ∉a. So ∀x₁…∀x_nϕ∉a. So $\forall x_{1}\dots \forall x_{n}\phi \not \in a{\downarrow ^{X}_{Y}}\subseteq a$. But then since ∀x₁…∀x_nϕ ∈RWQ_Y, $a{\downarrow ^{X}_{Y}}\not \in N_{Y}$.

Lifting::

Let $\overline {a}=\text {LQ}_{X}-a$ and let $c'=\{\phi \in \text {LQ}_{X\cup Y}:b\Vdash \phi \}$. Clearly c^′ is a theory and c^′⊇ b. Suppose $\phi \in c'\cap \overline {a}$. Let {x₁,…,x_n} be the free variables of ϕ that are not in Y. Then since $b\Vdash \phi$, $b\Vdash \forall x_{1}\dots \forall x_{n}\phi$. But ∀x₁…∀x_nϕ ∈ b ∩LQ_X∩Y = a ∩LQ_X∩Y. So $a\Vdash \forall x_{1}\dots \forall x_{n}\phi$, and thus by instantiation, $a\Vdash \phi$, which is impossible since $\phi \in \overline {a}$. It follows by the Lindenbaum Lemma that there is a c ⊇ c^′ such that $c\cap \overline {a}=\emptyset$. Thus, $c\downarrow ^{X\cup Y}_{X}\leq a$ and $b\leq c\downarrow ^{X\cup Y}_{Y}$.

Extension::

For each part, the proof is essentially a modification of the proof of one of the extension lemmas from Appendix B. We give only one of them, leaving the others to the reader.

Suppose R_Yabc and $d{\downarrow ^{X}_{Y}}=c$. Let $\overline {d}=\text {LQ}_{X}-d$, $e'=\{\phi \in \text {LQ}_{X}:a\Vdash \phi \}$ and let ${\Theta }=\left \{\alpha \in \text {LQ}_{X}:\alpha \to \beta \in e'\ \text { for some} \ \beta \in \overline {d}\right \}$. By familiar arguments, we see that Θ is closed under disjunction.

Suppose ϕ ∈ b ∩Θ. Then for some $\beta \in \overline {d}$, $a\Vdash \phi \to \beta$. Let x₁,…,x_n be the variables that are free in ϕ → β that are not in Y. Since a is a Y -theory and $a\Vdash \phi \to \beta$, $a\Vdash \forall x_{1}\dots \forall x_{n}(\phi \to \beta )$. Since ϕ ∈ b ⊆LQ_Y, it follows that $a\Vdash \phi \to \forall x_{1}\dots \forall x_{n}\beta$. But then since a is a Y -theory ϕ →∀x₁…∀x_nβ ∈ a. And since Rabc and ϕ ∈ b, it follows from this that ∀x₁…∀x_nβ ∈ c ⊆ d. It then quickly follows that β ∈ d, contradicting $\beta \in \overline {d}$. So b ∩Θ is empty.

Applying the Lindenbaum Lemma restricted to LQ_X, we then get that there is a prime X-theory f with b ⊆ f and f ∩Θ = ∅. It follows from definitions that R_Xe^′fd. Applying an analogue of Lemma 7 then gives the result.

Symmetry::

Following the discussion near Lemma 15 in [13], for any set of first-order wffs Σ, and any first-order wff ϕ, we write ${\Sigma }\Vdash ^{v,w}\phi$ when there is a sequence ϕ₁, ϕ₂,…,ϕ_n such that ϕ_n = ϕ and for all 1 ≤ i ≤ n, either

• ϕ_i ∈Σ, or

• For some j < i and k < i, ϕ_i = ϕ_j ∧ ϕ_k, or

• For some j < i, ν not free in Σ and free for x in ϕ_j(x), ϕ_i = ∀νϕ_j(x/ν), or

• For some j < i, ϕ_j is a v, w-variant of ϕ_i – that is, ϕ_i is ϕ_j with some occurrences of v replaced by w or with some occurrences of w replaced by v.

• For some j < i, ϕ_j → ϕ_i ∈RWQ.

The proof of the Lindenbaum Lemma generalizes to the use of $\Vdash ^{v,w}$. So if the hypotheses are met and we let ${\Delta }=\{\alpha : a\not \Vdash \alpha \}$, then the Lindenbaum Lemma gives a prime theory that satisfies the conclusion.

We are nearly done. What remains is to prove the truth-and-containment result.

Lemma 28

Let$\mathfrak {C}$bethe canonical model. Then ifa ∈ S_X,is X-coherent, andϕ ∈LQ_X,then$1{\in \mathfrak {C}^{a}_{X}}(\mathtt {va},\phi )$iffϕ ∈ aand$0{\in \mathfrak {C}^{a}_{X}}(\mathtt {va},\phi )$iff¬ϕ ∈ a.

Proof

By induction on complexity as usual. The base case and the cases involving the connectives are as they were in Appendix B. For the quantified cases, we argue as follows.

Suppose $1{\in \mathfrak {C}^{a}_{X}}(\mathtt {va},\forall v_{k}\psi (v_{k}))$. Then for some $Y\supsetneq X$ and i ∈ Y − X, for all b ∈ S_Y, if $b{\downarrow ^{Y}_{X}}=a$, then $1{\in \mathfrak {C}^{b}_{Y}}(\mathtt {va}^{v_{k}}_{v_{i}},\psi (v_{k}))$.

Now let c be a prime theory containing a. Since c ⊇ a, c_X ⊇ a. Thus, by Lemma 25, a ≤ c_X. So for some n ∈ N_X, Rnac_X. Suppose $b{\downarrow ^{Y}_{X}}=a$. It follows that $1{\in \mathfrak {C}^{b}_{Y}}(\mathtt {va}^{v_{k}}_{v_{i}},\psi (v_{k}))$. Also, by the Extension Condition, there is an m ∈ N_Y and d ∈ S_Y such that Rmbd and $d{\downarrow ^{Y}_{X}}=c_{X}$. Since m ∈ N_Y and Rmbd, b ≤ d. Thus, by Horizontal Heredity, $1{\in \mathfrak {C}^{d}_{Y}}(\mathtt {va}^{v_{k}}_{v_{i}},\psi (v_{k}))$. It follows by the Evaluation Lemma that $1{\in \mathfrak {C}^{d}_{Y}}(\mathtt {va},\psi (v_{i}))$. Thus, by Vertical Heredity, $1\in \mathfrak {C}^{c_{X}}_{X}(\mathtt {va},\psi (v_{i}))$. So by the inductive hypothesis, ψ(v_i) ∈ c_X. Thus ψ(v_i) ∈ c. Since c was arbitrary, by Lemma 19, ψ(v_i) ∈ a. Thus ∀v_kψ(v_k) ∈ a, as required.

On the other hand, suppose ∀v_kψ(v_k) ∈ a. Let $Y\supsetneq X$ and i ∈ Y − X. If $b{\downarrow ^{Y}_{X}}=a$, then a ⊆ b. So ∀v_kψ(v_k) ∈ b. Thus, by application of QA1, ψ(v_i) ∈ b. So by the inductive hypothesis, $1{\in \mathfrak {C}^{b}_{Y}}(\mathtt {va},\psi (v_{i}))$. So by the Evaluation Lemma, $1{\in \mathfrak {C}^{b}_{Y}}(\mathtt {va}^{v_{k}}_{v_{i}},\psi (v_{k}))$. Since b was arbitrary, $1{\in \mathfrak {C}^{a}_{X}}(\mathtt {va},\forall v_{k}\psi (v_{k}))$.

For the zero case, suppose $0{\in \mathfrak {C}^{a}_{X}}(\mathtt {va},\forall v_{k}\psi (v_{k}))$. Then for every $Y\supsetneq X$ and i ∈ Y − X there is a b ∈ S_Y so that $b{\downarrow ^{Y}_{X}}=a$ and $0{\in \mathfrak {C}^{b}_{Y}}(\mathtt {va}^{v_{k}}_{v_{i}},\psi (v_{k}))$. Thus, by the Evaluation Lemma, $0{\in \mathfrak {C}^{b}_{Y}}(\mathtt {va},\psi (v_{i}))$. So by the inductive hypothesis, ¬ψ(v_i) ∈ b. But by QA1, ¬ψ(v_i) →¬∀v_kψ(v_k) ∈RWQ. So since b is a theory, ¬∀v_kψ(v_k) ∈ b. And since ¬∀v_kψ(v_k) ∈LQ_X and $b{\downarrow ^{Y}_{X}}=a$, it follows that ¬∀v_kψ(v_k) ∈ a as required.

On the other hand, suppose ¬∀v_kψ(v_k) ∈ a. Let $Y\supsetneq X$, i ∈ Y − X, $b'=\{\beta \in \text {LQ}_{Y}:a,\neg \psi (v_{i})\Vdash \beta \}$, and $\overline {a}=\text {LQ}_{X}-a$. I claim that $b'\cap \overline {a}=\emptyset$; that is, that if β ∈ b^′∩LQ_X, then β ∈ a.

To see this, note that if $a,\neg \psi (v_{i})\Vdash \beta$, then there is α ∈ a so that $\alpha \land \neg \psi (v_{i})\Vdash \beta$. Thus by the Deduction Lemma, (α ∧¬ψ(v_i)) → β ∈RWQ. An instance of Fact 4 then gives that ¬β →¬(α ∧¬ψ(v_i)) ∈RWQ, from which A6 and Facts 2 and 7 give that ¬β → (¬α ∨ ψ(v_i)) ∈RWQ. Thus by QR1, so is ∀v_i(¬β → (¬α ∨ ψ(v_i))). On the assumption that β ∈LQ_X, it follows by instances of QA2, QA3, and Fact 2 that ¬β → (¬α ∨∀v_iψ(v_i)) ∈RWQ. Applying A6 and Facts 2 and 4, we then see that ¬(¬α ∨∀v_iψ(v_i)) → β ∈RWQ. The definition of ‘∨’ and instances of Fact 1 and A6 then give (α ∧¬∀v_iψ(v_i)) → β ∈RWQ. Thus, $a,\neg \forall v_{i}\psi (v_{i})\Vdash \beta$. But ¬∀v_iψ(v_i) ∈ a, so it follows that $a\Vdash \beta$. Thus since a is an X-theory and β ∈LQ_X, it follows that β ∈ a.

Since β ∈ b^′∩LQ_X only if β ∈ a, $\beta \cap \overline {a}=\emptyset$, the (restricted) Lindenbaum Lemma guarantees a prime Y -theory b ⊇ b^′ with $b\cap \overline {a}=\emptyset$. Since b ⊃ a by construction, $b{\downarrow ^{X}_{Y}}=a$. And since ¬ψ(v_i) ∈ b as well, the inductive hypothesis gives that $0{\in \mathfrak {C}^{b}_{Y}}(\mathtt {va}^{v_{k}}_{v_{i}}\psi (v_{k}))$. Thus $0{\in \mathfrak {C}^{a}_{X}}(\mathtt {va},\forall v_{k}\psi (v_{k}))$ as required. □

Theorem 4

IfϕisRWQ-valid,thenϕ ∈RWQ.

Proof

Suppose ϕ∉RWQ. Then since RWQ is normal and prime, there are normal prime X-theories that do not contain ϕ. So ϕ is not true in the canonical model and is thus not valid. □

4.5.1 D.5.1 Constant Domain Models

It’s straightforward to construct, from a varying-domain RWQ-model $C=\langle D,{\Omega },\delta ,\mathcal {M},\Downarrow \rangle$, an Ω-model C^cd = 〈D ∪Ω,δ,^′,⇓^′〉. ^′ and ⇓^′ are, essentially, just as they were before. The only difference is that ^′ now maps each X to the model $M_X^{\prime }$ that is just like M_X except that D_X = D ∪Ω. Because of the way the M_X are defined in the varying-domain case, each of the $M_X^{\prime }$ automatically satisfies the featurelessness condition.

Going the other direction is just as easy. So to prove that the constant-domain semantics is adequate (sound and complete) it suffices to prove that for all ϕ and C, $C\vDash \phi$ iff $C^{cd}\vDash \phi$. But this result, in turn, follows by an incredibly straightforward induction. So the constant domain semantics is adequate.