Revision Without Revision Sequences: Self-Referential Truth

Abstract

The model of self-referential truth presented in this paper, named Revision-theoretic supervaluation, aims to incorporate the philosophical insights of Gupta and Belnap’s Revision Theory of Truth into the formal framework of Kripkean fixed-point semantics. In Kripke-style theories the final set of grounded true sentences can be reached from below along a strictly increasing sequence of sets of grounded true sentences: in this sense, each stage of the construction can be viewed as an improvement on the previous ones. I want to do something similar replacing the Kripkean sets of grounded true sentences with revision-theoretic sets of stable true sentences. This can be done by defining a monotone operator through a variant of van Fraassen’s supervaluation scheme which is simply based on ω-length iterations of the Tarskian operator. Clearly, all virtues of Kripke-style theories are preserved, and we can also prove that the resulting set of “grounded” true sentences shares some nice features with the sets of stable true sentences which are provided by the usual ways of formalising revision. What is expected is that a clearer philosophical content could be associated to this way of doing revision; hopefully, a content directly linked with the insights underlying finite revision processes.

Appendices

Appendix A: The Duality Between Sieve and Jump

In the subsequent Corollary 1, we will prove an order-theoretic proposition about monotone operators on partially ordered sets from which Theorem 1 easily follows.

Let $\mathbb {P} = \langle P, \preceq \rangle$ and $\mathbb {Q} = \langle Q, \preceq \rangle$ be two partially ordered sets. We say that a function F : P → Q is antitone if and only if

$$p \preceq p^{\prime} \implies F(p^{\prime}) \preceq F(p), $$

for every p, p′ ∈ P. We say that a pair (F, G) of functions F : P → Q and G : Q → P is a Galois connection if and only if

• (1) Both F and G are antitone maps.

• (2) p ≼ G(F(p)) and q ≼ F(G(q)), for every p ∈ P and q ∈ Q.

Given a function f : P → P and an element p ∈ P, we say that

• p is f-sound if and only if p ≼ f(p).

• p is f-replete if and only if f(p) ≼ p.

• p is a fixed point of f if and only if f(p) = p.

The set of all fixed points of f will be denoted by fix(f).

Proposition 1

[24, thm. 3.16] Let (F, G) be a Galois connection and let Γ = G ∘ Fand Λ = F ∘ G. Then the mapsF : fix(Γ) → fix(Λ) andG : fix(Λ) → fix(Γ) are inverse bijections. Thus fix(Γ) = {G(q) | q ∈ Q} and fix(Λ) = {F(p) | p ∈ P}.

Lemma 1

LetF : P → QandG : Q → Pbe two antitonemaps and let Γ = G ∘ Fand Λ = F ∘ G. Then

• (1) Both Γ and Λ are monotone operators.

• (2) G ∘ Λ = Γ ∘ GandF ∘ Γ = Λ ∘ F.

• (3) For everyp ∈ Pandq ∈ Q: (a) ifp is Γ-sound,thenF(p) is Λ-replete;(b) ifq is Λ-replete,thenG(q) is Γ-sound;(c) ifp is Γ-replete,thenF(p) is Λ-sound;(d) ifq is Λ-sound,thenG(q) is Γ-replete.In particular, ifp is a fixed point of Γ, thenF(p) is a fixed point of Λ, and ifq is a fixed point of Λ, thenG(q) is a fixed point of Γ.

• (4) (F, G) is an antitone Galois connection between the sets fix(Γ) and fix(Λ).

Proof

• (1) Obvious, since both Γ and Λ are compositions of antitone maps. ⊣

• (2) G(Λ(q)) = G((F ∘ G)(q)) = (G ∘ F)(G(q)) = Γ(G(q)) and F(Γ(p)) = F((G ∘ F)(p)) = (F ∘ G)(F(p)) = Λ(F(q)). ⊣

• (3) (a) Let p be Γ-sound, i.e., p ≼ Γ(p). Hence, since F is antitone and by (2), Λ(F(p)) = F(Γ(p)) ≼ F(p), so F(p) is Λ-replete. (b) Similarly, if q is Λ-replete, i.e., Λ(q) ≼ q then, since G is antitone and by (2), G(q) ≼ G(Λ(q)) = Γ(G(q)), so G(q) is Γ-sound. (c) and (d) are proved by symmetric arguments. ⊣

• (4) Both F and G are antitone when restricted to fix(Γ) and fix(Λ), respectively. By (3), ran(F ↾ fix(Γ)) ⊆ fix(Λ) and ran(G ↾ fix(Λ)) ⊆ fix(Γ). We have only to show that the defining condition of antitone Galois connection holds.

Let p ∈ fix(Γ) and q ∈ fix(Λ). On one direction, assume p ≼ G(q). Hence q = Λ(q) = F(G(q)) ≼ F(p). On the other direction, assume q ≼ F(p). Hence p = Γ(p) = G(F(p)) ≼ G(q). □

Corollary 1

F andG are inverse bijections between fix(Γ) and fix(Λ).Hence they are order isomorphisms between fix(Γ) and fix(Λ)^op, the latter denoting the set fix(Λ) ordered by reversing ≼. In particular, if Γ has a least fixed pointlfp(Γ), thenF(lfp(Γ)) is the greatest fixed point of Λ and, conversely, if Λ has a greatest fixed pointgfp(Λ), thenG(gfp(Λ)) is the least fixed point of Γ.

Proof

By Lemma 1, (F′, G′) is an antitone Galois connection between fix(Γ) and fix(Λ), where F′ = F ↾ fix(Γ) and G′ = G ↾ fix(Λ). Moreover, G′ ∘ F′ = Γ ↾ fix(Γ) and F′ ∘ G′ = Λ ↾ fix(Λ) hence, by Proposition 1, F′ and G′ are inverse bijections. Since F is antitone, p ≼ p′ implies F(p′) ≼ F(p) for every p, p′ ∈ fix(Γ). Conversely, p, p′ ∈ fix(Γ) and F(p′) ≼ F(p) implies p = Γ(p) = GF(p) ≼ GF(p′) = Γ(p′) = p′. So, F and G are order isomorphisms between the sets fix(Γ) and fix(Λ), the latter ordered by reversing ≼.

Suppose $\bar {p} = \mathsf {{lfp}}({\Gamma })$ and let $\bar {q} = F(\bar {p})$. Since $\bar {p} \in \text{fix} ({\Gamma })$, $\bar {q} \in \text{fix} ({\Lambda } )$. Let q ∈ fix(Λ). Since G(q) ∈ fix(Γ) and $\bar {p} = \mathsf {{lfp}}({\Gamma })$ it follows $\bar {p} \preceq G(q)$, hence $q = {\Lambda } (q) = FG(q) \preceq F(\bar {p}) = \bar {q}$. Thus $\bar {q} = \mathsf {{gfp}}({\Lambda } )$, namely, F(lfp(Γ)) = gfp(Λ). The converse claim that G(gfp(Λ)) = lfp(Λ) follows by a symmetric argument. □

Let us now back to the proof of Theorem 1. Consider the set P of all partial hypotheses p and the set Q of all sets of (total) hypotheses $\mathcal {{H}}$, both ordered by inclusion. Clearly, the two maps J :p↦{h | p ⊆ h} and ${K :} {\mathcal {{H}}} \mapsto {\bigcap \{{\tau ^{\omega }(h)} \ \lvert \ {h \in \mathcal {{H}}} \}}$, defined in Section 3, are antitone maps between P and Q. Moreover, for the sieve Δ it holds Δ = J ∘ K, and for the jump σ^ω it holds σ^ω = K ∘ J. Hence, by Corollary 1, J and K are order isomorphisms between the set of all fixed points of σ^ω and the set of all fixed points of Δ (this latter ordered by reverse inclusion). Because we already know that both lfp(σ^ω) and gfp(Δ) exist, Corollary 1 yields that lfp(σ^ω) = K(gfp(Δ)), and that gfp(Δ) = J(lfp(σ^ω)). Therefore, Theorem 1 is proved.

Appendix B: Revision-Theoretic and Standard Supervaluation

We want to prove in this section that (a) for each σ-sound partial interpretation p (i.e., such that p ⊆ σ(p)) there exists the least fixed point of σ^ω above p, and (b) the least fixed point of σ^ω is exactly the least fixed point of σ^ω above the least fixed point of σ.

For a partial interpretation p and an operator 𝜃 on partial interpretations, let lfp(𝜃, p) denote the least fixed point of 𝜃 above p (when it exists). Under this notation, our goal becomes to prove the following

Theorem 2

For eachσ-sound partialinterpretation p, lfp(σ^ω, p) exists.Moreover,

$$\mathsf{{lfp}}(\sigma^{\omega}, \mathsf{{lfp}}(\sigma)) = \mathsf{{lfp}}(\sigma^{\omega}). $$

By an easy induction on ω we see that whenever p is σ-sound, p is σ^ω-sound too. Hence lfp(σ^ω, p) exists. The second part of Theorem 2 will follow from the dual fact that (a) for each fixed point q of σ^ω there exists the greatest fixed point of σ below q, denoted by gfp(σ, q), and (b) the maps d :p↦lfp(σ^ω, p) and e :q↦gfp(σ, q) form a monotone Galois connection¹ between the sets of all fixed points of σ and σ^ω ordered by inclusion.

To prove the existence of gfp(σ, q) for every q ∈ Fix(σ^ω) we define an auxiliary jump operator σ^∗ as follows:

$$\sigma^{*}(p) = \bigcap \{{\bigcap {\tau}^{\curvearrowright}(h)} \ \lvert \ {p \subseteq h} \}. $$

It is not difficult to check the following key properties of σ^∗:

• 1. Each σ^ω-replete q is σ^∗-replete too, namely, σ^ω(q) ⊆ q implies σ^∗(q) ⊆ q.

• 2. Fix(σ^∗) = Fix(σ).

From these two properties immediately follows that for each fixed point q of σ^ω (which obviously is also σ^ω-replete), by (1) there exists the greatest fixed point of σ^∗ below q which, by (2), coincides with gfp(σ, q).

Lemma 2

Letd : p ↦ lfp(σ^ω, p) ande :q↦gfp(σ, q). Then (d, e) is a monotone Galois connection between Fix(σ) and Fix(σ^ω), both ordered by inclusion.

Proof

Clearly, both d and e are monotone maps. Let p ∈ Fix(σ) and q ∈ Fix(σ^ω). By definition of e, e(d(p)) is the largest p′ ∈ Fix(σ) below d(p). Since p ⊆ d(p) and p ∈ Fix(σ), it follows p ⊆ e(d(p)). By definition of d, d(e(q)) is the least q′ ∈ Fix(σ^ω) above e(q). Since e(q) ⊆ q and q ∈ Fix(σ^ω), it follows e(d(q)) ⊆ q. □

We are now ready to prove the second part of Theorem 2, namely that lfp(σ^ω) = lfp(σ^ω, lfp(σ)). Let $\bar {p} = \mathsf {{lfp}}(\sigma )$ and $\bar {q} = \mathsf {{lfp}}(\sigma ^{\omega })$. By definition of $\bar {p}$ and e, $\bar {p} \subseteq e(\bar {q}) \subseteq \bar {q}$. By definition of $\bar {q}$ and d, and by Lemma 2, $\bar {q} \subseteq d(\bar {p}) \subseteq d(e(\bar {q})) \subseteq \bar {q}$. Hence $\bar {q} = d(\bar {p})$, i.e., lfp(σ^ω) = lfp(σ^ω, lfp(σ)).

Appendix C: Complexity of Revision-Theoretic Supervaluation

Theorem 3

The setVof all (Gödel codes of) sentences declared true bythe least fixed point of revision-theoretic supervaluationis${{\Pi }^{1}_{1}}$.

For the purposes of this section, let us back to the official definition of τ as an operator on subsets of ω. Let ${\tau }^{\curvearrowright }(X) = \{{\tau ^{n}(X)} \ \lvert \ {n \in \omega } \}$ denote the trajectory of τ starting with X. Accordingly, we have the following definitions:

• stab⁺(X) = {k ∈ ω | ∃m∀n ≥ m(k ∈ τⁿ(X))}.

• stab⁻(X) = {k ∈ ω | ∃m∀n ≥ m(k∉τⁿ(X))}.

• ${\Theta }^{+}(X^{+}, X^{-}) = \bigcap \{{\mathsf {{stab}}^{+}(X)} \ \lvert \ {X^{+} \subseteq X \And X \cap X^{-} = \emptyset } \}$.

• ${\Theta }^{-}(X^{+}, X^{-}) = \bigcap \{{\mathsf {{stab}}^{-}(X)} \ \lvert \ {X^{+} \subseteq X \And X \cap X^{-} = \emptyset } \}$.

• Θ(X⁺, X⁻) = (Θ⁺(X⁺, X⁻),Θ⁻(X⁺, X⁻)).

It is straightforward to see that

• 1. Θ is a monotone operator on partial interpretations.

• 2. If (Z⁺, Z⁻) denotes the least fixed point of Θ, then Z⁺ = V.

Moreover, it is clear that the definition of Θ fits the same template of the definition of the supervaluational jump operator J_vF in Burgess [3, p. 666], with the Tarskian operator J_T (Burgess’ notation for our τ) and its complement just replaced by the operators stab⁺ and stab⁻ defined above. It is well known that the relation {(X, n) | n ∈ J_T(X)} and its complement are ${{\Delta }^{1}_{1}}$. Therefore, by mimicking Burgess’ computation of the complexity of $0^{+}_{\mathsf {{vF}}}$ (Burgess’ notation for V^σ) in [3, p. 670], if we can show that both relations {(X, n) | n ∈stab⁺(X)} and {(X, n) | n ∈stab⁻(X)} are ${{\Delta }^{1}_{1}}$, then the upper bound of the complexity of V will result to be ${{\Pi }^{1}_{1}}$ as well the upper bound of the complexity of V^σ.

To prove that the relations {(X, n) | n ∈stab⁺(X)} and {(X, n) | n ∈stab⁻(X)} are ${{\Delta }^{1}_{1}}$ all we need to show is that the relation {(k, n, X) | k ∈ τⁿ(X)} is ${{\Delta }^{1}_{1}}$.

In order to emphasise the generality of the result we replace the Tarskian operator τ with a generic operator Ψ on subsets of ω and prove the following

Proposition 2

Let Ψ be an operator on subsets ofω. If Ψ is${{\Delta }^{1}_{1}}$, so is its iteration${\Psi }^{\curvearrowright }$.

We will prove Proposition 2 by a series of lemmata and remarks.

Let $\mathcal {P}(A)$ denote the set of all subsets of some set A. In general, we can code any function ${f} : {B} \rightarrow {\mathcal {P}(A)}$ by a binary relation R ⊆ B × A, by putting

$$R(y, x) \iff x \in f(y), $$

for all y ∈ B and x ∈ A. Indeed, for all y ∈ B, we have

$$f(y) = \{{x \in A} \ \lvert \ {R(y, x)} \}. $$

We call R the relation associated tof.

Lemma 3

Let${\Psi } : {\mathcal {P}(A)} \rightarrow {\mathcal {P}(A)}$. For allS ⊆ ω × Aand ∀i, j ∈ ω, the following are equivalent:

• (a) ∃Y (∀z(S(i, z) ⇔ z ∈ Y ) & ∀w(S(j, w) ⇔ w ∈Ψ(Y ))).

• (b) ∀Y (∀z(S(i, z) ⇔ z ∈ Y ) ⇒ ∀w(S(j, w) ⇔ w ∈Ψ(Y ))).

Proof

(a) ⇒ (b). Let Y ⊆ A be such that ∀z(S(i, z) ⇔ z ∈ Y ) holds and let Y ′ be a subset of A satisfying (a). Then, ∀z(z ∈ Y ′ ⇔ S(i, z) ⇔ z ∈ Y ), so Y = Y ′. Thus ∀w(S(j, w) ⇔ w ∈Ψ(Y )) follows.

(b) ⇒ (a). Assume (b) and define Y = {z ∈ A | S(i, z)}. Then ∀z(S(i, z) ⇔ z ∈ Y ) holds, so, by (b), also ∀w(S(j, w) ⇔ w ∈Ψ(Y ))) holds. □

We denote by R(i, j, S) the relation equivalently defined by either the condition (a) or (b) of Lemma 3. Clearly, if S is the relation associated to the sequence ${s} : {\omega } \rightarrow {\mathcal {P}(A)}$, then R(i, j, S) holds if and only if s(j) = Ψ(s(i)), for all i, j ∈ ω.

Remark 1

Let S be the relation associated to the sequence ${s} : {\omega } \rightarrow {\mathcal {P}(A)}$. Then the relation

$$R^{\prime}(i, j, S) \iff {\Psi}(s(i)) \subseteq s(j), $$

admits the following equivalent definitions

• (a) ∃Y (∀z(S(i, z) ⇔ z ∈ Y ) & ∀w(w ∈Ψ(Y ) ⇒ S(j, w))).

• (b) ∀Y (∀z(S(i, z) ⇔ z ∈ Y ) ⇒ ∀w(w ∈Ψ(Y ) ⇒ S(j, w))).

Lemma 4

We can give two equivalent explicit definitions of the relationx ∈ Ψⁿ(X) associated tothe trajectory${\Psi }^{\curvearrowright }(X)$ofX in Ψ as follows:

• 1. Q(n, x) ⇔ ∃S(∀y(S(0, y) ⇔ y ∈ X) & ∀i < nR(i, i + 1, S) &S(n, x)).

• 2. Q′(n, x) ⇔ ∀Y (Y = Ψⁿ(X) ⇒ x ∈ Y ).

Proof

(1) We will show, by induction on n, that ∀x ∈ A(Q(n, x) ⇔ x ∈Ψⁿ(X)).

Let n = 0. Suppose x ∈Ψ⁰(X) = X and define S = {〈k, y〉 | k = 0 &y ∈ X}. Then, ∀y(S(0, y) ⇔ y ∈ X holds by definition, ∀i < 0R(i, i + 1, S) vacuously holds and S(0, x) holds, since x ∈ X. Conversely, suppose there exists S such that ∀y(S(0, y) ⇔ y ∈ X and S(0, x). Then x ∈ X = Ψ⁰(X).

Let n = m + 1. Suppose x ∈Ψ^m+ 1(X) = Ψ(Ψ^m(X)). By the inductive hypothesis, there exists S′ such that ∀y ∈ A(Q(m, y) ⇔ y ∈Ψ^m(X)). Define Y = {z | S′(m, z)} and $S = {S^{\prime }} \! \upharpoonright \! {(m + 1)} \cup \{{\langle m + 1, z\rangle } \ \lvert \ {z \in {\Psi }(Y)} \}$. Then, ∀y(S(0, y) ⇔ y ∈ X) and R(i, i + 1, S) for all i < m hold since ${S^{\prime }} \! \upharpoonright \! {(m + 1)} \subseteq S$. Let i = m. Then, by definition of Y and S, ∀z(S(m, z) ⇔ S′(m, z) ⇔ z ∈ Y ) & ∀w(S(m + 1, w) ⇔ w ∈Ψ(Y )), hence R(m, m + 1, S) holds. By the inductive hypothesis, ∀z(z ∈ Y ⇔ S′(m, z) ⇔ z ∈Ψ^m(X)), hence Y = Ψ^m(X). Thus x ∈Ψ^m+ 1(X) ⇒ x ∈Ψ(Ψ^m(X)) ⇒ x ∈Ψ(Y ) ⇒ S(m + 1, x).

Suppose, conversely, that there exists S ⊆ ω × A such that ∀y(S(0, y) ⇔ y ∈ X) & ∀i < m + 1R(i, i + 1, S) &S(m + 1, x) holds.

Claim: Let P(T, k) be the condition ∀y(T(0, y) ⇔ y ∈ X) & ∀i < kR(i, i + 1, T). Then, ∀T′T′(P(T, k) &P(T′, k) ⇒ ∀i ≤ k∀y(T(i, y) ⇔ T′(i, y))).

Proof of the claim. By induction on k. If k = 0, then ∀y(T(0, y) ⇔ y ∈ X ⇔ T′(0, y)). Let k = j + 1. By the inductive hypothesis, ∀u(T(j, u) ⇔ T′(j, u)). Hence, from R(j, j + 1, T) and R(j, j + 1, T′) it follows:

∃Y (∀z(T(j, z) ⇔ z ∈ Y ) & ∀w(T(j + 1, w) ⇔ w ∈Ψ(Y ))) and ∃Y ′(∀z(T′(j, z) ⇔ z ∈ Y ′) & ∀w(T′(j + 1, w) ⇔ w ∈Ψ(Y ′))). Hence ∀z(z ∈ Y ⇔ T(j, z) ⇔ T′(j, z) ⇔ z ∈ Y ′), so Y = Y ′. Thus ∀w(T(j + 1, w) ⇔ w ∈Ψ(Y ) = Ψ(Y ′) ⇔ T′(j + 1, w)). ⊣

By the claim, we can prove that ∃S(P(S, m + 1) &S(m + 1, x) ⇒ x ∈Ψ^m+ 1(X)). By the hypothesis, there exists Y such that ∀z(S(m, z) ⇔ z ∈ Y ) & ∀w(S(m + 1, w) ⇔ w ∈Ψ(Y )). Since S(m + 1, x) holds, it follows x ∈Ψ(Y ). Thus, it remains to show that Y = Ψ^m(X). If y ∈ Y, then S(m, Y ). Since P(S, m) holds, by the inductive hypothesis y ∈Ψ^m(X). Conversely, let y ∈Ψ^m(X). By the inductive hypothesis, ∃S′(P(S′, m) &S′(m, y)). By the claim, ∀z(S(m, z) ⇔ S′(m, z)), so S(m, y) holds, hence y ∈ Y. Since Y = Ψ^m(X), x ∈Ψ(Y ) = Ψ(Ψ^m(X)) = Ψ^m+ 1(X). ⊣

(2) Immediate from the definition. □

Remark 2

The graph $Q = \{{\langle n, y\rangle } \ \lvert \ {y = {f}^{\curvearrowright }(x)(n)} \} = \{{\langle n, y\rangle } \ \lvert \ {y = f^{n}(x)} \}$ of the trajectory ${f}^{\curvearrowright }(x)$ admits the following inductive definition:

• (a) 〈0, x〉 ∈ Q,

• (b) if 〈m, z〉 ∈ Q then 〈m + 1, f(z)〉 ∈ Q.

• (c) Q is the intersection of all relations satisfying (1) and (2)

which can be converted into the following explicit definition:

$$\begin{array}{@{}rcl@{}} \langle n, y \rangle \in Q \iff \forall R ((\langle 0, x\rangle \in R \And \forall m, z (\langle m, z \rangle \in R \!\!&\implies&\!\! \langle m + 1, f(z) \rangle \in R)) \\ &\implies& \!\!\langle n, y\rangle \in R). \end{array} $$

Proof of Proposition 2

More precisely, we will prove the following statement. Let $D \subseteq {\mathcal {P}(\omega )} \times {\omega }$ and let Ψ(Y ) = {n ∈ ω | D(Y, n)} for all Y ⊆ ω. Let $Q \subseteq {\mathcal {P}(\omega )} \times {\omega ^{2}}$ be defined by

$$Q(X, n, k) \iff k \in {\Psi}^{n}(X), $$

for all n, k ∈ ω and X ⊆ ω. If D is ${{\Delta }^{1}_{1}}$, so is Q.

Given a relation S ⊆ ω × ω, define ${s} : {\omega } \rightarrow {\mathcal {P}(\omega )}$ by s(i) = {j ∈ ω | S(i, j)}, for all i ∈ ω.

Let R ⊆ ω × ω ×^ω×ω2 be the relation

$$R(i, j, S) \iff s(j) = {\Psi}(s(i)). $$

Since n ∈Ψ(Y ) ⇔ D(Y, n) and D is ${{\Delta }^{1}_{1}}$, then, by Lemma 3, R is ${{\Delta }^{1}_{1}}$. Let $P \subseteq {{}^{{{\omega } \times {\omega }}} {2}} \times {{\omega } \times {\mathcal {P}(\omega )}}$ be the relation

$$P(S, k, X) \iff \forall j (S(0, j) \iff j \in X) \And \forall i < k R(i, i + 1, S). $$

The relation P is ${{\Delta }^{1}_{1}}$ since R is. By Lemma 4, Q(X, n, k) holds if and only if ∃S(P(S, k, X) &S(n, k)) holds. Hence Q is ${{\Sigma }^{1}_{1}}$.

On the other hand, by Lemma 4, Q(X, n, k) holds if and only if ∀Y (Y = Ψⁿ(X) ⇒ k ∈ Y ). By Remark 2,

$$\begin{array}{@{}rcl@{}} &&Y = {\Psi}^{n}(X) \iff \\ &&~~~~~~~~\forall R ((\langle 0, X\rangle \in R \And \forall m, Z (\langle m, Z \rangle \in R \implies \langle m + 1, {\Psi}(Z) \rangle \in R)) \\ &&~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\implies \langle n, Y\rangle \in R). \end{array} $$

Hence Y = Ψⁿ(X) is ${{\Pi }^{1}_{1}}$, so also Q(X, n, k) is ${{\Pi }^{1}_{1}}$, and thus ${{\Delta }^{1}_{1}}$.

Since the relation Q(X, n, k) is ${{\Delta }^{1}_{1}}$, also the relation Y = Ψⁿ(X) is ${{\Delta }^{1}_{1}}$. For, Y = Ψⁿ(X) ⇔ ∀z(z ∈ Y ⇔ Q(X, n, z)). □

Now we can complete the proof of Theorem 3 as follows. Since τ is a ${{\Delta }^{1}_{1}}$ operator on subsets of ω, by Proposition 2 the ternary relations k ∈ τⁿ(X) and k∉τⁿ(X) are ${{\Delta }^{1}_{1}}$. It follows that the binary relations n ∈stab⁺(X) and n ∈stab⁻(X) are ${{\Delta }^{1}_{1}}$ too. Hence, as observed at the beginning of this section, by mimicking Burgess [3, p. 670], the set Z⁺ = V is ${{\Pi }^{1}_{1}}$.

Question 1

Is V a ${{\Pi }^{1}_{1}}$-complete set?²

Appendix D: Revision-Theoretic Supervaluation and Standard Revision

Theorem 4

lfp(σ^ω) is compatible withstab^#.

We will show that lfp(σ^ω) and stab^# are compatible as partial interpretations (namely, that they agree on the common part of their respective domains) by proving that they are compatible in the order-theoretic sense, namely that there exists a partial interpretation which extends both. This partial interpretation will be defined as the set of stabilities of a variant of revision based (as well as stab^#) on the concept of near stability.

The notion of revision sequence was defined by imposing on each limit stage of the sequence a coherence condition stated in terms of the notion of stability: h coheres with ${S} \! \upharpoonright \! {\beta }$ if and only if $\mathsf {{stab}}({S} \! \upharpoonright \! {\beta }) \subseteq h_{\beta }$. We can do the same replacing the notion of stability with the notion of near stability:

Definition 2

A revision^#sequenceS = 〈h_α | α ∈On〉 is an ordinal-length sequence of hypotheses satisfying the following two requirements:

• 1. h_α+ 1 = τ(h_α), for every α ∈On.

• 2. $\mathsf {{stab}}^{\#}({S} \! \upharpoonright \! {\beta }) \subseteq h_{\beta }$, for every β limit.

The partial interpretation yielded by using near stability in the above sense is the set

$$\mathsf{{stab}}^{\#\#} = \bigcap \{{\mathsf{{stab}}^{\#}(S)} \ \lvert \ {S \text{ is a revision\(^{\#}\) sequence}} \}$$

It is straightforward to see that, for every sequence S, stab(S) ⊆stab^#(S), hence every revision^# sequence is also a revision sequence and stab^# ⊆stab^##.

To prove that also lfp(σ^ω) ⊆stab^##, we will reformulate the partial interpretation stab^## as the least fixed point of a suitable monotone operator:

Definition 3

σ^## denotes the monotone operator defined by

$$\sigma^{\#\#}(p) = \bigcap \{{\mathsf{{stab}}^{\#}(S)} \ \lvert \ {p \subseteq S(0) \And S \text{ is a revision\(^{\#}\) sequence}} \}. $$

Obviously, σ^##(∅) = stab^##.

Given a revision sequence S, let Cf(S) denote the set of all hypotheses which occur cofinally many times in S. A key feature of near stability we will use in the following is that

$$\mathsf{{stab}}^{\#}(S) = \bigcap \{{\tau^{\omega}(h)} \ \lvert \ {h \in \mathsf{Cf}(S)} \}. $$

Lemma 5

stab^## = lfp(σ^##).

Proof

Let $\bar {p} = \sigma ^{\#\#}(\emptyset )$. By monotonicity, $\bar {p} \subseteq \sigma ^{\#\#}(\bar {p})$.

Conversely, let $\langle \phi , v\rangle \in \sigma ^{\#\#}(\bar {p})$. Hence 〈ϕ, v〉∈stab^#(S) for any revision^# sequence S such that $\bar {p} \subseteq S(0)$. For any sequence S and any ordinal α < lh(S), let S^α denote the finalsegment of S at α, namely the sequence S′ defined by

• 1. lh(S′) = lh(S) − α.

• 2. S′(ξ) = S(α + ξ), for every ξ < lh(S′).

Let S be any revision^# sequence. Since S is ordinal-length, there exists a limit ordinal γ such that Cf(S) = ran(S^γ): let g = S(γ). Since $\bar {p} \subseteq \mathsf {{stab}}^{\#}(S) = \bigcap \{{\tau ^{\omega }(h)} \ \lvert \ {h \in \mathsf {Cf}(S)} \}$ and g ∈Cf(S), it follows $\bar {p} \subseteq \tau ^{\omega }(g)$. Let δ = γ + ω. Since $\mathsf {{stab}}^{\#}({S} \! \upharpoonright \! {\delta }) = \tau ^{\omega }(g)$, by the coherence condition $\bar {p} \subseteq \tau ^{\omega }(g) \subseteq S(\delta )$. Since S^δ is a revision^# sequence and $\bar {p} \subseteq S^{\delta }(0) = S(\delta )$, it follows that 〈ϕ, v〉∈stab^#(S^δ) = stab^#(S). Hence, we have showed that for every revision^# sequence S, 〈ϕ, v〉∈stab^#(S). Therefore $\langle \phi , v\rangle \in \bar {p}$, thus $\sigma ^{\#\#}(\bar {p}) \subseteq \bar {p}$.

It follows that $\bar {p} = \sigma ^{\#\#}(\emptyset ) = \mathsf {{stab}}^{\#\#}$ is a fixed point of σ^##. Let q be any fixed point of σ^##. By monotonicity, stab^## = σ^##(∅) ⊆ σ^##(q) = q, hence stab^## = lfp(σ^##). □

Lemma 6

σ^ω(p) ⊆ σ^##(p), for everyσ^ω-soundp. In particular, everyσ^ω-soundp is also sound forσ^##.

Proof

Let p be σ^ω-sound and let S be a revision^# sequence such that p ⊆ S(0). We will show, by induction on γ, that p ⊆ S(γ) for every γ limit.

For γ = ω, $p \subseteq \sigma ^{\omega }(p) \subseteq \tau ^{\omega }(S(0)) = \mathsf {{stab}}^{\#}({S} \! \upharpoonright \! {\omega }) \subseteq S(\omega )$. Let γ = δ + ω. By the inductive hypothesis, p ⊆ S(δ). Hence, by definition of σ^ω, $p \subseteq \sigma ^{\omega }(p) \subseteq \tau ^{\omega }(S(\delta )) = \mathsf {{stab}}^{\#}({S} \! \upharpoonright \! {\gamma }) \subseteq S(\gamma )$. Finally, let γ be a limit of limits. By the inductive hypothesis, p ⊆ S(δ) for every δ ∈ γ ∩ Lim. Hence p ⊆ σ^ω(p) ⊆ τ^ω(S(δ)) for every δ ∈ γ ∩ Lim. Hence $p \subseteq \sigma ^{\omega }(p) \subseteq \bigcap \{{\tau ^{\omega }(S(\delta ))} \ \lvert \ {\delta \in \gamma \cap \text{Lim} } \} \subseteq \liminf \{{\tau ^{\omega }(S(\delta ))} \ \lvert \ {\delta \in \gamma \cap \text{Lim} } \} = \mathsf {{stab}}^{\#}({S} \! \upharpoonright \! {\gamma }) \subseteq S(\gamma )$.

Since p ⊆ S(γ) for every γ limit, it follows $\sigma ^{\omega }(p) \subseteq \bigcap \{{\tau ^{\omega }(S(\gamma ))} \ \lvert \ {\gamma \in \text{Lim} } \} \subseteq \liminf \{{\tau ^{\omega }(S(\gamma ))} \ \lvert \ {\gamma \in \text{Lim} } \} = \mathsf {{stab}}^{\#}(S)$. Hence p ⊆ σ^ω(p) ⊆ σ^##(p). □

Finally, we can conclude the proof of Theorem 4 as follows. From Lemma 6 it follows, by transfinite induction, that lfp(σ^ω) ⊆lfp(σ^##). Hence, by Lemma 5, lfp(σ^ω) ⊆stab^##. We already saw that stab^# ⊆stab^## too. Hence stab^## is a common upper bound of both lfp(σ^ω) and stab^# which, therefore, are order-theoretically compatible.