Abstract
According to the structured theory of propositions, if two sentences express the same proposition, then they have the same syntactic structure, with corresponding syntactic constituents expressing the same entities. A number of philosophers have recently focused attention on a powerful argument against this theory, based on a result by Bertrand Russell, which shows that the theory of structured propositions is inconsistent in higher order-logic. This paper explores a response to this argument, which involves restricting the scope of the claim that propositions are structured, so that it does not hold for all propositions whatsoever, but only for those which are expressible using closed sentences of a given formal language. We call this restricted principle Closed Structure, and show that it is consistent in classical higher-order logic. As a schematic principle, the strength of Closed Structure is dependent on the chosen language. For its consistency to be philosophically significant, it also needs to be consistent in every extension of the language which the theorist of structured propositions is apt to accept. But, we go on to show, Closed Structure is in fact inconsistent in a very natural extension of the standard language of higher-order logic, which adds resources for plural talk of propositions. We conclude that this particular strategy of restricting the scope of the claim that propositions are structured is not a compelling response to the argument based on Russell’s result, though we note that for some applications, for instance to propositional attitudes, a restricted thesis in the vicinity may hold some promise.
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References
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Acknowledgements
HL had the basic ideas for the consistency proofs in Section ??. GU had the basic idea for the inconsistency result in Section ??. PF wrote the paper, found a gap in HL’s argument for Theorem 3.11, and fixed it. All three of us are grateful to two anonymous reviewers for this journal, as well to audiences at (virtual) talks at USC, ACU and the 2021 Central APA meeting, especially Andrew Bacon, Cian Dorr, Stephen Finlay, Simon Goldstein, Jeremy Goodman, David Ripley, Avi Sommer, Sean Walsh, Isaac Wilhelm, and Juhani Yli-Vakkuri.
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Open access funding provided by University of Oslo (incl Oslo University Hospital).
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Appendices
Appendix A: Consistency
1.1 A.1 Soundness
To show soundness, we require two standard lemmas, showing that the interpretation of free variables is well-behaved.
Lemma A.1
In any csm, if a and b agree on the free variables in ε, then ⟦ε⟧a = ⟦ε⟧b.
Proof
If ε is a variable x, then a(x) = b(x), whence ⟦x⟧a = ⟦x⟧b. If ε is a constant c, then ⟦c⟧a = ⟦c⟧b by construction. The remaining cases follow routinely by induction hypothesis. □
This means that the assignment function is inert in the interpretation of any closed term ε. Consequently, we will omit it, and write simply ⟦ε⟧.
Lemma A.2
In any csm, \(\llbracket \eta [\bar {\varepsilon }/\bar {x}]\rrbracket ^{a}=\llbracket \eta \rrbracket ^{a[\llbracket \varepsilon _{1}\rrbracket ^{a}/x_{1},\dots ,\llbracket \varepsilon _{n}\rrbracket ^{a}/x_{n}]}\).
Proof
By induction on the structure of η.
- Atomic::
-
If η is x, then \(\llbracket x[\varepsilon /x]\rrbracket ^{a}=\llbracket \varepsilon \rrbracket ^{a}=\llbracket x\rrbracket ^{a[\llbracket \varepsilon \rrbracket ^{a}/x]}\). The case of constants is immediate by Lemma A.1.
- Application::
-
Routine by induction hypothesis.
- Abstraction::
-
Let η be \(\lambda \bar {y}.\varphi \). Since every εi is free for xi, \(\llbracket \lambda \bar {y}.\varphi [\bar {\varepsilon }/\bar {x}]\rrbracket ^{a}=\llbracket \lambda \bar {y}.(\varphi [\bar {\varepsilon }/\bar {x}])\rrbracket ^{a}\). This is:
$$\langle\bar{o}\mapsto\llbracket\varphi[\bar{\varepsilon}/\bar{x}]\rrbracket^{a[\bar{o}/\bar{y}]}_{1},\lambda\bar{y}+\llbracket\varphi[\bar{\varepsilon}/\bar{x}]\rrbracket^{a[\bar{y}]}_{2}\rangle$$By induction hypothesis and the fact that \(\bar {x}\) and \(\bar {y}\) are disjoint, this is:
$$\langle\bar{o}\mapsto\llbracket\varphi\rrbracket_{1}^{a[\llbracket\varepsilon_{1}\rrbracket^{a}/x_{1},\dots,\llbracket\varepsilon_{n}\rrbracket^{a}/x_{n}][\bar{o}/\bar{y}]},\lambda\bar{y}+\llbracket\varphi\rrbracket_{2}^{a[\llbracket\varepsilon_{1}\rrbracket^{a}/x_{1},\dots,\llbracket\varepsilon_{n}\rrbracket^{a}/x_{n}][\bar{y}]}\rangle$$
which is \(\llbracket \lambda \bar {y}.\varphi \rrbracket ^{a[\llbracket \varepsilon _{1}\rrbracket ^{a}/x_{1},\dots ,\llbracket \varepsilon _{n}\rrbracket ^{a}/x_{n}]}\), as required. □
Further, we note that the classical truth-conditions hold for primitive logical resources.
Lemma A.3
For every csm \(\mathfrak {M}\) and assignment function a:
- (1):
-
\(\mathfrak {M},a\vDash (\lambda \bar {x}.\varphi )\bar {\varepsilon }\) iff \(\mathfrak {M},a[\llbracket \varepsilon _{1}\rrbracket ^{a}/x_{1},\dots ,\llbracket \varepsilon _{n}\rrbracket ^{a}/x_{n}]\vDash \varphi \)
- (2):
-
\(\mathfrak {M},a\vDash \varphi \to \psi \) iff \(\mathfrak {M},a\vDash \varphi \) only if \(\mathfrak {M},a\vDash \psi \)
- (3):
-
\(\mathfrak {M},a\vDash \forall x^{t}\varphi \) iff \(\mathfrak {M},a[o/x]\vDash \varphi \) for all o ∈ Dt
Proof
- (1):
-
\(\mathfrak {M},a\vDash (\lambda \bar {x}.\varphi )\bar {\varepsilon }\)
iff \(@\in \llbracket \lambda \bar {x}.{\varphi \rrbracket ^{a}_{1}}(\langle \llbracket \varepsilon _{1}\rrbracket ^{a},\dots ,\llbracket \varepsilon _{n}\rrbracket ^{a}\rangle )\)
iff \(@\in \llbracket \varphi \rrbracket ^{a[\llbracket \varepsilon _{1}\rrbracket ^{a}/x_{1},\dots ,\llbracket \varepsilon _{n}\rrbracket ^{a}/x_{n}]}_{1}\)
iff \(\mathfrak {M},a[\llbracket \varepsilon _{1}\rrbracket ^{a}/x_{1},\dots ,\llbracket \varepsilon _{n}\rrbracket ^{a}/x_{n}]\vDash \varphi \).
- (2) and (3):
-
are routine.
□
Similar conditions hold for the defined logical connectives; we note three representative cases.
Lemma A.4
For every csm \(\mathfrak {M}\) and assignment function a:
- (1):
-
\(\mathfrak {M},a\nvDash \bot \)
- (2):
-
\(\mathfrak {M},a\vDash \neg \varphi \) iff \(\mathfrak {M},a\nvDash \varphi \)
- (3):
-
\(\mathfrak {M},a\vDash \varepsilon =\eta \) iff ⟦ε⟧a = ⟦η⟧a
Proof
Routine, using Lemma A.3. □
We can now establish soundness:
Proposition A.5
If ⊩ φ then φ is valid in every csm.
Proof
By a routine induction on the length of proofs, using the previous lemmas. □
1.2 A.2 Coarsenings
Given a coarsening \(\sim \), fix a type-indexed choice function γt on \(L_{t}/{\sim _{t}}\). That is, for each type t, let \(\gamma _{t}:L_{t}/{\sim _{t}}\to L_{t}\) such that γt(X) ∈ X, for all \(X\in L_{t}/{\sim _{t}}\). Assume further that for each variable x, \(\gamma ([x]_{\sim })=x\); there exist such functions since being a coarsening guarantees that x = y whenever \(x\sim y\). As usual, we omit the type indices. γ maps every equivalence class of terms under \(\sim \) to a designated representative. With this, we can define the result of replacing, in a given expression ε, any free variable x by the designated representative of the syntactic component of a(x). We notate this as follows:
\(\varepsilon [a]:=\varepsilon [\gamma (\pi _{2}(a(x_{1})))/x_{1},\dots ,\gamma (\pi _{2}(a(x_{n})))/x_{n}]\) (\(\bar {x}\) the free variables in ε)
Lemma A.6
In any csm, \({\llbracket \varepsilon \rrbracket ^{a}_{2}}=[\varepsilon [a]]_{\sim }\).
Proof
By induction on the complexity of ε.
- Atomic: :
-
If ε is a variable x, then \(\llbracket x{\rrbracket ^{a}_{2}}=\pi _{2}(a(x))=[\gamma (\pi _{2}(a(x)))]_{\sim }=[x[a]]_{\sim }\). If ε is a constant c, then \(\llbracket c{\rrbracket ^{a}_{2}}=[c]_{\sim }=[c[a]]_{\sim }\).
- Application: :
-
If ε is an application \(\xi \bar {\eta }\), then \(\llbracket \xi \bar {\eta }{\rrbracket ^{a}_{2}}={\llbracket \xi \rrbracket ^{a}_{2}}+\llbracket \eta _{1}{\rrbracket ^{a}_{2}}+\dots +\llbracket \eta _{n}{\rrbracket ^{a}_{2}}\). By IH, this is \([\xi [a]]_{\sim }+[\eta _{1}[a]]_{\sim }+\dots +[\eta _{n}[a]]_{\sim }=[\xi \bar {\eta }[a]]_{\sim }\).
- Abstraction: :
-
If ε is an abstraction \(\lambda \bar {y}.\varphi \), then \(\llbracket \lambda \bar {y}.{\varphi \rrbracket ^{a}_{2}}=\lambda \bar {y}+\llbracket \varphi \rrbracket ^{a[\bar {y}]}_{2}\). By IH, this is \(\lambda \bar {y}+[\varphi [a[\bar {y}]]]_{\sim }\). Let \(\bar {x}\) be the free variables of \(\lambda \bar {y}.\varphi \). Then \([\varphi [a[\bar {y}]]]_{\sim }\) may be specified as follows:
$$[\varphi[\gamma(\pi_{2}(a(x_{i})))/x_{1},\dots,\gamma(\pi_{2}(a[\bar{y}](y_{i})))/y_{1},\dots]_{\sim}$$Note that \(\pi _{2}(a[\bar {y}](y_{i}))=[y_{i}]_{\sim }\), so \(\gamma (\pi _{2}(a[\bar {y}](y_{i})))=y_{i}\). Thus the replacements on \(\bar {y}\) are vacuous; so \(\lambda \bar {y}+[\varphi [a[\bar {y}]]]_{\sim }=\lambda \bar {y}+[\varphi [\gamma (\pi _{2}(a(x_{i})))/x_{1},\dots ]_{\sim }=[(\lambda \bar {y}.\varphi )[a]]_{\sim }\).
□
Note that in the case of closed terms ε, the assignment function can be omitted, in which case we obtain \(\llbracket \varepsilon \rrbracket _{2}=[\varepsilon ]_{\sim }\).
Lemma A.7
For any consistent set of equations T:
- (1):
-
If \(\xi \sim ^{sT}\zeta \) then \(\xi \bar {\varepsilon }\sim ^{sT}\zeta \bar {\varepsilon }\).
- (2):
-
If \(\varepsilon _{i}\sim ^{sT}\eta _{i}\) then \(\xi \varepsilon _{1}\dots \varepsilon _{i}\dots \varepsilon _{n}\sim ^{sT}\xi \varepsilon _{1}\dots \eta _{i}\dots \varepsilon _{n}\).
- (3):
-
If \(\varphi \sim ^{sT}\psi \) then \(\lambda \bar {x}.\varphi \sim ^{sT}\lambda \bar {x}.\psi \).
Proof
(1) Assume \(\xi \sim ^{sT}\zeta \) is witnessed by expressions 𝜗, χ and κ such that ξ = 𝜗(χ) and ζ = 𝜗(κ). Without loss of generality, the designated free variable of 𝜗 may be assumed not to occur in \(\bar {\varepsilon }\). Then \(\xi \bar {\varepsilon }=\vartheta \bar {\varepsilon }(\chi )\) and \(\zeta \bar {\varepsilon }=\vartheta \bar {\varepsilon }(\kappa )\), whence \(\xi \bar {\varepsilon }\sim ^{sT}\zeta \bar {\varepsilon }\).
The cases of (2) and (3) are analogous, where in the case of (3), we assume the designated free variable of 𝜗 not to be among \(\bar {x}\). □
Lemma A.8
For any consistent set of equations T, \(\sim ^{T}\) is a coarsening.
Proof
It is immediate by construction that \(\sim _{t}\) is an equivalence relation. We check the three defining conditions.
For (1), we consider any distinct variables x≠y, and show that \(x\not \sim ^{T}y\). Assume \(x\sim ^{T}y\) for contradiction. Then T ⊩ x = y: In general, if ε = η ∈ T, then T ⊩ 𝜗(ε) = 𝜗(η). So if \(\xi \sim ^{iT}\zeta \), then T ⊩ ξ = ζ. By the reflexivity, symmetry and transitivity of = in ⊩, it follows also that if \(\xi \sim ^{T}\zeta \), then T ⊩ ξ = ζ. So, as claimed, it follows from \(x\sim ^{T}y\) that T ⊩ x = y, whence T ⊩∀x∀y(x = y) by the definition of entailment. But ⊩¬∀x∀y(x = y): for type 〈〉, this is witnessed by ⊩⊥≠⊤; for any other type, it is witnessed by \(\vdash (\lambda \bar {x}.\bot )\neq (\lambda \bar {x}.\top )\). Thus T is inconsistent, in contradiction to the assumption. So \(x\not \sim ^{T}y\).
For (2), assume \(\varepsilon _{1}\sim ^{T}\eta _{1},\dots ,\varepsilon _{n}\sim ^{T}\eta _{n}\) and \(\xi \sim ^{T}\zeta \). Then there are \(\bar {\vartheta },\bar {\bar {\chi }}\) such that
\(\xi =\vartheta ^{1}\sim ^{sT}\dots \sim ^{sT}\vartheta ^{m}=\zeta \), and
\(\varepsilon _{i}={\chi _{i}^{1}}\sim ^{sT}\dots \sim ^{sT}\chi _{i}^{l_{i}}=\eta \) for all i < n.
Then by Lemma A.7 (1 & 2),
So \(\xi \bar {\varepsilon }\sim ^{T}\zeta \bar {\eta }\), as required.
For (3), assume \(\varphi \sim ^{T}\psi \). Then there are \(\bar {\chi }\) such that
\(\varphi =\chi _{1}\sim ^{sT}\dots \sim ^{sT}\chi _{n}=\psi \).
Then by Lemma A.7 (3),
\(\lambda \bar {x}.\varphi =\lambda \bar {x}.\chi _{1}\sim ^{sT}\dots \sim ^{sT}\lambda \bar {x}.\chi _{n}=\lambda \bar {x}.\psi \).
So \(\lambda \bar {x}.\varphi \sim ^{T}\lambda \bar {x}.\psi \), as required. □
Lemma A.9
\(\sim ^{\alpha ,\eta }\) is a coarsening.
Proof
Standard models show that α,η is consistent. The claim therefore follows by Lemma A.8. □
Lemma A.10
\(\sim ^{\alpha ,\eta }\) is structural.
Proof
We show that if \(\xi \bar {\varepsilon }\sim ^{\alpha ,\eta }\zeta \bar {\chi }\), then \(\xi \sim ^{\alpha ,\eta }\zeta \) and \(\varepsilon _{i}\sim ^{\alpha ,\eta }\chi _{i}\) by induction on the length of the witnessing sequence. The case of a sequence of length 1 is immediate. So assume \(\xi \bar {\varepsilon }\sim ^{\alpha ,\eta }\zeta \bar {\chi }\) is witnessed by a sequence of length n + 1, with the n th element being 𝜗. Then \(\vartheta \sim ^{s\alpha ,\eta }\zeta \bar {\chi }\). Since no instance of α or η is an identity flanked by an application on either side, 𝜗 must be an application \(\nu \bar {\mu }\), with \(\nu \sim ^{s\alpha ,\eta }\zeta \) or ν = ζ, and \(\mu _{i}\sim ^{s\alpha ,\eta }\chi _{i}\) or μi = χi. Thus \(\nu \sim ^{\alpha ,\eta }\zeta \) and \(\mu _{i}\sim ^{\alpha ,\eta }\chi _{i}\). Further, by IH, \(\xi \sim ^{\alpha ,\eta }\nu \) and \(\varepsilon _{i}\sim ^{\alpha ,\eta }\mu _{i}\). The claim follows by transitivity of \(\sim ^{\alpha ,\eta }\). □
Lemma A.11
\(\sim ^{\alpha ,\eta }\) is safe.
Proof
We first show that for every instance χ = κ of α and η, ⟦χ⟧a = ⟦κ⟧a, considering the two coordinates separately.
To show \({\llbracket \chi \rrbracket _{1}^{a}}={\llbracket \kappa \rrbracket _{1}^{a}}\), we distinguish two cases. Consider first any instance of α: \(\llbracket \lambda \bar {y}.\varphi [\bar {y}/\bar {x}]{\rrbracket ^{a}_{1}}=\bar {o}\mapsto \llbracket \varphi [\bar {y}/\bar {x}]\rrbracket ^{a[\bar {o}/\bar {y}]}_{1}\); by Lemma A.2, this is \(\bar {o}\mapsto \llbracket \varphi \rrbracket ^{a[\bar {o}/\bar {y}][\bar {o}/\bar {x}]}_{1}\), which by Lemma A.1 is \(\bar {o}\mapsto \llbracket \varphi \rrbracket ^{a[\bar {o}/\bar {x}]}_{1}=\llbracket \lambda \bar {x}.{\varphi \rrbracket ^{a}_{1}}\). Consider now an instance of η: \(\llbracket \lambda \bar {x}.\varepsilon \bar {x}{\rrbracket ^{a}_{1}}=\bar {o}\mapsto \llbracket \varepsilon \bar {x}\rrbracket ^{a[\bar {o}/\bar {x}]}_{1}\), which by Lemma A.1 is \(\bar {o}{\mapsto \llbracket \varepsilon \rrbracket ^{a}_{1}}(\bar {o})={\llbracket \varepsilon \rrbracket ^{a}_{1}}\).
We now show \({\llbracket \chi \rrbracket _{2}^{a}}={\llbracket \kappa \rrbracket _{2}^{a}}\). Since χ = κ is an instance of α or η, χ[a] = κ[a] is an instance of α or η as well. Thus \(\chi [a]\sim ^{\alpha ,\eta }\kappa [a]\), whence \([\chi [a]]_{\sim ^{\alpha ,\eta }}=[\kappa [a]]_{\sim ^{\alpha ,\eta }}\). By Lemma A.6, it follows that \({\llbracket \chi \rrbracket ^{a}_{2}}={\llbracket \kappa \rrbracket ^{a}_{2}}\).
So, for every instance χ = κ of α and η, ⟦χ⟧a = ⟦κ⟧a. Therefore \(\llbracket \vartheta \rrbracket ^{a[\llbracket \chi \rrbracket ^{a}/x]}=\llbracket \vartheta \rrbracket ^{a[\llbracket \kappa \rrbracket ^{a}/x]}\), whence with Lemma A.2, ⟦𝜗(χ)⟧a = ⟦𝜗(κ)⟧a. It follows that if \(\xi \sim ^{i\alpha ,\eta }\zeta \), then ⟦ξ⟧a = ⟦ζ⟧a. Using the fact that identity is reflexive, symmetric and transitive, it follows by induction also that if \(\xi \sim ^{\alpha ,\eta }\zeta \), then ⟦ξ⟧a = ⟦ζ⟧a. □
Proposition A.12
Any safe csm based on a strongly structural coarsening validates Strong Closed Structure.
Proof
Let \(\mathfrak {M}\) be safe and based on a strongly structural coarsening \(\sim \). Assume \(\mathfrak {M},a\vDash \xi [\bar {\varepsilon }/\bar {x}]=\xi [\bar {\eta }/\bar {x}]\), for closed \(\bar {\varepsilon },\bar {\eta }\), with all \(\bar {x}\) free in ξ. Then \(\llbracket \xi [\bar {\varepsilon }/\bar {x}]{\rrbracket ^{a}_{2}}=\llbracket \xi [\bar {\eta }/\bar {x}]{\rrbracket ^{a}_{2}}\), so with Lemma A.6, \([\xi [\bar {\varepsilon }/\bar {x}][a]]_{\sim }=[\xi [\bar {\eta }/\bar {x}][a]]_{\sim }\). Let \(\bar {y}\) be the free variables distinct from those in \(\bar {x}\). Let \(\bar {z}\) be variables of the same types as \(\bar {x}\), which do not occur in γ(π2(a(yi))), for any i ≤ n. Then \(\xi [\bar {\varepsilon }/\bar {x}][a]\) is
By the choice of \(\bar {z}\), this is \(\xi ^{\prime }[\bar {\varepsilon }/\bar {z}]\), where \(\xi ^{\prime }\) is
Analogously, \(\xi (\bar {\eta })[a]\) is \(\xi ^{\prime }[\bar {\eta }/\bar {z}]\). So \(\xi ^{\prime }[\bar {\varepsilon }/\bar {z}]\sim \xi ^{\prime }[\bar {\eta }/\bar {z}]\). Since \(\sim \) is strongly structural, it follows that \(\varepsilon _{i}\sim \eta _{i}\). And as \(\mathfrak {M}\) is safe, it follows that ⟦εi⟧a = ⟦ηi⟧a, as required. □
Appendix B: Inconsistency
We use ‘\(\trianglelefteq \) ’ for the relation of being an initial sub-chain:
\({\trianglelefteq }:=\lambda RS.R\sqsubseteq S\wedge \forall p\forall q(Spq\to (q\prec \mathsf {ff}(R)\to p\prec \mathsf {ff}(R)))\)
We start by showing that \(\trianglelefteq \) totally orders chains:
Lemma A.13
\(\vdash ^{\prime }\mathsf {C}R\wedge \mathsf {C}S\to R\trianglelefteq S\vee S\trianglelefteq R\).
Proof
Assume CR and CS, i.e., that R and S are chains. Let qq be such that
qq = πp.p ≺ff(R) ∧ p ≺ff(S) ∧ R|p ≡ S|p.
We first show that qq is downward closed along R and S, in the following sense:
- (↓):
-
If Rpq or Spq, then q ≺ qq only if p ≺ qq.
Without loss of generality, consider any p,q such that Rpq and q ≺ qq. The claim is immediate if p = q, so assume p≠q. Since q ≺ qq, R|q ≡ S|q. So as R|qpp, also S|qpp, whence Spq. Note that this means that our assumptions are symmetric with respect to R and S. We establish the three conjuncts required for p ≺ qq:
- (i & ii):
-
Since Rpq, p ≺ff(R). By symmetry, p ≺ff(S).
- (iii):
-
We show that \(R|_{p}\sqsubseteq S|_{p}\). So assume R|prs. Then Rrs, Rsp and s≠p. By transitivity and antisymmetry of R, Rsq and s≠q. Thus R|qrs and R|qsp, and so S|qrs and S|qsp. Therefore Srs and Srq. Since r≠q, it follows that S|prs. Thus \(R|_{p}\sqsubseteq S|_{p}\), and by symmetry, \(S|_{p}\sqsubseteq R|_{p}\) as well. So R|p ≡ S|p, as required.
Thus p ≺ qq, which establishes (↓).
Next, we show the following claim:
- (∗):
-
\(\mathsf {ff}(R)\sqsubseteq qq\) or \(\mathsf {ff}(S)\sqsubseteq qq\).
Assume for contradiction that (∗) fails to hold. Then \(\mathsf {ff}(R)\not \sqsubseteq qq\), whence \(\pi p.p\prec \mathsf {ff}(R)\wedge p\nprec qq\) is non-empty. So as R is well-founded, there is an R-minimal p ≺ff(R) not among qq. Symmetrically, there is an S-minimal q ≺ff(R) not among qq.
We show that ff(R|p) ≡ qq: Consider first any r ≺ff(R|p). Then Rrp and r≠p, so by the minimality of p, it follows that r ≺ qq. Conversely, consider any r ≺ qq. Since \(p\nprec qq\), r≠p. And by (↓), Rpr fails to hold. But as both p and r are among ff(R), Rrp follows with the totality of R. Thus r ≺ff(R|p).
By symmetry, ff(S|q) ≡ qq as well, and so
ff(R|p) ≡ff(S|q).
Using Ext, we conclude:
χ(ff(R|p)) = χ(ff(S|q)).
Since p ≺ff(R), q ≺ff(S), and R and S are chains, we have:
p = χ(ff(R|p)) = χ(ff(S|q)) = q.
We show that R|p ≡ S|q follows as a consequence. Without loss of generality, assume for contradiction that there are r,s such that R|prs and not S|qrs. Then Rrs, Rsp and s≠p. So r,s ≺ff(Rp) ≡ff(Sq). Thus Srq and Ssq. Since not S|qrs was assumed it follows that not Srs. By the totality of S, this means that Ssr. So with transitivity and antisymmetry of S, we obtain Ssq and thus s = q. But this contradicts that p = q and s≠p. Thus, R|p ≡ S|q.
Therefore p (which is q) is such that p ≺ff(R), p ≺ff(S), and R|p ≡ S|p; therefore p ≺ qq. But this contradicts the choice of p as not among qq. This establishes (∗), i.e., that \(\mathsf {ff}(R)\sqsubseteq qq\) or \(\mathsf {ff}(S)\sqsubseteq qq\).
Assuming \(\mathsf {ff}(R)\sqsubseteq qq\), we show that \(R\trianglelefteq S\). To show \(R\sqsubseteq S\), assume Rpq. Then q ≺ qq, so q ≺ff(S). Therefore, if p = q, then Spq. This leaves only the case of p≠q to be considered. In this case, R|qpp. And since q ≺ qq, R|q ≡ S|q; thus S|qpp. So Spq, as required. Thus, \(R\sqsubseteq S\). So, to show \(R\trianglelefteq S\), assume Spq and q ≺ff(R). Then q ≺ qq, so by (↓), p ≺ qq, whence p ≺ff(R), as required. Thus if \(\mathsf {ff}(R)\sqsubseteq qq\), then \(R\trianglelefteq S\).
By symmetry, if \(\mathsf {ff}(S)\sqsubseteq qq\), then \(S\trianglelefteq R\). Since we have established (∗), i.e., that \(\mathsf {ff}(R)\sqsubseteq qq\) or \(\mathsf {ff}(S)\sqsubseteq qq\), it follows as claimed that \(R\trianglelefteq S\) or \(S\trianglelefteq R\). □
With this, we can show that M is itself a chain.
Lemma A.14
\(\vdash ^{\prime }\mathsf {CM}\).
Proof
We first verify WM, i.e., that M is a well-order, by checking the four defining conditions.
- Reflexivity::
-
Immediate by reflexivity of chains.
- Transitivity::
-
If Mpq and Mqr, then Rpq and Sqr for chains R and S. By Lemma B.1, \(R\sqsubseteq S\) or \(S\sqsubseteq R\). Thus, by the transitivity of chains, Rpr or Spr, whence Mpr.
- Antisymmetry::
-
Analogous to the case of transitivity, if Mpq and Mqp, then by Lemma B.1, there is a chain Q such that Qpq and Qqp, whence with the antisymmetry of chains, p = q.
- Well-foundedness::
-
Consider any \(pp\sqsubseteq \mathsf {ff}(\mathsf {M})\) such that there is some p ≺ pp. By definition of M, p ≺ff(R) for some chain R. Let qq be such that
qq = πq.q ≺ pp ∧ q ≺ff(R)
p ≺ qq, so qq is non-empty. Therefore, by the well-foundedness of R, there is an R-minimal q ≺ qq. We show that q is M-minimal among pp. So consider any r ≺ pp. If r ≺ff(R), then Rqr by the R-minimality of q, so Mqr. Otherwise, i.e., if \(r\nprec \mathsf {ff}(R)\), there is a chain S such that r ≺ff(S). Then \(S\not \sqsubseteq R\), so by Lemma B.1, \(R\trianglelefteq S\). Assume for contradiction that Mqr is not the case. Then Sqr also fails to be the case, whence by the totality of S, Srq. But q ≺ff(R), so with \(R\trianglelefteq S\), r ≺ff(R), contradicting the assumption otherwise. Thus Mqr, establishing that the M-minimality of q. Therefore M is well-founded.
Finally, we show that if p ≺ff(M), then p = χ(ff(M|p)). If p ≺ff(M), then p ≺ff(R) for some chain R. Then p = χ(ff(R|p)). We establish the following:
M|p ≡ R|p
\(R|_{p}\sqsubseteq \mathsf {M}|_{p}\) is immediate from the fact that R is a chain. So assume M|pqr. Then Mqr, Mrp and r≠p. By Lemma B.1, \(R\trianglelefteq \mathsf {M}\), since if \(\mathsf {M}\trianglelefteq R\) then M ≡ R. Thus q,r ≺ff(R). We know that Mpr cannot hold, whence Rpr cannot hold either. By totality of R, Rrp follows. It remains to establish Rqr: If q = r, this follows by reflexivity, so assume otherwise. Then Mrq fails, whence Rrq fails as well; by totality, Rqr follows. Thus R|pqr follows. This establishes M|p ≡ R|p.
If follows that ff(M|p) ≡ff(R|p), and so with Ext that p = χ(ff(R|p)) = χ(ff(M|p)) as required. □
We can now show that \(\infty \) is among the field of M.
Lemma A.15
\(\vdash ^{\prime }\infty \prec \mathsf {ff}(\mathsf {M})\)
Proof
Assume for contradiction that \(\infty \nprec \mathsf {ff}(\mathsf {M})\). Define:
\(\mathsf {M}^{+}:=\lambda pq.\mathsf {M}pq\vee (p\prec \mathsf {ff}(\mathsf {M})\wedge q=\infty )\vee (p=\infty \wedge q=\infty )\)
We show that M+ is a chain. First, we establish that it is a well-order:
- Reflexivity::
-
Immediate by construction.
- Transitivity::
-
Assume M+pq and M+qr. If \(r=\infty \), then M+pr, so assume otherwise, i.e., \(r\neq \infty \). Then Mqr, and as \(\infty \nprec \mathsf {ff}(\mathsf {M})\), \(q\neq \infty \) as well. Then Mpq, so Mpr by transitivity of M.
- Antisymmetry::
-
Assume M+pq and M+qp. Since \(\infty \nprec \mathsf {ff}(\mathsf {M})\), \(p=\infty \) iff \(q=\infty \). So assume neither p nor q is \(\infty \). Then Mpq and Mqp, whence p = q by antisymmetry of M.
- Well-foundedness::
-
Consider any non-empty \(pp\sqsubseteq \mathsf {ff}(\mathsf {M^{+}})\). If there is no p ≺ pp such that p ≺ff(M), then p ≺ pp iff \(p=\infty \). Then trivially, \(\infty \) is M+-minimal among pp. So assume otherwise, i.e., that there is some p ≺ pp such that p ≺ff(M). Let qq such that:
qq = πq.q ≺ pp ∧ q ≺ff(M)
Since p ≺ qq, qq is non-empty, so it follows from the well-foundedness of M that there is a q ≺ qq which is M-minimal among qq. We show that q is M+-minimal among pp. So consider any r ≺ pp. If r ≺M, then r ≺ qq, so Mqr by M-minimality of q, whence M+qr. Otherwise, \(r=\infty \), in which case, M+qr as well. Thus q is M+-minimal among pp as required.
Finally, we show that for all p ≺ff(M+), p = χ(ff(M+|p)). We distinguish two cases:
Case 1: p ≺ff(M). Then as M is a chain, p = χ(ff(M|p)). We show that \(\mathsf {ff}(\mathsf {M}|_{p})\equiv \mathsf {ff}(\mathsf {M}^{+}|_{p})\). \(\mathsf {M}\sqsubseteq \mathsf {M}^{+}\), so \(\mathsf {ff}(\mathsf {M}|_{p})\sqsubseteq \mathsf {ff}(\mathsf {M}^{+}|_{p})\). Conversely, consider any q ≺ff(M+|p). Then M+qp and q≠p, and as p ≺ff(M) and \(\infty \nprec \mathsf {ff}\mathsf {M})\), Mqp. So q ≺ff(M|p). Therefore, as claimed, \(\mathsf {ff}(\mathsf {M}|_{p})\equiv \mathsf {ff}(\mathsf {M}^{+}|_{p})\). By Ext, \(p=\chi (\mathsf {ff}(\mathsf {M}|_{p}))=\chi (\mathsf {ff}(\mathsf {M}^{+}|_{p}))\), as required.
Case 2: \(p\nprec \mathsf {ff}(\mathsf {M})\); then \(p=\infty \). For any p, p ≺ff(M) iff \(\mathsf {M}^{+}p\infty \) and \(p\neq \infty \), which is the case iff \(p\prec \mathsf {ff}(\mathsf {M}^{+}|_{\infty })\). Thus \(\mathsf {ff}(\mathsf {M})\equiv \mathsf {ff}(\mathsf {M}^{+}|_{\infty })\). By Ext, \(\infty =\chi (\mathsf {ff}(\mathsf {M}))=\chi (\mathsf {ff}(\mathsf {M}^{+}|_{\infty }))\), as required.
By construction, \(\mathsf {M}^{+}\infty \infty \). We have just shown M+ to be a chain, so it witnesses \(\mathsf {M}\infty \infty \). Thus \(\infty \prec \mathsf {ff}(\mathsf {M})\), contradicting our assumption to the contrary. □
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Fritz, P., Lederman, H. & Uzquiano, G. Closed Structure. J Philos Logic 50, 1249–1291 (2021). https://doi.org/10.1007/s10992-021-09598-5
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DOI: https://doi.org/10.1007/s10992-021-09598-5