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Studies.HollidayMandelkern2024

Holliday and Mandelkern (2024): The orthologic of epistemic modals #

This file formalizes Holliday and Mandelkern's possibility semantics for epistemic modals on their running example, the Epistemic Scale. The Scale is the epistemic frame of the two-world Boolean algebra of a coin flip, whose five possibilities run from knowing p through full uncertainty to knowing ¬p. The frame comes from the general construction Orthologic.epistemicFrame, so its compatibility and accessibility relations are derived rather than stipulated, and the paper's truth-value table follows by decision. On this model Wittgenstein sentences are contradictions while ◇¬p does not entail ¬p, and the classical laws the paper gives up fail at the point of full uncertainty.

Main results #

Implementation notes #

The Boolean algebra is Finset (Fin 2), so every claim about the Scale is decidable; the general truth conditions of embedded propositions are Orthologic.mem_core_embed and Orthologic.mem_diamond_embed. The paper contrasts its symmetric Wittgenstein's Law with the order asymmetry of dynamic semantics, whose side is Veltman1996.consistent_might_neg beside Veltman1996.not_consistent_up_might_neg; no theorem here conjoins the two.

TODO #

The Epistemic Grid (Example 4.34), the logic EO+ with its completeness theorem, and the remaining principles of Proposition 5.12 are not formalized.

References #

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Coin is the two-world Boolean algebra of the coin flip (Example 5.3).

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    Poss is the type of possibilities of the epistemic frame of Coin.

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      frame is the Epistemic Scale's compatibility frame, with accessibility access.

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        The possibility x1 = ({0}, {0}) knows p.

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          The possibility x2 = ({0}, {0, 1}) settles p without knowing it.

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            The possibility x3 = ({0, 1}, {0, 1}) is fully uncertain.

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              The possibility x4 = ({1}, {0, 1}) settles ¬p without knowing it.

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                The possibility x5 = ({1}, {1}) knows ¬p.

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                  theorem HollidayMandelkern2024.card_poss :
                  Fintype.card Poss = 5

                  Two worlds give five possibilities, the first row of Table 1.

                  theorem HollidayMandelkern2024.compat_iff (x y : Poss) :
                  frame.compat x y ↔ x = y ∨ (x, y) ∈ [(x1, x2), (x2, x1), (x2, x3), (x3, x2), (x3, x4), (x4, x3), (x4, x5), (x5, x4)]

                  Compatibility is the path x1 — x2 — x3 — x4 — x5 (Figure 7 and Example 5.3).

                  theorem HollidayMandelkern2024.access_iff (x y : Poss) :
                  (x, y) ∈ Orthologic.Possibility.access ↔ x = y ∨ (x, y) ∈ [(x2, x1), (x2, x3), (x4, x3), (x4, x5)]

                  x2 accesses x1 and x3, x4 accesses x3 and x5, and every possibility accesses itself (Figure 12 and Example 5.3).

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                  P is the proposition that the coin lands 0, the embedding of {0}.

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                    instance HollidayMandelkern2024.instDecidablePredPossMemSetP :
                    DecidablePred fun (x : Poss) => x ∈ P
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                    instance HollidayMandelkern2024.instDecidablePredPossMemSetNP :
                    DecidablePred fun (x : Poss) => x ∈ nP
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                    instance HollidayMandelkern2024.instDecidablePredPossMemSetBP :
                    DecidablePred fun (x : Poss) => x ∈ bP
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                    instance HollidayMandelkern2024.instDecidablePredPossMemSetBnP :
                    DecidablePred fun (x : Poss) => x ∈ bnP
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                    instance HollidayMandelkern2024.instDecidablePredPossMemSetDP :
                    DecidablePred fun (x : Poss) => x ∈ dP
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                    instance HollidayMandelkern2024.instDecidablePredPossMemSetDnP :
                    DecidablePred fun (x : Poss) => x ∈ dnP
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                    Example 4.33: the truth values of Figure 13 #

                    theorem HollidayMandelkern2024.mem_P (x : Poss) :
                    x ∈ P ↔ x = x1 ∨ x = x2
                    theorem HollidayMandelkern2024.mem_nP (x : Poss) :
                    x ∈ nP ↔ x = x4 ∨ x = x5
                    theorem HollidayMandelkern2024.mem_bP (x : Poss) :
                    x ∈ bP ↔ x = x1
                    theorem HollidayMandelkern2024.mem_bnP (x : Poss) :
                    x ∈ bnP ↔ x = x5
                    theorem HollidayMandelkern2024.mem_dP (x : Poss) :
                    x ∈ dP ↔ x = x1 ∨ x = x2 ∨ x = x3
                    theorem HollidayMandelkern2024.mem_dnP (x : Poss) :
                    x ∈ dnP ↔ x = x3 ∨ x = x4 ∨ x = x5
                    theorem HollidayMandelkern2024.mem_dP_inter_dnP (x : Poss) :
                    x ∈ dP ∩ dnP ↔ x = x3

                    ◇p ∧ ◇¬p holds only at the point of full uncertainty.

                    □p ∨ □¬p holds exactly where something is known.

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                    instance HollidayMandelkern2024.instDecidablePredPossMemSetCoeFinset (s : Finset Poss) :
                    DecidablePred fun (x : Poss) => x ∈ ↑s
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                    The compatibility frame has ten regular subsets (Example 4.11).

                    The desiderata of §2 on the Scale #

                    theorem HollidayMandelkern2024.wittgenstein :
                    nP ∩ dP = ∅ ∧ P ∩ dnP = ∅

                    Wittgenstein sentences of both orders are contradictions ((2) and (3)), by Proposition 4.27 applied to p and to ¬p, using that ¬¬p = p for the regular p.

                    At the coin whose outcome is unknown, ◇¬p does not entail ¬p (§1) and p does not entail □p (§2). In a Boolean algebra wittgenstein would force both entailments (Orthologic.wittgensteinLaw_iff_diamondHom_compl_le).

                    EO does not derive ◇¬p ⊢ ¬p: the Scale's regular propositions form an epistemic ortholattice (Proposition 4.27), so by soundness (Theorem 3.26) the derivation would make x3, which settles ◇¬p, settle ¬p (§1).

                    EO does not derive p ⊢ □p: at x2, which settles p without knowing it (§2).

                    At x3 the conjunction (p ∨ ¬p) ∧ (◇p ∧ ◇¬p) holds but its distribution (p ∧ ◇¬p) ∨ (¬p ∧ ◇p) does not ((10) and Example 4.33).

                    Disjunctive syllogism fails (13), since p ∨ □¬p is a logical truth and ¬□¬p holds at x3 while p does not.

                    The regular propositions of the Scale are not orthomodular ((21) and Example 3.20), since p entails ◇p but ◇p does not entail p ∨ (¬p ∧ ◇p), which fails at x3.

                    Nor are they modular, since a modular ortholattice is orthomodular ([HM24] footnote 4), and so they are not distributive.

                    theorem HollidayMandelkern2024.pseudocomplementation_fails :
                    ¬∀ (a b : frame.Regular), Disjoint a b → b ≤ aᶜ

                    Orthonegation is not pseudocomplementation (Example 3.20), since p ∧ ◇¬p = ⊥ although ◇¬p ≰ ¬p.

                    De Morgan's law makes ◇p ∧ ◇¬p equivalent to ¬(□¬p ∨ □p) (29).

                    The disjunction □¬p ∨ □p is not a logical truth (30).

                    The diamond of p does not collapse to p (Theorem 5.7.4).

                    Example 4.42: levelwise classicality #

                    Level0 indexes the Boolean level B0 = {∅, p, ¬p, ⊤}.

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                      Level1 indexes the modal level B1, the eight propositions of Example 3.33.

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                        theorem HollidayMandelkern2024.level1_closed :
                        (∀ (l : Level1), Orthologic.IsRegular frame l.set) ∧ (∀ (l : Level1), ∃ (m : Level1), ∀ (x : Poss), x ∈ Orthologic.orthoNeg frame l.set ↔ x ∈ m.set) ∧ ∀ (l m : Level1), ∃ (n : Level1), ∀ (x : Poss), x ∈ l.set ∩ m.set ↔ x ∈ n.set

                        B1 is closed under orthonegation and meet, and its members are regular.

                        theorem HollidayMandelkern2024.stratified_level0 (l m : Level0) :
                        (∃ x ∈ l.set, ∃ y ∈ m.set, frame.compat x y) → ∃ z ∈ l.set, z ∈ m.set

                        Within B0, compatible witnesses of two propositions yield a common witness, so B0 is Boolean by the criterion of Proposition 4.39.

                        theorem HollidayMandelkern2024.stratified_level1 (l m : Level1) :
                        (∃ x ∈ l.set, ∃ y ∈ m.set, frame.compat x y) → ∃ z ∈ l.set, z ∈ m.set

                        Within B1 the criterion of Proposition 4.39 holds too, so B1 is an eight-element Boolean algebra.

                        theorem HollidayMandelkern2024.not_stratified_across_levels :
                        x2 ∈ P ∧ x3 ∈ dnP ∧ frame.compat x2 x3 ∧ ∀ (z : Poss), z ∉ P ∩ dnP

                        Across levels the criterion fails, since x2 settles p, x3 settles ◇¬p, the two are compatible, and nothing settles both.