Modal force: existential (◇) or universal (□).
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A modal disjunction: conjunction of modal propositions.
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A single disjunct is true iff its modal claim holds. ◇: ∃w ∈ A, B(w). □: ∀w ∈ A, B(w).
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A modal disjunction is true iff every disjunct's modal claim holds.
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- disj.holds = List.all disj fun (d : Semantics.Modality.Disjunction.Disjunct W) => d.holds
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Exhaustivity: C ⊆ ∪(Aᵢ ∩ Bᵢ). All background worlds are covered by some disjunct's modal cell.
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- Semantics.Modality.Disjunction.exhaustivity C disj = ∀ (w : W), C w = true → (List.any disj fun (d : Semantics.Modality.Disjunction.Disjunct W) => d.cell w) = true
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Disjointness for binary disjunctions: cells don't overlap. (Aᵢ ∩ Bᵢ) ∩ (Aⱼ ∩ Bⱼ) = ∅ for i ≠ j.
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Non-triviality: Aᵢ ≠ ∅. Each modal domain is non-empty.
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- Semantics.Modality.Disjunction.nonTriviality disj = ∀ d ∈ disj, Core.Proposition.FiniteWorlds.worlds.any d.domain = true
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Free choice: for a modal disjunction, each disjunct's modal claim holds individually.
Geurts §3 Case #1/2: "It may be here or it may be there" → each individual "may" claim holds. This is immediate from the structure: the disjunction IS a conjunction of modal claims.
Disjointness gives exclusive reading.
If cells are disjoint and world w is in cell 1, it is not in cell 2. This is Geurts §5: exclusive 'or' from Disjointness, NOT scalar implicature.
Exhaustivity + Disjointness → each C-world in exactly one cell.
Default domain binding: by default, each modal domain equals the background C. The hearer tries A = C first, and only restricts if constraints force it (Geurts p. 394).
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With default binding and existential force, truth = each disjunct is possible w.r.t. C. This is the basic free choice structure.
Construct a Geurts existential disjunction from two presuppositional propositions: domains = presuppositions, contents = assertions.
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The overall presupposition of a Geurts disjunction from PrProps is p.presup ∨ q.presup — matching PrProp.orFlex.
The assertion of a Geurts disjunction from PrProps matches orFlex: (p.presup ∧ p.assertion) ∨ (q.presup ∧ q.assertion).
When presuppositions conflict (p ∧ q = ⊥), the Geurts domains are automatically disjoint — the Disjointness constraint is satisfied for free.
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Background C: it's either here or there (not elsewhere).
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Disjunct 1: □(here) over domain {here}.
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Disjunct 2: □(there) over domain {there}.
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"It must be here or it must be there" with domain restriction. Exhaustivity+Disjointness force A = {here}, A' = {there}.
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The disjunction holds: □(here) over {here} ∧ □(there) over {there}.
Exhaustivity: bgHereOrThere ⊆ {here} ∪ {there}.
Disjointness: {here} ∩ {there} = ∅.
Key prediction: "It must be here or it must be there" does NOT entail "it must be here". The necessity over the full background fails.
"It may be here or it may be there" with default domain binding.
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The disjunction holds: ◇(here) w.r.t. C ∧ ◇(there) w.r.t. C.
Free choice: ◇(here) holds individually.
Free choice: ◇(there) holds individually.