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Linglib.Studies.Anaphora.AbneyKeshet2025

Abney & Keshet (2025): PIP Compositional Operations #

@cite{abney-keshet-2025} @cite{keshet-abney-2024}

The Glossa companion paper to @cite{keshet-abney-2024} enriches PIP with explicit set-based compositional operations: sigma evaluation (Σxφ), set-based generalized quantifiers (EVERY/SOME as set inclusion/intersection), three-argument modals with modal bases, and the FX type-lifting operation for restricted variables.

This file verifies these operations on finite models and demonstrates the paper's predictions about quantificational subordination, strong donkey anaphora, modal subordination, and the exists/sigma bridge.

Worked Examples #

  1. Sigma evaluation — Σx(farmer(x)) = {alice, bob} on a finite model
  2. Sigma set algebra — Σx(φ∧ψ) = Σxφ ∩ Σxψ instantiated
  3. GQ over sigma sets — EVERY(Σf farmer, Σf (farmer∧buyer)) via setEvery
  4. Quantificational subordination — Σx(farmer(x) ∧ bought-donkey(x)) ⊆ Σx(farmer(x))
  5. Exists↔sigma bridge — ∃x(farmer(x)) ↔ (Σx(farmer(x))).Nonempty
  6. Strong donkey — summation pronouns as exhaustive sigma sets
  7. FX thematic roles — ↑AGENT accumulates assertions
  8. Paycheck pronouns — sigma set varies with external free variable
  9. Modal subordination — MUST/MIGHT on two-world intensional model

Four-entity domain: two farmers and two donkeys.

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      Single-world model (extensional fragment).

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          • One or more equations did not get rendered due to their size.

          Σx(farmer ∧ donkey) = Σx(farmer) ∩ Σx(donkey) — no entity is both.

          Σx(farmer ∨ donkey) = Σx(farmer) ∪ Σx(donkey) — all four entities.

          Verify: the farmer∧donkey intersection is empty.

          EVERY(Σf farmer, Σf (farmer ∧ boughtDonkey)) — via setEvery.

          "Every farmer bought a donkey" modeled as: the set of farmers is a subset of the set of farmer-buyers. This connects to the GQ bridge in Bridges.lean via setEvery_eq_pipEvery.

          SOME(Σf donkey, Σf farmerBody) — no donkey is a farmer. This demonstrates setSome negation: the intersection is empty.

          Quantificational subordination:

          "Every farmer bought a donkey. He vaccinated it."

          The subordinate quantifier's sigma set (farmers who bought a donkey and vaccinated it) is a subset of the main quantifier's sigma set (farmers). This follows from sigma_subordination.

          All farmer-buyers are indeed farmers (pointwise verification).

          ∃x(farmer(x)) is true (at least one farmer exists).

          The sigma bridge: ∃x(farmer(x)) ↔ nonempty sigma set.

          ∀x(farmer(x)) is false (donkeys are not farmers).

          Strong donkey anaphora via summation pronouns (paper's §4.3 analysis):

          "Every farmer who bought a donkey vaccinated them."

          The pronoun "them" is a summation pronoun: its denotation is the sigma set of all farmer-buyers, not a single witness. As noted in PIP.Composition, summation pronoun denotation IS sigmaEval — the distinction from simple pronouns is presuppositional (PL vs SG), handled by PIP.Bridges.plural.

          Both alice and bob are in the sigma set — exhaustive, not a single witness.

          The sigma set has 2+ elements → plural presupposition satisfied.

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            ↑AGENT(buy)(alice) = true — alice is an agent of buying.

            ↑AGENT(buy)(spot) = false — spot is not an agent.

            Iterated FX: ↑PATIENT(↑AGENT(buy))(spot) = false — agent fails for spot.

            Paycheck pronouns (paper's §4.2, Karttunen 1969, Jacobson 2000):

            "Almost every girl brought the diorama she made to class. Very few of them forgot it at home."

            The pronoun "it" is a summation pronoun whose antecedent formula D ≡ DIORAMA([d]) ∧ MADE(x, d) contains a free variable x bound by a different quantifier at the pronoun's use site vs its antecedent's. The sigma set ΣdD varies with x: when x = alice, it denotes alice's diorama(s); when x = bob, bob's diorama(s).

            This is the defining property of paycheck pronouns. PIP handles them via summation pronouns with formula labels that contain free variables. Dynamic logics struggle with this because they only store individuals already quantified over, not formulas with free variables.

            Paycheck body: D ≡ DIORAMA([d]) ∧ MADE(x, d) with x as free variable. The sigma set ΣdD depends on who x is.

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              When x = alice, ΣdD = {spot} (alice's diorama).

              When x = bob, ΣdD = {rex} (bob's diorama).

              The paycheck property: the sigma set varies with the external free variable. This is what distinguishes paycheck pronouns from ordinary anaphora — the pronoun's denotation depends on who the binder is.

              Each maker's sigma set is a singleton — satisfying the SG presupposition.

              Two-world model for modal subordination.

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                  Modal subordination (paper's §4.5, Roberts 1987 wolf example):

                  "A wolf might come in. It would eat you first."

                  MIGHT(β_w, ΣwW) where W ≡ (WOLF([x]) ∧ ENTERS(x)) MUST(β_w, ΣwW, ΣwE) where E ≡ (W ∧ TIM([t]) ∧ EATS(x,t))

                  The nuclear scope of the first sentence is stored in label W. The restriction of the second sentence's MUST is anaphoric to W — this is modal subordination, parallel to quantificational subordination.

                  Modal base: actual can access alt, and alt accesses itself.

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                    Worlds where a wolf enters (only alt).

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                      Worlds where the wolf eats Tim (also only alt).

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                        MIGHT(β, ⊤, wolfEnters) — a wolf might enter. The modal base intersected with the wolf-enters set is nonempty.

                        MUST(β, wolfEnters, wolfEatsTim) — it would eat Tim. Modal subordination: the restriction wolfEnters (from the first sentence's label W) constrains MUST. Every wolf-entry world is also a wolf-eats-Tim world.

                        Modal duality instantiated: ¬MUST(β, ⊤, wolfEnters) because the wolf doesn't enter at actual.

                        Modal duality: ¬MUST ↔ MIGHT(complement). The wolf doesn't necessarily enter, so it's possible it doesn't.