Rooth-Partee Conditionals: Empirical Data #
@cite{sharvit-2025} @cite{rooth-partee-1982}Theory-neutral data for @cite{sharvit-2025} "Rooth-Partee Conditionals."
The puzzle #
(85) "If Mia is penniless or proud of her money, Sue shouldn't lend her any."
Two readings:
(86a) if-over-∃: If [penniless ∨ proud-of-money], shouldn't-lend. Single conditional with disjunctive antecedent.
(86b) ∀-over-if: For each P ∈ {penniless, proud-of-money}: if P(m), shouldn't-lend. Conjunction of two conditionals.
"Proud of her money" presupposes "Mia has money." Empirically the readings
are Strawson-equivalent — speakers cannot distinguish them. Under K/P
(material conditional filtering) they are NOT Strawson-equivalent because
or^{K/P}(penniless, proud) is undefined at penniless-worlds, causing those
worlds to drop from the ∃-reading's quantification domain.
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All worlds in the model.
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"Mia has money" — the presupposition of "proud of her money."
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- Phenomena.Presupposition.Studies.Sharvit2025.hasMoney Phenomena.Presupposition.Studies.Sharvit2025.W.proudNotLend = true
- Phenomena.Presupposition.Studies.Sharvit2025.hasMoney Phenomena.Presupposition.Studies.Sharvit2025.W.proudLend = true
- Phenomena.Presupposition.Studies.Sharvit2025.hasMoney x✝ = false
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"Mia is penniless" — presuppositionless.
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- Phenomena.Presupposition.Studies.Sharvit2025.penniless Phenomena.Presupposition.Studies.Sharvit2025.W.pennyNotLend = true
- Phenomena.Presupposition.Studies.Sharvit2025.penniless Phenomena.Presupposition.Studies.Sharvit2025.W.pennyLend = true
- Phenomena.Presupposition.Studies.Sharvit2025.penniless x✝ = false
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"Mia is proud of her money" — presupposes hasMoney.
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"Sue shouldn't lend Mia any money" — presuppositionless.
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Presuppositionless version of penniless as PrProp.
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The disjunction "penniless or proud" under orFilter is UNDEFINED
at penniless-worlds. This is the root cause of K/P's failure: orFilter's
filtering condition requires "proud of money" to be defined when
"penniless" is true, but penniless entails ¬hasMoney.
The disjunction IS defined at proud-worlds.
Under K/P, the ∃-reading uses orFilter for the antecedent disjunction.
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Under K/P, the ∀-reading is the conjunction of two K/P conditionals.
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K/P ∃-reading is UNDEFINED at penniless-worlds.
K/P ∃-reading IS defined at proud-worlds.
K/P ∀-reading: the first conditional (if penniless, shouldntLend) is always defined.
K/P ∀-reading presup = hasMoney (from the second conditional).
K/P ∃-reading assertion excludes penny-worlds from quantification.
K/P* conditional: if penniless, shouldntLend.
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K/P* conditional: if proud-of-money, shouldntLend.
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K/P* conditional with penniless: always defined.
K/P* conditional with proud-of-money: defined iff hasMoney.
K/P* ∃-reading.
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K/P* ∀-reading.
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Both K/P* readings have presupposition = hasMoney.
Both K/P* readings have identical assertions.
Strawson equivalence of ∃ and ∀ readings under K/P*.