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Linglib.Phenomena.Presupposition.Studies.Sharvit2025

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:

"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.

Worlds for sentence (85). Cross-product of Mia's financial status (penniless vs has-money-and-proud) with whether Sue lends.

  • pennyNotLend : W
  • pennyLend : W
  • proudNotLend : W
  • proudLend : W
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      All worlds in the model.

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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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            Penniless entails ¬hasMoney: the presuppositions conflict.

            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.

            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 presup = hasMoney (from the second conditional).

                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 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.