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Linglib.Phenomena.FreeChoice.Studies.Fox2007

Fox 2007: Free Choice via Recursive Exhaustification @cite{fox-2007} #

The original grammatical account of free-choice permission. Recursive application of the covert exhaustivity operator exh (Innocent Exclusion) over the Sauerland alternatives {◇(p∨q), ◇p, ◇q, ◇(p∧q)} derives the FC inference without Innocent Inclusion.

The five-world model below (ModalW) makes each ◇-formula correspond to which propositional situations are accessible from the evaluation world, so that the four Sauerland alternatives have non-trivial truth values and the IE algorithm has something to compute over.

Modal worlds named by which ◇-propositions they make true.

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      Sauerland alternatives under ◇: {◇(p∨q), ◇p, ◇q, ◇(p∧q)}.

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        Layer 1: Exh(◇(p∨q)) = ◇(p∨q) ∧ ¬◇(p∧q). Only ◇(p∧q) is innocently excludable; ◇p and ◇q cannot be IE-excluded because excluding ◇p while keeping ◇(p∨q) forces ◇q, and conversely.

        Exhaustifying ◇p (relative to the same alternatives) excludes ◇q.

        Exhaustifying ◇(p∧q) is vacuous: it entails every other alternative, so nothing is non-weaker.

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        Layer-2 alternatives: {Exh(φ) : φ ∈ C} — the exhaustified versions of the original Sauerland alternatives.

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          Free Choice: double exhaustification yields ◇p ∧ ◇q ∧ ¬◇(p∧q).

          Exh(C')[Exh(C)(◇(p∨q))] excludes the layer-2 alternatives Exh(◇p) = ◇p ∧ ¬◇q and Exh(◇q) = ◇q ∧ ¬◇p. Negating both (given the layer-1 prejacent ◇(p∨q) ∧ ¬◇(p∧q)) forces both ◇p and ◇q to hold.

          FC entails both disjuncts hold: the speaker has permission for each disjunct individually.