Kernel Semantics for Epistemic Modals #
@cite{von-fintel-gillies-2010} @cite{von-fintel-gillies-2021}
@cite{von-fintel-gillies-2010} kernel semantics. Epistemic modals carry an
evidential presupposition that the prejacent is not directly settled by the
kernel K ⊆ B_K. The central result (entailment_settling_gap) is that B_K can
entail φ without K directly settling it, making the presupposition non-trivial.
The 2021 paper extends the analysis to can't, exposing a dilemma for the weakness Mantra: no assignment of force to can't simultaneously explains its evidential distribution (Observation 4) and its incompatibility with it's possible (Observation 5). Kernel semantics resolves this because can't φ = must(¬φ) is simultaneously strong and evidentially constrained.
Helpers #
Kernel structure (@cite{von-fintel-gillies-2010} Def 4) #
A kernel: a set of direct-information propositions with B_K = ⋂K.
- props : List (BProp Attitudes.Intensional.World)
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B_K = ⋂K.
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Consistency: B_K ≠ ∅.
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φ follows from K iff B_K ⊆ ⟦φ⟧.
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Induce an EpistemicFlavor for Kratzer modal evaluation.
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- k.toEpistemicFlavor = { evidence := k.toModalBase }
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World-dependent kernel assignment K^c(w).
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Settling (@cite{von-fintel-gillies-2010} Def 5) #
K directly settles P iff some X ∈ K entails P or is incompatible with P (Implementation 1).
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Refine a partition along φ-boundaries (@cite{von-fintel-gillies-2010} subject matter).
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The partition S_K generated by K (@cite{von-fintel-gillies-2010} Def 7).
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K settles P via partition iff P is a union of cells in S_K (Implementation 2).
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What both implementations share: settling implies entailment. If K directly settles φ (Implementation 1), then B_K ⊆ ⟦φ⟧ or B_K ⊆ ⟦¬φ⟧. If X ∈ K entails φ then B_K ⊆ X ⊆ ⟦φ⟧; if X excludes φ then B_K ⊆ X ⊆ ⟦¬φ⟧.
Partition settling implies entailment: if S_K settles φ then B_K ⊆ ⟦φ⟧ or B_K ⊆ ⟦¬φ⟧. All worlds in B_K agree on every X ∈ K (they all satisfy every X), so they occupy a single cell of S_K. If that cell is φ-uniform, B_K inherits the uniformity.
Modal operators (@cite{von-fintel-gillies-2010} Defs 5–6) #
⟦must φ⟧: presupposes K doesn't settle φ; asserts B_K ⊆ ⟦φ⟧.
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⟦might φ⟧: presupposes K doesn't settle φ; asserts B_K ∩ ⟦φ⟧ ≠ ∅.
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⟦can't φ⟧ = must(¬φ).
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- Semantics.Modality.kernelCant k φ = Semantics.Modality.kernelMust k fun (w : Semantics.Attitudes.Intensional.World) => !φ w
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World-dependent ⟦must φ⟧.
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World-dependent ⟦might φ⟧.
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Core properties #
Must is strong: when defined, must φ = universal necessity over B_K.
T axiom: must φ entails φ when B_K is realistic (w ∈ B_K).
Duality: might φ ↔ ¬(must ¬φ) in assertion content.
Empty kernel: nothing is settled, so must is always defined.
Bridge to Kratzer.lean #
Kernel must assertion = Kratzer simple necessity.
Kernel must assertion = full Kratzer necessity with empty ordering.
This extends kernelMust_eq_simpleNecessity by connecting through
kernelMust_eq_simpleNecessity: kernel must = simple necessity.
Kernel must assertion = Kratzer necessity evaluated via EpistemicFlavor.
This bridges the kernel (@cite{von-fintel-gillies-2010}) to Kratzer's parametric semantics: kernel must is exactly Kratzer necessity with the kernel's epistemic flavor.
World-dependent kernel must = world-dependent Kratzer necessity.
kernelMustW kb φ asserts necessity (kb.toModalBase) emptyBackground φ
at each world.
Mastermind example (@cite{von-fintel-gillies-2010} pp. 365–366) #
w0 = red, w1 = blue, w2 = green, w3 = unknown. K = {redOrBlue, notRed}, B_K = {w1}.
Non-equivalence of the two implementations (@cite{von-fintel-gillies-2010} §7, p. 379) #
The explicit and partition-based implementations of "directly settles" are
non-equivalent. Implementation 1 settles supersets of K-propositions that
Implementation 2 misses; Implementation 2 settles propositions determined
jointly by K-propositions that no single proposition settles. Both, however,
imply entailment (see explicit_implies_entailment and
partition_implies_entailment above).
Counterexample (Impl 1 ↛ Impl 2): K = {red}, φ = redOrBlue. red ⊆ redOrBlue so Impl 1 settles, but S_K = {{w0},{w1,w2,w3}} and the cell {w1,w2,w3} straddles redOrBlue.
Counterexample (Impl 2 ↛ Impl 1): K = mastermindK, φ = blue. No single X ∈ K entails or excludes blue, but S_K = {{w0},{w1},{w2,w3}} resolves blue into a union of cells.
Entailment does not imply partition settling: B_K ⊆ ⟦φ⟧ does not guarantee S_K settles φ. Same witness as explicit_not_implies_partition: K = {red} entails redOrBlue (B_K = {w0} ⊆ ⟦redOrBlue⟧) but the partition cell {w1,w2,w3} straddles redOrBlue.
Deep theorems (@cite{von-fintel-gillies-2010}) #
Entailment-settling gap: B_K can entail φ without K settling it. This gap makes the evidential presupposition non-trivial: must φ can be simultaneously defined and true.
Indirectness ≠ weakness: three independent cases show indirectness and assertion strength are orthogonal dimensions.
Modus ponens with must (@cite{von-fintel-gillies-2010} Arg 4.3.1): the argument form "if φ, must ψ; φ; ∴ ψ" is valid under realistic B_K.
Must-perhaps contradiction (@cite{von-fintel-gillies-2010} Arg 4.3.2): must φ ∧ might ¬φ is contradictory. When B_K ⊆ ⟦φ⟧, we have B_K ∩ ⟦¬φ⟧ = ∅.
Direct evidence infelicity: when K settles φ, must φ is undefined.
Settling is monotone: more propositions in K means more is settled.
Can't dilemma (@cite{von-fintel-gillies-2021} §4, Observations 4–5) #
The Mantra faces a dilemma: no assignment of force to can't simultaneously explains its evidential distribution (paralleling must) and its incompatibility with it's possible that φ. Kernel semantics resolves this because can't φ = must(¬φ) is strong AND evidentially constrained.
Can't–might exclusion (@cite{von-fintel-gillies-2021} Obs 5): when can't φ holds (B_K ⊆ ⟦¬φ⟧), might φ is false (B_K ∩ ⟦φ⟧ = ∅).
Can't entails negation (@cite{von-fintel-gillies-2021} via S2): when B_K is realistic and can't φ is defined and true, φ(w) = false.
Can't evidential parallelism (@cite{von-fintel-gillies-2021} Obs 4): can't φ has the same presupposition structure as must(¬φ).
Can't dilemma resolved: a single kernel simultaneously exhibits evidentiality, strength, and might-exclusion. Witness: K = {redOrBlue, notRed}, φ = notBlue. Then can't(notBlue) = must(blue), which is defined (blue unsettled), true (B_K ⊆ ⟦blue⟧), and excludes might(notBlue).
Direct evidence blocks can't, paralleling must: when K settles ¬φ, can't φ has presupposition failure.
can't φ and must(¬φ) are intensionally identical.
Prior information state and nandao-Q felicity #
@cite{zheng-2025} shows Mandarin nandao-Qs are evidence-driven, not bias-driven.
The felicity condition has three parts: (i) evidence in K supports the prejacent,
(ii) K conflicts with prior expectations U, (iii) the prejacent is not directly
settled in K. Condition (iii) is the same presupposition as kernelMust.
Prior information state (beliefs, norms, desires) before encountering evidence. Distinct from the kernel (direct evidence) and the CG (shared).
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B_U = ⋂U: worlds compatible with the prior information state.
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K and U are incompatible: B_K ∩ B_U = ∅ (the evidence is unexpected).
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Evidence p raises the probability of φ: P(φ|p) > P(φ). Computed via cross-multiplication to avoid rationals.
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Some proposition in K raises the probability of φ (condition i).
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- k.evidenceSupports φ = k.props.any fun (x : BProp Semantics.Attitudes.Intensional.World) => Semantics.Modality.evidenceRaises x φ
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Nandao-Q felicity (@cite{zheng-2025}, condition 11 for polar questions): (i) some evidence in K raises P(φ), (ii) K and U are incompatible (evidence is unexpected), (iii) φ is not directly settled in K.
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Nandao condition (iii) = presupposition of kernelMust.
Pure inquiry: nandao is felicitous without epistemic bias (@cite{zheng-2025} ex. 3). The three conditions suffice; no prior belief about φ is required.
Dripping raincoat scenario (@cite{zheng-2025} exx. 2–3, 5) #
w0 = raining, w1 = sprinkler (wet but not rain), w2 = dry, w3 = unknown. K = {wearingRaincoat}: direct evidence that someone entered with a wet coat. U = {expectDry}: prior expectation of no rain (normative or doxastic).
"Nandao waimian xiayu-le ma?" is felicitous with a dripping raincoat. P(rain|coat) = 1/2 > P(rain) = 1/4; K ∩ U = ∅; rain unsettled by K.
Without evidence, nandao is infelicitous (@cite{zheng-2025} ex. 5 ctx 2).
When evidence is expected (K compatible with U), nandao is infelicitous (@cite{zheng-2025} ex. 6 ctx 2).
Nandao reuses the mastermind kernel: "must blue" and "nandao blue?" share condition (iii). When must is defined, nandao's indirectness condition holds.