Bar-Lev and Fox 2020: Free choice, simplification, and Innocent Inclusion #
Exhaustification with Innocent Inclusion returns the cell of the partition induced by the
alternatives whenever that cell is consistent. Free-choice disjunction ◇(a ∨ b), whose
alternatives are ◇a, ◇b and ◇(a ∧ b), strengthens to ◇a ∧ ◇b ∧ ¬◇(a ∧ b); plain a ∨ b,
whose conjunctive alternative is the conjunction of the other two, only denies it. only
presupposes the includable alternatives, so free choice under only projects. Free choice
under a universal quantifier, its negative counterpart, the existential-over-universal case
and most with a disjunctive restrictor are one computation over eight alternatives.
Simplification of disjunctive antecedents is the same strengthening over a variably strict
conditional: it fails when the consequent is one of the disjuncts, and an or both
antecedent, exhaustified by Hurford's constraint, turns the conjunctive simplification from
excluded into included.
Main results #
freeChoice,simpleDisjunction: the modal and the plain disjunction.II_fcAlts,isInnocentlyExcludable_fcAlts_iff: what free choice includes and excludes.support_only_fcAlts: free choice underonlyis presupposed.posExt_only_eq: without symmetric alternativesonlyagrees with the exclusiveFocus.Particles.only.universalFreeChoice,negativeUniversalFreeChoice,freeChoiceOverUniversal,simplificationMost: instances ofExhaustification.exhIEII_quantified.sda,sda_consequent_disjunct,orBoth: simplification, its failure, and the switches.
References #
Free choice and simple disjunction #
The alternatives of ◇(a ∨ b) replace the disjunction by its disjuncts and their
conjunction.
Equations
- BarLevFox2020.fcAlts R a b = {R.preimage (a ∪ b), R.preimage a, R.preimage b, R.preimage (a ∩ b)}
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Free choice needs a world permitting only a, one permitting only b, and one permitting
each but not both.
- only_left : ∃ w ∈ R.preimage a, w ∉ R.preimage b
- only_right : ∃ w ∈ R.preimage b, w ∉ R.preimage a
- not_both : ∃ w ∈ R.preimage a ∩ R.preimage b, w ∉ R.preimage (a ∩ b)
Instances For
Given the witnesses, ◇(a ∨ b) strengthens to ◇a ∧ ◇b ∧ ¬◇(a ∧ b), free choice.
The includable alternatives of ◇(a ∨ b) are the prejacent and the disjunct
alternatives.
The only excludable alternative of ◇(a ∨ b) is the conjunctive one.
Without the modal the conjunctive alternative is the conjunction of the disjunct alternatives, so exhaustification denies it and includes neither disjunct.
The cell of simple disjunction is contradictory, so cell identification does not apply.
only presupposes the innocently includable alternatives and asserts that the innocently
excludable ones are false (32).
Equations
- BarLevFox2020.only ALT φ = {w : W | ∀ r ∈ Exhaustification.II ALT φ, w ∈ r}.indicator fun (w : W) => SignType.ofProp (∀ (q : Set W), Exhaustification.IsInnocentlyExcludable ALT φ q → w ∉ q)
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Free choice under only is presupposed, since only ◇(a ∨ b) presupposes ◇a and ◇b.
Where the prejacent is the only includable alternative and the excludable ones are exactly
those it does not entail, only is true at the same worlds as the exclusive
Focus.Particles.only on the entailment scale, the entry (29b) that Innocent Inclusion
revises.
Simplification of disjunctive antecedents #
The alternatives of (p ∨ q) → r replace the antecedent's disjunction by its disjuncts
and their conjunction.
Equations
- BarLevFox2020.sdaAlts ord p q r = {Conditional.closestImp ord (p ∪ q) r, Conditional.closestImp ord p r, Conditional.closestImp ord q r, Conditional.closestImp ord (p ∩ q) r}
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Disjunctive antecedents simplify. Over total similarity preorders, given a world where only
the first simplification holds, one where only the second does, and one where both hold but the
conjunctive one fails, (p ∨ q) → r strengthens to (p → r) ∧ (q → r) ∧ ¬((p ∧ q) → r).
Simplification fails when the consequent is one of the disjuncts (71), since the other simplification is the only contingent alternative and so is excluded.
The alternatives of (Exh(p ∨ q) ∨ (p ∧ q)) → r, the or both antecedent parsed with an
embedded exhaustifier by Hurford's constraint (80), are the antecedent's three cells.
Equations
- BarLevFox2020.orBothAlts ord p q r = {Conditional.closestImp ord (p ∪ q) r, Conditional.closestImp ord (p \ q) r, Conditional.closestImp ord (q \ p) r, Conditional.closestImp ord (p ∩ q) r}
Instances For
With the antecedent's cells as alternatives nothing is excludable and, given a world
verifying each cell's conditional alone and one verifying all three, everything is included,
so (p ∨ q) → r asserts the conjunctive conditional it denied under sdaAlts (82).
The switches #
A world of [CZC18]'s scenario records whether each switch is up, and the wiring of the light.
- up₁ : Bool
- up₂ : Bool
- light : Bool → Bool → Bool
Instances For
Equations
- One or more equations did not get rendered due to their size.
Instances For
Equations
- BarLevFox2020.instFintypeSwitch = Fintype.ofEquiv ((_ : Bool) × (_ : Bool) × (Bool → Bool → Bool)) BarLevFox2020.Switch.proxyTypeEquiv
Worlds with a different wiring are farther than any with the same; among the latter, distance is the number of switches in another position.
Equations
Instances For
Switch A is down.
Equations
- BarLevFox2020.Switch.down₁ = {w : BarLevFox2020.Switch | w.up₁ = false}
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Switch B is down.
Equations
- BarLevFox2020.Switch.down₂ = {w : BarLevFox2020.Switch | w.up₂ = false}
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The light is off.
Equations
- BarLevFox2020.Switch.off = {w : BarLevFox2020.Switch | w.light w.up₁ w.up₂ = false}
Instances For
Both switches up, and the light on exactly when the switches agree.
Equations
- BarLevFox2020.Switch.actual = { up₁ := true, up₂ := true, light := fun (x y : Bool) => x == y }
Instances For
If switch A or switch B were down, the light would be off (76) is true in the scenario, since its strengthening asserts both simplifications and denies the conjunctive one.
If switch A or switch B or both were down, the light would be off (78) is false in the scenario, since its strengthening asserts the conjunctive conditional.
Free choice under quantifiers #
The alternatives of ∀x(Px ∨ Qx) (41) replace the disjunction by its disjuncts and by
their conjunctive counterpart B, and the universal by the existential.
Equations
- BarLevFox2020.universalAlts P Q B = {⋂ (x : D), P x ∪ Q x, ⋂ (x : D), P x, ⋂ (x : D), Q x, ⋂ (x : D), B x, ⋃ (x : D), P x ∪ Q x, ⋃ (x : D), P x, ⋃ (x : D), Q x, ⋃ (x : D), B x}
Instances For
Given a world where every individual has P and none Q, one the other way round, one where
each has exactly one and both occur, and one where every individual has P and Q but none
B, ∀x(Px ∨ Qx) strengthens to ∀x Px ∧ ∀x Qx ∧ ¬∃x Bx, universal free choice (44). With
Px = ◇px and Bx = ◇(px ∧ qx) this is (36).
The alternatives of ¬∃x □(px ∧ qx), no student is required to solve both (46), replace
the conjunction by its conjuncts and their disjunction, and no by not every.
Equations
- One or more equations did not get rendered due to their size.
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The alternatives of ¬∃x □(px ∧ qx) stand in the entailment pattern of universal free
choice, so with the corresponding worlds it strengthens to ¬∃x □px ∧ ¬∃x □qx ∧ ∀x □(px ∨ qx),
negative universal free choice (47).
The alternatives of ◇∀x(px ∨ qx) (55).
Equations
- One or more equations did not get rendered due to their size.
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Given the corresponding worlds, free choice with the existential modal over the universal
(57), [Nou17]'s case, strengthens ◇∀x(px ∨ qx) to ◇∀x px ∧ ◇∀x qx ∧ ¬◇∃x(px ∧ qx),
although ◇∀ does not distribute over disjunction.
Simplification with most #
Most P S says that more than half of P lies in S (86).
Equations
- BarLevFox2020.Most P S = (P.card < 2 * (P ∩ S).card)
Instances For
Overlaps P S says that P and S meet.
Equations
- BarLevFox2020.Overlaps P S = (P ∩ S).Nonempty
Instances For
The alternatives of most(P ∪ Q)(S) (87) replace the disjunctive restrictor by its
disjuncts and their intersection, and most by some.
Equations
- One or more equations did not get rendered due to their size.
Instances For
Given a world where most of P ∪ Q and most of P are in S but nothing of Q is, one the
other way round, one where most of P ∪ Q but of neither P nor Q is, and one where most of
each is but nothing of P ∩ Q, most(P ∪ Q)(S) strengthens to
most(P)(S) ∧ most(Q)(S) ∧ ¬some(P ∩ Q)(S), simplification with most (89).