Kratzer (1977) — Premise semantics worked example #
@cite{kratzer-1977} @cite{kratzer-2012}
A concrete formalization of the New Zealand judgments scenario from §1.3
of @cite{kratzer-1977} (= Chapter 1 of @cite{kratzer-2012}), exercising
the API of Core.Logic.Intensional.Premise.
The scenario #
@cite{kratzer-1977} imagines a country whose entire common law consists of three judgments:
- Murder is a crime. — call this proposition
p. - Deer are personally responsible for damage they inflict on young
trees. — call this proposition
q. - Deer are not personally responsible for damage they inflict on young
trees. — proposition
¬q.
The premise set A = [p, q, ¬q] is inconsistent (its intersection
is empty: no world makes q and ¬q both true). Yet our intuitions
about modal claims relativized to what the New Zealand judgments
provide are crisp:
- (7) "Murder must be a crime." — TRUE
- (8) "It must be that murder is not a crime." — FALSE
- (9) "It is possible that deer are responsible." — TRUE
- (10) "It is possible that deer are not responsible." — TRUE
- (14) "It is possible that murder is not a crime." — FALSE
Kratzer's original Defs 5–6 (necessity = consequence, possibility =
compatibility) collapse on inconsistent A: by ex falso quodlibet,
everything follows from A, so must p and must ¬p are both
true, and nothing is compatible with A, so can q and can ¬q
are both false. Defs 7–8 — quantifying over the consistent subsets of
A and asking for an extension that supports the conclusion — recover
the intuitive predictions.
What this study is #
A Phenomenon-layer integration test for Core.Logic.Intensional.Premise:
it picks the worked example Kratzer uses to motivate the revision from
Defs 5–6 to Defs 7–8 and verifies, by structural proofs over the four-world
frame, that the formalization gets each prediction right and that the
unrevised definitions fail in exactly the way Kratzer says they do.
§1. The model #
A four-world frame, indexed by Fin 4, that distinguishes the two
contingent dimensions of the scenario: whether murder is a crime
(p) and whether deer are responsible (q). All four combinations
are represented so that every singleton premise — p, q, ¬q, ¬p
— is individually consistent.
| world | p (murder is a crime) | q (deer responsible) |
|---|---|---|
w₀ | true | true |
w₁ | true | false |
w₂ | false | true |
w₃ | false | false |
"Murder is a crime." True at w₀ and w₁.
Equations
Instances For
"Deer are personally responsible for damage they inflict on young trees."
True at w₀ and w₂.
Equations
Instances For
The premise set A of @cite{kratzer-1977} §1.3 — what the New Zealand
judgments provide: the three rulings, taken together.
Equations
Instances For
The constant modal restriction: at every world, the premise set is A.
The scenario abstracts away from world-to-world variation in the
judgments.
Instances For
§2. A is inconsistent #
The premise set contains both q and its negation, so its
intersection is empty.
§3. Unrevised Defs 5–6 give paradoxical predictions #
These two theorems are the formal counterpart of Kratzer's diagnosis
of the original definitions: they trivialize over an inconsistent A.
Under the original Def 5 (mustInView), the inconsistent A
entails every proposition — including ¬p. This is the paradox
of ex falso quodlibet that motivates the revision to Def 7.
Under the original Def 6 (canInView), nothing is compatible with
the inconsistent A — including q, which intuitively is
possible in view of the judgments.
§4. Revised Defs 7–8 give the intuitive predictions #
Each theorem corresponds to a sentence number from @cite{kratzer-1977} §1.3, with the predicted truth value Kratzer argues for.
The proofs proceed by enumerating the consistent sublists of A and,
for each one, exhibiting an extension that witnesses (or refutes) the
target consequence.
Membership helpers for the four worlds #
[p] is a sublist of [p, q, negQ].
[q] is a sublist of [p, q, negQ].
[negQ] is a sublist of [p, q, negQ].
The empty list is a sublist of A.
Witness selection for the revised must/can theorems #
For any consistent sublist B of A, exhibit a consistent sublist
C of A with B ⊆ C and p ∈ C. We use [p, q] if B does not
contain negQ, and [p, negQ] otherwise.
(7) "Murder must be a crime" — TRUE under Def 7. The extension
[p, q] (or [p, ¬q]) of any consistent subset of A entails p.
(8) "It must be that murder is not a crime" — FALSE under Def 7.
Apply Def 7 at B = [p]; any consistent extension C ⊇ [p] of A
contains p, but negP cannot follow from C because C admits
a witness world (0 if q ∈ C, else 1) where p holds, hence
negP fails.
(9) "It is possible that deer are responsible" — TRUE under Def 8.
Witness: take B = [q]; every consistent extension already contains q,
so adding q preserves consistency.
(10) "It is possible that deer are not responsible" — TRUE under Def 8.
Symmetric to (9), with witness B = [¬q].
(14) "It is possible that murder is not a crime" — FALSE under Def 8.
Every consistent subset of A extends to one containing p, and
adding ¬p to such a set is never consistent.
§5. The contrast in one place #
Two theorems pinning the bug-vs-fix asymmetry: the revised definitions agree with intuition exactly where the original definitions trivialize.
Def 5 wrongly accepts must ¬p; Def 7 correctly rejects it.
Def 6 wrongly rejects can q; Def 8 correctly accepts it.