Katzir 2007: Structurally-Defined Alternatives (End-to-End) #
@cite{katzir-2007}
Katzir, R. (2007). Structurally-defined alternatives. Linguistics and Philosophy, 30(6), 669–690.
Unified Tree Demonstration #
This file demonstrates that a single Tree Cat String supports both:
- Structural operations (PF-level):
leafSubstgenerates scalar alternatives by same-category word substitution - Compositional interpretation (LF-level):
evalTreecomputes truth conditions via FA, PM, and Predicate Abstraction
One tree, two interfaces — the Y-model made concrete.
The Argument #
- Build φ = "some student sleeps" as
Tree Cat Stringwith QR - Generate φ' = "every student sleeps" via
leafSubst(Det substitution) - Interpret both: ⟦φ⟧ = true, ⟦φ'⟧ = false → asserting φ implicates ¬φ'
- Show φ contains no
ConjP/NegP→ symmetric alternative "some but not all" cannot be generated structurally
This is Katzir's solution to the symmetry problem: structural constraints on alternatives prevent the symmetric alternative from being generated, licensing the scalar implicature.
"Some student sleeps" after QR, with UD-grounded categories:
[S [DP [Det some] [N student]] [₁ [S [t₁:NP] [VP [V sleeps]]]]]
Equations
- One or more equations did not get rendered due to their size.
Instances For
Scalar alternative: substitute "some" → "every" at Det position. This is Katzir's core operation (def 19, substitution): replace a terminal with a same-category item from the substitution source. Both "some" and "every" are Det terminals in the lexicon.
Equations
Instances For
"Some student sleeps" is true: John is a student and sleeps.
The scalar alternative "every student sleeps" is false: Mary is a student but doesn't sleep.
The two readings differ: genuine scalar inference. Asserting "some" when "every" was available implicates ¬"every".
The symmetry problem: for any stronger alternative φ' = "every", there exists a symmetric alternative φ'' = "some but not all" which is also stronger. Naïve exhaustivity would predict no implicature.
Katzir's solution: φ'' requires ConjP and NegP structure, which cannot be generated from L(φ) = lexicon ∪ subtrees(φ) because the source tree φ contains neither category.
φ contains no ConjP anywhere in its structure.
φ contains no NegP anywhere in its structure.
None of φ's subtrees contain ConjP either.