What this involves for choice sentences, syntactically speaking, is that all choice sentences are analyzed as wide disjunctions (the contrast here with Simons couldn't be greater.) So the following inference does follow for Zimmermann:
Might(A)
Therefore, Might (A v B)
...it's just that the conclusion is not a parse that occurs in natural language. As far as I can tell, this gives [[or]] quite a complicated semantic entry. It is going to be a recursive, raised-type thing, but the basis for the recursion will be:
[[or]] = \lambda p . \lambda q . might p and might q.
In order to figure out how the recursion goes, I'll need to learn more about Montague lifts.
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