Topic: HUMAN SEMANTICS
Continuation from previous post
.Comments of na excerpt from page 7:
(Huitink, 2004) is the most recent attempt to solve the puzzle of anankastic conditionals. Huitink argues that if there are several non-conflicting goals at stake and several ways to achieve the goal in the antecedent, the anankastic reading cannot obtain. So anankastic sentences are false in such cases. However, they can be predicted true under the analysis in (Fintel and Iatridou, 2004). The scenario that should make the argument clear is the following:
(16) a. To go to Harlem, you can take the A train or the B train.
b. You want to go to Harlem.
c. You want to kiss Ruud van Nistelrooy (Dutch soccer star).
d. Ruud van Nistelrooy is on the A train.
The designated goal analysis would predict that the Harlem sentence is true at least in its ought-version:
(17) If you want to go to Harlem you ought to take the A train.
What we get is that in the best Harlem worlds, i.e. the worlds in which you kiss Ruud, you take the A train. This follows from the facts in the described scenario. So the sentence is true but it shouldn't, because the A train is not the necessary condition for going to Harlem in (16).
Well I have two objections against saying that (17) is not true in the scenario above.
The first objection is that intuitively (17) is true in the scenario described, accordingly to my judgment as speaker of human languages.
The second objection comes from the application of simple Logic. Assume two propositions in the described situation
Proposition 1. x must take train A=O(A)
Proposition 2. x must take train B=O(B)
Now consider that the disjunction containing the two is true:
(DISJ) T(O(A) OR O(B))
For this conjunction to be true, there are three possibilities:
a. T(O(A)), T(O(B));
b. T(O(A)), F(O(B));
c. F(O(A)), T(O(B)).
But, in such case, (17) is true in two of the possibilities. And if the first possibility is the case in the described scenario, then (17) is true.
Thus, if any formal semantic analysis predicts that (17) is true in the described scenario, such analysis is correct.
Now, remember that the Classic Maxim O(P)-->P of earlier deontic logic is not valid. So the impossibility of x to take both trains at the same time does not constitute a real problem. By claiming that O(A) is, true one does not claim that x takes train A, nor O(B) implies that X takes train B.
Posted by Tony Marmo
at 04:46 GMT
Updated: Sunday, 5 December 2004 14:21 GMT