tag:blogger.com,1999:blog-8027844571839885250.post5612218253298050372..comments2022-03-25T07:20:12.468-04:00Comments on Matters of Substance: Dispositions and Interferences: the PaperGabriele Contessahttp://www.blogger.com/profile/13607158011908969169noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-8027844571839885250.post-52562527924828115692013-10-17T13:13:46.713-04:002013-10-17T13:13:46.713-04:00Thanks, Matthews V! My first reaction would be to ...Thanks, Matthews V! My first reaction would be to say that the issue you raise is an epistemological one (and an interesting one) but analyses of disposition ascriptions are trying to answer a semantical/metaphysical question. In any case, it seems that trying to determine whether something is an antidote to D, we have to determine if o has D in the first place and we might never be able to find out if D is necessarily (or at least always) interfered with. That doesn't mean, however, that there is no fact of the matter as to whether o has D (see the section of the paper on necessary interferences).Gabriele Contessahttps://www.blogger.com/profile/13607158011908969169noreply@blogger.comtag:blogger.com,1999:blog-8027844571839885250.post-41524247642210328152013-10-12T16:17:57.107-04:002013-10-12T16:17:57.107-04:00Great paper. One question, though:
Suppose we wan...Great paper. One question, though:<br /><br />Suppose we want to test an object O to see if it possesses the disposition D - contrafactually(if S, then M). We also want to test A1 and check if this is an antidote to D.<br /><br />Test #1: We put O under S & A1. O doesn't manifest M.<br />Test #2: We put O under S & ~A1. O doesn't manifest M either.<br /><br />In this case I would feel compelled to say O isn't disposed to manifest M if subject to S in the absence of A1 - which is the same as saying that A1 isn't an antidote to D. But that would be too easy. Maybe O didn't manifest M because there was another hidden antidote in the active: A2. So we could set up test #3:<br /><br />Test #3: O is subject to S & ~A1 & ~A2. O doesn't manifest M.<br /><br />So what conclusion can we draw at this point? Either A1 and A2 are antidotes to D and there is still another antidote A3 on the active, or they're not antidotes to D and O just isn't disposed to M when S. But how can we distinguish those two alternatives unless we already know the whole extension of the set of every antidote-to-D?Matthews Vnoreply@blogger.com