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B entails that a conjunction of determinate truths is determinate

October 26, 2010

I know it’s been quiet for a while around here. I have finally finished a paper on higher order vagueness which has been  a long time coming, and since I expect it to be in review for quite a while longer I decided to put it online. (Note: I’ll come back to the title of this post in a bit, after I’ve filled in some of the background.)

The paper is concerned with a number of arguments that purport to show that it is always a precise matter whether something is determinate at every finite order. This would entail, for example, that it was always a precise matter whether someone was determinately a child at every order, and thus, presumably, that this is also a knowable matter. But it seems just as bad to be able to know things like “I stopped being a determinate child at every order after 123098309851248 nanoseconds from my birth” as to know the corresponding kinds of things about being a child.

What could the premisses be that give such a paradoxical conclusion? One of the principles, distributivity, says that a (possibly infinite) conjunction of determinate truths is determinate, the other, B, says p \rightarrow \Delta\neg\Delta\neg p. If \Delta^* p is the conjunction of p, \Delta p, \Delta\Delta p, and so on, distributivity easily gives us (1) \Delta^*p \rightarrow\Delta\Delta^* p. Given a logic of K for determinacy we quickly get \Delta\neg\Delta\Delta^*p \rightarrow\Delta\neg\Delta^* p, which combined with \neg\Delta^* p\rightarrow \Delta\neg\Delta\Delta^* p (an instance of B) gives (2) \neg\Delta^* p\rightarrow\Delta\neg\Delta^* p. Excluded middle and (1) and (2) gives us \Delta\Delta^* p \vee \Delta\neg\Delta^* p, which is the bad conclusion.

In the paper I argue that B is the culprit.* The main moving part in Field’s solution to this problem, by contrast, is the rejection of distributivity. I think I finally have a conclusive argument that it is B that is responsible, and that is that B actually *entails* distributivity! In other words, no matter how you block the paradox you’ve got to deny B.

I think this is quite surprising and the argument is quite cute, so I’ve written it up in a note. I’ve put it in a pdf rather than post it up here, but it’s only two pages and the argument is actually only a few lines. Comments would be very welcome.

* Actually a whole chain of principles weaker than B can cause problems, the weakest which I consider being \Delta(p\rightarrow\Delta p)\rightarrow(\neg p \rightarrow \Delta\neg p), which corresponds to the frame condition: if x can see y, there is a finite chain of steps from y back to x each step of which x can see.

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6 comments

  1. Incidentally, the argument goes through in Field’s non-classical logic.


  2. Hi Andrew,

    Good stuff!

    Just to be clear: I take it that accepting distributivity is not something Field can do, given his semantics?

    Cheers
    Sam


  3. Well there’s no reason a particular semantics would rule out something that might contingently true; but it’s ruled out in the sense that you can’t have it with a naïve truth theory.

    My point was that the failure of distributivity is what allows Field to deny precision over what’s determinate at every order.


  4. Sorry, I meant his semantics for a naive truth theory.

    I suppose I was try to understand the dialectic a bit better, since Field sets the problem up of iterated determinately operators (in STFP) for his truth theory without appeal to B, so that in a sense Field’s solution there (that distributivity fails) isn’t a solution to the problem you raise (which uses B).

    Or have I missed something?

    Sam


  5. Well the problem as I originally understood it (that you could prove \Delta\Delta^* p \vee \Delta\neg\Delta^* p) used both B and distributivity of determinacy over infinite conjunction (as well as some fairly uncontroversial background conditions.) Now I see that distributivity is redundant in that argument.

    Field’s doesn’t mention B, but as far as I can see he has all the principles I needed to get the argument running so B must fail in his logic. (Technically he doesn’t deal with infinitary conjunction. However he does want his theory to have `omega-models’, so infinitary conjunction could presumably be simulated using quantification.)

    By the way the dialectic is slightly different if you’re a non-classical logician anyway, as the problematic conclusion in that case is only \Delta^* p \vee \neg\Delta^* p. I had in my mind a classical logician who wanted to give up distributivity.


  6. So it is finished here but the continued way is here :
    http://www.paris-philo.com

    Happy fun for you ;)



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