## Indeterminacy and knowledge

November 21, 2008

What do people think of this principle: determinate implication preserves indeterminacy? Formally1:

• $\Delta(p \rightarrow q) \rightarrow (\nabla p \rightarrow \nabla q)$

If this principle is ok, and we accept that factivity of knowledge is determinate, it seems we can make trouble for the epistemicist, ignorance view of vagueness. That is, given:

• $\Delta(Kp \rightarrow p)$

we can infer that $\nabla p \rightarrow \nabla Kp$: whenever p is indeterminate, it is indeterminate whether you know p. This, I take it, is incompatible with (determinate) ignorance concerning p.

[1 Note that, although this looks similar, it’s not quite the same as $\Box (p \rightarrow q) \rightarrow (\Diamond p \rightarrow \Diamond q)$, which is a theorem of the weakest normal modal logic, K. $\nabla$ and $\Delta$ don’t stand in the same relation as $\Diamond$ and $\Box$.]

1. Hmm, I just thought of a counterexample!

p: Fred is bald
q: Fred = Fred

I guess I should think things through more thoroughly before posting! (Which pretty much sums up this blog :-p.)

det(p implies q) implies [indet(q) implies indet(p)]

3. But presumably ‘Fred is bald’ does imply ‘Fred is Fred’? Also, I would have thought that epistemicists, of the Williamson sort, would reject the principle for its use of indeterminacy?

4. Oh, wait, my suggested principle won’t work because we could have indet(q) and det(~p), even if det(p implies q).

Sam – right, but the indeterminacy of ‘Fred is bald’ doesn’t imply the indeterminacy of ‘Fred is Fred’.

5. Richard: yeah, it looks like your amendment doesn’t work either. But also, it’s not clear you can get that its indeterminate whether you know p, if p is vague (which was the reason I was interested in these principles.)

Sam: epistemicists don’t deny the existence of vagueness! In fact, their logic of indeterminacy ends up looking very similar to the supervaluationists. (Of course, they might object to me using the *word* “indeterminacy” to describe this phenomenon, I don’t know.)

6. Sorry, I should have been more precise. I meant det(‘fred is bald’ implies ‘fred is fred’) and so the ‘implies’ strengthening wont work. And I’m sure there is a difference between vagueness and indeterminacy (at least I’m sure there is an important distinction to be made which might be made using these terms, especially if you are an epistemicist). But maybe you disagree here.

Anyway, if you want to use them as synonymous here you obviously get a similar problem, i.e. ~vague(‘fred is bald’ implies ‘fred is fred’) implies vague(‘fred is bald’) etc etc will come out false if ‘fred is bald’ is vague. All of this is a bit useless since you already said it doesn’t work, but I feel I should defend myself 🙂

cheers

7. Good to see you again yesterday. Here are the notes from the Kit Fine talk I was telling you about. They are fairly sketchy but should give you some idea of what he’s up to:

http://philosophywiki.org/main/A_New_Approach_to_Vagueness

Feel free to play around with the web site, by the way. It’s something I’ve been working on for a while and would be interested to know what you think of it.

8. Andrew – actually, I thought your specific implementation was an instance of my suggested principle, not yours.

You want to infer from det (q implies Kq) to [indet(q) implies indet(Kq)]. So the general schema you’re appealing to is (letting ‘p’ stand for ‘Kq’]:

det(p implies q) implies [indet(q) implies indet(p)]

Now, as noted above this won’t work because one might hold indet(q) with det(~p) — which is to say, det(~Kq). So the general schema is false, and we can see why in this particular instance too.

But technically, from your originally proposed principle, you can’t even make the original inference you wanted. All you get is the converse: if it’s indeterminate whether you know q, then it’s indeterminate whether q.

Right?

9. Typo: You want to infer from det (Kq implies q) to [indet(q) implies indet(Kq)].

10. Hi Richard,

You wrote:

“But technically, from your originally proposed principle, you can’t even make the original inference you wanted. All you get is the converse: if it’s indeterminate whether you know q, then it’s indeterminate whether q.”

I think you can. By determinacy of factivity you get $\Delta(\neg p \rightarrow \neg Kp)$. Since by definition $\nabla p := \neg \Delta p \wedge \neg \Delta \neg p$ we have $\nabla p \equiv \nabla \neg p$. Applying the rule I stated to $\Delta(\neg p \rightarrow \neg Kp)$ to get $\nabla \neg p \rightarrow \nabla \neg Kp$ then just substitute $\nabla \neg p$ for $\nabla p$ and same for Kp…

You can also infer it from your principle (I was wrong earlier.)

11. Ah, clever! Thanks for the explanation.

12. No worries. Thanks for posting!