## Tuesday, April 30, 2013

### Vagueness and grounding

1. If p is a precification of q and p is true, then p grounds q.
2. If q is vaguely true, then q has a true precification.
3. So, every proposition that is vaguely true is grounded.
If we could add the thesis that every grounded proposition is non-fundamental (which in another post I argued against), we could conclude that all vaguely true propositions are non-fundamental.  But even without that thesis, groundedness is evidence of non-fundamentality.  So vagueness is evidence of non-fundamentality.

Brian Cutter said...

Where p is indeterminate, both p and ~p will have true precisifications. So both p and ~p are grounded?

Alexander R Pruss said...

Maybe the thing to say is that (p grounds q) is at least vaguely true in 1? Then we only get to conclude that vague propositions are vaguely grounded.
But perhaps they are grounded on the same precisifications on which they are true. So they are definitely not ungroundedly true.
But this is not quite right on the standard supervaluationist setting. For p grounds q on precisification u of q iff p grounds u. But then p doesn't ground q on precisification q as that would require q to ground q. Maybe this forces a revision of the standard supervaluationist semantics?

Heath White said...

It seems to me that in the indeterminate case, the precisification that grounds p and the one that grounds ~p are the same thing. That is, that Heath has 200 hairs precisifies, and therefore grounds, both of the vague propositions that he is bald and that he is not bald.

So why not say both are grounded?

Alexander R Pruss said...

The kinds of precifications I had in mind were ones like: bald → having less than 20 hairs, bald → having less than 201 hairs; bald → having less than 21 linear inches of hair, etc. And these don't ground ~q.

Of course, they have analogues like: nonbald → having at least 20 hairs, nonbald → having at least 201 hairs; nonbald → having at least 21 linear inches of hair. And these might be thought to ground ~q.

No doubt Heath having less than 201 hairs is grounded, at least in part (the other part might be done by some mathematical relations), in Heath having 200 hairs (if he does). But that's a second step, and it's a touch more controversial that this can always be done.

Alexander R Pruss said...

That said, I could also see a case that Heath having less than 201 hairs doesn't ground Heath being bald, but grounds Heath being at least vaguely bald.

Brian Cutter said...

Perhaps we can avoid the sting of having cases where both p and ~p are grounded by eliminating the requirement that grounding relations hold only between true propositions. Or, better: instead of grounding, we could introduce a notion of quasi-grounding, which is like grounding except it doesn't require the truth of its relata. So, e.g., A quasi-grounds (A or B) even when A isn't true. Then we can define (full) grounding as follows: {A1, A2,...} fully grounds B iff {A1, A2, ...} quasi-grounds B, and A1, A2,... and B are true. The appropriate reformulation of (1) in terms of quasi-grounding would be: If p is a precisification of q, then p quasi-grounds q. This of course will have the result that there are instances of q such that q and ~q are both quasi-grounded. But where q is indeterminate, it will not determinately be the case that q is grounded, and it will not determinately be the case that ~q is grounded, and it will determinately *not* be the case that (q is grounded and ~q is grounded).

Alexander R Pruss said...

Dan Johnson did a dissertation on "ontological explanation" which was rather like your quasi-grounding, though probably with different judgments about cases.

On this proposal, I wonder if we'll get something like this:
Definitely true: x has no hairs quasi-grounds x is bald
Definitely true: x has a million hairs quasi-grounds x is not bald
Neither definitely true nor definitely false: x has 200 hairs quasi-grounds x is bald.

Brian Cutter said...

"On this proposal, I wonder if we'll get something like this:
Definitely true: x has no hairs quasi-grounds x is bald
Definitely true: x has a million hairs quasi-grounds x is not bald
Neither definitely true nor definitely false: x has 200 hairs quasi-grounds x is bald."

The first two definitely sound alright. As for the third, I'm not sure. Let p1 be a precisification of (x is bald) which means something like "x has between 0 and 199 hairs, and let p2 be a precisification of (x is bald) which means something like "x has between 0 and 201 hairs." Plausibly, (x has 200 hairs) quasi-grounds p2, but not p1. So (x has 200 hairs) quasi-grounds some but not all of the precisifications of (x is bald). But that may not be sufficient for us to say that it's indeterminate whether the former quasi-grounds the latter, for reasons you mention in your first comment above. Here's an argument for the claim that it's definitely true that (x has 200 hairs) quasi-grounds (x is bald).
1. It's definitely true that: (x has 200 hairs) quasi grounds p2.
2. It's definitely true that p2 quasi-grounds (x is bald). (Because vague propositions are always [definitely] quasi-grounded in their precisifications.)
3. Therefore, it's definitely true that (x has 200 hairs) quasi-grounds (x is bald). (Given the transitivity of definite quasi-grounding.)

Eeh, it looks like this will have the result that some (non-contradictory) propositions will (definitely) quasi-ground both q and ~q, for some vague propositions q. But maybe that's not a bad result?

Alexander R Pruss said...

Here's a plausible thesis:
Definitely: if (p quasi-grounds q) and p is true, then q.

But now we don't want to allow having less than 201 hairs to quasi-ground being bald.

Brian Cutter said...

Hm.. Good point. Here's an alternative approach, which doesn't appeal to anything like quasi-grounding, which I think I prefer: Any vague proposition q is *partly* grounded in each of its true precisifications. If q is super-true, then q is *fully* grounded in the set of all its precisifications. If q is neither super-true nor super-false, then q is partly grounded (in each of its true precisifications) but not fully grounded (because not all of its precisifications are true). (Here we have another reason not to define partial grounding in the usual way in terms of full grounding!) This delivers the intuitively correct result about vaguely true propositions. They are only "partly" true, not "fully" true, so they ought to be partly, but not fully, grounded.

If we reformulate your argument by replacing (1) with:
1*. If p is a precisification of q and p is true, then p *partly* grounds q,
then we get the conclusion that vaguely true propositions are partly grounded. And partial grounding is at least evidence for non-fundamentality.