Strength of theories in Lewis's account of laws

I've learned a lot at the Lewis workshop, which was also enjoyable in every other respect. One thing I've learned is that my views about theory strength in Lewis's account of laws were rather naive.

Lewis defines a law of nature as a consequence of the best theory, where what makes a theory good is simplicity, strength, and fit (of assigned probabilities to actual occurrences). I claimed that objective standards for strength aren't hard to find: one could, for instance, use something like number and diversity of excluded possibilities (with a meaningful measure for 'number', these two criteria might coincide). But in the discussions, it turned out that this doesn't work, for at least two reasons.

First, let w1 be a (deterministic) world with lots of As that are B where "all As are B" is a fundamental law. Let w2 be like this world except that there are no As. As Humeans, we don't want "all As are B" to be a law at w2. Instead, the fundamental laws at w2 might well be the fundamental laws at w1 minus "all As are Bs". But we can't get that result if we collect all true theories of w1 and w2 respectively and then select the one that is best in terms of simplicity (defined in terms of syntactical complexity in a suitable language) and strength (defined as above). For the laws of w1 are also true at w2 and vice versa, so the theory that does better in one world must also do better in the other.

The upshot is that strength has to be world-relative: the w1-laws are stronger with respect to w1 than with respect to w2, which is why they lose to a simpler theory in w2, but not in w1. I'm not sure how to define this world-relative notion of strength. Intuitively, what matters is something like the number of positive instances of a law at the world, but of course "all non-Bs are non-As" has many positive instances at w2, so that doesn't really work.

Second, take a world w with three fundamental, indeterministic laws: L1, L2 and L3. Consider the same theory without L3: why isn't this the best theory of w? It is certainly simpler, and it fits just as well as the original one. So the problem can only be that it is weaker. But how so? A probabilistic law does not exclude anything, so the simpler theory excludes exactly the same possibilities as the original theory. Again, I'm not sure how to account for this.

Comments

# on 06 July 2006, 09:56

Hey Wo,

Nice set of problems. I'll have to think a bit more about the one with which you end. I have two questions and a comment about the first problem, though. (i) If I were a Humean, why should I want or need to deny that "All As are B" is a law at worlds in which there are no As? After all, this law supervenes upon the global distribution of fundamental local qualities. So, I'm not sure why you say this. Is there some other commitment you have in mind? (ii) Call the lawbook true of w1 L, and a lawbook which includes every law in L except "All As are B" L-minus. If L and L-minus agree upon all matters of fact in w2, shouldn't L-minus count as the simpler theory for the (perhaps naive) reason that L-minus is just as strong and syntactically "nice" as L while containing fewer fundamental laws? (iii) Maybe the world-relative notion of strength, if the Humean ends up needing it after all, should be fixed with respect to the set of fundamental local qualities instantiated at that world (or regions of that world). Just a thought.

# on 06 July 2006, 11:54

Hi Alex,

Re i): if there are no As, I don't see how a Humean could assume that "all As are B" is a fundamental law, given that there are so many other laws fitting the world, like "all As are not-B". Surely not all vacuous generalizations are fundamental laws. But what Humean fact could single out "all As are B"?

Maybe one could say that whenever there are no As at a world, "there are no As" is a fundamental law. That doesn't sound too bad. Then all those vacuous generalizations would be derivative laws.

Re ii): I agree that L-minus is simpler than L, so that if they were equally strong, L-minus would win. But this is the problem: how can I say that L-minus is equally strong as L? After all, L rules out many possibilities L-minus doesn't rule out; moreover, with respect to w1, L should trump L-minus in virtue of its superior strength.

I don't quite understand your suggestion (iii): what is needed, I think, is an explanation of strength that begins like "L1 is stronger than L2 with respect to w iff (or to the degree that) --", which allows that L is much stronger than L-minus w.r.t. w1, but not much stronger w.r.t. w2.

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