## Friday, April 1, 2011

### A consequence argument

Write pMq for the claim that if p were to hold, q might hold. Let Np be the claim that that p is true and there is no agent x and action A such that Can(x,A) and Does(x,A)M~p. I.e., p is true and no one can act in such a way that p might be false.

Let L be the laws, and P be the state of the world in the distant past. Plausibly:
1. If pM(q or r), then pMq or pMr.
2. If pMq and q entails r, then pMr.
3. If determinism holds, then L&P entails p for any true proposition p about the present or future.
4. NL
5. NP
It follows from (1)-(5) that:
1. If determinism holds, then Np for any true proposition p about the present or future.
So if determinism is true, nothing is even chancily up to anyone.

The argument that (6) follows from (1)-(5) uses this theorem:

Theorem: If (1) and (2) hold, then if q&r entails p and if Nq and Nr, then Np.

To see that (6) follows from (1)-(5) given the Theorem, just let q be L and let r be P.

The proof of the Theorem is pretty easy, too. Suppose q&r entails p and suppose Nq and Nr. For a reductio, suppose ~Np. Then there is an action A and agent x such that Can(x,A) and Does(x,A)M~p. But ~p entails (~q or ~r) by contraposition and De Morgan. So, Does(x,A)M(~q or ~r) by (2). And so by (1) we have Does(x,A)M~q or Does(x,A)M~r, and hence either ~Nq or ~Nr, which contradicts our assumptions. The proof is complete.

Now, (1) and (2) follow from the official David Lewis semantics for might-conditionals (as well as from some modified versions of the semantics) so at least we have good reason to think that there won't be any really easy and uncontroversial counterexamples to (1) and (2). Besides, they are both pretty plausible.