Friday, August 21, 2009

Circularity and regress

I am not entirely convinced of the arguments below. But they are fun.

Say that p is explanatorily prior to q provided that p contributes to some explanation of q. I shall assume that explanatory priority only holds between true propositions. Let us suppose that:

  1. No contingent proposition is explanatorily prior to itself.
Thus, anybody who thinks that explanatory priority among contingent propositions could be circular—for instance, p prior to q, q prior to r, and r prior to p—is committed to denying the transitivity of explanatory priority.

Now, let us posit two plausible formal principles for explanatory priority:

  1. If p is prior to q, and r is prior to s, then p&q is prior to q&s.
  2. If p and q are conjunctions, and differ only in the order of the conjuncts, and p is prior to r, then q is prior to r.
For a reductio, suppose a circle of size three (the argument works in general): p is prior to q, q is prior to r and r is prior to p, where all three are contingent and true. Using (1) twice, we conclude that p&q&r is prior to q&r&p. Using (3), we conclude that q&r&p is prior to q&r&p, which contradicts (1), since the conjunction of contingent truths is a contingent truth.

Now suppose that a backwards infinite regress of contingently true propositions is possible:

  1. ... p−3 prior to p−2, p−1 prior to p0.
Then a doubly infinite regress is surely also possible:
  1. ..., p−3 prior to p−2, p−1 prior to p0, p0 prior to p1, p1 prior to p2, p2 prior to p3, ....
(Here is a quick way to go from (4) to (5): let pn be the proposition that 1+1=3 or pn−1 for n>0. Then, pn−1 is explanatorily prior to pn.) Suppose that all the propositions in the regress are distinct (otherwise, there is a circularity, which has already been ruled out). Suppose that the infinitary analog of (2) holds:
  1. If A and B are two sets of propositions, with a one-to-one correspondence c between the members of A and those of B, such that c(a) is prior to a for every member a of A, and if p is the conjunction of all the members of A, and q is the conjunction of all the members of B, then q is prior to p.
Now, let A be the set of all the pn for n an integer (positive, negative or zero). Let B=A. Let c(pn)=pn−1. The antecedents in (6) are satisfied. Hence, the conjunction of all the members of B is prior to the conjunction of all the members of A. But A=B. Thus, we have a violation of (1), once again.

The weakness of this argument is that (6) seems less plausible than the finite version (2).

An argument against (2) (and hence also against (6)) is that (2) implies that p can be prior to q even though p and q have a conjunct in common. (Suppose p is prior to q and q is prior to r; then p&q is prior to q&r.) And that might be implausible.

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