I'm reading chapter 11 of The Blackwell Guide to the Philosophy of Language by Frank Jackson, and once he touched upon Lewis' triviality results he writes:
However, this definitely doesn't look that simple to me. Could you break down this proof line by line and show why it holds?
The argument can be outlined as follows (I presume it is not hard to supply philosophical and mathematical details):
Ex hypothesi, P(A → B) = P(B | A) = P(A & B) / P(A).
By the rule of total probability, we have
(*) P(A → B) = P(A → B | B) P(B) + P(A → B | ~B) P(~B)
where the propositions B and ~B are complementary (instead, one may take any relevant set of mutually exclusive and collectively exhaustive propositions).
Consider multiple conditioning (i.e., “conditionalising”): Suppose we have P(A | B) and want to put a further condition C; that is, we seek the probability of a proposition A given B, given C. In ill-formed notation (just to illustrate the idea!), what we want is P(A | B | C). Properly, this is P(A | B & C) which is equal to P(A & B & C) / P(B & C) – you can easily check these by the conditional probability formula.
We apply multiple conditioning to P(A → B) with B and ~B:
P(A → B | B) = P( B | A & B)
P(A → B | ~B) = P( B | A & ~B)
Substituting into (*), we get
P(A → B) = P(B | A & B) P(B) + P(B | A & ~B) P(~B)
Observe that
P(B | A & B) = P(B & A & B) / P(A & B) = P(A & B) / P(A & B) = 1
P(B | A & ~B) = P(B & A & ~B) / P(A & ~B) = P(⊥) / P(A & ~B) = 0
So, we get P(A → B) = P(B | A) = P(B)
Therefore, the equation P(A & B) = P(A)P(B) holds true and propositions A and B are independent, while, according to the argument, the probability of B has to be dependent on the probability of A in general.
