Well, since I was horribly wrong when I thought I saw a flaw in the math, let me instead look at the conclusions, and maybe I won't be horribly wrong :D

if p(A|B) > p(A), then |p(A v B|B) – p(A v B)| > |p(A v ~B|B) – p(A v ~B)|

That is, to the extent that B increases the probability of A, it does so by increasing the probability of A v B more than it decreases the probability of A v ~B. However, since A v B is a logical consequence of B to begin with, the increase in probability is a purely deductive inference.

This is not what the equation says above. Yes, p(AvB|B)=1. But there's another term on that side too: -p(AvB), which has to be mentioned. If it could be ignored, then B would increase the probability of A for any choice of A and B!

How's this: to the extent that B increases the probability of A, p(A v ~B) - p(A v ~B|B) is less than 1 - p(A v B).

Not deductive, I know, but accurate.

Not only do logical consequences of A which are independent of B not increase in probability, they may actually decrease in probability.

This is a bit circular. If something is really independent of B, it will not change at all if we condition on B. Nearly all things that are logical consequences, though, aren't independent. Maybe the author had some example in mind while writing?