I'm back and there's been no response, so I'll be specific. Starting from
p(A v B|B) – p(A v B) + p(A v ~B|B) – p(A v ~B) = .15
Using p(X v Y) = p(X) + p(Y) - p(XY), we get
.15 = p(A|B) + p(B|B) - p(AB|B) - p(A) - p(B) + p(AB) + p(A|B) + p(~B|B) - p(A~B|B) - p(A) - p(~B) + p(A~B)
= p(A|B) + 1 - p(A) - p(A) - p(B) + p(AB) + p(A|B) - p(A) - p(~B) + p(A~B)
= 2 p(A|B) - 2 p(A)
= twice the thing you started from, which is bad.
It looks like you simplified p(AB|B) as p(A), but in fact p(AB|B)=p(ABB)/p(B)=p(AB)/p(B)=p(A|B). (I made a similar mistake earlier.)
I get p(A|B) + p(B|B) - p(AB|B) - p(A) - p(B) + p(AB) + p(A|B) + p(~B|B) - p(A~B|B) - p(A) - p(~B) + p(A~B)
= p(A|B) + 1 - p(A|B) - p(A) - p(B) + p(AB) + p(A|B) + 0 - 0 - p(A) - 1 + p(B) + p(A~B)
= - p(A) - p(B) + p(AB) + p(A|B) - p(A) + p(B) + p(A~B)
= - p(A) + p(AB) + p(A|B) - p(A) + p(A~B)
= p(A|B) -2p(A) +p(AB) + p(A~B)
= p(A|B) -2p(A) + p(A)
= p(A|B) - p(A)
But this is quod erat demonstrandum.
In 1983 Karl Popper and David Miller published an argument to the effect that probability theory could be used to disprove induction. Popper had long been an opponent of induction. Since probability theory in general, and Bayes in particular is often seen as rescuing induction from the standard objections, the argument is significant.
It is being discussed over at the Critical Rationalism site.