Yes, even without the extra condition. Let a = P(A), b = P(B), c = P(A & B).
P(B|A) > P(B) is equivalent to c > ab.
P(~B|~A) > P(~B) is equivalent to 1-a-b+c > (1-a)(1-b) = 1 - a—b + ab, which is equivalent to c > ab, which is the hypothesis.
As a check that the conventional definition of P(B|A)=0 when P(A)=0 doesn’t affect things, if P(A)=0, P(A)=1, P(B)=0, or P(B)=1, then P(B|A) = P(B), making the antecedent false and the proposition trivially true.
Yes, even without the extra condition. Let a = P(A), b = P(B), c = P(A & B).
P(B|A) > P(B) is equivalent to c > ab.
P(~B|~A) > P(~B) is equivalent to 1-a-b+c > (1-a)(1-b) = 1 - a—b + ab, which is equivalent to c > ab, which is the hypothesis.
As a check that the conventional definition of P(B|A)=0 when P(A)=0 doesn’t affect things, if P(A)=0, P(A)=1, P(B)=0, or P(B)=1, then P(B|A) = P(B), making the antecedent false and the proposition trivially true.