You’ve introduced a third piece of information C: observing the result of a computation that B indeed divides A. With that information, I have P(A|C) = P(A & B|C) = 1. I also have P(A) < P(A | C). But without C, I still have P(A) ≥ P(A & B), not the opposite.
I am having trouble with your second paragraph. Certainly, P(A | B&C) = P( A & B | C) = 1. But when I do not know C, P(A | B) = 1 and P(A & B) < 1.
You’ve introduced a third piece of information C: observing the result of a computation that B indeed divides A. With that information, I have P(A|C) = P(A & B|C) = 1. I also have P(A) < P(A | C). But without C, I still have P(A) ≥ P(A & B), not the opposite.
I am having trouble with your second paragraph. Certainly, P(A | B&C) = P( A & B | C) = 1. But when I do not know C, P(A | B) = 1 and P(A & B) < 1.