Thread:TheHumanAmbassador/@comment-26907577-20200128211255/@comment-32182236-20200204232536

Here goes nothing.

P=AB/AB+CD

Given:B=D [B is my shorthand for P(E|H), and D is my shorthand for P(E|¬H). Likewise, A is my shorthand for P(H), and C is my shorthand for P(¬H). And of course, P here is equivalent to P(H|E).]

I'll substitute D with B.

P=AB/AB+CB

AB+CB share a common factor, B, so we can pull the B out like this:

P=AB/B(A+C)

Now, A+C=1, so AB+CB=B.

P=AB/B

P=A

Yep, the theorem holds.