Expanding expression based on conditional probability

Question

The following expression (1) is supposedly "trivially" equal to expression (2), and hence the details are not shown in the example. But of course, I can't get it right.

(1) p(A|B,C)*p(B|C)
(2) p(A,B|C) = p(A|C)*p(B|C) (either side of this equation is equivalent to (1), whichever is easier)

I suspect the answer will make me look pretty stupid, but I am stuck. Why/how are they equivalent?

Answer 1

The reason it is deemed trivial is because it is trivial that $$\Pr(A|B)*\Pr(B) = \Pr(A,B)$$ Now, imagine that we restrict the probability space to $C$. Then the equality holds also when all the events are conditioned on $C$.

Answer 2

I think I've got it...

1) p(A | B, C) * p(B|C)
2) p(A, B, C) * (1 / p(B, C) ) * p(B|C)
3) p(A, B, C) * (1 / p(B, C) ) * p(B, C) * (1/ p(C) )
4) [ p(A, B, C) * (1 / p(C) ) ]  * ( 1/ p(B, C) ) * p(B, C)
5) p(A, B | C) * p(B, C) * ( 1/ p(B, C) )
6) p(A, B | C)

Please let me know if this looks correct. Thank you. (Does that really seem trivial?)