Probability Independence to reduce terms.

Question

Is the following probability reduction correct?

p(A,B|C,D) = p(A,B|C), where (A,B) is independent of D

Is it because of the following prove:

p(A,B|C,D) 
= p(A,B,C,D) / p(C,D)
= p(D)*p(A,B,C) / p(C)*p(D)
= p(A,B,C) / p(C)
= P(A,B|C)

If so, then isn't it (A,B,C) is independent of D rather than (A,B) is independent of D?

Answer 1

To get a simple counterexample, throw a fair coin twice, call $C$ the event that the first throw was a head, $D$ the event that the second throw was a head and $A=B$ the event that both throws coincide.

Then $A$ is independent of $D$ while $P(A\mid C,D)=1$ and $P(A\mid C)=\frac12$.

On the other hand, if $D$ is independent of $(A,B,C)$, the proof you suggest is correct and shows indeed that $P(A,B\mid C,D)=P(A,B\mid C)$.