If $p(A,B,C,D) = p(A,B) \cdot p(C,D)$, then is $p(A \mid B,C,D) = p(A \mid B)$? Given the discrete random variables $A,B,C,$ and $D$, if
$$
p(A = a,B = b,C = c,D = d) = p(A = a,B = b) \cdot  p(C = c,D = d) \ \forall a,b,c,d
$$
then is
$$
p(A = a \mid B = b,C = c,D = d) = p(A = a \mid B = b)
$$
and
$$
p(B = b \mid A = a,C = c,D = d) = p(B = b \mid A = a)
$$
for all $a,b,c,$ and $d$?

\begin{align}
p(A = a \mid B = b,C = c,D = d) &= \frac{p(A = a,B = b,C = c,D = d)}{\sum_{a} p(A = a,B = b,C = c,D = d)} \\
&= \frac{p(A = a,B = b) \cdot p(C = c,D = d)}{\sum_{a} p(A = a,B = b) \cdot p(C = c,D = d)} \\
&= \frac{p(A = a,B = b) \cdot p(C = c,D = d)}{p(C = c,D = d) \cdot p(B = b)} \\
&= \frac{p(A = a,B = b)}{p(B = b)} \\
&= p(A = a \mid B = b)
\end{align}
The proof for
$$
p(B = b \mid A = a,C = c,D = d) = p(B = b \mid A = a)
$$
follows in a similar way.
