Is $B \perp\kern-5pt\perp C | A $ for the two graphical models? I would like to know whether $B \perp\kern-5pt\perp C | A $ in the following two graphical models and would like to know if my reasoning is correct:

For the left graphical model, which is a Belief Network, here's how I deduce it:
$P(B,C|A) \propto \sum_{D,E} P(B)P(A|B)P(C|A)P(D|B,C)P(E|D) = \underbrace{P(B)P(A|B)}_{f(B)}\underbrace{P(C|A)}_{f(C)}$
Since the probability factors into a product of functions of $B$ and $C$, we can say that they are independent given $A$.
For the right graphical model, which is a Markov Network, I use the following method:

*

*Remove all edges from $A$

*Check if there is a path leading from $B$ to $C$
So we remove all edges from $A$ and we can see that we have a path, $B-D-C$ and hence $B$ and $C$ are not independent given $A$.
Is my reasoning correct? Also, for the Belief network deduction, is there a faster way to see this? My reasoning, without writing anything, would be to just say that since $A$ is not a collider for $B$ and $C$, then they are independent. Would this be correct reasoning?

Your arguments both for the BN and the MN are correct, provided those graphs are perfect maps for your two distributions (which are then necessarily different).
The formula-free explanation for the BN is correct, too, although I would add that the path $B\to D\leftarrow C$ is blocked because of $D$ being a collider.
