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

For the left model (Belief Network), here's my work:
$$
\begin{align}
P(C,D|A) &\propto \sum_{B,E} P(B)P(A|B)P(C|A)P(D|B,C)P(E|D)\\
    &=\sum_B P(B)P(A|B)P(C|A)P(D|B,C).
\end{align}
$$
We can't factorize it anymore. So this means that we can't factorize the probability into a product of functions of $f(C)$ and $f(D)$, so $C$ and $D$ are not 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 $C$ to $D$
We can see that if we remove the edges from $A$, we still have the path $C-D$ left, so this means that they are not independent given $A$.
Is this correct? To me it seems like for the left network, it's not correct, since $A$ is not a collider for $C$ and $D$, so they should be independent given $A$.

$\newcommand{\npperp}{\not\perp\kern-8pt\perp}$
In your reasoning for the BN, you state "We can't factorize it anymore." But this would have to be proven.
Nevertheless, it is indeed true that $C\npperp D | A$, and the graph-based, d-separation, explanation would go like this: the path $C\to D$ is (trivially) d-separation open. There is no node between $C$ and $D$ that could block this path. The conditioning on $A$ would only have a chance of blocking this path if it was on this path, which it is not.
Your procedure and conclusion for the MN are correct.
