Finding an unconditional joint probability in a Bayesian Belief Network I have a Bayesian network as drawn in the picture:

We can see that $B$ and $C$ are conditionally independent given $A$. My goal is to find $P(B\cap C)$. My first thought was to use the Law of Total Probability:
\begin{align}
P(B\cap C) &= P (B\cap C\cap A) + P(B\cap C\cap \neg A)\\ 
&= P(B\cap C \mid A)P(A) + P(B\cap C \mid \neg A)P(\neg A).
\end{align}
Using conditional independence, we should obtain
\begin{align}
P(B\cap C \mid A) = P(B \mid A)P(C\mid A)=(0.4)(0.6) = 0.24.
\end{align}
However, when trying to calculate $P(B\cap C \mid \neg A)$, I know that I can't necessarily perform a similar product, as conditional independence given $A$ does not imply conditional independence given $\neg A$. This is where I am stuck. I thought about utilizing the Markov blanket of $A$ or even involving $D$ and $E$, as in a Bayesian network it is very easy to calculate
\begin{align}
P(A\cap B\cap C\cap D\cap E) &= P(A)P(B\mid A)P(C\mid A)P(D\mid B\cap C)P(E\mid C)\\
&=(0.8)(0.4)(0.6)(0.8)(0.3)=0.04608.
\end{align}
However, I am not sure if this is a useful route to take. Any help or comments are greatly appreciated!
 A: It is true that
$$ P\left(B \cap C \mid \neg A \right) = P\left( B \mid \neg A \right)\times P\left( C \mid \neg A \right). $$
The link that you attached (about conditional independence given $A$ not implying conditional independence given $\neg A$) is not relevant here. In that link, $A$, $B$ and $C$ are events.
 Whereas the Bayesian network tells you about dependencies between random variables.
To spell it out, let $X_A$ be the random variable that takes the value $1$ if $A$ is true, and takes the value $0$ if $A$ is false. Let $X_B$ and $X_C$ be defined in the same way for $B$ and $C$. The Bayesian network diagram is a statement about these random variables: it says that the random variables $X_B$ and $X_C$ are conditionally independent given $X_A$, i.e.
$$ P\left( X_B, X_C \mid X_A \right) = P\left( X_B \mid X_A \right) \times P\left( X_C \mid X_A \right).$$
Thus
\begin{align}  P\left(B \cap C \mid \neg A \right) & = P\left( X_B = 1, X_C = 1 \mid X_A = 0\right) \\ & = P\left( X_B = 1 \mid X_A = 0 \right) \times P\left( X_C = 1\mid X_A = 0\right)\\ & = P\left( B \mid \neg A \right)\times P\left( C \mid \neg A \right) . \end{align}
