Dependency of non descendant in Bayesian Network I am learning Probabilistic Graphical Models from a book by Daphne Koller and the basic definition of a Bayesian Network says this:

I tried to make a counter example to this and am confused if the definition takes into the consideration a following model.

Consider here the random variable Grade G . I would refer all variables with their first letter.
P(G | D, I, R , P)   !=   P(G | D, I, R ) 
(Independence of variable G from a non decescendant is violated)
Intuitively if we know that we got a very high recommendation(R) and did very poorly in the project(P) then this will definitely improve the probability that we got a good grade(G), given intelligence and difficulty , compared to no evidence given about the project(P). But according to the basic definition this should not be true as independence over non descendant is violated here.
Can anyone tell me where I am wrong?
 A: The 'non-descendant' in this case is the Project node.
What you're doing here is conditioning on R. This doesn't match with the statement
given in the text, since that statement is concerned with conditioning on
the parents of G.
It might interest you to know that this is an example of conditional
dependence.
Consider the following network:
$$ A \rightarrow C \leftarrow B $$
where $ A $ and $ B $ are the two parent nodes of $ C $.
It is true that
$$
P(A, B) = P(A)P(B)
$$
in that they are independent. Consequently, $ P(A \mid B) = P(A) $ and vice versa
for $ B $.
However, $ P(A, B \mid C) \neq P(A \mid C)P(B \mid C) $. Intuitively, if $ C$ happens then we get information
about the combined influence that both $ A $ and $ B $ had on $ C $;
as such, knowledge about $ A $ will influence our knowledge about $ B $.
To further specify this, let
$$
A \sim \text{Bernoulli}(0.1)
$$
$$
B \sim \text{Bernoulli}(0.1)
$$
$$
C \mid A, B \sim \text{Bernoulli}(0.1 + (A \times B) 0.9)
$$
Now we'll show that $ P(A, B \mid C) \neq P(A \mid C)P(B \mid C) $
(which implies that $ A $ and $ B $ are not independent given $ C $).
\begin{equation}
\begin{aligned}
P(A=1, B=1 \mid C=1)
& = \frac{P(C=1 \mid A=1, B=1)P(A=1, B=1)}{P(C=1)} \\
& = \frac{1 \cdot P(A=1)P(B=1)}{P(C=1)} \\
& = \frac{0.01}{P(C=1)}
\end{aligned}
\end{equation}
whereas:
\begin{equation}
\begin{aligned}
P(A=1 \mid C=1)
& = \frac{P(C =1 \mid A = 1)P(A = 1)}{P(C=1)} \\
& = \frac{[P(C = 1 \mid A=1, B=1)P(B=1) + P(C = 1 \mid B = 0, A = 1)P(B = 0)]P(A = 1)}{P(C = 1)} \\
& = \frac{[P(B=1) + 0.1 \cdot P(B = 0)]P(A = 1)}{P(C = 1)} \\ 
& = \frac{[0.1 + 0.1 \cdot 0.9]P(A = 1)}{P(C = 1)} \\ 
& = \frac{0.019}{P(C = 1)} \\ 
& = P(B = 1 \mid C = 1) \text{ by symmetry} \\ 
\end{aligned}
\end{equation}
Now since $$ P(C = 1) = P(C = 1 \mid A = 1)P(A = 1) + P(C = 1 \mid A = 0) P(A = 0) = 0.029 $$ we have
$$
P(A =1, B = 1\mid C = 1) = \frac{0.01}{0.029} \approx 0.3448
$$
whereas
$$
P(A = 1 \mid C=1)P(B = 1 \mid C=1) = \frac{0.019^2}{0.029^2} \approx 0.4293
$$
