How to prove conditional independence symmetry $X\perp Y | Z$? In probabilistic graphical modeling, conditional independence $X\perp Y | Z$ means $P(X,Y|Z)=P(X|Z)P(Y|Z)$.
How to prove it?
EDIT: this post might help your understanding
https://math.stackexchange.com/questions/832370/proving-conditional-independence
 A: Check on Dawid, 1980 for a more detailed explanation, but generally  proving conditional independence depense on the form of your join probability distribution. You must show that the variables are independent after being conditioned on. For example consider the following joint distribution:    
$P(x,y,z)=P(x|z)P(y|z)P(z)$    
The distribution of $x$ and $y$ can be obtained my marginalizing out $z$. The distribution of $x$ and $y$ is dependent as it will not factor , symbolically:
$P(x,y)=\sum_{z} P(x|z)P(y|z)p(z) \neq P(x)P(y) .$
When conditioning on $z$ one can show the distribution factors and is independent:
$P(x,y|z)=P(x,y,z)/P(z)= (P(x|z)P(y|z)P(z))/P(z)=P(x|z)P(y|z)$    
Similarity for this second distribution one can show that $x$ and $y$ are conditionally independent for the second joint distribution:
$P(x,y,z)=P(y|z)P(z|x)P(x)$
The marginal for $x$ and $y$ is given by: 
$P(x,y)=\sum_{z} P(y|z)P(z|x)P(x)=P(x) \sum_{z} P(y|z)P(z|x)$
$=P(x) \sum_{z} P(y,z|x)=P(x)P(y|x) \neq P(x)P(y) .$
The variables $x$ and $y$ are dependent but when conditioning on $z$  they become independent.
$P(x,y|z)=P(y|z)P(z|x)P(x)/P(z)=P(y|z)(P(z,x)/P(z))=P(y|z)P(x|z)$.
