Consider the following definition:
Definition (Pearl 2009)
Let $V = {V_1, V_2, \ldots }$ be a finite set of variables.
Let $P(\cdot)$ be a joint probability function over the variables in $V$, and let $X$, $Y$, $Z$ stand for any three subsets of the variables in $V$.
The sets $X$ and $Y$ are said to be conditionally independent given $Z$ if
\(P(x | y, z) = P(x|z)\) whenever $P(y,z) > 0$.
Denoting this ternary relation as \((X \mathrel{\unicode{x2AEB}} Y \vert Z)\), whe have the following properties below.
The first property is symmetry.
\[(X \mathrel{\unicode{x2AEB}} Y \vert Z) \implies (Y \mathrel{\unicode{x2AEB}} X \vert Z)\]While this is a form of symmetry, it is not the same as strongly symmetric when describing any given $n$-ary relation. If it were, we would have to accept propositions such as \((X \mathrel{\unicode{x2AEB}} Y \vert Z) \implies (Y \mathrel{\unicode{x2AEB}} Z \vert Y)\). But this can not be so for the same type of reason as to why $Pr(A\vert B) \neq Pr(B \vert A)$ for at least some probability measure $Pr$ on events $A$ and $B$.
The second property is decomposition,
\((X \mathrel{\unicode{x2AEB}} (Y, W) \vert Z) \implies (Y \mathrel{\unicode{x2AEB}} X \vert Z)\),
Pearl 2009 seems to use $YW$ instead of $(Y,W)$, giving a false impression that we might be considing some kind of multiplication. See Notational confusion about conditional independence in Pearl 2009 for more information.
The third property is weak union.
\[(X \mathrel{\unicode{x2AEB}} (Y, W) \vert Z) \implies (Y \mathrel{\unicode{x2AEB}} X \vert (Z,W))\]The fourth property is contraction.
\[(X \mathrel{\unicode{x2AEB}} Y | Z) \land (X \mathrel{\unicode{x2AEB}} W | (Z,Y)) \implies (X \mathrel{\unicode{x2AEB}} (Y,W) \vert Z)\]Finally the fifth property is intersection.
\[(X \mathrel{\unicode{x2AEB}} W \vert (Z, Y)) \land (X \mathrel{\unicode{x2AEB}} Y \vert (Z, W)) \implies (X \mathrel{\unicode{x2AEB}} (Y,W) \vert Z)\]