Notational confusion about conditional independence in Pearl 2009 First I read this definition which introduces $X$, $Y$ and $Z$ as sets of random variables.

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 $P(y,z) > 0$.

Pearl then introduces Dawid's notation:

$(X \mathrel{\unicode{x2AEB}} Y | Z)_P \iff P(x | y, z) = P(x | z)$

That's all fine as far as I understand, but I got confused by the decomposition property of conditional independence:
$$(X \mathrel{\unicode{x2AEB}} YW | Z)_P \implies (X \mathrel{\unicode{x2AEB}} Y | Z)_P$$
The symbol $W$ is not defined, nor is putting it to the right of $Y$ in the form $YW$. I'm not sure if there is a set operation implied here. What does this notation mean?
 A: I think $X, Y, Z, W$ should all be interpreted as random variables, and "$YW$" is not the product of $Y$ and $W$, but the random vector $(Y, W)$.  Therefore, the expression means:
\begin{align}
(X \mathrel{\unicode{x2AEB}} (Y, W)|Z)_P \implies (X \mathrel{\unicode{x2AEB}} Y|Z)_P,
\end{align}
i.e., if $X$ and $(Y, W)$ are conditionally independent given $Z$, then $X$ and $Y$ are conditionally independent given $Z$.  This interpretation is supported by the following statement which appears in the second paragraph after the expression:

The decomposition axiom asserts that if two combined items of information are judged irrelevant to $X$, then each separate item is irrelevant as well.

The mathematical proof for this property should also be straightforward: $(X \mathrel{\unicode{x2AEB}} (Y, W)|Z)_P$ means that for any $A_1 \in \sigma(X), A_2 \in \sigma(Y, W)$, it holds that (see this thread for a rigorous definition of conditional independence):
\begin{align}
P(A_1A_2|\sigma(Z)) = P(A_1|\sigma(Z))P(A_2|\sigma(Z)). \tag{1}
\end{align}
Since $\sigma(Y) \subset \sigma(Y, W)$, $(1)$ holds for any $A_2 \in \sigma(Y)$, hence $(X \mathrel{\unicode{x2AEB}} Y | Z)_P$.
