How to show these two conditional independence relationships? If $X \perp Y|Z $, and $U=h(X)$, then $X \perp Y|(Z,U) $
If $X \perp Y|Z $, and $X \perp W|(Y,Z) $, then $X \perp (W,Y)|Z $
The book doesn't give any explanation. So I am confused.
 A: To show above relationships, the following theorem is useful(c.f., Y. S. Chow & H. Teicher, Probability Theory, 3Ed, Springer Verlag, 1997, p.230, Theorem 7.3.1)
Theorem If  $\mathscr{G}_i,i=1,2,3,$ are $\sigma$-algebras of events,
then conditioinal independence of $\mathscr{G}_1$ and $\mathscr{G}_2$ is eqivalent to any one
of the following:
(i) For all $A_1\in\mathscr{G}_1 $,
\begin{equation*}
 \mathsf{P}(A_1|\sigma(\mathscr{G}_2\cup \mathscr{G}_3))=\mathsf{P}(A_1|\sigma(\mathscr{G}_3)), 
\end{equation*}
(iv) for every $\sigma(\mathscr{G}_1\cup \mathscr{G}_3) $-measurable function $X$ with $|\mathscr{E}X|\le \infty$,
\begin{equation*}
 \mathsf{E}[X|\sigma(\mathscr{G}_2\cup \mathscr{G}_3)]=\mathsf{E}[X| \mathscr{G}_3]. 
\end{equation*}
Now we derive the first relationship. Since $U=h(X)$, then $\sigma(U,X)=\sigma(X) $. From
$X\perp Y|Z (\iff \mathsf{E}[1_B(Y)|X,Z]=\mathsf{E}[1_B(Y)|Z])$ and $\sigma(U,X)=\sigma(X) $ get
\begin{equation*}
 \mathsf{E}[1_B(Y)|X,U,Z]=\mathsf{E}[1_B(Y)|Z], \quad \forall B\in\mathscr{B}(\mathbb{R}).  \tag{1}
\end{equation*}
Taking $\mathsf{E}[\; \cdot \; |U,Z]$ for above equation,
\begin{equation*}
 \mathsf{E}[1_B(Y)|U,Z]=\mathsf{E}[1_B(Y)|Z], \quad \forall B\in\mathscr{B}(\mathbb{R}). \tag{2}
\end{equation*}
Comparing (1) and (2) get
\begin{gather*}
 \mathsf{E}[1_B(Y)|X,U,Z]=\mathsf{E}[1_B(Y)|U,Z], \quad \forall B\in\mathscr{B}(\mathbb{R})\\
 \hphantom{\text{ by  (iv)}}\Downarrow  \text{ by (iv)} \\
 X\perp Y|(Z,U)
\end{gather*}
For the second relationship, since $X\perp W|(Y,Z)$ and (iv)
\begin{align*}
 \mathsf{E}[1_B(X)|Z,Y,W]&=\mathsf{E}[1_B(X)|Z,Y]\\
 (\text{by } X\perp Y|Z) \qquad &=\mathsf{E}[1_B(X)|Z]
\end{align*}
Comparing the lefthandside and righthandside of above equalities get  $X\perp (W,Y)|Z$ from (iv).
A: $\newcommand{\classG}{\mathscr{G}}$
It is really helpful to first clarify the formal mathematical definition of conditional independence. Oddly enough, many texts/papers chose to extensively use this important concept (including adopting "$\perp$" notations as you listed in the question and making statements such as "$X$ and $Y$ are independent given $Z$") before giving its rigorous definition. The reference Probability Theory: Independence, Interchangeability, Martingales mentioned in @JGWang's answer is an exception. To complement his answer, I would like to quote the definition of conditional independence as follows:

Let $\classG$ be a $\sigma$-algebra of events and $\{\classG_n, n \geq 1\}$ a sequence of classes of events. Two such classes $\classG_1$ and $\classG_2$ are said to be conditionally independent given $\classG$ if for all $A_i \in \classG_i, i = 1, 2$,
\begin{align*}
P(A_1A_2|\classG) = P(A_1|\classG)P(A_2|\classG), \quad \text{a.c.}
\end{align*}
More generally, the sequence $\{\classG_n, n \geq 1\}$ is declared conditionally independent given $\classG$ if for all choices of $A_m \in \classG_{k_m}$, where $k_i \neq k_j$ for $i \neq j, m = 1, \ldots, n$, and $n = 2, 3, \ldots$,
\begin{align*}
P(A_1A_2\cdots A_n|\classG) = \prod_{i = 1}^nP(A_i|\classG), \quad \text{a.c.}
\end{align*}
Finally, a sequence $\{X_n, n \geq 1\}$ of random variables is called conditionally independent given $\classG$ if the sequence of classes $\classG_n = \sigma(X_n), n \geq 1$, is conditionally independent given $\classG$.

This definition, of course, is measure-theoretic.  It conveys two vital takeaways that how advanced probability theory defines conditional independence within a unified framework:

*

*It begins with defining the conditional independence of classes of events given a $\sigma$-algebra (such method is also shared with defining unconditional independence). The conditional independence of random variables/vectors given a random variable/vector is then naturally defined as the independence of their corresponding generated $\sigma$-algebras.  For example, $X \perp Y|Z$ really means $\sigma(X)$ and $\sigma(Y)$ are independent given $\sigma(Z)$.

*It heavily relies on the measure-theoretic concept of conditional probability (or equivalently, conditional expectation).

The greatest advantage of these two points is that separate treatments for continuous random variables, discrete random variables, and even a hybrid of continuous and discrete random variables, which are usually done in elementary probability theory (the elementary definition of conditional independence typically requires that the conditional joint pdf/pmf factorizes to marginal ones, which is very tedious and sometimes infeasible), are no longer needed.  A rigorous proof to your questions based on the above cited definition has been provided by @JGWang.
That said, for a heuristic proof that only uses the definition $P(A|B) = P(A \cap B)/P(B)$ (given $P(B) > 0$), you can assume that all the random variables under consideration are discrete. For example, to show that $X \perp Y|Z $ and $X \perp W|(Y,Z)$ imply $X \perp (W,Y)|Z$, just verify:
\begin{align*}
 & P(X = x, W = w, Y = y | Z = z) \\
=& P(X = x, W = w | Y = y, Z = z) P(Y = y | Z = z) \\
\overset{*}{=}& P(X = x | Y = y, Z = z)P(W = w | Y = y, Z = z)P(Y = y | Z = z) \\
\overset{**}{=}& P(X = x | Z = z)P(W = w, Y = y | Z = z),
\end{align*}
where we used $X \perp W | (Y, Z)$ to derive "$\overset{*}{=}$" and $X \perp Y | Z$ to derive "$\overset{**}{=}$".
