Interchanging random variables with the same distribution Let $X, W, Y, Z$ be $\mathbb{R}$-valued random variables on the probability space $(\Omega, \mathcal{F}, \mathbb{P})$ such that $Y, Z, W$ are i.i.d. and $\sigma(X)$ is independent of $\sigma(W, Y, Z)$. Let $f : \mathbb{R}^2 \rightarrow \mathbb{R}$ be a measurable function and $B \in \mathcal{B}(\mathbb{R})$.

Is it true that
$$
\mathbb{P} (f(X, Y) \in B, f(X, Z) \in B) = \mathbb{P}(f(X, Y) \in B, f(X, W) \in B).
$$

The probability on the left-hand side depends on the joint law of $X$, $Y$, $Z$, which is given by the joint law of $X$ and $(Y, Z)$. Since $X$ is independent of $(Y, Z)$, the joint law is given by the product of the two laws. But $(Y, Z)$ has the same law as $(Y, W)$, and so the joint law of $X, Y, Z$ is the same as that of $X, Y, W$. Can this be used to show the equality?
 A: The statement is true. By Theorem $20.3$ of Probability and Measure and the independence assumption,
\begin{align}
& P[f(X, Y) \in B, f(X, Z) \in B] = \int_{-\infty}^\infty P[f(x, Y) \in B, f(x, Z) \in B]\mu(dx), \\
& P[f(X, Y) \in B, f(X, W) \in B] = \int_{-\infty}^\infty P[f(x, Y) \in B, f(x, W) \in B]\mu(dx),
\end{align}
where $\mu$ is the distribution of $X$.
It is therefore sufficient to show $P[f(x, Y) \in B, f(x, Z) \in B] = P[f(x, Y) \in B, f(x, W) \in B]$ for any fixed $x \in \mathbb{R}^1$, which is the consequence of (use the i.i.d. assumption):
\begin{align*}
& P[f(x, Y) \in B, f(x, Z) \in B] = P[f(x, Y) \in B]P[f(x, Z) \in B] \\ 
=& P[f(x, Y) \in B]^2 \\
=& P[f(x, Y) \in B]P[f(x, W) \in B] = P[f(x, Y) \in B, f(x, W) \in B].
\end{align*}

More details of applying Theorem $20.3$: basically identify the $X$ in the question as the $X$ in the theorem, and identify $(Y, Z)$ (or $(Y, W)$) in the question as the $Y$ in the theorem.  Argue as follows:
Define the mapping $g: \mathbb{R}^3 \to \mathbb{R}^2$ by $g(x, y, z) = (f(x, y), f(x, z))$. Clearly $g$ is measurable $\mathscr{R}^3/\mathscr{R}^2$ and $[f(X, Y) \in B, f(X, Z) \in B] = [g(X, Y, Z) \in B \times B]$. Therefore,
\begin{align}
& P[f(X, Y) \in B, f(X, Z) \in B] = P[g(X, Y, Z) \in B \times B] \\
=& P[(X, Y, Z) \in g^{-1}(B \times B)] \\
=& \int_{-\infty}^\infty P[(x, Y, Z) \in g^{-1}(B \times B)]\mu(dx) \\ 
=& \int_{-\infty}^\infty P[f(x, Y) \in B, f(x, Z) \in B]\mu(dx). \\[2em]
 & P[f(X, Y) \in B, f(X, W) \in B] = P[g(X, Y, W) \in B \times B] \\
=& P[(X, Y, W) \in g^{-1}(B \times B)] \\
=& \int_{-\infty}^\infty P[(x, Y, W) \in g^{-1}(B \times B)]\mu(dx) \\ 
=& \int_{-\infty}^\infty P[f(x, Y) \in B, f(x, W) \in B]\mu(dx). 
\end{align}
P.S. Using the notation I introduced above, I think your (verbal) argument in the question actually works too (and gives an even shorter proof), as it fully takes advantage of the fact $(X, Y, Z) \overset{d}{=} (X, Y, W)$ under given assumptions. If we denote their common distribution on $(\mathbb{R}^3, \mathscr{R}^3)$ as $\eta$ (it can be easily seen that $\eta = \mu \times \nu \times \nu$, which is a product measure of the distribution $\mu$ of $X$ and the distribution of $\nu \times \nu$ of $(Y, Z)$ (or $(Y, W)$)), and   note that $g^{-1}(B \times B)$ is a Borel set in $\mathscr{R}^3$, it follows that
\begin{align}
 & P[f(X, Y) \in B, f(X, Z) \in B] = P[(X, Y, Z) \in g^{-1}(B \times B)] \\
=& \eta(g^{-1}(B \times B)) \\
=& P[(X, Y, W) \in g^{-1}(B \times B)] = P[f(X, Y) \in B, f(X, W) \in B]. 
\end{align}
