The law of iterated expectations tells us that ${\bf E}\big [{\bf E}[X\, |\, Y]\big ]={\bf E}[X]$. Suppose that we want apply this law in a conditional universe, given another random variable $Z$, in order to evaluate ${\bf E}[X \, |\, Z]$. Then:
${\bf E}\big [{\bf E}[X\, |\, Y,Z] \, |\, Z\big ]={\bf E}[X\, |\, Z]$
I'm not sure how to apply the Law of Iterated Expectations to show this relationship is true. Initially, I thought I could do this: $ {\bf E}[X\, |\, Y,Z] = {\bf E}[X]$
But by blindly applying the formula from the inside out, I get an incorrect result, so I'm missing some of the behind-the-scenes reasoning:
${\bf E}\big [{\bf E}[X]\ |\, Z\big ]$
${\bf E}\big [{\bf E}[X]] = {\bf E}[X]$
Another way I tried to think about this is that it seems like $ {\bf E}[X\, |\, Y,Z] $ is the same as ${\bf E}[X\, |\, Y,Z] | Z$. But then I get to the same place as above where I'm not sure how to deal with conditioning on multiple random variables. It appears that $ {\bf E}[X\, |\, Y,Z] \neq {\bf E}[X]$
Let $\sigma(Y,Z)$ be the $\sigma$-algebra generated by $(Y,Z)$ and $\sigma(Z)$ the $\sigma$-algebra generated by $Z$. I will also assume that $X,Y: \Omega \to \mathbb R$ are real random variables defined on $\Omega$ (although this can be generalized). It then holds that
$$\sigma(Z) \subset \sigma(Y,Z)$$
One can see this as follows:
Let $A \in \sigma(Z)$, then $Z(A) \in \mathcal B(\mathbb R)$ and $Z(A) \times \mathbb R \in \mathcal B(\mathbb R^2)$. Then $A$ is the inverse image of $Z(A) \times \mathbb R$ under $(Z,Y)$ and hence $A \in \sigma(Y,Z)$.
Your identity is now simply the tower property for conditional expectations.