Suppose $X, Y, Z$ are random variables. By Tower rule, $E(X) = E(E(X|Y))$. Is $E(X) = E(E(X|Y, Z))$? Suppose $X, Y, Z$ are random variables. By Tower rule (iterated expectations), $$E(X) = E(E(X|Y))$$
My question is, is $$E(X) = E(E(X|Y, Z))?$$
My attempt (assuming $X, Y, Z$ are discrete r.v.):
\begin{align*}
E(E(X|Y, Z)) &= \sum_y\sum_zE(X|Y=y,Z=z)P(Y = y,Z=z)\\
&= \sum_y \sum_z \sum_x xP(X=x|Y=y,Z=z)P(Y=y,Z=z)\\
&= \sum_y \sum_z \sum_x x \frac{P(Y=y,Z=z|X=x)P(X=x)}{P(Y=y,Z=z)}P(Y=y,Z=z)\\
&= \sum_y \sum_z \sum_x x P(Y=y,Z=z|X=x)P(X=x)\\
&=  \sum_x x P(X=x)\sum_y \sum_zP(Y=y,Z=z|X=x)\\
&= \sum_x xP(X=x)\\
&= E(X)
\end{align*}

Yes, it is correct because these formulas are also correct for random vectors (so groups of random variable). This means $Y,Z$ can be represented by a vector $\mathbf{V}$, and the following formula can be applied:
$$\mathbb E[X]=\mathbb E[\mathbb E[X|\mathbf{V}]]$$
