Does $\text E[\text E[X\mid\mathcal F]\mid\mathcal G]=\text E[\text E[X\mid\mathcal G]\mid\mathcal F]$ hold when $\mathcal G\not\subseteq\mathcal F$? Let $X$ be a random variable on a measurable space $(\Omega,\mathcal A)$ and $\mathcal F$, $\mathcal G$ two sub-$\sigma$-algebras of $\mathcal A$. Assume that $$Y:=\operatorname E\left[X\mid\mathcal G\right]$$ is $\mathcal F$-measurable. I'm curious whether or not $(\ast)$ in $$Y=\operatorname E\left[Y\mid\mathcal F\right]\stackrel{(\ast)}=\operatorname E\left[\operatorname E\left[X\mid\mathcal F\right]\mid\mathcal G\right]\tag 1$$
holds. I know, that if $\mathcal G\subseteq\mathcal F$, then $(1)$ is just the tower property of conditional expectation.

I've tried the following, but I'm not sure if I've made a mistake: Let $Z:=\operatorname E\left[Y\mid\mathcal F\right]$. Then, $$Z=\operatorname E\left[Z\mid\mathcal A\right]\stackrel{\text{tower property}}=\operatorname E\left[\operatorname E\left[Z\mid\mathcal A\right]\mid\mathcal G\right]\stackrel{\text{tower property}}=\operatorname E\left[Z\mid\mathcal G\right]\;.\tag 2$$ Why do I think that this might be wrong? Well, I don't understand why we should otherwise force $\mathcal G\subseteq\mathcal F$ in the statement of the tower property, since we could always switch to $\mathcal A$ in an intermediate steplike in $(2)$.

Hint:
Suppose $X$ can have two states, state $A$ and state $B$, with values 0 and 1 respectively. Consider $\mathcal F$ and $\mathcal G$ to be related to $A$ and $B$. Try to make it as simple as possible, so you don't have to take some complicated expectation.
