Take $(\Omega,F,P)$ where
$\Omega =\begin{Bmatrix} \omega_{1,1}&\omega_{1,2} &\omega_{1,3} \\ \omega_{2,1}&\omega_{2,2} &\omega_{2,3} \\ \omega_{3,1}&\omega_{3,2} &\omega_{3,3} \end{Bmatrix}$ and $F=2^\Omega$.
Let $A_1$ a sub $\sigma$-field generated by: $\left \{ \omega_{1,1},\omega_{1,2},\omega_{1,3} \right \}, \left \{ \omega_{2,1},\omega_{2,2},\omega_{2,3} \right \}, \left \{ \omega_{3,1},\omega_{3,2},\omega_{3,3} \right \}$
and $A_2$ a sub $\sigma$-field generated by: $\left \{ \omega_{1,1} \right \}, \left \{ \omega_{1,2},\omega_{2,2},\omega_{2,1} \right \}, \left \{\omega_{1,3},\omega_{2,3},\omega_{3,1},\omega_{3,2},\omega_{3,3} \right \}$
If define a random variable $X$ as $X(\omega_{i,j})=i+j$, can we say that:
$E[E[X|A_1]|A_2]=E[E[X|A_2]|A_1]?$
HINT
I think that the best way of thinking about this simple yet overcomplicated (discrete) problem is as follows.
Consider two random variables $Y$ and $Z$ with the following definitions:
$$Y=\begin{cases} a&\text{if}&\omega\in\{ \omega_{1,1},\omega_{1,2},\omega_{1,3} \}\\ b&\text{if}&\omega\in\{ \omega_{2,1},\omega_{2,2},\omega_{2,3} \}\\ c&\text{if}&\omega\in\{ \omega_{3,1},\omega_{3,2},\omega_{3,3} \},\end{cases}$$
$$Z=\begin{cases} u&\text{if}&\omega\in\left \{ \omega_{1,1} \right \}\\ v&\text{if}&\omega\in\left \{ \omega_{1,2},\omega_{2,2},\omega_{2,1} \right \}\\ w&\text{if}&\omega\in\left \{\omega_{1,3},\omega_{2,3},\omega_{3,1},\omega_{3,2},\omega_{3,3} \right \}\end{cases}$$ where $a,b,c$ and $u,v,w$ are distinct reals.
With this notation, and because of the definition of the conditional expectation, for every $\omega\in\Omega$, $E[X\mid A_1],\ E[X\mid A_2] $ as functions of $\omega$ can be calculates as
$$E[X\mid A_1\ ](\omega)=E[X\mid Y=y\ ]$$ and $$E[X\mid A_2\ ](\omega)=E[X\mid Z=z\ ]$$ where $Y(\omega)=y$ and $Z(\omega)=z$.
By this the calculations are lead back to methods known from elementary probability theory. Now, we have $S=E[\ X\mid A_1]$ and $R=E[\ X\mid A_2\ ]$ as functions of $\omega$. Then use the similar thread of thoughts when calculating
$$E[S\mid Z\ ]\text{ and } E[R\mid Y\ ].$$
EDIT
Let's get one step forward. A question: Who said that $P(\{\omega_{i,j}\})=\frac19$? In the general case, these probabilities are all different. Let me use the following notation: $p_{ij}=P(\{\omega_{i,j}\}), \ i,j\in\{1,2,3\}$.
Now, let's calculate $E[X\mid A_1]$ as a function of $\omega$. First, for $\omega_{1,1},\omega_{1,2}, \text{ and } \omega_{1,3}$. For all of these lementary events $Y(\omega)=a$.
$$E[X\mid A_1](\omega_{1,1})=E[X\mid A_1](\omega_{1,2})=E[X\mid A_1](\omega_{1,3})=$$ $$=E[X\mid Y=a]=$$ $$=2P(X=2\mid Y=a)+3P(X=3\mid Y=a)+4P(X=4\mid Y=a)=$$ $$=2\frac{p_{11}}{p_{11}+p_{12}+p_{13}}+3\frac{p_{12}}{p_{11}+p_{12}+p_{13}}+4\frac{p_{1,3}}{p_{11}+p_{12}+p_{13}}.$$
The same way
$$E[X\mid A_1](\omega_{2,1})=E[X\mid A_1](\omega_{2,2})=E[X\mid A_1](\omega_{2,3})=$$ $$=E[X\mid Y=b]=$$ $$=2P(X=2\mid Y=b)+3P(X=3\mid Y=b)+4P(X=4\mid Y=b)=$$ $$=3\frac{p_{21}}{p_{21}+p_{22}+p_{23}}+4\frac{p_{22}}{p_{21}+p_{22}+p_{23}}+5\frac{p_{2,3}}{p_{21}+p_{22}+p_{23}}.$$
and
$$E[X\mid A_1](\omega_{3,1})=E[X\mid A_1](\omega_{3,2})=E[X\mid A_1](\omega_{3,3})=$$ $$=E[X\mid Y=c]=$$ $$=2P(X=2\mid Y=c)+3P(X=3\mid Y=c)+4P(X=4\mid Y=c)=$$ $$=4\frac{p_{31}}{p_{31}+p_{32}+p_{33}}+5\frac{p_{3,2}}{p_{31}+p_{32}+p_{33}}+6\frac{p_{33}}{p_{31}+p_{32}+p_{33}}.$$
If $p_{ij}$ are $\frac19$ then $E[X\mid A_1]$ as a random variable (a function of $\omega$ can be pictured easily over:
$$\Omega=\begin{Bmatrix} \omega_{1,1}&\omega_{1,2} &\omega_{1,3} \\ \omega_{2,1}&\omega_{2,2} &\omega_{2,3} \\ \omega_{3,1}&\omega_{3,2} &\omega_{3,3} \end{Bmatrix}$$
like this
$$\begin{Bmatrix} 3&3 &3 \\ 4&4 &4 \\ 5&5 &5 \end{Bmatrix}.$$
Could you do $E[X\mid A_2]$ now?