Let $(\Omega,\mathcal A,\operatorname P)$ be a probability space, $(E,\mathcal E)$ be a measurable space, $(W_n)_{n\in\mathbb N_0}$ be an $(E,\mathcal E)$-valued process on $(\Omega,\mathcal A,\operatorname P)$, $(N_t)_{t\ge0}$ be a $\mathbb N_0$-valued process on $(\Omega,\mathcal A,\operatorname P)$ and $$X_t:=W_{N_t}\;\;\;\text{for }t\ge0.$$
Let $\mathcal F^Z$ denote the filtration generated by any process $Z$. How can we show that $$\mathcal F^X_t=\sigma\left(\left\{A_1\cap A_2\cap\left\{N_t=n\right\}:n\in\mathbb N\text{ and }(A_1,A_2)\in\mathcal F^N_t\times\mathcal F^W_n\right\}\right)\tag1$$ for all $t\ge0$?
This should be provable by showing that the set inside the $\sigma$-operator on the right-hand side is a Dynkin system. But even then it's not immediate that equality holds in $(1)$.
Maybe it's useful to note that we've clearly got $$\left.\sigma(X_t)\right|_{\left\{\:N_t\:=\:n\:\right\}}=\left.\sigma(W_n)\right|_{\left\{\:N_t\:=\:n\:\right\}}\tag2$$ for all $t\ge0$ and $n\in\mathbb N_0$.