If $W$ is a Markov chain and $N$ is a Poisson process, then $left(W_{N_t}right)_{tge0}$ is Markov

Let $(Omega,mathcal A,operatorname P)$ be a probability space, $(E,mathcal E)$ be a measurable space, $(W_n)_{ninmathbb N_0}$ be a time-homogeneosu Markov chain on $(Omega,mathcal A,operatorname P)$ with transition kernel $kappa$ and $(N_t)_{tge0}$ be a Poisson process on $(Omega,mathcal A,operatorname P)$ with intensity $lambda>0$ independent of $W$.

I would like to conclude that $$X_t:=W_{N_t};;;text{for }tge0$$ is a time-homogeneous Markov process on $(Omega,mathcal A,operatorname P)$ with transition semigroup$^1$ $left(e^{t(kappa-lambda)}right)_{tge0}$.

The idea is begin{equation}begin{split}operatorname Pleft[X_{s+t}in Bmidmathcal F^X_sright]&=operatorname Pleft[Y_{N_s+(N_{s+t}-N_s)}in Bmidmathcal F^X_sright]\&=sum_{n=0}^inftyoperatorname Pleft[N_{s+t}-N_s=n,Y_{N_s+n}in Bmidmathcal F^X_sright]\&=sum_{n=0}^inftyoperatorname Pleft[N_{s+t}-N_s=nright]operatorname Pleft[Y_{N_s+n}in Bmidmathcal F^X_sright]\&=e^{-lambda t}sum_{n=0}^inftyfrac{(lambda t)^n}{n!}operatorname Pleft[Y_{N_s+n}in Bmidmathcal F^X_sright].end{split}tag1end{equation} for all $Binmathcal E$ and $s,tge0$.

However, in order for the third equality in $(1)$ to hold, we need that $N_{s+t}-N_s$ is independent of $mathcal F^X_s$ for all $s,tge0$.

Intuitively, this seems to be obvious, since $(N_t)_{tge0}$ is independent of $(W_n)_{ninmathbb N}$ and $N_{s+t}-N_s$ is independent of $mathcal F^N_s$ for all $s,tge0$, but how can we prove it rigorously?

It’s easy to see that $$left.sigma(X_s)right|_{{:N_s:=:n:}}=left.sigma(W_n)right|_{{:N_s:=:n:}}tag3$$ for all $sge0$ and $ninmathbb N$. So, maybe the desired claim follows from the local property of conditional expectation.

EDIT: Let $$mathcal E_t:=bigcup_{ninmathbb N_0}left{A_1capleft{N_t=nright}cap A_2:ninmathbb Ntext{ and }(A_1,A_2)inmathcal F^N_ttimesmathcal F^W_nright}$$ for $tge0$. It’s easy to see that

1. $emptysetinmathcal E_t$ for all $tge0$;
2. $mathcal E_t$ is closed under finite intersections for all $tge0$;
3. $N_{s+t}-N_s$ is independent of $mathcal E_s$ for all $s,tge0$.

All these properties together imply that $N_{s+t}-N_s$ is independent of $sigma(mathcal E_s)$ for all $s,tge0$. So, it would be sufficient to show that $sigma(mathcal E_s)=mathcal F^X_s$ for all $sge0$.