Summing two independent Markov Chains on a discrete-time and space environment I have the following problem on my Stochastic Processes problem set:

Let $(X_n)_{n \in \mathbb{N}}$ and $(Y_n)_{n \in \mathbb{N}}$ be two independent Markov Chains taking values on $I_{X}$ and $I_{Y}$ respectively, both countable sets. Also, let $(Z_{n})_{n \in \mathbb{N}}$ be iid random variables taking values on $\{0,1\}$. Now, define $W_n = X_n \mathbb{1}_{\{Z = 0\}} + Y_n \mathbb{1}_{\{Z = 1\}}$. Is $W_n$ a Markov Chain?

I understand that I have to show that 
$\mathbb{P}(W_{n+1} = j| W_{n} = i, W_{n-1} = i_{n-1},..., W_0 = i_0) = \mathbb{P}(W_{n+1} = j| W_{n} = i)$,
for any states $j,i,i_{n-1}, ..., i_{0} \in I_{X}\cup I_{Y}$. I am stuck right now, but I have the impression that this result is false as stated.
For example, if $I_X$ and $I_Y$ were disjoint sets, observing that $W_n = i$ for some $i \in I_X$ gives me some information about $X_n$, although no information about $Y_n$. But if $i \in I_{X}\cup I_{Y}$, I can get "confused" since I cannot attribute this realization to one or another variable.
On the other hand, $Z_{n}$ is such a simple process and both $X_{n}$ and $X_{n}$ are Markov Chains, so the dependence on the past is somewhat limited. 
I have no idea how to answer this question. I do know, however, that the sum of independent Markov Chains is not necessarily another Markov Chain, but I don't see how this result can help me here.
Thanks a lot in advance!!
 A: You're right about the case of disjoint $I_X$ and $I_Y$, and that's all you need. The question isn't asking whether $W_n$ might be a Markov chain under some circumstances (the answer is clearly yes, e.g. if $Z_n\equiv0$), but whether $W_n$ is necessarily a Markov chain, given the assumptions, so your counterexample is enough to disprove this. If $W_n\in I_X$, then it gives you no information about $Y_{n-1}$, but unless $Z_n\equiv0$ or $Y_{n+1}$ is independent of previous states, the next state $W_{n+1}$ depends on $Y_{n-1}$.
Edit in response to the comment:
A concrete counterexample is given by $Z_n$ uniformly distributed on $\{0,1\}$ and $X_n$ and $Y_n$ symmetric random walks on disjoint copies $\mathbb Z_X$ and $\mathbb Z_Y$ of $\mathbb Z$ starting at the respective origins at $n=0$. Assume that $Z_1=1$ and $Z_2=0$, so $W_1=y_1\in\mathbb Z_Y$ and $W_2=x_2\in\mathbb Z_X$. Now 
$$P(W_3=a\mid W_1=y_1,W_2=x_2)=
\begin{cases}
\frac14&a\in\{y_1,x_2+1_X,x_2-1_X\}\\
\frac18&a\in\{y_1+2_Y,y_1-2_Y\}\\
0&\text{otherwise}
\end{cases}
$$
whereas e.g. $P(W_3=3_Y\mid W_2=x_2)=\frac1{16}$.
