The statement is as follows:
$(\mathbf{Y} \perp \mathbf{W} \mid \mathbf{Z})$ and $(\mathbf{X} \perp \mathbf{Y},\mathbf{W} \mid \mathbf{Z})$ imply $(\mathbf{X},\mathbf{W} \perp \mathbf{Y} \mid \mathbf{Z})$.
So far, I only managed to develop them;
$(\mathbf{Y} \perp \mathbf{W} \mid \mathbf{Z})\Leftrightarrow P(\mathbf{Y}, \mathbf{W} \mid \mathbf{Z}) = p(w,y\mid z) = p(y\mid z)p(w \mid z)$
$(\mathbf{X} \perp \mathbf{Y},\mathbf{W} \mid \mathbf{Z}) \Leftrightarrow P(\mathbf{X} , \mathbf{Y},\mathbf{W} \mid \mathbf{Z}) = p(x,y,w\mid z) = p(x\mid z)p(y,w\mid z)$ and given the statement above, I imagine I can further develop it to = $p(x\mid z) p(y\mid z) p(w\mid z)$
However, I am super confused, and have no idea how I can get to the implication stated (if it is true) that
$$(\mathbf{X},\mathbf{W} \perp \mathbf{Y} \mid \mathbf{Z}) \Rightarrow P(X,Y,W\mid Z) = p(x,y,w\mid z) = p(x,w\mid z) p(y\mid z)$$
or even how I could go about disproving this... Am I allowed to simply regroup terms in the multiplication of conditional probabilities and say that from the second statement:
$$p(x\mid z) p(y\mid z) p(w\mid z) = p(x,w\mid z) p(y\mid z)$$ and then say it is as expected?
I have no idea what knowledge is needed to solve this.. I was thinking perhaps some sort of marginalization, but even then, I don't know in what way to apply it..
Thanks for any possible help. Please consider $\perp$ as independent (not sure how to make it happen here)