Show that x and y are independent in this joint distribution

[endeavor]

Given that the joint probability Pr(w,x,y,z) over four variables factorizes as
$Pr(w,x,y,z) = Pr(w) Pr(z|y) Pr(y|x,w)Pr(x)$
show that x is independent of w by showing that Pr(x,w) = Pr(x)Pr(w).

Attempt: if we simply assume Pr(x,w) = Pr(x)Pr(w), then:
$\begin{align}<br /> Pr(w,x,y,z) &= Pr(w) Pr(z|y) Pr(y|x,w) Pr(x)\\<br /> &\stackrel{?}= Pr(x,w) Pr(z|y) Pr(y|x,w)\\<br /> &\stackrel{?}= Pr(z|y) Pr(w,x,y)<br /> \end{align}$

But can we say the last line equals Pr(w,x,y,z)? I think this problem can be solved by simply applying Bayes' rule several times, but I can't seem to wrap my head around it.

 

[hikaru1221]

What we need to prove is P(w,x) = P(w)P(x). P(w,x) can be computed easily by integrating (or summing) P(w,x,y,z) over all y and z

 

[endeavor]

Ah, I see. So we have:
$\begin{align}<br /> Pr(x,w) &= \iint Pr(w,x,y,z)\,dydz \\<br /> &= \iint Pr(w)Pr(z|y)Pr(y|x,w)Pr(x)\,dydz \\<br /> &= Pr(w)Pr(x) \iint Pr(z|y)Pr(y|x,w)\,dydz \\<br /> &= Pr(w)Pr(x) <br /> \end{align}$