Does $x \perp (y,z)$ imply $x \perp y \mid z$? Does $x \perp (y,z)$ imply $x \perp y \mid z$, where $\perp$ denotes stochastic independence?
I was told it is true and the following is the proof (which I believe is wrong):
We want to show that $p(x,y,z) = p(x)p(y,z)$ implies $p(x,y \mid z)=p(x \mid z) p(y \mid z)$.
then:
$$p(x,y \mid z) = \frac{p(x,y,z)}{p(z)} = p(x) \frac{p(y,z)}{p(z)} = p(x)p(y \mid z)$$
QED...Except that, that is not what conditional independence means. The correct conclusion should have arrived at: $p(x \mid z) p(y \mid z)$.
1) Is this proof incorrect? (as I suspect it is)
2) If it is incorrect, I still don't know what the right answer is. I feel its false, but have been unable to produce a counter example. Any ideas anyone?
3) If its true, does someone have an intuitive explanation of why its correct? Or a different mathematical proof that is more intuitive/clear?

The proof is correct and it shows the desired result. 
To see this, sum on $y$ the identity $p(x,y\mid z)=p(x)p(y\mid z)$, valid for every $(x,y,z)$. One gets $p(x\mid z)=\sum\limits_yp(x,y\mid z)=p(x)\sum\limits_yp(y\mid z)=p(x)$ for every $(x,z)$ hence $p(x)=p(x\mid z)$ for every $(x,z)$. 
Thus, the identity $p(x,y\mid z)=p(x)p(y\mid z)$ for every $(x,y,z)$ implies $p(x,y\mid z)=p(x\mid z)p(y\mid z)$ for every $(x,y,z)$, as desired.
