About conditional expectation Can someone give me some hints on the following problem? Many thanks!!

Let $x$, $y$, and $z$ be integrable random variables on a probability space $(X,\Sigma, \mathbb{P})$. Show that if both $x$ and $y$ are independent of $z$, then $\mathbb{E}(x\,|\,y)=_{\text{a.s.}}\mathbb{E}(x\,|\,y,z)$.(Here $\mathbb{E}(x\,|\,y,z)$ stands for $\mathbb{E}(x\,|\,\sigma\{y,z\})$, where $\sigma\{y,z\}$ is the smallest $\sigma$-algebra generated by $y$ and $z$.)
If the hypothesis is that $x$ and $z$ are independent and that $y$ and $z$ are independent, the result does not hold and the usual counterexample applies: assume that $x$ and $y$ are i.i.d. symmetric Bernoulli $\pm1$ random variables and that $z=xy$.

Then $x$ and $y$ are independent hence $E[x\mid y]=E[x]=0$, while $x=yz$ hence $E[x\mid y,z]=x$.
Hint: the conditional expectation is the (almost) unique function satisfying...