Factorization of conditional expectation I have to proof this. And I have no idea how to do it.

Let (X,U) and (Y,V) independent random vectors, s.t. X and Y are real valued r.v.s with finite expectations. Show that then:

E[X.Y|(U,V)]=E[X|U]E[Y|V]

I would really appreciate it, If any body could also give me an example for it that I would understand this better. Thank you.
Let's call the event $(U,V)=(H,K) $ as A, the event $U=H$ as B and the event $V=K$ as C, we have:

$E[XY|(U,V)=(H,K)]=E[XY|A]=\frac{E[XY \cdot 1_{A}]}{P(A)}=\frac{E[XY \cdot 1_B \cdot 1_C]}{P(B)P(C)}$ because $U,V$ are independent. Then because the two vectors are independent:

$\frac{E[XY \cdot 1_B \cdot 1_C]}{P(B)P(C)}=\frac{E[X \cdot 1_B ]E[Y \cdot 1_C]}{P(B)P(C)}=\frac{E[X \cdot 1_B ]}{P(B)}\frac{E[Y  \cdot 1_C]}{P(C)}=E[X|B]E[Y|C]=E[X|U=H]E[Y|V=K].$ $1_{A}(x)=1, \forall x \in A \; 1_{A}(x)=0, \forall x \notin A.$
Let's call the event $(U,V)=(H,K) $ as A, the event $U=H$ as B and the event $V=K$ as C, we have:

$E[XY|(U,V)=(H,K)]=E[XY|A]=\frac{E[XY \cdot 1_{A}]}{P(A)}=\frac{E[XY \cdot 1_B \cdot 1_C]}{P(B)P(C)}$ because $U,V$ are independent. Then because the two vectors are independent:

$\frac{E[XY \cdot 1_B \cdot 1_C]}{P(B)P(C)}=\frac{E[X \cdot 1_B ]E[Y \cdot 1_C]}{P(B)P(C)}=\frac{E[X \cdot 1_B ]}{P(B)}\frac{E[Y  \cdot 1_C]}{P(C)}=E[X|B]E[Y|C]=E[X|U=H]E[Y|V=K].$ $1_{A}(x)=1, \forall x \in A \; 1_{A}(x)=0, \forall x \notin A.$