Suppose $R_1$ is a partial order on $A_1$, $R_2$ is a partial order on $A_2$, and $A_1 \cap A_2 = \emptyset$.
Prove that $R_1 \cup R_2 \cup (A_1 \times A_2)$ is antisymmetric on $A_1 \cup A_2$.
Proof: Suppose $(x,y) \in R_1 \cup R_2 \cup (A_1 \times A_2) $ and $(y,x) \in R_1 \cup R_2 \cup (A_1 \times A_2) $ with $x,y \in A_1 \cup A_2$.
Case $(x,y)\in R_1 \cup R_2$:
SubCase1: $(y,x) \in R_1 \cup R_2 $. Because $(x,y)\in R_1 \cup R_2$ and since we know $R_1 \cup R_2$ is antisymmetric, it follows that x=y.
SubCase2: $(y,x) \in (A_1 \times A_2)$. Then $y \in A_1$ and $x \in A_2$. Because $A_1 \cap A_2 = \emptyset$, $y \notin A_2$ and $x \notin A_1$.
At this point , I don't know how to show if given $(y,x) \in (A_1 \times A_2)$ and $(x,y)\in R_1 \cup R_2$ hold, then x=y, if it's even possible to show.