Order relation on two Disjunct sets For the following relation we wish to prove that it is an order relation, thus we wish to prove reflexivity, transitivity and anti-symmetry.
Suppose $ (A,\sqsubseteq_A)$  and $ (B,\sqsubseteq_B)$ are partially ordered sets. If $A$ and $B$ are disjunct, we define the new relation $\sqsubseteq$ on $A\cup B$ as follows:
$x \sqsubseteq y$  if
($x,y \in A$ and $x \sqsubseteq_A y$
or $x,y \in B$ and $x \sqsubseteq_B y$)
or if ( $x \in A$ and $y \in B$.)
I figured that for reflexivity we state this for a disjunct set $A$ and $B$ so they do not have elements in common, so $a\in A$ and $a\in B$ is always false. Moreover if $a,a \in A$ we know since $ (A,\sqsubseteq_A)$  is a poset, $ a\sqsubseteq_A a$. similar argument when in B.
 Thus $ a\sqsubseteq a $ is always  true. But how do I go about the other two properties? It feels a bit... tricky
 A: (Anti-Symmetry)
Assume $x\leq y$ and $y\leq x$
Then 
(1) $x,y\in A$ or 
(2) $x,y\in B$ 
and it can not be that 
(3) $x\in A$ and $y\in B$ or 
(4) $x\in B$ and  $y\in A$

Assume (1) $x,y\in A$
Then $x=y$ since $\leq_A$ is a partial order.

Assume (2) $x,y\in B$
Then $x=y$ since $\leq_B$ is a partial order.

(3) is impossible since then $x\in A$ and $x\in B$ contradicts $A$ and $B$ are disjunct
(4) is impossible (same reason as for (3))

(Transitivity)
Assume $x\leq y$ and $y\leq z$
Then 
(1) $x,y,z\in A$ or
(2) $x,y,z\in B$ or
(3) $x\in A$ and $y,z\in B$ or
(4) $x,y\in A$ and $z\in B$
and it can not be that 
(5) $x,z\in A$ and $y\in B$ or
(6) $x,z\in B$ and $y\in A$ or
(7) $x\in B$ and $y,z\in A$ or
(8) $x,y\in B$ and $z\in A$

Assume (1) $x,y,z\in A$
Then $x\leq z$ since $\leq_A$ is a partial order

Assume (2) $x,y,z\in B$
Then $x\leq z$ since $\leq_B$ is a partial order

Assume (3) $x\in A$ and $y,z\in B$
Then $x\leq z$ since $x\in A$ and $z\in B$

Assume (4) $x,y\in A$ and $z\in B$
Then $x\leq z$ since $x\in A$ and $z\in B$

(5) or (6) or (7) or (8) are impossible since this would contardict that $A$ and $B$ are disjunct
