I want to prove the following theorem and have some questions and doubts about it:
Let $(A_1, R_1)$ and $(A_2, R_2)$ be two well-ordered posets with $A_1$ and $A_2$ disjoint. Now, consider the new relation $R$ defined on $A_1 \cup A_2$ defined as follows: $xRy$ if $x,y \in A_1$ and $xR_1y$ , or $x,y \in A_2$ and $xR_2y$, or $x \in A_1$ and $x \in A_2$. Prove that $(A_1 \cup A_2, R)$ is a poset and that it is well-ordered.
First of all, it's really strange how is that possible to have $x \in A_1$ and $x \in A_2$ if $A_1 \cap A_2 = \emptyset$. I first thought it was a typo and it had to be like $x \in A_1$ and $y \in A_2$. But in that case it seems that the relation can't be antisymmetric. Therefore I thought that it was like "always-false" preposition and just ignored it. It actually allowed me to prove that the relation is a partial order on $A_1 \cup A_2$.
But now I'm wondering how it can be a well-order if we can have a set $S = \{ x, y \}$, where $x \in A_1 $ and $y \in A_2$. Since $S \subseteq A_1 \cup A_2 $ and if I show that $S$ is not empty, then we can see that $S$ doesn't have the least element in respect to $R$, so that $R$ can't be a well-order.
Could someone please tell me what I'm missing here? Or is there really something wrong with the theorem?
UPDATE:
There is a typo in the theorem, so that it should be defined the $R$ as follows:
$xRy$ if $x,y \in A_1$ and $xR_1y$ , or $x,y \in A_2$ and $xR_2y$, or $x \in A_1$ and $y \in A_2$.
And my assumptions that the relation defined in this way can't be antisymmetric was wrong.