This is an exercise from Velleman's "How To Prove It":
Suppose $R$ is a partial order on $A$ and $S$ is a partial order on $B$. Define a relation $L$ on $A \times B$ as follows: $$L = \{((a,b), (a',b')) \in (A \times B) \times (A \times B) | aRa' \wedge (a=a' \rightarrow bSb')\}$$ Show that $L$ is a partial order on $A \times B$. If both $R$ and $S$ are total orders, will $L$ also be a total order?
I was able to prove that that $L$ is a partial order, but I just want to make sure that my proof for the second part is correct.
Proof: Let $(a,b)$ and $(a',b')$ be arbitrary elements of $A \times B$. If $((a,b), (a',b')) \in L$, then we are done. So now suppose that $((a,b), (a',b')) \notin L$. This means that either $(a, a') \notin R$ or $a = a' \wedge (b, b') \notin S$. We consider these two cases separately.
Case 1. $(a, a') \notin R$. Since $R$ is a total order on $A$, either $aRa'$ or $a'Ra$. Since $(a, a') \notin R$, $a'Ra$. Since $R$ is reflexive and $(a, a') \notin R$, we must have $a \neq a'$. It immediately follows that $a' = a \rightarrow b'Sb$. Therefore, $((a',b'),(a,b)) \in L$.
Case 2. $a = a' \wedge (b,b') \notin S$. Since $S$ is a total order on $B$ and $b,b' \in B$, $bSb'$ or $b'Sb$. Since $(b, b') \notin S$, we have $b'Sb$. Therefore, $a'=a \rightarrow b'Sb$. Since $a=a'$ and $R$ is reflexive, $(a', a) \in R$. So it follows that $((a', b'), (a,b)) \in L$.
In either case, $((a',b'),(a,b)) \in L$. Since $(a,b)$ and $(a', b')$ were arbitrary, $L$ is a total order on $A \times B$. $\square$