Trying to understand question in exercise about relations I'm working on my own through How to Prove It by Daniel J. Velleman and I am trying to understand what is being asked for exercise 9 in section 4.4:
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: $L = \{ ((a,b),(a',b')) \in (A \times B) \times (A \times B) ~ | ~ aRa', ~ \text{and if} ~ a = a' \text{then} ~  bSb'\}$.
Specifically, I am trying to understand the definition of the relation $L$.
Must $a = a'$ for any pair of ordered pairs to belong to $L$?
As a specific example, let $A = \{1, 2\}$ and $B = \{3, 4\}$. Therefore, $A \times B = \{(1,3), (1,4), (2,3), (2,4)\}$. Let $R = \{(1,1), (2,2), (1,2) \}$ and $B = \{ (3,3), (4,4), (3,4) \}$.
Now in this specific example is $L = \{ ((1,3), (1,3)), ~((1,4),(1,4)), ~ ((1,3),(1,4)), ~ ((2,3),(2,3)), ~ ((2,4),(2,4)), ~ ((2,3),(2,4))\}$?

Update: Thanks everyone for the helpful replies to my question.
A little more context to my question:
One of the reasons I am having difficulty with the definition of $L$ is when trying to use it when showing that $L$ is a partial order on $A \times B$.
The definition of $L$ has a logical form of $P \land (O \implies Q)$, where $P$ is $aRa'$, $O$ is $a = a$, and $Q$ is $bSb'$.
Now to show, for example, that $L$ is a reflexive relation on $A \times B$ we must show that $\forall (a, b) \in A \times B ((a,b),(a,b)) \in L$. To do this, let $(a,b)$ be arbitrary and suppose $(a,b) \in A \times B$. Thus, $a \in A$ and $aRa$. This shows the $P$ part of $P \land (O \implies Q)$ from above.
Now we must show that $a = a \implies bSb$. The method we are shown to do this in the book is to assume the antecedent and prove the consequent. So assume $a = a$, but this does not tell us anything about $bSb$. So I'm not sure what to do from here.
Perhaps I should open a new question for the stuff in the update part of this question?

After reviewing the comments below and thinking about it more here is another attempt to show that $L$ is a reflexive relation on $A \times B$:
Let $(a,b)$ be arbitrary and suppose $(a,b) \in A \times B$. Thus $a \in A$ and because $R$ is a partial order on $A$, then $aRa$. Now assume $a = a$. We know that since $(a,b) \in A \times B$ then $b \in B$. Since $S$ is a partial order on $B$, then $bSb$. Therefore, if $a = a$ then $bSb$. Since $aRa$ and if $a = a$ then $bSb$, then $((a,b),(a,b)) \in L$. Since $(a,b)$ was arbitrary we can conclude $L$ is a reflexive relation on $A \times B$. $\square$
 A: Let $x := ((a,b)\times(a',b'))$. Whether $x \in L$, depends on whether $aRa'$ and whether $bSb'$.  There are 4 possibilities for the latter two. Let's make a table.
$\begin{array}{r|cc}
& bSb' & \lnot bSb' \\ \hline
aRa' & & \\
\lnot aRa'
\end{array}$
The definition for $L$ is an and condition which starts with $aRa'$, so we can exclude the $\lnot aRa'$ cases.
$\begin{array}{r|cc}
& bSb' & \lnot bSb' \\ \hline
aRa' &  & \\
\lnot aRa' & x \notin L & x \notin L
\end{array}$
Further, the definition for $L$ states that in the case where $aRa'$, we also need to consider whether $a=a'$, so we need to split the $aRa'$ case.
$\begin{array}{r|cc}
& bSb' & \lnot bSb' \\ \hline
a=a',aRa' &  & \\
a \ne a',aRa' &  & \\
\lnot aRa' & x \notin L & x \notin L
\end{array}$
The second part of the conditional says if $a=a'$ then $bSb'$, so we can fill in the top row.
$\begin{array}{r|cc}
& bSb' & \lnot bSb' \\ \hline
a=a',aRa' & x\in L & x \notin L \\
a \ne a',aRa' &  & \\
\lnot aRa' & x \notin L & x \notin L
\end{array}$
Finally, the if conditional is true when the hypothesis is false, that is when $a \ne a'$, so $aRa'$ makes the first half of the conditional true and $a \ne a'$ makes  the second half true, and we finish out the table.
$\begin{array}{r|cc}
& bSb' & \lnot bSb' \\ \hline
a=a',aRa' & x\in L & x \notin L \\
a \ne a',aRa' & x\in L & x\in L \\
\lnot aRa' & x \notin L & x \notin L
\end{array}$
A: I would interpret it as follows: The relation goes as follows ((a,b),(a',b')) where aRa', except if a= a'. If a= a', then only the following relations exist ((a,b), (a',b')) where aRa' and bSb'.
A: 
Must =′ for any pair of ordered pairs to belong to ?

No. You can have pairs $(a,b)\times (a',b')$ where $a\neq a'$ as long as $aRa'$.
Using your example, since $(1,2)\in R$, then (1,3)x(2,4)$\in L$.
