How to determine the truth set of a conditional predicate "If P(x), then Q(x)"? My question is related to the truth table of the conditional operator "implies". i.e. "$r: p \to q$".
\begin{array}{c c | c}
p & q & p \to q\\ 
\hline
T & T & T\\
T & F & F\\
F & T & T\\
F & F & T
\end{array}
Consider this question in an example: 
Let $Q(x,y)$ be the predicate "If $x<y$, then $x^2<y^2$" with both $x$ and $y$ real numbers. Find the truth set of $Q(x,y)$.
Should I consider the cases where "$x>y$" and "$x=y$"? Should these pairs be included in the truth set? It confuses me because according to the truth table or definition of conditional operator, the statement "If p, then q" is true when the proposition p is false. However to my mathematical intuition whenever we provide a condition, it means we are talking about q assuming p is correct.
 A: EDITED
The truth set is  
$$\{(x,y):x<y,|x|\ge|y|\}^c=\left[\{(x,y):x<y\}\cap \{(x,y):|x|\ge|y|\} \right]^c=$$
$$= \{(x,y):x\ge y\}\cup \{(x,y):|x|<|y|\}.$$
That is, the truth set is the complement of the set where the implication is false according to the truth table. ($p(x,y)$ is true and $q(x,y)$ is false; $x<y$ and $x^2\ge y^2$.) 
It may be surprising that the truth set is simply the union of the sets where $p(x,y)$ is false and $q(x,y)$ is true.
It is easy to check all the possible combinations in the figure below. (Color red indicates that the premise is false, that is, there is no need to further investigation.)

If this example goes against your intuition then think of the following implication: Let $R=$ "it is raining." $W=$ "the grass is wet".
$$R \rightarrow W.$$
Is this false just because it is not raining right now?
It is worth to take a look at this question. (You will find there further explanation regarding the naive set theoretic version of implication of $A \rightarrow B=A^c\cup B$)
