When there is a proposition $(P\rightarrow Q)$, which row in the truth table of $\rightarrow $ should I use? I solved one question in a book of analysis, and although I used an informal method to check it, I'd like to know more about what should be done. The question was the following:

  
*
  
*$A\subset X$ and $ B \subset X$;
  
*If $A\subset Y$ and $ B\subset Y$, then $X\subset Y$.
Prove that $X=A\cup B$.

Looking at the second line, we have an implication. If $Q$, then $P$. And it's truth table is the following:
$$\begin{matrix}
{P}&{Q}&{}&{P\rightarrow Q}\\ 
{0}&{0}&{}&{1}\\ 
{0}&{1}&{}&{1}\\ 
{1}&{0}&{}&{0}\\ 
{1}&{1}&{}&{1}
\end{matrix}$$
Then $Q\equiv A\subset Y, B\subset Y$ and $P\equiv X\subset Y$, to answer it, what line should I take from that truth table? I did by thinking that $P=1$ and $Q=1$, but there are other choices too. Is it possible to obtain the same answer with any choices of $P$ and $Q$? I've been thinking in the other ways but assuming that $P=1=Q$ seems more natural. I guess that it should be valid for every row, but I'm not sure.
 A: If I understand you correctly, you want to know the following:

I want to prove
$$(A\subset X \land B \subset X) \land ((A\subset Y \land B\subset Y) \rightarrow X\subset Y)\implies X=A\cup B $$
since the proposition $P \implies Q$ is formally true in the cases
$$\begin{matrix}
{P}&{Q}&{}&{P\implies Q}\\ 
{0}&{0}&{}&{1}\\ 
{0}&{1}&{}&{1}\\ 
{1}&{1}&{}&{1}
\end{matrix}$$
If I take $$P=(A\subset X \land B \subset X) \land ((A\subset Y \land B\subset Y) \rightarrow X\subset Y)$$ $$Q=(X=A\cup B)$$
Can I use any of the cases above?

The answer is no, here is a counter-example:
$A = \{1,2\}$, $B=\{3,4\}$, $X =\{1,2,3,4,5\}$ and $Y=\{2,3,4\}$
we have $A \subset X$, $B \subset X$, $A \not \subset Y$, $B \subset Y$, $X \not \subset Y$ and $X \neq A \cup B$
In this case both $A\subset X \land B \subset X$ and $(A\subset Y \land B\subset Y) \rightarrow X\subset Y$ would be true, so we should be able to prove the implication since we are in one of the cases above, but it doesn't work.
Here is how I explain it:
The proposition $P \implies Q$ is $\textbf{formally}$ true in all the cases above, but when we are trying to prove something we don't interpret it this way, it's more like a "construction", you have to show that $Q$ "depends on" $P$, and this case is only the last one in the table. Think of it as sets, you want to show that $Q \subset P$, it's always true if $Q = \emptyset$, but this one doesn't count.
