Elementary proof technique used in set theory. Let A,B be sets. Prove that  if $x \notin B$ and $A\subseteq B$, then $x \notin A$.
I'm just wondering if I can prove it like this:
$x\notin B \to x\notin A$ is the counter-positive version of:
If $x\in A \to x\in B$. Since $A \subseteq B$, $\forall x\in A, x\in B$.
This proves that $x\notin B \to x\notin A$. Is this sufficient?
 A: $x \in A \subseteq B$ is equivalent to $x \in A \Rightarrow x \in B$. Contrapositive is $x \notin B \Rightarrow x \notin A$.
A: The contrapositive of $P\implies Q$ is indeed $\bar Q\implies \bar P$
The two propositions are logically equivalent: $(\bar Q\implies \bar P)\equiv(\bar{\bar Q}\lor \bar P)\equiv(Q\lor \bar P)\equiv(P\implies Q)$

But we can wonder, giving a context $C$, if the contrapositive still holds under this form:


*

*Do we have $\left((C\land P)\implies Q\right)\equiv\left((C\land \bar Q)\implies \bar P\right)$ ?



In fact it is easier to work with implications, since :
$\left(C\implies(P\implies Q)\right)\equiv \left(C\implies(\bar Q\implies\bar P)\right)$
Although
$\begin{array}{ll}
\left((C\land P)\implies Q\right)
&\equiv\left(\overline{(C\land P)}\lor Q\right)\\
&\equiv\left((\bar C\lor \bar P)\lor Q\right)\\
&\equiv\left(\bar C\lor(\bar P\lor Q)\right)\\
&\equiv\left(C\implies(P\implies Q)\right)
\end{array}$
Thus the expected "extended" contraposition with context holds.

Now coming back to your problem, we effectively have
$\left((A\subset B\ \ \land\ \ x\in A)\implies x\in B\right)\quad\equiv\quad\left((A\subset B\ \ \land\ \ x\notin B)\implies x\notin A\right)$
Actually, we often do this "factorization" of the context intuitively, and with reasons since it is true.
