Suppose $A\setminus B \subseteq C\cap D$ and $x\in A$. Prove that $x\notin D \rightarrow x\in B$

Question

Suppose $x\in A \land x\notin B$. Then $A\setminus B \subseteq C\cap D$ and thus $x\in D$. Therefore, $x\notin D \rightarrow x\in B$.

I understand the proof when i take the contra positive of what im trying to prove. But I dont see how $x\notin D \rightarrow x\in B$. Lets try to prove it without the contrapositive as i think this will help me see how $x\notin D \rightarrow x\in B$.

Suppose $x\notin D$, then $x\notin C\cap D$. I dont see how we can assume anything about B from here, especially that $x\in B$. What am i missing?

Answer 1

Here is how you would prove with the assumption that $x\notin D$.

Proof: Suppose that $x\notin D$. This implies that $x\notin C\cap D$. Given that $A\setminus B \subseteq C\cap D$, we know that $x\notin A\setminus B$. As if $x\in A\setminus B$, then $x\in C\cap D$. The element $x\notin A\setminus B$ only if $x\notin A$ or $x\in B$. We are already given that $x\in A$. From this we conclude that $x\in B$.

Answer 2

Another possible approach based on the contrapositive. \begin{align*} A\cap B^{c}\subseteq C\cap D & \Longleftrightarrow (C\cap D)^{c}\subseteq(A\cap B^{c})^{c}\\\\ & \Longleftrightarrow C^{c}\cup D^{c}\subseteq A^{c}\cup B \end{align*}

Consequently, one may deduce that \begin{align*} x\not\in D & \Rightarrow x\in D^{c}\\\\ & \Rightarrow x\in C^{c}\cup D^{c}\\\\ & \Rightarrow x\in A^{c}\cup B\\\\ & \Rightarrow (x\in A^{c})\vee(x\in B) \end{align*}

Once $x\in A$, this means that $x\not\in A^{c}$. Therefore $x\in B$, just as desired.

Hopefully this helps!

Answer 3

From $D\ \supseteq\ C\cap D\ \supseteq\ A\setminus B$, we obtain $$A\setminus D\ \subseteq\ A\setminus(C\cap D)\ \subseteq\ A\setminus(A\setminus B)\ =\ A\cap B\ \subseteq\ B.$$ Therefore $$x\in A\ \wedge\ x\notin D \ \Longrightarrow\ x\in A\setminus D \ \Longrightarrow\ x\in B.$$