Prove that if $A \setminus B ⊆ C ∩ D$ and $x \in A$ and $x \not \in D$ then $x \in B$

Question

Suppose $A \setminus B ⊆ C ∩ D$ and $x \in A$. Prove that if $x \not \in D$ then $x \in B$.

We can rewrite $A \setminus B ⊆ C ∩ D$ using logical connectives as follows: $(x \in A \land x \not \in B) \implies (x \in C \land x \in D)$

It can be seen that if $x \in A$ and $x \not \in B$ then x must be in $D$.

Given that $x \not \in D$. In this case, either $x \not \in A$ or $x \in B$.

Given that $x \in A$. Therefore, $x \in B$.

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Apparently, more complete way to describe statement $A \setminus B ⊆ C ∩ D$ would be $\forall x \big((x \in A \land x \not \in B) \implies (x \in C \land x \in D)\big)$

Is it OK to omit quantifier ($\forall$) like I did? And in general, is the proof above correct? Any suggestions for improvement will be welcome!

Answer 1

The proof is correct, there's no need to improve anything unless you want to be very formal. You can omit the quantifier $\forall$, because if something is true for all $x$, then it's particularily true for the one $x$ that you're given.

Answer 2

If $x \notin D$ then $x \notin C \cap D$, so $x \notin A\setminus B$ by the given inclusion. But we already know $x \in A$ so $x \notin B$ would make it an element of $A \setminus B$ which it is not, so $x \in B$ must hold.