An exercise asks me to prove the following: $$A\cap B \subseteq \overline{A \triangle B}$$
Unlike most other exercises, this one implies a symmetric difference, of which I am unfamiliar in this kind of proofs. There was little I could do, here:
The statement can be rewritten as the following: $$A\cap B \subseteq \overline{(A-B)\cap (B-A)}$$ $$A\cap B \subseteq \overline{(A-B)}\cap \overline{(B-A)}$$ $$A\cap B \subseteq (\overline{A} - \overline{B}) \cap (\overline{B} - \overline{A})$$ I rewrote it because the symmetric difference doesn't seem "primitive" enough for me to operate with. Then my proof begins: $$x \in A \cap B \implies x\in A \land x \in B$$ $$\implies x \in (A \cap B) \land x \in (B \cap A)$$ $$\implies (x \in A \land x \in B) \land (x \in B \land x \in A)$$ And then, I got stuck. I don't see how can $(x \in A \land x \in B) \land (x \in B \land x \in A)$ become what I needed at all.