1. The problem statement, all variables and given/known data [tex]\forall x \in S <-> \exists x \not \in S[/tex] 3. The attempt at a solution The statement is clearly false. I will try to show that by the proof of contradiction. Let [tex]P: \forall x \in S[/tex] and [tex]Q: \exists x \not\in S [/tex] The negation of Q is [tex]negQ: \forall x \not\in S [/tex] and the negation of P is [tex]negP: \exists x \in S [/tex]. negQ means that all x are not in S, while negP means that there is one x in S. This is a contradiction. Therefore, the original statement must be false. Is my proof correct?