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soopo
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Homework Statement
[tex]\forall x \in S <-> \exists x \not \in S[/tex]
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?
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