Proof of a simple logic statement

[soopo]

Homework Statement

\forall x \in S <-> \exists x \not \in S

The Attempt at a Solution

The statement is clearly false.
I will try to show that by the proof of contradiction.

Let
P: \forall x \in S
and
Q: \exists x \not\in S

The negation of Q is
negQ: \forall x \not\in S
and the negation of P is
negP: \exists x \in S.

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?

 

[CompuChip]

Your statement is a bit odd.
"For all x in S, is equivalent to there exists an x such that there exists an S"
What is on the right hand side? What does x exist in and there exist x and S such that what?In first-order logic, your statement is not a well-formed formula

 

[soopo]

"For all x in S, is equivalent to there exists an x such that there exists an S"
What is on the right hand side? What does x exist in and there exist x and S such that what?

The statement should be "For all x in S, is equivalent to there does not exist an x in S."

The rest of the proof is also now as I want.

 

[soopo]

Your statement is a bit odd.
"For all x in S, is equivalent to there exists an x such that there exists an S"
What is on the right hand side? What does x exist in and there exist x and S such that what?


In first-order logic, your statement is not a well-formed formula

I agree with you. The statement is still false.

Perhaps, the initial statement should be
\forall x \in S <=> \neg(\exists x \not \in S)

It means that: all x in S is equivalent with the negation of the statement that
there exists x not in S.