(adsbygoogle = window.adsbygoogle || []).push({}); 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?

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# Homework Help: Proof of a simple logic statement

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