(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?

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Proof of a simple logic statement

**Physics Forums | Science Articles, Homework Help, Discussion**