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$$ \forall A. \exists a,b. a \neq b \land a,b \in P(A) $$

My intuition tells me it is false, because given $A=\emptyset$, then $P(\emptyset) = \{\emptyset\}$, so $a=b=\emptyset$.
I proceeded to proving it and negated it to:

$$ \exists A. \forall a,b. a = b \lor a,b \notin P(A) $$

And then I ran into trouble, because I can't prove that $\forall a,b. a=b$, nor that $\forall a,b. a,b \notin P(\emptyset)$.

I think I could assume $a,b \in P(A)$, and then prove $a=b$ by contradiction, but I'm curious, why didn't the negation I wrote above work for me?

Asaf Karagila
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MasterMastic
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1 Answers1

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This is what happens when you abbreviate.

Recall that $a,b\in\mathcal P(A)$ is really a shorthand for $a\in\mathcal P(A)\land b\in\mathcal P(A)$. I think that now you can finish the negation properly.

Asaf Karagila
  • 393,674
  • Could you go a bit further? I wind up with needing to prove $\forall a,b. a \notin P(\emptyset) \lor b \notin P(\emptyset)$ instead, and it still doesn't seems possible. Is it? – MasterMastic Nov 25 '14 at 19:45
  • You forgot the part the subformula $a=b$ there, which will solve this problem. – Asaf Karagila Nov 25 '14 at 20:10
  • I didn't see it relevant, because the negation I have is $\exists A. \forall a,b. a=b \lor (a \notin P(A) \lor b \notin P(A))$. I can't prove $\forall a,b. a=b$, so I disregarded it. The negated statement I wrote here does not get me closer. I would love to get some more. – MasterMastic Nov 25 '14 at 23:23
  • Can you prove that given any two sets, they are either equal, or at least one of them is not the empty set? – Asaf Karagila Nov 26 '14 at 04:12