Question

what is the negation of ∃!x P(x) ?

Answer 1

There are none, or there are two or more.

Answer 2

\[
[\forall x \quad \lnot P(x)]\lor[\exists x,y \quad x\neq y\land P(x)\land P(y)]
\]

Answer 3

oh ok.

Answer 4

What were you expecting?

Answer 5

I thought it was the only second condition

Answer 6

what about for all x, p(x) ? is this also not ∃!x P(x) ?

Answer 7

That won't work if there is only one \(x\) in our universal set.

Answer 8

:o

Answer 9

And if there were more than one, it would be included by the second condition.

Answer 10

ahh i see. So it wouldn't work either in the case x being the only element in the universal set if the negation is the only the second condition, right?

Answer 11

huhm... nvm. I don't even know what I meant by that :D