How can I get the negation of $\exists!$ (unique existential quantification)?

Question

How can I get the negation of $\exists!$ (unique existential quantification)? if it's $\forall$, So if I wanna re-negate the last one, I'll get $\exists$ but it's not the same as what we started with! Did I do something wrong here?

For example: $$P : \exists! x\in \mathbb{R} \text{ such that } x^2 = 0$$ it means $\exists x \in \mathbb{R}\text{ such that } x^2 = 0\wedge x $ is unique, So the negation is $\forall x \in \mathbb{R}\space x^2 \ne 0 \vee x \text{ isn't unique}$ is this a false statement?

Answer 1

I think the best thing to do is unpack what $\exists !$ means.

$\exists ! x \phi(x)$ is shorthand for $$\exists x (\phi(x) \wedge \forall y (\phi(y) \rightarrow y=x))$$

so negating this gives $$\forall x(\neg \phi(x) \vee \exists y(\phi(y) \wedge y \ne x))$$ which is to say: either no $x$ satisfies $\phi$ or there is a $y$ distinct from $x$ which satisfies $\phi$.

This is of the form $\forall x (\neg A \vee B)$, which is the same as $\forall x(A \to B)$, so we could write $$\forall x(\phi(x) \to \exists y(\phi(y) \wedge y \ne x))$$ which is to say: if $x$ satisfies $\phi$ then there is a $y$ distinct from $x$ which does too. This is closer to how I'd think intuitively about the negation of $\exists!$.

Unfortunately there isn't a very succinct way of writing it.

Answer 2

Take $\exists! x : E(x)$ if you want to negate that, you have to take $\exists x : E(x) \Rightarrow \exists y\neq x : E(y)$.

Answer 3

The symbol $\exists !$ means "there exists a unique", and is not really a unit, it carries two conditions: existence and uniqueness. The negation of $A$ and $B$ is not $A$ or not $B$, in symbols:

$$\lnot (A \wedge B) = \lnot A \vee \lnot B.$$

The negation of "there exists" is "there does not exist". The uniqueness assumes existence, and its negation is plurality. So it seems to me that the negation is

There does not exist, or there exist many, such that...

Answer 4

Just write it out, unfold definitions and keep pushing the negation inwards

$\neg \exists ! x, P(x)$

$\neg \exists x, (P(x) \wedge (\forall y, P(y) \to y=x))$

$\forall x, \neg (P(x) \wedge (\forall y, P(y) \to y=x))$

$\forall x, \neg P(x) \vee \neg (\forall y, P(y) \to y=x)$

$\forall x, \neg P(x) \vee \exists y, \neg (P(y) \to y=x)$

$\forall x, \neg P(x) \vee \exists y, P(y) \wedge y\not = x$

You can read this as saying that for each $x$ we must not have $P(x)$ or there's some $y$ satisfying $P(y)$ which isn't $x$. So that could happen if there's absolutely nothing satisfying $P(x)$, or if there's multiple things satisfying it.