TITLE: What is dual to "There exists unique?"
QUESTION [21 upvotes]: I know that "for all" $(\forall)$ and "there exists" $(\exists)$ are dual, in the sense that $$\neg \forall \neg = \exists,\quad \neg \exists \neg = \forall$$
What is dual to "there exists unique"? In other words, how should we interpret $$\neg (\exists !) \neg$$
???

REPLY [10 votes]: This answer doesn't really differ from Clive Newstead's answer, but arrives at essentially the same place via a slightly different approach.
Note that 

$( \exists x ) ( \phi (x) )$ holds iff $\{ x : \phi (x) \} \neq \varnothing$;
$( \forall x ) ( \phi (x) )$ holds iff $\{ x : \neg \phi (x) \} = \varnothing$.

Another set of dual quantifiers are $\exists^\infty$ and $\forall^\infty$ which satisfy

$( \exists^\infty x ) ( \phi (x) )$ holds iff $\{ x : \phi (x) \}$ is infinite (i.e., is not finite);
$( \forall^\infty x ) ( \phi (x) )$ holds iff $\{ x : \neg \phi (x) \}$ is finite.

So by analogy, since, 

$( \exists ! x )( \phi (x) )$ holds iff $| \{ x : \phi (x) \} | = 1$

this would imply that the natural definition of $\forall !$ would be

$ ( \forall ! x ) ( \phi (x) )$ holds iff $| \{ x : \neg \phi (x) \} | \neq 1$.