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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 $$\neg (\exists !) \neg$$

???

Asaf Karagila
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goblin GONE
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3 Answers3

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I suppose you could define $\forall!$ by $$\forall!x \psi(x) \leftrightarrow \forall x[\psi(x) \vee \exists y(\neg \psi(y) \wedge y \ne x)]$$ ("Either $\psi$ holds here or it fails somewhere else too" / "Either $\psi$ is true everywhere or there are at least two counterexamples")

Then $\exists ! = \neg \forall ! \neg$ and $\forall ! = \neg \exists ! \neg$ in the same sense that $\exists = \neg \forall \neg$ and so on: really $\exists = \neg \forall \neg$ is just shorthand for $\exists x \phi(x) \leftrightarrow \neg \forall x \neg \phi(x)$.

Why? Well, $\exists ! x\ \phi(x)$ is shorthand for $$\exists x [\phi(x) \wedge \forall y (\phi(y) \to y=x)]$$ Dualising the quantifiers gives $$\neg \forall x \neg [\phi(x) \wedge \neg \exists y\neg(\phi(y) \to y=x)]$$ which is equivalent to $$\neg\forall x[\neg \phi(x) \vee \exists y (\phi(y) \wedge y \ne x)]$$ So we have $$\exists ! x \phi(x) \leftrightarrow \neg \forall ! x \neg \phi(x)$$ And so on.

Clive Newstead
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    So basically, $\forall ! x \phi(x)$ is saying that, okay, either $\phi(x)$ holds for all $x$, or it fails for at least two distinct $x$? – goblin GONE May 10 '13 at 15:30
  • Yep.${}{}{}{}{}$ – Clive Newstead May 10 '13 at 15:31
  • Hmmmmm. How would one actually say $\forall ! x \phi(x)$? The sentence "For all unique x we have $\phi(x)$" is clearly incorrect. – goblin GONE May 10 '13 at 15:33
  • I don't think it has a snazzy name like $\exists !$ does - I can't see it being much use other than to observe that it's dual to $\exists !$. (I'll warn you that I made up the notation so it might clash with another established notation.) – Clive Newstead May 10 '13 at 15:35
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    Maybe you could read $\forall ! x \psi(x)$ as "there is not a unique counterexample to $\psi$", though that's almost the same as reading $\neg \exists ! x \neg \psi(x)$ out loud. – Clive Newstead May 10 '13 at 15:37
  • Thanks. I'll let you know if I find a more direct way of saying it. – goblin GONE May 10 '13 at 15:41
  • Somethings's wrong here, for example the string doesn't say "or there are at least two counterexamples" as in the $\forall x$ expression, $\psi$ is only negated for the thing which isn't $x$. – Nikolaj-K Sep 10 '13 at 15:13
  • @NickKidman: For given $x$ we have $\psi(x) \vee \exists y (\neg \psi(y) \wedge y \ne x)$. Thus if there is some $x$ such that $\psi(x)$ fails, then this expression means that there is also a $y$ distinct from $x$ such that $\psi(y)$ fails. Another way of writing the expression is $$\forall x(\neg \psi(x) \to \exists y(\neg \psi(y) \wedge y \ne x))$$ which holds by the law $(A \to B) \leftrightarrow (\neg A \vee B)$. Thus if there is one counterexample to $\psi$ then there is another one too. – Clive Newstead Sep 10 '13 at 15:50
  • @CliveNewstead: You're right. – Nikolaj-K Sep 10 '13 at 17:00
  • I agree; $\lnot \exists ! \lnot$ is $\forall x \psi(x)\lor \exists y(\lnot \psi(y)\land \lnot y=x)$, i.e. $\exists x \lnot \psi(x) \rightarrow \exists y(\lnot \psi(y)\land \lnot y=x)$. So its reading must be : if there is something such that $\psi(x)$ fails, then that thing is not unique. – Mauro ALLEGRANZA Jan 08 '14 at 15:38
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    "Einmal ist keinmal." – moteutsch Aug 09 '17 at 23:17
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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 $| \{ x : \neg \phi (x) \} | \neq 1$.
user642796
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If you have a formula $\phi$ for which you proved that $\exists ! x, \phi(x)$ (i.e. $\exists x, \phi(x) \land \forall y, \phi(y) \implies y=x$), then you have :

$(\exists x. \phi(x) \land \psi(x)) \iff (\forall x. \phi(x) \implies \psi(x))$, and so the connective $\exists x.\phi(x) \land \cdot$ is equivalent to its dual $\forall x. \phi(x) \implies \cdot$

So the "let $x$ be the unique object satisfying $\phi$ in ..." connective (you could write it $\nabla_\phi x. \psi(x)$) is self-dual, and behaves rather "neutrally". For example it "commutes" with almost everything : you can push it inside of or pull it out of formulas easily (as long as you don't break $\phi$ by pulling out of quantifiers over variables appearing in $\phi$)

mercio
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