# Negating the uniqueness quantifier

I am trying to negate ##\exists ! x P(x)##, which expanded means ##\exists x (P(x) \wedge \forall y (P(y) \rightarrow y=x))##. The negation of this is ##\forall x (\neg P(x) \lor \exists y (P(y) \wedge y \ne x))##. How can this be interpreted in natural language? Is it logically equivalent to the statement that either 0 or more than 1 value of x satisfies P(x)?

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I am trying to negate ##\exists ! x P(x)##, which expanded means ##\exists x (P(x) \wedge \forall y (P(y) \rightarrow y=x))##. The negation of this is ##\forall x (\neg P(x) \lor \exists y (P(y) \wedge y \ne x))##. How can this be interpreted in natural language? Is it logically equivalent to the statement that either 0 or more than 1 value of x satisfies P(x)?
Yes, because there exists exactly one are two statements: existence and uniqueness, so both can be violated in the negation. This means either none exists at all or if, then more than one.

There is exactly one odd prime. is wrong, so the contrary is true: there are either more than one odd primes (true) or all primes are even (wrong). But true or wrong is true.

There is exactly one even prime greater than ##2##. is wrong, so the contrary is true: there are either more than one even prime greater than ##2## (wrong) or all primes greater than ##2## are odd (true). Again wrong or true is true.

The case true and true cannot occur, since either there are many or none at all, which cannot both be true. So the negation of ##\exists !## can be expressed by an exclusive or, sometimes noted as ##\dot{\vee}##.

Math_QED