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I'm trying to express $∃!x : (P(x))$ in a different way. i want to know how to express it with the other quantifiers.

Here is what I have tried:

$∃!x(P(x)) = ∃x : (P(x) \wedge ∀y, y≠x (\neg P(y)))$

What I'm trying to say is that there exists only one $x$ such that $x$ is true and for all $y$ that are not $x$, $y$ is false. $x$ and $y$ share the same universe of discourse. I'm new to representing this mathematically. Any help is appreciated.

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    $\begingroup$ Try to express what you are trying to say better, it doesn't make sense. $\endgroup$
    – Rebellos
    Commented May 5, 2016 at 17:42
  • $\begingroup$ just updated it $\endgroup$
    – Andrew Kor
    Commented May 5, 2016 at 17:42
  • $\begingroup$ $\exists x\forall y\ P(y)\rightarrow (y=x)$ $\endgroup$ Commented May 5, 2016 at 17:50
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    $\begingroup$ @FedericaMaggioni That would also be true if P is never true of any x. $\endgroup$
    – user4894
    Commented May 5, 2016 at 18:00
  • $\begingroup$ @CharalamposFilippatos ∃!x(P(x))=∃x:(P(x)∧∀y,y≠x(¬P(y))) does that make more sense? $\endgroup$
    – Andrew Kor
    Commented May 5, 2016 at 18:28

3 Answers 3

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$$\neg\left(\exists x: P(x) \right)$$ is equivalent to $$\forall x: \neg P(x)$$

Is this what you wanted to show?

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First, there exists some $x$ such that $P(x)$:

$\exists x P(x)$

Then if there is some other $y$ such that $P(y)$, it must be the case that $y = x$:

$\exists x P(x) \land \forall y (P(y) \implies y = x)$

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$$∃!x(P(x)) = [∃x P(x)] \ \wedge ∀x\ \forall y [P(x) \wedge P(y) \implies x=y\ ]. $$ Here the first clause says there is at least one, while the second says there is at most one, $x$ for which $P(x)$ is true.

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