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Negating uniqueness

  1. Oct 18, 2005 #1
    I think I've got this one, I'd just like someone to check my work
    Negate the statement [tex] (\exists! x \in S) P(x) [/tex]

    Since [tex] (\exists ! x \in S) P(x) \Longleftrightarrow \{(\exists x \in S) (P(x) \} \wedge \{(\forall x,y \in S) [P(x) \wedge P(y) \longrightarrow x = y \} [/tex]

    The negation would be [tex] \sim (\exists ! x \in S) P(x) \Longleftrightarrow \{(\forall x \in S) \sim P(x)\} \wedge \{\exists x,y \in S) \sim [P(x) \wedge P(y) \longrightarrow x = y \} [/tex]

    Does this look correct?
    Last edited: Oct 18, 2005
  2. jcsd
  3. Oct 20, 2005 #2


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    Flip something upside down.
    [tex]\neg(P \wedge Q) \Leftrightarrow (\neg P \vee \neg Q)[/tex]
  4. Oct 21, 2005 #3


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    Sorry, my comment may not have been clear, and I read that you're teaching yourself (:cool:), so I'll try to explain just to be safe. (I use grouping symbols a bit differently; I think it makes more sense this way.)

    [tex](\exists ! x \in S)(Px)[/tex]

    has two parts, existence and uniqueness.

    [tex]\mbox{Existence: } (\exists x \in S)(Px)[/tex]
    [tex]\mbox{Uniqueness: } (\forall x, y \in S)((Px \wedge Py) \rightarrow (x = y))[/tex]

    Call the existence statement E and the uniqueness statement U. The right-hand side of your definition of [itex](\exists ! x \in S)(Px)[/itex] can then be stated as

    [tex](E \wedge U)[/tex]

    And its negation is

    [tex](\neg[E \wedge U]) \Leftrightarrow (\neg E \vee \neg U)[/tex]

    So just plug back in the definitions of E and U to get the right-hand side of [itex]\neg[(\exists ! x \in S)(Px)][/itex].

    [tex](\neg[(\exists x \in S)(Px)]) \ \vee \ (\neg[(\forall x, y \in S)((Px \wedge Py) \rightarrow (x = y))])[/tex]

    which you can further simplify by "distributing" the negations. Here are some handy http://people.cornell.edu/pages/ps92/414/LogicalOpLogicQuantifiers.pdf [Broken] (PDF); see page 4 for a shorter way to write a statement of unique existence.
    Last edited by a moderator: May 2, 2017
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