Negate Uniqueness: Check Correctness

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The discussion focuses on the correct negation of the statement "there exists a unique x in S such that P(x)." The original statement is broken down into two components: existence and uniqueness. The proposed negation is examined, leading to the conclusion that it can be expressed as the disjunction of the negation of existence and the negation of uniqueness. A clarification is provided on how to properly apply logical operations to achieve the correct form. The conversation emphasizes the importance of understanding the structure of logical statements for accurate negation.
Jeff Ford
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I think I've got this one, I'd just like someone to check my work
Negate the statement (\exists! x \in S) P(x)

Since (\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 \}

The negation would be \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 \}

Does this look correct?
 
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Jeff Ford said:
I think I've got this one, I'd just like someone to check my work
Negate the statement (\exists! x \in S) P(x)
Since (\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 \}
The negation would be \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 \}
Does this look correct?
Flip something upside down.
\neg(P \wedge Q) \Leftrightarrow (\neg P \vee \neg Q)
 
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.)

(\exists ! x \in S)(Px)

has two parts, existence and uniqueness.

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

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

(E \wedge U)

And its negation is

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

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

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

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