# Negating uniqueness

1. Oct 18, 2005

### Jeff Ford

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?

Last edited: Oct 18, 2005
2. Oct 20, 2005

### honestrosewater

Flip something upside down.
$$\neg(P \wedge Q) \Leftrightarrow (\neg P \vee \neg Q)$$

3. Oct 21, 2005

### honestrosewater

Sorry, my comment may not have been clear, and I read that you're teaching yourself (), 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 [Broken] (PDF); see page 4 for a shorter way to write a statement of unique existence.

Last edited by a moderator: May 2, 2017