# Unique existence quantifier equivalent to what?

1. Jul 25, 2012

### Aziza

According to my book,
$(\exists!x)P(x)$ is equivalent to $(\exists x)P(x)\wedge(\forall y)(\forall z)[P(y)\wedge P(z)\Rightarrow y=z]$

But I don't see why the variable z is necessary. Wouldn't the following also be correct but shorter and easier to understand:

$(\exists x)P(x)\wedge(\forall y)(P(y)\Rightarrow y=x)$

??

2. Jul 25, 2012

### Stephen Tashi

I don't know how the book's notation indicates the scope of quantifiers. Your way requires that $\exists x$ has a longer scope than the book's way:

$(\exists x)\{ P(x) \wedge (\forall y)\{ P(y) \Rightarrow y = x) \} \}$

Your way is also equivalent to unique existence. The book's way is how unique existence is often proven in mathematical systems. For example, to prove the identity element of a Group is unique, one argues that, by definition of a Group, an identity element of the Group exists. Then one shows that if two elements of the Group are both identity elements then they are equal to each other.

3. Jul 28, 2012

### xxxx0xxxx

Because you statement says: for all y, if y is Jack, then Jack is Jill, whereas the correct statement says for all y and for all z, if y is Jack and z is Jack then Jack is Jack.

4. Jul 28, 2012

### micromass

His statement is perfectly fine. The two statements in the OP are equivalent. He only nees to be careful about the scope of the quantifiers.

5. Jul 29, 2012

### Aziza

What does it mean that it has a longer scope? My book didn't talk about scope so far

6. Jul 29, 2012

### Stephen Tashi

A variable such as "x" may mean one thing on one page of a math book or in one function of a computer program and it may mean something entirely different on another page or in another funciton. The "scope" of a quantifier of such as $\exists x$ is, roughtly speaking, the expressions where the 'x' referred to by that quantifier stands for the same thing.

For example, the statement

Everyone is mortal and there exists a person who is happy

could be symbolized as

$\{(\forall x)M(x)\} \wedge \{(\exists x) H(x)\}$

The 'x' in the left hand side of the wedge means something different than the 'x' on the right hand side of the wedge. The "scope" of the $\forall x$ only includes $M(x)$,

7. Jul 29, 2012

### Aziza

ohh ok, so that's why you put the extra brackets around my statement, thanks!