Translating Quantified statements.

[Bashyboy]

Homework Statement

Let L(x, y) be the statement “x loves y,” where the do main for both x and y consists of all people in the world. Use quantifiers to express each of these statements.

g) There is exactly one person whom everybody loves.

Homework Equations

The Attempt at a Solution

I couldn't determine the answer myself, so I looked to the answer key for aid. According to the answer key, \exists x[(\forall y L(y,x) \wedge \forall z ( \forall w L(w,z)) \implies z = x))] is the proper translation. However, I the introduction of the variable w is extraneous, that the answer to be simplified roughly to \exists x \forall y [(L(y,x) \wedge \forall z L(y,z)) \implies z = x)] Would you care to share your opinion?

 

[haruspex]

According to the answer key, \exists x[(\forall y L(y,x) \wedge \forall z ( \forall w L(w,z)) \implies z = x))] is the proper translation. However, I the introduction of the variable w is extraneous, that the answer to be simplified roughly to \exists x \forall y [(L(y,x) \wedge \forall z L(y,z)) \implies z = x)]

I disagree. First, let's make the first form a bit clearer (not sure what the standard binding order is, and a parenthesis is unmatched):
\exists x[(\forall y L(y,x)) \wedge (\forall z ( \forall w L(w,z) \implies z = x))] .
In your version, the implication at the end has the wrong relationship to L(y,x). Worse, what do you think this translates to: \forall z L(y,z)?

 

[verty]

Some authors define the quantifier $\exists!$ to mean there exists uniquely, there exists one and only one. It translates this way:

$\exists! x P(x)$ <==> $\exists x (P(x) \wedge \forall y (P(y) → y = x))$ where y is not a free variable of P.

You should verify that the answer given is just a version of $\exists! x \forall y L(y,x)$.