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Translating Quantified statements.

  1. Jun 21, 2013 #1
    1. The problem statement, all variables and given/known data
    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.

    2. Relevant equations



    3. 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, [itex]\exists x[(\forall y L(y,x) \wedge \forall z ( \forall w L(w,z)) \implies z = x))][/itex] is the proper translation. However, I the introduction of the variable w is extraneous, that the answer to be simplified roughly to [itex]\exists x \forall y [(L(y,x) \wedge \forall z L(y,z)) \implies z = x)][/itex] Would you care to share your opinion?
     
  2. jcsd
  3. Jun 21, 2013 #2

    haruspex

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    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):
    [itex]\exists x[(\forall y L(y,x)) \wedge (\forall z ( \forall w L(w,z) \implies z = x))][/itex] .
    In your version, the implication at the end has the wrong relationship to L(y,x). Worse, what do you think this translates to: [itex] \forall z L(y,z)[/itex]?
     
  4. Jun 22, 2013 #3

    verty

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    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)##.
     
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