Predicate calculus and use of the form there exists exactly one

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Main Question or Discussion Point

Predicate calculus and use of the form "there exists exactly one"

Given the following utterance does the analysis necessarily follow. Is there something wrong with it or would it be deemed a correct analysis.

“Everyone has exactly one best friend”

∀x( if x is a person then there exists exactly one y such that x has a best friend y)
F(x, y) = “x has a best friend y” Pe(x) = “x is a person”
∀x(Pe(x) → ∃!y(F(x, y))
∀x(Pe(x) → ∃y(F(x, y) & ~∃z((y ≠ z) & F(x, z))
∀x(Pe(x) → ∃y(F(x, y) & ∀z((y = z) v ~F(x, z))
∀x(Pe(x) → ∃y(F(x, y) & ∀z((y = z) v ~F(x, z))
∀x(Pe(x) → ∃y(F(x, y) & ∀z((y ≠ z) → ~F(x, z)) = “Everyone has exactly one best friend”
 

Answers and Replies

  • #2
Stephen Tashi
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would it be deemed a correct analysis.
What do you mean by "analysis"? Is what you gave supposed to be a series of steps, each following from the other? Or is it a collection of possible answers, each one to be marked correct or incorrect?
 
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What do you mean by "analysis"? Is what you gave supposed to be a series of steps, each following from the other? Or is it a collection of possible answers, each one to be marked correct or incorrect?
Yes its supposed to be a series of steps. By analysis I mean is the following formula ∀x(Pe(x) → ∃y(F(x, y) & ∀z((y ≠ z) → ~F(x, z)) where F(x, y) = “x has a best friend y” and Pe(x) = “x is a person”, equivalent to saying "everyone has exactly one best friend".
 
  • #4
Stephen Tashi
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Whether its a correct analysis will depend on whether you can justify the steps using whatever assumptions or theorems that your course materials employ. Different textbooks may use different axioms and theorems, so I can't evaluate whether your analysis is correct.

I do agree that each individual step is equivalent to statement you began with (assuming that [itex] \exists ! [/itex] is defined the way that lets to go from step 1 to step 2).
 

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