Exogenist
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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”
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”