A Quirky Function

Re:A Quirky Function by Double_Cast on Saturday June 25, 2016 @02:02PM (#11977264)

  0^0
  (\be.eb)(\sz.z)(\sz.z)
      [b := (\sz.z)]
  (\e.e(\sz.z))(\sz.z)
      [e := (\sz.z)]
  ((\sz.z)(\sz.z))
      [s := (\sz.z)]
  ((\z.z))

Lambda Calculus returns (\z.z) AKA the Identity Function. Which kinda sounds like 1, except 1 is usually represented as (\sz.sz). Which means 0^0 is NaN? I wonder if division by zero also gives (\z.z), but I don't know what division in Lambda Calculus looks like.

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Re:A Quirky Function by kmill on Saturday June 25, 2016 @06:44PM (#11978353)

I'm pretty sure (\sz.sz) and (\z.z) are the same function: (\sz.(\z.z)sz)=(\sz.sz).

It would make sense that 0^0 for church numerals is 1, since this is exponentiation to the natural (defined as recursive multiplication, with the base case x^0=1).

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