4 ms·
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 Calculu
by Double_Cast 10y ago
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.
- kmill 10y agoI'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).