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In Church encoding of the natural numbers in lambda calculus raising zero to the power zero gives the answer zero. Does anybody know of an encoding where the answer is 1?

William Pearson
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1 Answers1

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(λm.λn.n m) (λf.λx.x) (λf.λx.x)
⇒   (λn.n (λf.λx.x)) (λf.λx.x)
⇒   (λf.λx.x) (λf.λx.x)
⇒   λx.x

Another different calculation ((\m.\n. ((n (\m.\n.\f. (m (n f)))) f x))(\f.\x.x)(\f.\x.x)):

(λm.λn.n (λi0.λi1.λi2.i0 [i1 i2]) f x) (λi0.λi1.i1) (λi0.λi1.i1)
⇒   (λn.n (λm.λi0.λi1.m [i0 i1]) f x) (λi0.λi1.i1)
⇒   (λi0.λi1.i1) (λm.λn.λi0.m (n i0)) f x
⇒   (λi0.i0) f x
⇒   f x

Calculations done using: http://www.cburch.com/lambda/index.html

Loreno Heer
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    Which is not a Church numeral at all, is it? So, can we conclude that $0^0$ is undefined in the context of Church encoding? – Daniel R Jan 30 '15 at 12:59
  • @DanielR I am slightly confused by it myself now. I calculated it again using a different approach, which does indeed result in 1. – Loreno Heer Jan 30 '15 at 13:02
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    $\lambda x.x$ and $1$ are equivalent under $\eta$-expansion – mercio Jan 30 '15 at 13:03
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    I'm confused. In "Constructive Foundations for Functional Languages" http://www.academia.edu/456528/CONSTRUCTIVE_FOUNDATIONS_FOR_FUNCTIONAL_LANGUAGES it says λx.x is the zero object – William Pearson Jan 30 '15 at 13:18
  • Ping @mercio. $;$ – Daniel R Jan 30 '15 at 13:48
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    the zero object is $\lambda f. \lambda x. x$. Meanwhile, $\lambda x. x =\alpha \lambda f.f =\eta \lambda f. (\lambda x. f x) = 1$ – mercio Jan 30 '15 at 15:36
  • thanks for the clarification. Just found "For example, the combinators (λy.λx.yx) and (λy.y) are observationally equivalent" on another answerlink. Out of interest could you not also eta expand to λf.(λx.x) f which I really don't understand? why dont you put as answer so I can accept it? – William Pearson Jan 30 '15 at 16:26
  • @William When you address a user who is not the author of the answer or question, you can add a @ and the user name. This way, the user is made aware of your comment by getting a notification in their inbox. This may help your attempt to reach out to mercio. – Daniel R Jan 31 '15 at 11:04
  • @mercio sorry, new here. please see comment before last – William Pearson Jan 31 '15 at 20:12