
An answer would probably depend on the reductions order you've chosen. Would that do? (\e -> e (\u -> e (\v -> u))) (\f -> \x -> f x) -- all variables have different names, see? = (\f -> \x -> f x) (\u -> (\f -> \x -> f x) (\v -> u)) = \x -> (\u -> (\f -> \x -> f x) (\v -> u)) x = \x -> (\f -> \x -> f x) (\v -> x) = \x -> \x -> (\v -> x) x -- Ouch! = \x -> \x -> x = \x -> id where the real answer is \x -> (\f -> \x -> f x) (\v -> x) = \x -> (\f -> \y -> f y) (\v -> x) = \x -> \y -> (\v -> x) y = \x -> \y -> x = \x -> const x On 21 Jun 2009, at 20:53, Andrew Coppin wrote:
OK, so I'm guessing there might be one or two (!) people around here who know something about the Lambda calculus.
I've written a simple interpretter that takes any valid Lambda expression and performs as many beta reductions as possible. When the input is first received, all the variables are renamed to be unique.
Question: Does this guarantee that the reduction sequence will never contain name collisions?
I have a sinking feeling that it does not. However, I can find no counter-example as yet. If somebody here can provide either a proof or a counter-example, that would be helpful.
_______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe