> Unfortunately (or suprisingly), even this does not work
> as long as we interpret f = g as f x = g x at the top-level:
>
> f5 (.) g 3 ->* g (g 3)
> -> incr (g 3) (with rule g x = incr x)
> -> (g 3) + 1
> -> (decr 3) + 1 (with rule g x = decr x)
> ->* 3
Hmmm, I should have thought of that!
Anyway, here are two more cases. Given the definitions
coin = 0
coin = 1
f6 h g = h g g
f7 h g = h g0 g0 where g0 = g
what are the solutions of:
f6 (\x y -> x + y) coin
f7 (\x y -> x + y) coin
I suppose that 1 is a possible solution in the former case, while it isn't in the latter. (Or will the use of the parameter g for both arguments of h in f6 already introduce sharing?)
Regards
Wolfgang
--
Wolfgang Lux Phone: +49-251-83-38263
Institut fuer Wirtschaftinformatik FAX: +49-251-83-38259
Universitaet Muenster Email: lux_at_helios.uni-muenster.de
Received on Mi Dez 02 1998 - 17:49:00 CET