More algebraically, including 'or' for symmtry: and xs = foldr (&&) True xs or xs = foldr (||) False xs The True and False are the (monoid) identities with respect to && and || : True && x == x x && True == x False || x == x x || False == x And so an empty list, if defined at all, should be the identity: and [] = True or [] = False In english: "and returns false" when "is any element of the list false?" is yes "or returns true" when "is any element of the list true?" is yes Matthew Brecknell wrote:
Dan Weston wrote:
Here, "any path" means all paths, a logical conjunction:
and [True, True] = True and [True ] = True and [ ] = True
Kim-Ee Yeoh wrote:
Hate to nitpick, but what appears to be some kind of a limit in the opposite direction is a curious way of arguing that: and [] = True.
Surely one can also write
and [False, False] = False and [False ] = False and [ ] = False ???
No. I think what Dan meant was that for all non-null xs :: [Bool], it is clearly true that:
and (True:xs) == and xs -- (1)
It therefore makes sense to define (1) to hold also for empty lists, and since it is also true that:
and (True:[]) == True
We obtain:
and [] == True
Since we can't make any similar claim about the conjuctions of lists beginning with False, there is no reasonable argument to the contrary.