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.
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