[Haskell-cafe] Wrong Answer Computing Graph Dominators

[Matthew Brecknell]

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.