Tim Newsham quotes somebody /I didn't follow this thread!/:
I assume you mean then that it is a valid proof because it halts for
*some* argument? Suppose I have:
thm1 :: (a -> a) -> a
thm1 f = let x = f x in x
There is no f for which (thm1 f) halts (for the simple reason that _|_ is
the only value in every type), so thm1 is not a valid theorem.
=================
Since, as I said, I didn't follow you, it would be indecent of me to try to
be clever now, but the statement above (that there is no f for which thm1
halts) is false *IN HASKELL*. Try f = const 1.
Unless I missed some important point elsewhere...
Jerzy Karczmarczuk
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