[Haskell-cafe] Tutorial: Curry-Howard Correspondence

[jerzy.karczmarczuk at info.unicaen.fr]

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