Re: [Haskell] Correct interpretation of the curry-howard isomorphism

2004-04-23 Thread JP Bernardy
Hi, --- Bruno Oliveira [EMAIL PROTECTED] wrote: coerce :: a - b coerce x = undefined As an obvious consequence, Haskell type system would be unsound. Actually, in the isomrphism, undefined is a proof that any theorem is correct. Somehow it can represent an assumption that a subtheorem is

Re: [Haskell] Correct interpretation of the curry-howard isomorphism

2004-04-23 Thread Lennart Augustsson
JP Bernardy wrote: I'd say, check what any primitive 'proves' before using it. Besides that, calling other functions is ok. Except for general recursion. coerce :: a - b coerce = coerce -- Lennart ___ Haskell mailing list [EMAIL PROTECTED]

Re: [Haskell] Correct interpretation of the curry-howard isomorphism

2004-04-23 Thread Bruno Oliveira
Hi! Thanks Connor, I enjoyed your answer, it was very illustrative. but then, it would be too easy to write this in haskell: coerce :: a - b coerce x = undefined As an obvious consequence, Haskell type system would be unsound. So, I assumed that this would be a wrong interpretation.

Re: [Haskell] Correct interpretation of the curry-howard isomorphism

2004-04-23 Thread Andre Pang
On 23/04/2004, at 10:56 PM, Bruno Oliveira wrote: but then, it would be too easy to write this in haskell: coerce :: a - b coerce x = undefined As an obvious consequence, Haskell type system would be unsound. So, I assumed that this would be a wrong interpretation. This is the part of your