On Thu, 2009-01-01 at 03:50 +0000, raeck@msn.com wrote:

I am afraid I am still confused.

...

foo [] = ... foo (x:xs) = ... There is an implied: foo _|_ = _|_ The right side cannot be anything but _|_. If it could, then that would imply we could solve the halting problem:

in a proof, how I could say the right side must be _|_ without defining foo _|_ = _|_ ?

This definition is taken care of for you by the definition of Haskell pattern matching. If the first equation for a function has a pattern other than * a variable or * a lazy pattern (~p) for a given argument, then supplying _|_ for that argument /must/ (if the application is total) return _|_. By rule. (We say the pattern is strict, in this case).

and in the case of

...

bad () = _|_ bad _|_ = ()

Note that these equations (which are not in the right form for the Haskell equations that define Hasekll functions) aren't satisfied by any Haskell function!

mean not every function with a _|_ input will issue a _|_ output,

True --- but we can say a couple of things: * For all Haskell functions f, if f _|_ is an application of a constructor C, then f x is an application of C (to some value), for all x. [1] * For all Haskell functions f, if f _|_ is a lambda expression, then f x is a lambda expression, for all x. The only other possibility for f _|_ is _|_. (Do you see why bad above is impossible?)

so we have to say what result will be issued by a _|_ input in the definitions of the functions if we want to prove the equvalence between them?

You have to deduce what the value at _|_ will be.

However, in the case of map f _|_ , I do believe the result will be _|_ since it can not be anything else, but how I could prove this? any clue?

Appeal to the semantics of Haskell pattern matching. If you like, you can de-sugar the definition of map a little, to get map = \ f xn -> case xn of [] -> [] x:xn0 -> f x : map f xn0 And then you know that case _|_ of [] -> ... ... = _|_ whatever you fill in for the ellipses. (Do you see why this *must* be part of the language definition?)

ps, the definition of map does not mention anything about _|_ .

The behavior of map f _|_ is fixed by the definition of Haskell pattern matching. jcc

Am Donnerstag, 1. Januar 2009 04:50 schrieb raeck@msn.com:

I am afraid I am still confused.

...

foo [] = ... foo (x:xs) = ... There is an implied: foo _|_ = _|_ The right side cannot be anything but _|_. If it could, then that would imply we could solve the halting problem:

in a proof, how I could say the right side must be _|_ without defining foo _|_ = _|_ ? and in the case of

Because _|_ is matched against a refutable pattern ([], in this case), so when foo is called with argument _|_, the runtime tries to match it against []. For that, it must reduce it far enough to know its top level constructor, which by definition of _|_ isn't possible, so the pattern match won't terminate, hence foo _|_ is a nonterminating computation, i.e. _|_.

...

bad () = _|_ bad _|_ = ()

You can't do that. You can only pattern-match against patterns, which _|_ isn't.

mean not every function with a _|_ input will issue a _|_ output, so we have to say what result will be issued by a _|_ input in the definitions of the functions if we want to prove the equvalence between them?

If you match against an irrefutable pattern (variable, wildcard or ~pattern), the matching succeeds without evaluating the argument, so then you can have functions which return a terminating value for nonterminating arguments: lazyMap ~(x:xs) = [[],[x],xs] *LazyTest> lazyMap undefined [[],[*** Exception: Prelude.undefined *LazyTest> lazyMap [] [[],[*** Exception: PrintPer.hs:28:0-28: Irrefutable pattern failed for pattern (x : xs) *LazyTest> take 1 $ lazyMap undefined [[]]

However, in the case of map f _|_ , I do believe the result will be _|_ since it can not be anything else, but how I could prove this? any clue?

ps, the definition of map does not mention anything about _|_ .

As above, evaluation of map f _|_ tries to match _|_ against [], which doesn't terminate.

Thanks Raeck