[Haskell-beginners] Re: [Haskell-cafe] bottom case in proof by induction

[Jonathan Cast]

On Thu, 2009-01-01 at 03:50 +0000, raeck at 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