[Haskell-cafe] bottom case in proof by induction

[Derek Elkins]

On Wed, 2008-12-31 at 22:08 -0600, Jonathan Cast wrote:
> 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!

This isn't just a quirk of Haskell semantics.  bad is not computable.
Period.