I guess nontermination is a problem (e.g. if one or both functions fail to terminate for some values, equality will be undecidable).
/Jonas On 14 April 2010 08:42, Ashley Yakeley <ash...@semantic.org> wrote: > On Wed, 2010-04-14 at 16:11 +1000, Ivan Miljenovic wrote: >> but the only way you can "prove" it in >> Haskell is by comparing the values for the entire domain (which gets >> computationally expensive)... > > It's not expensive if the domain is, for instance, Bool. > > -- > Ashley Yakeley > > _______________________________________________ > Haskell-Cafe mailing list > Haskell-Cafe@haskell.org > http://www.haskell.org/mailman/listinfo/haskell-cafe > _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
