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
> bad () = _|_
> bad _|_ = ()
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?
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 _|_ .
Thanks
Raeck
From: Luke Palmer
Sent: Wednesday, December 31, 2008 10:43 PM
To: Max.cs ; raeck(a)msn.com
Subject: Re: [Haskell-cafe] bottom case in proof by induction
On Wed, Dec 31, 2008 at 3:28 PM, Max.cs <max.cs.2009(a)googlemail.com> wrote:
thanks Luke,
so you mean the _|_ is necessary only when I have defined the pattern _|_ such as
foo [] = []
foo _|_ = _|_
foo (x:xs) = x( foo xs )
-- consider non-termination case
That is illegal Haskell. But another way of putting that is that whenever you do any pattern matching, eg.:
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:
halts () = True
halts _|_ = False
Because _|_ is the denotation of a program which never halts.
To do it a bit more domain-theoretically, I'll first cite the result that every function has a fixed point. That is, for every f, there is a function fix f, where fix f = f (fix f). (The fix function is actually available in Haskell from the module Data.Function). Then let's consider this bad function:
bad () = _|_ -- you can't write _|_ in Haskell, but "undefined" or "let x = x in x" mean the same thing
bad _|_ = ()
Then what is fix f? Well, it either terminates or it doesn't, right? I.e. fix f = () or fix f = _|_.
Taking into account that fix f = f (fix f):
If it does: fix f = () = f () = _|_, a contradiction.
If it doesn't: fix f = _|_ = f _|_ = (), another contradiction.
>From a mathematical perspective, that's why you can't pattern match on _|_.
>From an operational perspective, it's just that _|_ means "never terminates", and we can't determine that, because we would try to run it "until it doesn't terminate", which is meaningless...
Luke
