[Haskell-beginners] Re: [Haskell-cafe] bottom case in proof by
lrpalmer at gmail.com
Wed Dec 31 17:19:26 EST 2008
2008/12/31 <raeck at msn.com>
> Dear all,
> Happy New Year!
> I am learning the Induction Proof over Haskell, I saw some proofs for the
> equivalence of two functions will have a case called 'bottom' but some of
> them do no have. What kind of situation we should also include the bottom
> case to the proof? How about the functions do not have the 'bottom' input
> such as:
> foo  = 
> foo (x:xs) = x : (foo xs)
Okay, I'm not sure what you mean by bottom. You could either mean the base
case, or you could mean bottom -- non-terminating inputs -- as in domain
Let's say you wanted to see if foo is equivalent to id.
id x = x
We can do it without considering nontermination, by induction on the
structure of the argument:
First, the *base case*: empty lists.
foo  = 
id  = 
Just by looking at the definitions of each.
Now the inductive case. Assume that foo xs = id xs, and show that foo
(x:xs) = id (x:xs), for all x (but a fixed xs).
foo (x:xs) = x : foo xs
foo xs = id xs by our the induction hypothesis, so
foo (x:xs) = x : id xs = x : xs
And then just by the definition of id:
id (x:xs) = x : xs
And we're done.
Now, maybe you meant bottom as in nontermination. In this case, we have to
prove that they do the same thing when given _|_ also. This requires a
deeper understanding of the semantics of the language, but can be done
First, by simple definition, id _|_ = _|_. Now let's consider foo _|_. The
Haskell semantics say that pattern matching on _|_ yields _|_, so foo _|_ =
_|_. So they are equivalent on _|_ also. Thus foo and id are exactly the
See http://en.wikibooks.org/wiki/Haskell/Denotational_semantics for more
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