If above is true, I am confused why we have to distinguish the terms which
have NF and be in NF? isn't the terms have NF will eventually become in NF?
or there are some way to avoid them becoming in NF?
Another way to think about it: what about terms which have no NF?
And given both kinds of
hi,
What is the different between 'in beta normal form' and 'has beta normal
form' ? Does the former means the current form of the term is already in
normal form but the latter means that it is not a normal form yet and can be
reduced to be normal form? Like \x.x is in NF and (\x.x) (\x.x) has
Given the Y combinator, Y (\x.x) has no normal form.
However, (\a. (\x. x)) (Y (\x. x)) does have a normal form; (\x. x).
But it only reduces to that normal form if you reduce the (\a. ...)
redex, not if you reduce its argument. So depending on evaluation
order you might not reach a normal form.
Thus, if you use normal-order evaluation (which Haskell uses), you
will inevitably reach the normal form if and only if it exists at all.
So, if all you care about is the normal form (if you don't care about
computation time), then terms that *have* an NF and terms that *are*
in NF are
On Sat, Mar 21, 2009 at 07:29:05PM +, Algebras Math wrote:
If above is true, I am confused why we have to distinguish the terms which
have NF and be in NF? isn't the terms have NF will eventually become in NF?
or there are some way to avoid them becoming in NF?
Spoken like a mathematician
Actually I was also going to provide exactly the same example, but
hesitated to :)
2009/3/21 Antti-Juhani Kaijanaho antti-juh...@kaijanaho.fi:
On Sat, Mar 21, 2009 at 07:29:05PM +, Algebras Math wrote:
If above is true, I am confused why we have to distinguish the terms which
have NF and
