[Haskell-cafe] about beta NF in lambda calculus
Eugene Kirpichov
ekirpichov at gmail.com
Sat Mar 21 15:59:24 EDT 2009
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 indistinguishable for you in Haskell.
2009/3/21 Ryan Ingram <ryani.spam at gmail.com>:
> 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.
>
> -- ryan
>
> 2009/3/21 Algebras Math <algebras2009 at googlemail.com>:
>> 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 NF?
>>
>> 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?
>>
>> Thanks
>>
>> Alg
>>
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--
Eugene Kirpichov
Web IR developer, market.yandex.ru
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