The Coq mailing list

[Thorsten Altenkirch <Thorsten.Altenkirch AT nottingham.ac.uk>]

Why not just using the Church encoding:

Definition Mu (f : Type -> Type) :=
    forall x, (f x -> x) -> x

I need to know that f is a functor if I want to implement fold. I don¹t
see why this is a problem?

Is Type still impredicative in Coq? I guess this works only for Type(0)
because if you have several impredicative levels then Girard¹s paradox
applies. 

I don¹t think it is a good idea to encode types using this impredicative
trickery: it is conceptually rather heavy (do I need to understand what it
means to quantify over all types to understand the natural numbers?)and
practically not very useful. Also you need to assume parametricity to
derive any induction principle.

Thorsten



On 30/03/2016 16:44, "John Wiegley" 
<johnw AT newartisans.com>
 wrote:

>>>>>> Igor Zhirkov 
>>>>>> <igorjirkov AT gmail.com>
>>>>>>  writes:
>
>> I wonder whether it is possible to define the Mu  combinator in Coq in a
>> general way.
>
>Hi Igor,
>
>In many cases you can use a "Mendler-style Church encoding" of Mu:
>
>  Definition Mu (f : Type -> Type) :=
>    forall r, (forall x, (x -> r) -> f x -> r) -> r.
>
>Note that you only need to know f is a Functor at reduction time.
>
>However: this "Mu" may be "larger" in CiC than f (Mu f), which could have
>implications for what you're doing.
>
>-- 
>John Wiegley                  GPG fingerprint = 4710 CF98 AF9B 327B B80F
>http://newartisans.com                          60E1 46C4 BD1A 7AC1 4BA2





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