[Haskell-cafe] forall & ST monad

[Kim-Ee Yeoh]

There's a lot to chew on (thank you!), but I'll just take something 
I can handle for now.


Dan Doel wrote:
> 
> An existential:
> 
>     exists a:T. P(a)
> 
> is a pair of some a with type T and a proof that a satisfies P (which has
> type 
> P(a)). In Haskell, T is some kind, and P(a) is some type making use of a.
> That 
> doesn't mean that there is only one a:T for which P is satisfied. But it 
> *does* mean that for any particular proof of exists a:T. P(a), only one
> a:T is 
> involved. So if you can open that proof, then you know that that is the 
> particular a you're dealing with, which leads to the problems in the 
> grandparent.
> 

re: Constructivity and the opening of a proof

A form of the theorem that the primes are infinite goes
"Given a finite set of primes, there's a prime bigger than any of them."

The usual proof is constructive since factorization is algorithmic,
but I don't see why a priori, applications of this theorem on a given input
should always yield the same prime when more than one factor exists.

Is non-deterministic choice forbidden in constructive math? A cursory
google seems to suggest that if not, it's at least a bête noire to some.

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