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[Jean-Francois Monin <jean-francois.monin AT imag.fr>]

Additional exercise: prove the following stronger version

Theorem identical : forall (X: Type) (f: X->X) n, 
  (fun x => ant f x n) = (fun x => ant' f x n).

This can be done without axiom (in Coq 8.4).
Warning: there is a funny trick to use somewhere, so to speak.

The above statement is somewhat weird. It is better to consider:

Theorem identical2 : forall (X: Type) (f: X->X) n, 
  ant2' f n = ant2 f n. 

with obvious definitions for ant2 and ant2' such that
ant2 f n   =  fun x => ant f x n
ant2' f n  =  fun x => ant' f x n


JF


On Sat, Nov 10, 2012 at 02:54:57PM +0100, Kevin Sullivan wrote:
> My students presented two solutions to the simple problem of applying a
> function f n times to an argument x (iterated composition). The first says
> apply f to the result of applying f n-1 times to x; the second,  apply f n-1
> times to the result of applying f to x. I challenged one student to prove 
> that
> the programs are equivalent. This ought to be pretty easy based on the
> associativity of composition. Your best solution?
> 
> Fixpoint ant {X: Type} (f: X->X) (x: X) (n: nat) : X :=
> match n with
> | O => x
> | S n' => f (ant f x n')
> end.
> 
> Fixpoint ant' {X: Type} (f: X->X) (x: X) (n: nat) : X :=
> match n with
> | O => x
> | S n' => ant' f (f x) n'
> end.
> 
> Theorem equiv: forall (X: Type) (f: X->X) (x: X) (n: nat),
> (ant f x n) = (ant' f x n).
> Proof.  admit.
> 
> 

-- 
Jean-Francois Monin