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[Arnaud Spiwack <aspiwack AT lix.polytechnique.fr>]

forall X:Type, forall (A B:X->Prop), (forall x, A x <-> B x) -> forall P:X->Prop, exists x, A x /\ Px <-> exists x, B x /\ P x.

On 5 September 2013 22:20, t x <txrev319 AT gmail.com> wrote:

[We're using propositional equivalence through all this.]

Here is a simple lemma. (I don't need help proving it. I need help making it more abstract).

(1)

  Lemma lem__Z_split:
    forall (l r: Z) (P: Z -> Prop),
    (exists i, l <= i < r /\ P i) =
    (P l \/ (exists i, l +1 <= i < r)).

Using propositional equivalence, this can be proved easily using the Z destructuring of {a = b} + {a < b} + {a > b}.

However, I want to formulate a more general theorem, something like this:

(2)

  forall (A B : Sets) (*  -- not necessairly the same "Set" as Coq's "Set" *),

    A = B (* in the sense that for all x, x in A <-> x in B *),

      exists x, x in A /\ P x = exists x, x in B /\ P x

Then, all I need is to show that the "set" defined by l <= i < r is equiv to teh set defined by i = l \/ (l+1 < i < r).

I'm having trouble formulating (2) in Coq. Can someone show me sample code (or point me at some chapter I should read?)

Thanks!