How to destruct a exist in goal for coq

Question

This is a bit generic so feel free to ask for details, I simplified to make the problem easier to communicate.

Say my goal is

let (r,_) := f x in Q r

where fx has type {u | P u} {u | P u} .

I want to "destruct" this so that I have Q r as the goal with P r as a hypothesis. What is the best way to achieve this ? In the past, I have achieved what I wanted by

pose (f x).

and then simplifying.

Per request here is some simplified code.

Parameter T:Type.
Parameter P:T -> Prop.

Axiom A : {t:T|P t}.

Definition myvar:T.
  destruct A.
  exact x.
Defined.
Theorem B : P myvar.

Answer 1

unfold myvar; destruct A.

will solve your goal in this particular case.

In general, the destruct tactic can be used with with any term and it will try to abstract the term out and destruct it. However, note that this may fail sometimes, particularly when using dependent types.

The reason is that destruct uses the gallina match underneath, and in Coq, pattern matching may lose some typing information if not done carefully, search for "Convoy Pattern" for more information.

Answer 2

You can name the 'outputs' of a destruct tactic if you want:

(* Stupid definitions to create a minimal example *)
Parameter A: Set.
Parameter P Q: A -> Prop.
Definition f (x: A) : {r | P r }.
Admitted.

Goal (forall x, let (r, _) := f x in Q r).
intro x.
destruct f as [r hr]. (* you goal has now r: A and hr : P r as premises *)
Abort.

Edit: more info after comment. If you don't name them, Coq will do it automatically for you using a scheme based on xi variables ( x0, x1, ... ). You can only force the naming of the first variable if you don't care about the name of the second part using destruct f as [r].