Coq: destruct (co)inductive hypothesis without losing information - coq

Question

Consider the following development:
Require Import Relation RelationClasses.
Set Implicit Arguments.
CoInductive stream (A : Type) : Type :=
| scons : A -> stream A -> stream A.
CoInductive stream_le (A : Type) {eqA R : relation A}
`{PO : PartialOrder A eqA R} :
stream A -> stream A -> Prop :=
| le_step : forall h1 h2 t1 t2, R h1 h2 ->
(eqA h1 h2 -> stream_le t1 t2) ->
stream_le (scons h1 t1) (scons h2 t2).
If I have a hypothesis stream_le (scons h1 t1) (scons h2 t2), it would be reasonable for the destruct tactic to turn it into a pair of hypotheses R h1 h2 and eqA h1 h2 -> stream_le t1 t2. But that's not what happens, because destruct loses information whenever doing anything non-trivial. Instead, new terms h0, h3, t0, t3 are introduced into the context, with no recall that they are respectively equal to h1, h2, t1, t2.
I would like to know if there is a quick and easy way to do this kind of "smart destruct". Here is what i have right now:
Theorem stream_le_destruct : forall (A : Type) eqA R
`{PO : PartialOrder A eqA R} (h1 h2 : A) (t1 t2 : stream A),
stream_le (scons h1 t1) (scons h2 t2) ->
R h1 h2 /\ (eqA h1 h2 -> stream_le t1 t2).
Proof.
intros.
destruct H eqn:Heq.
remember (scons h1 t1) as s1 eqn:Heqs1;
remember (scons h2 t2) as s2 eqn:Heqs2;
destruct H;
inversion Heqs1; subst; clear Heqs1;
inversion Heqs2; subst; clear Heqs2.
split; assumption.
Qed.

Answer 1

Indeed, inversion basically does what you want, however as Arthur pointed out it is a bit unstable, mainly due to the different congruence steps.
Under the hood, inversion just calls a version of destruct, but remembering some equalities first. As you have well discovered, pattern matching in Coq will "forget" arguments of constructors, except if these are variables, then, all the variables under the scope of the destruct will be instantiated.
What does that mean? It means that in order to properly destruct an inductive I : Idx -> Prop, you want to get your goal of the form: I x -> Q x, so that destructing the I x will also refine the x in Q. Thus, a standard transformation for an inductive I term and goal Q (f term) is to rewrite it to I x -> x = term -> Q (f x). Then, destructing I x will get you x instantiated to the proper index.
With that in mind, it may be a good exercise to implement inversion manually using the case: tactic of Coq 8.7;
From Coq Require Import ssreflect.
Theorem stream_le_destruct A eqA R `{PO : PartialOrder A eqA R} (h1 h2 : A) (t1 t2 : stream A) :
stream_le (scons h1 t1) (scons h2 t2) ->
R h1 h2 /\ (eqA h1 h2 -> stream_le t1 t2).
Proof.
move E1: (scons h1 t1) => sc1; move E2: (scons h2 t2) => sc2 H.
by case: sc1 sc2 / H E1 E2 => h1' h2' t1' t2' hr ih [? ?] [? ?]; subst.
Qed.
You can read the manual for more details, but basically with the first line, we create the equalities we need; then, in the second we can destruct the term and get the proper instantiations solving the goal. A good effect of the case: tactic is that, contrary to destruct, it will try to prevent us from destructing a term without first bringing its dependencies into scope.

Answer 2

Calling destruct will not directly give you what you want. You need to use inversion instead.
Theorem stream_le_destruct : forall h1 h2 t1 t2,
stream_le (scons h1 t1) (scons h2 t2) ->
h1 <= h2 /\ (h1 = h2 -> stream_le t1 t2).
Proof.
intros.
inversion H; subst; clear H.
split; assumption.
Qed.
Unfortunately, the inversion tactic is quite ill behaved, as it tends to generate a lot of spurious equality hypotheses, making it hard to name them consistently. One (somewhat heavyweight, admittedly) alternative is to use inversion only to prove a lemma like the one you did, and apply this lemma in proofs instead of calling inversion.