coqssreflect

ssreflect inversion, I need two equations instead of one

I have next definitions (code can be compiled):

From mathcomp Require Import all_ssreflect.

Set Implicit Arguments.
Set Asymmetric Patterns.

Unset Strict Implicit.
Unset Printing Implicit Defensive.

Inductive val : Set := VConst of nat | VPair of val & val.
Inductive type : Set := TNat | TPair of type & type.

Inductive tjudgments_val : val -> type -> Prop :=
| TJV_nat n :
    tjudgments_val (VConst n) TNat
| TJV_pair v1 t1 v2 t2 : 
    tjudgments_val v1 t1 ->
    tjudgments_val v2 t2 ->
    tjudgments_val (VPair v1 v2) (TPair t1 t2).

And I would like to prove the following lemma:

Lemma tjexp_pair v1 t1 v2 t2 (H : tjudgments_val (VPair v1 v2) (TPair t1 t2)) :
  tjudgments_val v1 t1 /\ tjudgments_val v2 t2.
Proof.
  case E: _ _ / H => // [v1' t1' v2' t2' jv1 jv2].
  (* case E: _ / H => // [v1' t1' v2' t2' jv1 jv2]. *)

But it looks to me like I need both of them together. How to?

Solution

  • There is a way of using inversion's power with ssreflect tactics.

    Derive Inversion tjudgments_val_inv with (forall v t, tjudgments_val v t).

    You can use it with elim/tjudgments_val_inv: H.

    The proof is straightforward after this.