注射策略可以修改最終目標，還是添加無關緊要的假設？

Question

考慮以下開發，這是Adam Chlipala的simplHyp的孤立部分：

(** Fail if H is in context *)
Ltac notInCtx H := assert H; [ assumption | fail 1 ] || idtac.

Ltac injectionInCtx :=
  match goal with
  (* Is matching on G strictly necessary? *)
  | [ H : ?F ?X = ?F ?Y |- ?G ] =>
    (* fail early if it wouldn't progress *)
    notInCtx (X = Y); 
    injection H;
    match goal with
      (* G is used here *)
      | [ |- X = Y -> G ] =>
        try clear H; intros; try subst
    end
  end.

Goal forall (x y : nat), S x = S y -> x = y.
  intros x y H.
  injectionInCtx.
  exact eq_refl.
Qed.

請參閱內聯注釋G在開始時就已匹配，並最終用於驗證最終目標是否保持不變。 這是否排除了injection H可能改變目標或添加無關緊要的假設的可能性？

Answer 1

我認為您不能修改G ，但是您可以提出一個假設，根據該假設， injection將產生多個等式。

我們定義injectionInCtx2 ，它與injectionInCtx相同，只是它不使用G

Ltac injectionInCtx2 :=
  match goal with
  | [ H : ?F ?X = ?F ?Y |- _ ] =>
    (* fail early if it wouldn't progress *)
    notInCtx (X = Y);
    injection H;
    match goal with
    | [ |- X = Y -> _ ] =>
      try clear H; intros; try subst
    end
  end.

Definition make_pair {A} (n:A) := (n, n).

Goal forall (x y : nat), make_pair x = make_pair y -> x = y.
Proof.
  intros x y H.
  (* [injection H] gives [x = y -> x = y -> x = y] *)
  Fail injectionInCtx.
  injectionInCtx2.
  reflexivity.
Qed.