Question

我注意到符号可以区别对待。例如，<只是常规定义的表示法，而unfold "<"的确如下例所示：

Theorem a : 4 < 5.
Proof.
    unfold "<".

但是，<=是与le类型相关联的表示法，由于某种原因unfold "<="不起作用，如下例所示：

Theorem a : 4 <= 5
Proof.
   unfold "<=".

以Unable to interpret "<=" as a reference失败。

我可以使用一些ltac命令将4 <= 5转换为le 4 5吗？

Answer 1

This happens because < is interpreted as lt which is a definition (here):

Definition lt (n m:nat) := S n <= m.

You can achieve the same effect with unfold lt.

In the same manner <= means le, but you cannot unfold le, because it is a type constructor. The manual says that you can unfold only a defined transparent constant or local definition.

The upshot here is that you don't unfold notations, you unfold the definitions those notations refer to.

Answer 2

添加到Anton的答案：如果您已经知道如何定义符号并且只想让它在目标中可见，您可以执行类似

的操作

Definition make_visible {X} (f : X) := f.

Notation "` f" := (make_visible f) (at level 5, format "` f").

Tactic Notation "unfold" "notation" constr(f) :=
  change f with (`f).
Tactic Notation "fold" "notation" constr(f) :=
  unfold make_visible.

Theorem a : 4 <= 5.
Proof.
  unfold notation le.
  fold notation le.

（编辑：我的第一个解决方案是Definition make_visible {X} (f : X) := (fun _ => f) tt.，但正如安东指出的那样，这更容易。）

在ltac中展开符号

2 个答案: