在Coq中完全忽略

Question

我找到了我想研究的一個有趣的邏輯定理的源代碼 。 但是，當我在CoqIDE中運行它時，它一開始就卡住了。

Inductive Term: Set :=
   K: Term |
   S: Term |
   app: Term -> Term -> Term. 

Inductive one_step: Term -> Term -> Prop :=
   redk: (m, n: Term) (one_step (app (app K m) n) m) |
   reds: (m, n, p: Term) (one_step (app (app (app S m) n) p) (app (app m p) (app n p))) |
   redl: (m, m', n: Term) (one_step m m') -> (one_step (app m n) (app m' n)) |
   redr: (m, n, n': Term) (one_step n n') -> (one_step (app m n) (app m n')). 

由於The reference m was not found in the current environment. ，因此one_step定義失敗The reference m was not found in the current environment.

我知道缺少的術語是forall ，但是沒有它，原始代碼是如何運行的？ 是否需要加載一些模塊以使其隱式？

編譯代碼所需的其他模塊列表也將非常有幫助。

Answer 1

自7.4版以來，該語法已進行了一些更改，該版本已通過此開發進行了測試。 這是一個適用於8.4的版本：

Inductive Term: Set :=
   K: Term |
   S: Term |
   app: Term -> Term -> Term.

Inductive one_step: Term -> Term -> Prop :=
| redk (m n: Term) : one_step (app (app K m) n) m
| reds (m n p: Term) : one_step (app (app (app S m) n) p) (app (app m p) (app n p))
| redl (m m' n: Term) : one_step m m' -> one_step (app m n) (app m' n)
| redr (m n n': Term) : one_step n n' -> one_step (app m n) (app m n').