Question

我找到了一个我想要解决的有趣逻辑定理的source code。但是当我在CoqIDE中运行它时，它会在最开始时卡住。

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')).

one_step的定义失败，因为The reference m was not found in the current environment.

我理解缺少的术语是forall，但原始代码是如何在没有它的情况下运行的？是否需要加载一些模块以使其隐含？

编译代码所需的其他模块列表也非常有用，如果有的话。

Answer 1

自版本7.4以来，语法发生了一些变化，这是使用此开发进行测试的版本。这是一个适用于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').

在Coq中省略forall

1 个答案: