軟件基礎：適用於……策略

Question

我嘗試apply ... with ... Pierce的“軟件基礎”中的策略在apply ... with ...上運行一些簡單的示例。

看來本書中的示例對我不起作用：

Theorem trans_eq: forall (X: Type) (n m o: Type),
                    n = m -> m = o -> n = o.
Proof.
  intros X n m o eq1 eq2. rewrite -> eq1. rewrite -> eq2. reflexivity.
Qed.

Example trans_eq_example' : forall (a b c d e f : nat),
     [a;b] = [c;d] ->
     [c;d] = [e;f] ->
     [a;b] = [e;f].
Proof.
  intros a b c d e f eq1 eq2.
  (* If we simply tell Coq apply trans_eq at this point,
     it can tell (by matching the goal against the
     conclusion of the lemma) that it should instantiate X
     with [nat], n with [a,b], and o with [e,f].
     However, the matching process doesn't determine an
     instantiation for m: we have to supply one explicitly
     by adding with (m:=[c,d]) to the invocation of
     apply. *)
  apply trans_eq with (m:=[c;d]). apply eq1. apply eq2. Qed.

trans_eq_example'失敗，並出現以下錯誤：

trans_eq_example' < apply trans_eq with (m:=[c;d]).
Toplevel input, characters 6-30:
> apply trans_eq with (m:=[c;d]).
>       ^^^^^^^^^^^^^^^^^^^^^^^^
Error: Impossible to unify "?1707 = ?1709" with "[a; b] = [e; f]".

有關Coq版本的其他信息：

coqtop -v
The Coq Proof Assistant, version 8.4pl4 (July 2014)
compiled on Jul 27 2014 23:12:44 with OCaml 4.01.0

如何解決此錯誤？

Answer 1

問題不是apply而是您先前代碼中的錯字。 trans_eq的定義應為：

Theorem trans_eq: forall (X:Type) (n m o: X), n = m -> m = o -> n = o.

請注意， nmo的類型應為X ，而不是Type 。