Question

I was trying out the examples from the Coq documentation Software Foundations (http://www.cis.upenn.edu/~bcpierce/sf/current/Induction.html#lab40) when I noticed that to solve the example give in the link:

Theorem andb_true_elim1 : ∀b c : bool,
  andb b c = true → b = true.
Proof.
  intros b c H.
  destruct b.
  Case "b = true". (* <----- here *)
    reflexivity.
  Case "b = false". (* <---- and here *)
    rewrite ← H.
    reflexivity.
Qed.

we are destructing b which appears on the left of c for "andb". However to do the exercise that follows in the book i.e.

Theorem andb_true_elim2 : ∀b c : bool,
  andb b c = true → c = true.
Proof.
  (* FILL IN HERE *) Admitted.

I found the same method does not help. "reflexivity" did not help which I assume is due to the position of 'b' in the subgoal. Eventually I completed the proof in the following manner:

Theorem andb_true_elim2 : forall b c:bool,
andb b c = true -> c = true.
Proof.
intros b c H.
destruct c.
Case "c=true".
reflexivity.
Case "c=false".
rewrite <- H.
destruct b.
reflexivity.
reflexivity.
Qed.

Can someone explain why the above code worked and the following dint?

Theorem andb_true_elim2 : forall b c:bool,
andb b c = true -> c = true.
Proof.
intros b c H.
destruct c.
Case "c=true".
reflexivity.
Case "c=false".
rewrite <- H.
reflexivity.

Answer 1

最后一次证明不起作用的原因是andb定义的方式以及Coq简化此类定义的方式（reflexivity在目标中执行一些简化）。< / p>

以下是Coq。{/ p>中andb的定义

Definition andb (b1 b2:bool) : bool := if b1 then b2 else false.

请注意，我们会根据 first 参数做出决定。因此，当您在第一个参数上使用destruct然后使用reflexivity时，您可以让Coq减少andb b c（到false或{ {1}}）。

由于c的不对称定义， second 参数的案例分析不起作用，因为Coq无法减少函数体内的andb表达式

Using reflexivity in Coq

1 个答案: