如何证明从典型类型到“Prop”的所有函数 P、Q，“forall a, b, P(a) or Q(b) 成立”当且仅当“forall a, P(a), or, forall b, Q (b)，持有”？
[英]How to prove for all functions P, Q from typical type to `Prop`, “forall a, b, P(a) or Q(b) holds” iff “forall a, P(a), or, forall b, Q(b), holds”?
问题描述
Lemma forall_P_Q_a_b_notPa_or_notPb_iff_forall_a_notPa_or_forall_b_notPb : forall (T : Type) (P Q : T->Prop),
(forall a b, P a \/ Q b) <-> ((forall a, P a) \/ (forall b, Q b))
.
Proof.
intros. split; intros.
- admit.
- destruct H; auto.
Admitted.
I proved the one side easily, but couldn't find a way to prove the other side.
我很容易证明了一方，但找不到证明另一方的方法。 edestruct didn't work for me due to the scope.由于edestruct ，edestruct 对我不起作用。 How should I prove the theorem?我应该如何证明这个定理？
1 个解决方案
解决方案1
3 已采纳 2021-02-20 15:25:07
I believe this requires classical logic:我相信这需要经典逻辑：
Require Import Coq.Logic.Classical.
Lemma forall_P_Q_a_b_notPa_or_notPb_iff_forall_a_notPa_or_forall_b_notPb : forall (T : Type) (P Q : T->Prop),
(forall a b, P a \/ Q b) <-> ((forall a, P a) \/ (forall b, Q b))
.
Proof.
intros T P Q. split; intros H.
- destruct (classic (forall a, P a)) as [HP|HP]; auto. (* Classical reasoning *)
destruct (not_all_ex_not _ _ HP) as [a Ha]. (* Also here *)
right; intros b.
specialize (H a b). tauto.
- destruct H; auto.
Qed.
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