[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”?
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.
我很容易证明了一方，但找不到证明另一方的方法。 由于edestruct ，edestruct 对我不起作用。 我应该如何证明这个定理？
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.
在 Coq 中如何证明或证伪 `forall (PQ : Prop), (P -> Q) -> (Q -> P) -> P = Q.`？
[英]How or is that possible to prove or falsify `forall (P Q : Prop), (P -> Q) -> (Q -> P) -> P = Q.` in Coq?
如何用给定的假设证明排除中间（forall PQ: Prop, (P -> Q) -> (~P \/ Q)）？
[英]How can I prove excluded middle with the given hypothesis (forall P Q : Prop, (P -> Q) -> (~P \/ Q))?
如何证明所有人（pq：Prop），〜p->〜（（p-> q）-> p） 使用coq
[英]How to prove forall (p q:Prop), ~p->~((p ->q) ->p). using coq
如何证明 (forall x, P x /\\ Q x) -> (forall x, P x)
[英]How to prove (forall x, P x /\ Q x) -> (forall x, P x)
在 Coq 中提供示例，其中 (AB: Prop), P: Prop -> Type，这样 A <-> B，但不能用 PB 替换 PA
[英]Provide example in Coq where (A B: Prop), P: Prop -> Type, such that A <-> B, but one cannot replace P A with P B
你如何阅读 coq 量词`forall P: Set -> Prop`？
[英]How do you read the coq quantifier `forall P: Set -> Prop`?
我怎样才能证明forall ab，a <=？ b = true -> a <=? S b = 在 Coq 中为真
[英]How can I proove forall a b, a <=? b = true -> a <=? S b = true in Coq
声明:本站的技术帖子网页，遵循CC BY-SA 4.0协议，如果您需要转载，请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.