Created
May 29, 2011 17:54
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(∃x.P(x)) → Q ⇔ ∀x.(P(x) → Q)の証明
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Lemma L0 : forall (A : Type) (P : A -> Prop) (Q : Prop), | |
((exists x, P x) -> Q) -> (forall x, P x -> Q). | |
Proof. | |
intros. | |
apply H. | |
exists x. | |
apply H0. | |
Qed. | |
Lemma L1 : forall (A : Type) (P : A -> Prop) (Q : Prop), | |
(forall x, P x -> Q) -> ((exists x, P x) -> Q). | |
Proof. | |
intros. | |
case H0. | |
apply H. | |
Qed. | |
Theorem L2 : forall (A : Type) (P : A -> Prop) (Q : Prop), | |
((exists x, P x) -> Q) <-> (forall x, P x -> Q). | |
Proof. | |
intros. | |
unfold iff. | |
split. | |
apply L0. | |
apply L1. | |
Qed. |
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