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Partial proof that Axiom of Choice --> Law of Excluded Middle in Coq
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Require Import Coq.Logic.ClassicalChoice. | |
Definition two (n : nat) : Prop := n = 0 \/ n = 1. | |
Lemma zero_in_two : two 0. | |
Proof. | |
unfold two. left. reflexivity. | |
Qed. | |
Lemma one_in_two : two 1. | |
Proof. | |
unfold two. right. reflexivity. | |
Qed. | |
Definition A (n : nat) (P : Prop) (x : two n) : Prop := n = 0 \/ P. | |
Lemma zero_in_A : forall (P : Prop), A 0 P zero_in_two. | |
Proof. | |
intros. | |
unfold A. | |
left. reflexivity. | |
Qed. | |
Definition B (n : nat) (P : Prop) (x : two n) : Prop := n = 1 \/ P. | |
Lemma one_in_B : forall (P : Prop), B 1 P one_in_two. | |
Proof. | |
intros. | |
unfold B. | |
left. reflexivity. | |
Qed. | |
(* What should this be? *) | |
Definition R (n : nat) (n' : nat) (P : Prop) (x : two n) (y : two n') (a : A n P x) (b : B n' P y) : Prop := (n = 0 /\ n' = 1) \/ P. | |
(* How to prove this? Is this actually the inhabitation statement? *) | |
Lemma inhabited : forall (n : nat) (n' : nat) (P : Prop) (x : two n) (y : two n'), | |
(forall a : A n P x, exists b : B n' P y, R n n' P x y a b). | |
Admitted. | |
Theorem choiceABR : forall (n : nat) (n' : nat) (P : Prop) (x : two n) (y : two n'), | |
exists f : A n P x -> B n' P y, (forall a : A n P x, R n n' P x y a (f a)). | |
intros. | |
apply choice. | |
apply inhabited. | |
Qed. | |
Theorem lem : forall (P : Prop), P \/ ~P. | |
Admitted. |
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