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I am currently confused about how to prove the following theorem:

Theorem excluded_middle2 : 
 (forall P Q : Prop, (P -> Q) -> (~P \/ Q)) -> (forall P, P \/ ~P).

I am stuck here:

Theorem excluded_middle2 : 
  (forall P Q : Prop, (P -> Q) -> (~P \/ Q)) -> (forall P, P \/ ~P).
Proof.
  intros.
  evar (Q : Prop).
  specialize H with (P : Prop) (Q : Prop).

I know that it's impossible to simply prove the law of excluded middle in coq, but I really want to know with this given theorem is it possible to prove the law of excluded middle?

Mark Rotteveel
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Pikapikapi Kaori
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1 Answers1

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Yes, you can. One way, using ssreflect, is as follows (there are probably shorter ways):

Lemma orC P Q : P \/ Q -> Q \/ P.
Proof. by case; [right | left]. Qed.

Theorem excluded_middle2 : 
 (forall P Q : Prop, (P -> Q) -> (~ P \/ Q)) -> (forall P, P \/ ~ P).
Proof.
move=> orasimply P.
have pp : P -> P by [].
move: (orasimply P P pp).
exact: orC.
Qed.
Pierre Jouvelot
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