如何利用Coq中包含“ forall”的假设？

Question

I am trying to prove the equivalence of P \\/ Q and ~ P -> Q , under the assumption of Excluded Middle,

Theorem eq_of_or :
  excluded_middle ->
  forall P Q : Prop,
    (P \/ Q) <-> (~ P -> Q).

where the Excluded Middle is the following.

Definition excluded_middle := forall P : Prop, P \/ ~ P.

Actually, the proof of one direction does not require the Excluded Middle. In my attempt at proving the other direction, I get stuck when I am trying to utilize the Excluded Middle among the hypotheses,

Proof.
  intros EM P Q. split.
  { intros [H | H]. intros HNP. 
    - unfold not in HNP. exfalso.
      apply HNP. apply H.
    - intros HNP. apply H. }
  { intros H. unfold excluded_middle in EM.
    unfold not in EM. unfold not in H.
  }

where the current environment is the following:

1 subgoal
EM : forall P : Prop, P \/ (P -> False)
P, Q : Prop
H : (P -> False) -> Q
______________________________________(1/1)
P \/ Q

I understand that under such circumstance, what we need to do next is to do something like the "case analysis" of P, including the use of tactics left and right , if my proof makes sense till now.

Thanks in advance for any advice and suggestion!

Answer 1

You can instantiate EM : forall P : Prop, P \\/ ~ P with any proposition (I instantiated it with P below and destructed it immediately), since EM is essentially a function that takes an arbitrary proposition P and returns a proof of either P or ~ P .

Theorem eq_of_or' :
  excluded_middle ->
  forall P Q : Prop, (~ P -> Q) -> P \/ Q.
Proof.
  intros EM P Q.
  destruct (EM P) as [p | np].     (* <- the key part is here *)
  - left. apply p.
  - right.
    apply (H np).
    (* or, equivalently, *)
    Undo.
    apply H.
    apply np.
    Undo 2.
    (* we can also combine two `apply` into one: *)
    apply H, np.
Qed.