Reputation: 35
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!
Upvotes: 3
Views: 3722
Reputation: 15404
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.
Upvotes: 2