How to prove (p -> q) -> (~ p \/ q) in Coq

Question

I am trying to prove (p -> q) -> (~ p / q) in Coq using the Axiom:

Axiom tautology : forall P:Prop, P \/ ~ P.

I am trying to convert ~ p / q into ~ p / p by applying p -> q. So do something like this:

Theorem Conversion: forall (p q: Prop),(p -> q) -> (~ p \/ q).
Proof.
  intros p q.
  intros p_implies_q.
  (do something here, change ~p\/q into ~p\/p)
  apply tautology...

But I don't know how can I do this. And if there is a better way to do this, please tell me. Thanks!.

Answer 1

One way to use your tautology is with the tactic destruct . This allows you reduce to the cases where p is true and where p is not true.

Axiom tautology : forall P:Prop, P \/ ~ P.

Theorem Conversion: forall (p q: Prop),(p -> q) -> (~ p \/ q).
Proof.
  intros p q.
  intros p_implies_q.
  destruct (tautology p) as [p_true | p_not_true].
  - (* prove ~p \/ q using p *)
  - (* prove ~p \/ q using ~p *)
Qed.

Can you see how to prove ~p \\/ q in each of those cases?