How to prove the following in Coq?

Question

How to prove the following in Coq?

(p->q)->(~q->~p)

Here is what I started with:

Lemma work : (forall p q : Prop, (p->q)->(~q->~p)).
Proof.
  intros p q.
  intros p_implies_q not_q.
  intros p_true.

Answer 1

Lemma work : (forall p q : Prop, (p->q)->(~q->~p)).
Proof.
  intros p q.
  intros p_implies_q not_q.
  intros p_true.
  apply not_q.
  apply p_implies_q.
  auto.
Qed.

Some remarks:

• whenever you have a goal of the form A -> B , you can use intros H command to add H: A into your list of premises and be left with a goal B .

• ~P is a syntactic sugar for P -> False . Thus, if your goal is ~P , then intros H will add H: P to your premises list and reduce the goal to False

• If your goal is Q and you have a premise H: P -> Q , executing command apply H will change your goal to P

• Consecutive intros commands can be merged into one, so the proof can be shortened to

  Proof. intros pq p_implies_q not_q p_true. apply not_q. apply p_implies_q. auto. Qed.