(A -> B) /\ (B -> C) -> (A -> C) in Coq?

Question

I am learning Coq via the book software foundations, and have trouble proving the following lemma (which I require to prove other theorems.)

Lemma if_trans : 
  forall (P Q R: Prop),
    (P -> Q) /\ (Q -> R) -> (P ->R).
Proof.
  intros P Q R [E1 E2]. 
Admitted.

This is where I get stuck. I can't destruct on propositions, and I although I can apply E2 in E1 , it generates two subgoals (I don't understand why), and the second subgoal is logically incorrect to prove. Please help. Also I know only the following tactics:

simpl , reflexivity, symmetry, destruct, induction, apply, replace, .. in , exfalso, discriminate, injection, split, left , right, intros, unfold, assert, rewrite.

Q2: Another question on somewhat similar lines. I need to use the above proved lemma for proving other theorems. So suppose I have two hypothesis H1: P -> Q and H2: Q -> R and the goal is P -> R . How can I use the above lemma to prove the goal in this case ie how can I combine H1 and H2 into a single hypothesis H : (P ->Q ) /\\ (Q -> R) ?

Answer 1

You're trying to prove the following:

Lemma if_trans :
  forall (P Q R : Prop),
    (P -> Q) /\ (Q -> R) -> (P -> R).

but you only introduce P, Q, R the proof of P -> Q and that of Q -> R so it leaves you to prove P -> R . In the same way you can use intro p to get p : P as an extra assumption and then prove R .

To summarise you have

P, Q, R : Prop
E1 : P -> Q
E2 : Q -> R
p : P
===============
R

after the tactic

intros P Q R [E1 E2] p.

(notice the extra p ).

Perhaps then it will appear clearer how to prove it.

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When you use apply E2 in E1 it basically sees that E1 proves Q under the assumption that P holds, so it applies E2 : Q -> R in it to obtain R and asks on the side that you provide evidence for P .

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For your second question, if you apply your lemma, it will ask for (P -> Q) /\\ (Q -> P) which you can prove with split and then assumption . You should not try to combine P -> Q and Q -> R into (P -> Q) /\\ (Q -> P) though, but if really you must you can use pose proof (conj H1 H2) as H .