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I'm very new at coq.(I'm reading now Poly section in Software Foundation)

In Basics section, they define ble_nat function that is x <= y, then I want to prove transitive law about this, like:

Notation "x =< y" := (ble_nat x y) (at level 50, left associativity) : nat_scope.

Theorem ble_trans: forall (n m o:nat),
   n =< m = true -> m =< o = true -> n =< o = true.
Proof.
(* proof *)

But I could not prove this by using simpl, destruct, induction, rewrite or apply tactic.

I googled and found out there is already proved library of this, but I could not found out code.

How would I prove this ?

french_cruller
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  • Can you explain what did you tried or where are you stuck in proof. – Divyanshu Ranjan Jan 28 '22 at 13:42
  • sorry.. somehow, I proved transitivity, but change theorem to `forall (n m:nat), n =< m =true -> exists o, m =< o = true -> n =< o = true.` actually, I am also unfamiliar with those logics, so my first lemma was wrong hypothesis? – french_cruller Jan 28 '22 at 14:46

1 Answers1

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To prove forall (n m : nat), n =< m =true -> exists o, m =< o = true -> n =< o = true, it is enough to show o := S m satisfy existential quantifier.

Theorem bleS : forall (n m: nat), n =< m = true -> n =< S m = true.
Proof.
  intros n.
  induction n.
  + intros m H. reflexivity.
  + intros m H. destruct m.
    - simpl in H. discriminate.
    - simpl. simpl in H. apply IHn. exact H.
Qed.

Theorem ble_trans_ex: forall (n m :nat),
   n =< m = true -> exists o, m =< o = true -> n =< o = true.
Proof.
  intros n m H1.
  apply ex_intro with (x := S m).
  intros H2. apply bleS. exact H1.
Qed.
Divyanshu Ranjan
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