Coq's proof #Coq

Question

I try to solve this proof but I don't find how to it. I have two subgoals but I don't even know if it's correct.

Here the lemma that I trid to solve with this but I'm stuck :

2 subgoals

a, b : Nat

H : Equal (leB a b) True

______________________________________(1/2)

Equal match b with

| Z => False

| S m' => leB a m'

end (leB a b) / Equal (leB b (S a)) (leB a b)

______________________________________(2/2)

Equal (leB (S a) b) True / Equal (leB b (S a)) True

Inductive Bool : Type :=
          True : Bool | False : Bool.

Definition Not(b : Bool) : Bool :=
          Bool_rect (fun a => Bool)
                     False
                     True
                     b.



Lemma classic : forall b : Bool, Equal b (Not (Not b)).
Proof.
intro.
induction b.
simpl.
apply refl.
simpl.
apply refl.
Qed.

Definition Equal(T : Type)(x y : T) : Prop :=
           forall P : T -> Prop, (P x) -> (P y).

Arguments Equal[T].
(* Avec certaines versions Arguments Equal[T] *)

Lemma refl : forall T : Type, forall x : T, Equal x x.
Proof.
intros.
unfold Equal.
intros.
assumption.
Qed.

Fixpoint leB n m : Bool :=
  match n, m with
    | Z, _ => True
    | _, Z => False
    | S n', S m' => leB n' m'
  end.

Your question doesn't provide a definition to `Equal` or `Bool`. Also you need to tell us what you tried and why you are stuck if we are to help you. — Théo Winterhalter, Jan 01 '22 at 13:56
@I edit my questions thanks. I'm stuck with 2 sub-goals and I don't know how to continue. — Mervin Macky, Jan 01 '22 at 14:33
You're still not giving the definitions, and you're not explaining why you are stuck. How would you do it on paper and why doesn't it work for you in Coq? — Théo Winterhalter, Jan 01 '22 at 18:52

score 1 · Accepted Answer · answered Jan 02 '22 at 16:59

1

First, don't introduce all variables in the beginning with intros. You will get a too weak induction hypothesis. Just introduce a.

Then in each branch, consider the different cases of b with the destruct tactic. It will simplify your goal and you can see if it is the left or the right side of goal that is true, and use your refl lemma to finish the goal.

The last case require that you use your induction hypothesis, and it is here that it is important that it holds for all b, not just one specific b.

Also, you didn't provide a definition for you Nat type, I guess it is something like this:

Inductive Nat := Z | S (n:Nat).

Here is a proof.

Lemma Linear : forall a b, (Equal (leB a b) True) \/ (Equal (leB b a) True). 
  Proof.
induction a. 
- intros b. destruct b; simpl.
    + left. apply refl. 
    + left. apply refl.
- intros b. destruct b; simpl.
  + right. apply refl.
  + destruct (IHa b) as [Hleft | Hright].
    ++ left. apply Hleft.
    ++ right. apply Hright.
Qed.

While it may not be as insightful, you can also use tactics that try these steps to get a shorter proof.

induction a; destruct b; firstorder.

will also prove your lemma.

answered Jan 02 '22 at 16:59

larsr

5,447
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38

Hello, thanks for your help. I understand better now. I will try your advices to others lemma – Mervin Macky Jan 04 '22 at 21:40
Hello, it's me again. I just don't understand the meaning of "-" and "++". Have a nice day – Mervin Macky Jan 05 '22 at 13:24
They are just indentation "bullets" that show where a sub-proof starts and ends. They are not needed, but makes the proof script more readable. Its somewhat like brackets around code blocks in C or white space indentation in python, but they start with a "bullet", you can use both minus-signs, pluses or asterisks. So for example, the first tactic `induction` creates two sub-goals. With the - bullet you can easily see where the proof of the first goal starts and ends. Destruct also creates two subgoals, marked with +. etc. – larsr Jan 05 '22 at 17:10
Thank you for your quick answer. Good evening :) – Mervin Macky Jan 05 '22 at 21:04

Coq's proof #Coq

1 Answers1