Coq将功能扩展为目标的一部分是什么意思？

Question

I was trying to solve the following theorem and got stuck at the last simpl. :

Lemma nonzeros_app : forall l1 l2 : natlist,
  nonzeros (l1 ++ l2) = (nonzeros l1) ++ (nonzeros l2).
Proof.
  intros l1 l2. induction l1 as [| n' l' IHl'].
  -simpl. reflexivity.
  -simpl. 
Qed.

at that point Coq changes the goal from:

1 subgoal (ID 170)

  n' : nat
  l', l2 : natlist
  IHl' : nonzeros (l' ++ l2) = nonzeros l' ++ nonzeros l2
  ============================
  nonzeros ((n' :: l') ++ l2) = nonzeros (n' :: l') ++ nonzeros l2

to:

1 subgoal (ID 185)

  n' : nat
  l', l2 : natlist
  IHl' : nonzeros (l' ++ l2) = nonzeros l' ++ nonzeros l2
  ============================
  match n' with
  | 0 => nonzeros (l' ++ l2)
  | S _ => n' :: nonzeros (l' ++ l2)
  end =
  match n' with
  | 0 => nonzeros l'
  | S _ => n' :: nonzeros l'
  end ++ nonzeros l2

which seems completely mysterious to me. What does it mean when Coq just copy pastes the definition of a function into my goal? What do I even do with this?

---

Context of Question:

Someone told me that the solution is:

Lemma nonzeros_app : forall l1 l2 : natlist,
  nonzeros (l1 ++ l2) = (nonzeros l1) ++ (nonzeros l2).
Proof.
  intros l1 l2. induction l1.
  - simpl. reflexivity.
  - simpl. { induction n.
             - ...
             - ... }
Qed.

which made me want to understand why they use induction on n since it feels it would never occur to me to use induction there. So I am asking, why? But I realized that before I could ask that why I didn't even understand the proof state before that since it just seemed to copy paste a function to a proof state (which makes no sense to me). So before I asked why use induction there I have to ask what does the proof state before that, maybe that would yield light into why induction on n .

Answer 1

I am assuming that you have defined nonzeros in the following way (or similar):

Require Import List.
Import ListNotations.


Definition natlist := list nat.

Fixpoint nonzeros (l : natlist) :=
  match l with
  | [] => []
  | 0 :: xs => nonzeros xs
  | x :: xs => x :: nonzeros xs
  end.

So that nonzeros is recursive with structural decreasing on l . Coq's simpl tactic employs a heuristic in which it unfolds definitions of fixpoints if they are applied to a term that has a constructor as the head symbol. In your case, eg, nonzeros (n' :: l') , the constant nonzeros is followed by a term formed by a constructor, Cons (= :: ). Coq performs a so-called "delta reduction", replacing the occurrence of nonzero with its definition. Since that definition is a match , you get a match as your new term. Further substitutions do simplify it a bit, but cannot eliminate the two cases: one for zero head and one for nonzero head.

Same happens to the occurrence nonzeros ((n' :: l') ++ l2) , which is first simplified to nonzeros (n' :: (l' ++ l2)) , so that again, the head of the argument is Cons .

If you want to avoid exposing match expressions when simplifying, you can put the following directive after the definition of nonzeros :

Arguments nonzeros l : simpl nomatch.

This specifically tells simpl to avoid expanding a term if this will end up exposing a match at the location of the change.

As for the induction used by your friend here: it is applied to force a case split over n' , so that each case ( n' = 0 , n' = S _ ) can be handled separately. Indeed, induction is not needed here. A simple case split ( case n' ) will do the same.