如何在 Coq 中证明 a*b*c=a*(b*c)？

Question

Im trying to prove the above question. I have been given a definition of an induction:

Definition nat_ind 
  (p : nat -> Prop)
  (basis : p 0)
  (step : forall n, p n -> p (S n)) :
    forall n, p n := fix f n :=
      match n return p n with
      | 0 => basis
      | S n => step n (f n)
      end.

This is my attempt, but don't know how to finish

Goal forall a b c, a * b * c = a * (b * c).
Proof. 
 apply nat_ind.
  - intros a b c. revert a.
    apply (nat_ind (fun a => a * b * c = a * (b * c))); simpl.
    + reflexivity.
    + intros. f_equal. intros. 

Answer 1

After your very first nat_ind invocation, if you look at your goal, you see that Coq did not do the right thing at all!

______________________________________(1/3)
forall a b c : nat, a * b * c = a * (b * c)
______________________________________(2/3)
nat ->
(forall a b c : nat, a * b * c = a * (b * c)) ->
forall a b c : nat, a * b * c = a * (b * c)
______________________________________(3/3)
nat

What happened here is that it made a guess for your motive p , and decided to unify it with fun (_: nat) => <YOUR_WHOLE_GOAL> , a function which given any nat would give your goal... Yes, this is silly!

One way to nudge it into doing the induction on a is by explicitly forcing it to do so, with:

apply nat_ind with (n:= a)

(where the n matches the name used in your definition of nat_ind )

After this, you get the much more reasonable goals:

______________________________________(1/2)
forall b c : nat, 0 * b * c = 0 * (b * c)
______________________________________(2/2)
forall n : nat,
(forall b c : nat, n * b * c = n * (b * c)) ->
forall b c : nat, S n * b * c = S n * (b * c)

where indeed a has been replaced by 0 and S n respectively.

[EDIT: I guess this does not quite answer your question as you had gotten your way to the same point with the second induction call...]

To solve your goal, it will help a lot to have a property about distributivity of multiplication over addition:

forall n m p, (n + m) * p = n * p + m * p

All of these, as well as what you're trying to prove, already exists in Coq. Is this homework? Are you just training?