Definition manual_grade_for_plus_comm_informal : option (nat*string) := None.
(** [] *)

(** **** Exercise: 2 stars, standard, optional (eqb_refl_informal)  

    Write an informal proof of the following theorem, using the
    informal proof of [plus_assoc] as a model.  Don't just
    paraphrase the Coq tactics into English!

    Theorem: [true = n =? n] for any [n].

    Proof: (* FILL IN HERE *)

    [] *)

(* ################################################################# *)
(** * More Exercises *)

(** **** Exercise: 3 stars, standard, recommended (mult_comm)  

    Use [assert] to help prove this theorem.  You shouldn't need to
    use induction on [plus_swap]. *)

Theorem plus_swap : forall n m p : nat,
  n + (m + p) = m + (n + p).
Proof.
  (* FILL IN HERE *) Admitted.

(** Now prove commutativity of multiplication.  (You will probably
    need to define and prove a separate subsidiary theorem to be used
    in the proof of this one.  You may find that [plus_swap] comes in
    handy.) *)

Theorem mult_comm : forall m n : nat,
  m * n = n * m.
Proof.
  (* FILL IN HERE *) Admitted.
(** [] *)

(** **** Exercise: 3 stars, standard, optional (more_exercises)  

    Take a piece of paper.  For each of the following theorems, first
    _think_ about whether (a) it can be proved using only
    simplification and rewriting, (b) it also requires case
    analysis ([destruct]), or (c) it also requires induction.  Write
    down your prediction.  Then fill in the proof.  (There is no need
    to turn in your piece of paper; this is just to encourage you to
    reflect before you hack!) *)

Check leb.

Theorem leb_refl : forall n:nat,
  true = (n <=? n).
Proof.
  (* FILL IN HERE *) Admitted.

Theorem zero_nbeq_S : forall n:nat,
  0 =? (S n) = false.
Proof.
  (* FILL IN HERE *) Admitted.

Theorem andb_false_r : forall b : bool,
  andb b false = false.
Proof.
  (* FILL IN HERE *) Admitted.

Theorem plus_ble_compat_l : forall n m p : nat,
  n <=? m = true -> (p + n) <=? (p + m) = true.
Proof.
  (* FILL IN HERE *) Admitted.

Theorem S_nbeq_0 : forall n:nat,
  (S n) =? 0 = false.
Proof.
  (* FILL IN HERE *) Admitted.

Theorem mult_1_l : forall n:nat, 1 * n = n.
Proof.
  (* FILL IN HERE *) Admitted.

Theorem all3_spec : forall b c : bool,
    orb
      (andb b c)
      (orb (negb b)
               (negb c))
  = true.
Proof.
  (* FILL IN HERE *) Admitted.

Theorem mult_plus_distr_r : forall n m p : nat,
  (n + m) * p = (n * p) + (m * p).
Proof.
  (* FILL IN HERE *) Admitted.

Theorem mult_assoc : forall n m p : nat,
  n * (m * p) = (n * m) * p.
Proof.
  (* FILL IN HERE *) Admitted.
(** [] *)

(** **** Exercise: 2 stars, standard, optional (eqb_refl)  

    Prove the following theorem.  (Putting the [true] on the left-hand
    side of the equality may look odd, but this is how the theorem is
    stated in the Coq standard library, so we follow suit.  Rewriting
    works equally well in either direction, so we will have no problem
    using the theorem no matter which way we state it.) *)

Theorem eqb_refl : forall n : nat,
  true = (n =? n).
Proof.
  (* FILL IN HERE *) Admitted.
(** [] *)

(** **** Exercise: 2 stars, standard, optional (plus_swap')  

    The [replace] tactic allows you to specify a particular subterm to
   rewrite and what you want it rewritten to: [replace (t) with (u)]
   replaces (all copies of) expression [t] in the goal by expression
   [u], and generates [t = u] as an additional subgoal. This is often
   useful when a plain [rewrite] acts on the wrong part of the goal.

   Use the [replace] tactic to do a proof of [plus_swap'], just like
   [plus_swap] but without needing [assert (n + m = m + n)]. *)
(* etc... *)