From mathcomp Require Import all_ssreflect.

Set Implicit Arguments.
Unset Strict Implicit.
Unset Printing Implicit Defensive.

Section Nombre.

(* Convertir une liste d'entiers en un entier                                 *)

Definition l2nat l := foldl (fun v i => i + 10 * v) 0 l.

Lemma l2nat_rcons l n : l2nat (rcons l n) = l2nat l * 10 + n.
Proof.
rewrite /l2nat -[rcons _ _]revK rev_rcons foldl_rev.
by rewrite /= -foldl_rev revK addnC mulnC.
Qed.

Compute l2nat [:: 1; 2; 3].

(* Convertir un entier en une liste d'entiers                                 *)

Fixpoint nat2l2 m n :=
   if n < 10 then [:: n] else
   if m is m1.+1 then rcons (nat2l2 m1 (n %/ 10)) (n %% 10) else [::n].

Compute nat2l2 100 123.

Definition nat2l n := nat2l2 n n.

Compute nat2l 123.

Lemma nat2l2K m n : n < 10 ^ m -> l2nat (nat2l2 m n) = n.
Proof.
elim: m n => [[]|m IH n nLD] //=.
case: leqP => _ /=; last by rewrite /l2nat /= addnC.
rewrite l2nat_rcons IH -1?divn_eq //.
rewrite -(ltn_pmul2r (_ : 0 < 10)) // -expnSr (leq_ltn_trans _ nLD) //.
by rewrite {2}(divn_eq n 10) leq_addr.
Qed.

Lemma nat2l2_eq m1 m2 n :
  n < 10 ^ m1 ->  n < 10 ^ m2 -> nat2l2 m1 n = nat2l2 m2 n.
Proof.
wlog : m1 m2 / m1 <= m2 => [H nLm1 nLm2|].
  case: (leqP m1 m2)=> [m1Lm2|m2Lm1]; first by apply: H.
  by apply: sym_equal; apply: H => //; apply: ltnW.
elim: m1 m2 n => [[|m2] [] |] //= m1 IH [|m2] //= n.
rewrite ltnS => m1Lm2 nLm1 nLm2.
case: leqP=> // _; congr rcons.
apply: IH => //.
  rewrite -(ltn_pmul2r (_ : 0 < 10)) // -expnSr (leq_ltn_trans _ nLm1) //.
  by rewrite {2}(divn_eq n 10) leq_addr.
rewrite -(ltn_pmul2r (_ : 0 < 10)) // -expnSr (leq_ltn_trans _ nLm2) //.
by rewrite {2}(divn_eq n 10) leq_addr.
Qed.

Lemma nat2lK n : l2nat (nat2l n) = n.
Proof. by apply: nat2l2K; apply: ltn_expl. Qed.

(******************************************************************************)
(*                                                                            *)
(*              Modélisation des différentes propositions                     *)
(*                                                                            *)
(******************************************************************************)

(* Chiffres décroissants                                                      *)

Definition prop1 l := (1 \in l) == sorted (fun x y => x >= y) l.

Compute prop1 [:: 4; 3; 2; 1; 1; 0].
Compute prop1 [:: 7; 5; 2; 2; 1; 0].

(* Au moins deux chiffres impairs                                             *)

Definition prop2 l := (2 \in l) == (count odd l >= 2).

Compute prop2 [:: 5; 2; 6; 1; 0].
Compute prop2 [:: 1; 1; 3; 2].

(* Tous les chiffres différents                                               *)

Definition prop3 l := (3 \in l) == uniq l.

Compute prop3 [:: 3; 1; 2].
Compute prop3 [:: 4; 4; 3].

(* Quatrième chiffre en partant de la gauche est pair                         *)

Definition prop4 l := (4 \in l) == ~~ odd (nth 0 l 3).

Compute prop4 [:: 4; 3; 1; 6; 0].

(* Produit non divisible par 5                                                *)

Definition prop5 l := (5 \in l) == ~~ (5 %| foldr muln 1 l).

Compute prop5 [:: 4; 2].
Compute prop5 [:: 3; 7].

(* Trois chiffres impairs à la suite                                          *)

Fixpoint todd l := if l is a :: l1 then
                     if l1 is b :: c :: l2 then
                              [&& odd a, odd b & odd c] || todd l1
                     else false
                   else false.

Definition prop6 l := (6 \in l) == todd l.

Compute prop6 [:: 1; 3; 7; 6].
Compute prop6 [:: 6; 2; 1; 1; 1; 3; 4].

(* Est un nombre premier                                                      *)

Definition prop7 l := (7 \in l) == prime (l2nat l).

Compute prop7 [:: 7].
Compute prop7 [:: 3; 7].

(* Pas plus 2 nombres pairs à la suite                                        *)

Fixpoint teven l := if l is a :: l1 then
                     if l1 is b :: c :: _ then
                              [&& ~~ odd a, ~~ odd b & ~~ odd c] || teven l1
                     else false
                   else false.

Definition prop8 l := (8 \in l) == ~~ teven l.

Compute prop8 [:: 8; 2; 2].
Compute prop8 [:: 4; 8; 2; 1].

(* Produit des impairs est un carré parfait                                   *)

Definition perfect_square n :=
   all (fun p => ~~ odd p.2) (prime_decomp n).

Compute perfect_square 47.

Definition iprod l := foldr (fun i v => if odd i then i * v else v) 1 l.

Compute iprod [:: 9; 2; 3].

Definition prop9 l := (9 \in l) == perfect_square (iprod l).

(* Un élément distinct est la somme des autres                                *)

Definition prop0 l := (0 \in l) ==
                        (let sum := foldr addn 0 l in
                         ~~ odd sum && (count (pred1 sum./2) l == 1)).

Compute prop0 [:: 5; 7; 2; 0; 0].
Compute prop0 [:: 1; 0; 1; 3; 1].
Compute prop0 [:: 4; 4; 0].

(* Toutes les propositions                                                    *)

Definition check_prop n :=
  let l := nat2l n in
  [&& prop1 l, prop2 l, prop3 l, prop4 l, prop5 l,
      prop6 l, prop7 l, prop8 l, prop9 l& prop0 l].

(* Une version plus rapide                                                    *)
Definition fcheck_prop n :=
  let l := nat2l n in
  if prop9 l then if prop8 l then if prop0 l then if prop6 l then if prop5 l
  then if prop4 l then if prop3 l then if prop2 l then if prop1 l then prop7 l
  else false else false else false else false else false else false else false
  else false else false.

Lemma Pfcheck_prop n : fcheck_prop n = check_prop n.
Proof.
rewrite /check_prop /fcheck_prop.
case: prop9 => //; case prop8 => //; case: prop0; rewrite ?andbF //.
case: prop6 => //; case prop5 => //; case: prop4; rewrite ?andbF //.
case: prop3 => //; case prop2  => //; case: prop1 => //; case: prop7 => //.
Qed.

(* Le résultat                                                                *)

Definition result := l2nat [:: 9; 4; 2; 2; 1; 0].

(* Le résultat vérifie les 10 propositions                                    *)

Fact Presult : check_prop result.
Proof. by vm_compute. Qed.

(******************************************************************************)
(*                                                                            *)
(*     Tester tous les nombres plus petits que 942210                         *)
(*                                                                            *)
(******************************************************************************)

(* Test à zero sur une liste                                                  *)

Definition is_zero l := all (pred1 0) l.

Lemma is_zero_rcons l a : is_zero (rcons l a) = (is_zero l && (a == 0)).
Proof.
elim: l a => [a| b l IH a] /=; first by rewrite andbT.
by rewrite IH andbA.
Qed.

Lemma head_nat2l2_eq0 m n :
   n < 10 ^ m -> (nth 1 (nat2l2 m n) 0  == 0) == (n == 0).
Proof.
elim: m n => [[]|m IH n nLD] //.
rewrite [nat2l2 _ _]/=.
case: leqP=> [DLn|nLN //].
rewrite nth_rcons.
have->: 0 < size (nat2l2 m (n %/ 10)).
  by case: (m) => /=; case: (_ < 10) => // s; rewrite size_rcons.
rewrite (eqP (IH _ _)); last first.
  rewrite -(ltn_pmul2r (_ : 0 < 10)) // -expnSr (leq_ltn_trans _ nLD) //.
  by rewrite {2}(divn_eq n 10) leq_addr.
by move: (DLn); rewrite -divn_gt0 //; case: (n) DLn => // n1; case: (_ %/ _).
Qed.

Lemma is_zero_nat2l2 m n : n < 10 ^ m -> is_zero (nat2l2 m n) == (n == 0).
Proof.
elim: m n => [[]|m IH n nLD] //=.
case: leqP=> [DLn|nLN /=]; last by rewrite andbT.
rewrite is_zero_rcons (eqP (IH _ _)); last first.
  rewrite -(ltn_pmul2r (_ : 0 < 10)) // -expnSr (leq_ltn_trans _ nLD) //.
  by rewrite {2}(divn_eq n 10) leq_addr.
by rewrite {3}(divn_eq n 10) addn_eq0 muln_eq0 orbF.
Qed.

(* Remplacer tous les chiffres par des 9                                    *)

Definition to_nine l := map (fun i : nat => 9) l.

(* Décrémenter de 1 une liste                                               *)

Fixpoint decrr l :=
  if l is a :: l1 then
     if is_zero l1 then
        if a == 0 then to_nine l else a.-1 :: to_nine l1
     else a :: decrr l1
  else [::].

Definition decr l :=
  if is_zero l then [:: 0] else
  let l1 := decrr l in
  if l1 is 0 :: a :: l2 then a :: l2 else l1.

Compute decr [:: 1; 2; 3].
Compute decr [:: 1].

Lemma decrr_rcons l a :
  decrr (rcons l a) =
     if (a == 0) then rcons (decrr l) 9 else rcons l a.-1.
Proof.
elim: l => //= b l IH.
rewrite is_zero_rcons; case: (boolP (is_zero l)) => [zZ|zNZ] /=; last first.
  by rewrite {}IH; case: (a =P 0) => //.
case: (b =P 0) => [-> | bNZ] /=.
  by rewrite /to_nine map_rcons {}IH; case: (_ == _).
by rewrite /to_nine map_rcons {}IH; case: (_ == _).
Qed.

Lemma Pdecrr m n : n < 10 ^ m -> 0 < n ->
  let l1 := decrr (nat2l2 m n) in
  (if l1 is 0 :: a :: l2 then a :: l2 else l1)  = nat2l2 m n.-1.
Proof.
elim: m n => [[]|m IH [|n] nLm _] //=.
case: (leqP 10  n.+1) => [DLm|] /= ; last first.
  case: (n) nLm => // n1 nLm nLD.
  by rewrite (leq_ltn_trans _ nLD).
rewrite decrr_rcons.
case: (_ =P 0) => [modZ|modNZ] /=; last first.
  have : n %% 10 < 10 by apply: ltn_pmod.
  rewrite leq_eqVlt ltnS => /orP[/eqP nE10 | mL9].
    by case: modNZ; rewrite -addn1 -modnDml addn1 nE10.
  have /= := head_nat2l2_eq0 nLm.
  rewrite ltnNge DLm /=.
  move: DLm mL9; rewrite ltnS leq_eqVlt => /orP[/eqP<-|] //.
  rewrite ltnNge -ltnS => /negPf-> mL9.
  have [-> -> /=]: (n.+1 %/ 10) = (n %/ 10) /\ (n.+1 %% 10) = (n %% 10).+1.
    have ->: n.+1 = n %/ 10 * 10 + (n %% 10).+1.
      by rewrite addnS -divn_eq.
    rewrite divnMDl // [_.+1 %/ _]divn_small ?addn0 //.
    by rewrite modnMDl  // modn_small.
  by case: nat2l2=> [|[|a] l] //=; case: modn.
have nLSD : n.+1 %/ 10 < 10 ^ m.
  rewrite -(ltn_pmul2r (_ : 0 < 10)) // -expnSr (leq_ltn_trans _ nLm) //.
  by rewrite {2}(divn_eq n.+1 10) leq_addr.
move: DLm modZ; rewrite ltnS leq_eqVlt => /orP[/eqP<-|DLm] //=.
  by case: (m).
rewrite leqNgt ltnS DLm /= => modD0.
have posD : 0 < n %/ 10 by rewrite divn_gt0.
have modD9 : n %% 10 = 9.
  move: modD0.
  have : n %% 10 < 10 by apply: ltn_pmod.
  rewrite leq_eqVlt eqSS ltnS => /orP[/eqP-> //| DD].
  by rewrite -addn1 -modnDml modn_small addnC.
have posSD : 0 < n.+1 %/ 10.
  by rewrite divn_gt0 // (leq_trans DLm).
rewrite modD9; move: (IH _ nLSD posSD).
have -> /=: (n.+1 %/ 10) = (n %/ 10).+1.
  by rewrite {1}[n](divn_eq n 10) -addnS modD9 divnMDl ?addn1.
case: decrr => [|[[HH |a1 l1 <-]|a l <-]] //=.
  by case: (m) => [|m1] //=; case: (_ < _) => //; case: nat2l2.
have nLD : n %/ 10 < 10 ^ m.
  by apply/(leq_ltn_trans _ nLSD)/leq_div2r.
have := nat2l2K nLD; rewrite -HH /l2nat/= muln0.
by case: (_ %/ _) posD.
Qed.

Lemma Pdecr n : decr (nat2l n) = nat2l n.-1.
Proof.
suff {n}H : forall m n, n < 10 ^ m -> decr (nat2l2 m n) = nat2l2 m n.-1.
  suff->: nat2l n.-1 = nat2l2 n n.-1; first by apply/H/ltn_expl.
  apply: nat2l2_eq; first by apply: ltn_expl.
  by apply: leq_trans (leq_ltn_trans (leq_pred _) _) (ltn_expl _ _).
move=> [|m] [|n] // nLD.
rewrite /decr (eqP (is_zero_nat2l2 _)) //=.
by apply: (Pdecrr nLD).
Qed.

(* Une version plus efficace pour vérifier les propositions                   *)

Definition check_propl l :=
  if prop9 l then
    if prop8 l then
     if prop0 l then
     if prop6 l then
     if prop5 l then
     if prop4 l then
     if prop3 l then
     if prop2 l then
     if prop1 l then prop7 l else false
     else false else false else false else false else false else false
     else false else false.

Lemma Pcheck_propl n : check_prop n = check_propl (nat2l n).
Proof.
rewrite /check_prop /check_propl.
case: prop9 => //; case prop8 => //; case: prop0; rewrite ?andbF //.
case: prop6 => //; case prop5 => //; case: prop4; rewrite ?andbF //.
case: prop3 => //; case prop2  => //; case: prop1 => //; case: prop7 => //.
Qed.

(* Vérifier en décrémentant progressivement                                   *)
Fixpoint check_less l n :=
  let l1 := decr l in
  if n is n1.+1 then check_propl l1 || check_less l1 n1
                         else false.

Lemma Pcheck_less n m :
  check_less (nat2l n) m = false ->  forall k, n - m <= k < n -> ~ check_prop k.
Proof.
elim: m n => [m _ k|m IH [|n]] //=.
  by rewrite subn0 leqNgt; case: (_ <= _).
rewrite Pdecr /= subSS -Pcheck_propl.
case: (boolP (check_prop _)) => //= /negP cP /IH H k.
rewrite [k < _]leq_eqVlt eqSS ltnS; case: eqP=> [->|kDl] //=.
by exact: H.
Qed.

(* Tous les nombres plus petits ne vérifient pas les 10 propositions          *)

Fact Lresult : forall m, m < result -> ~ check_prop m.
Proof.
move=> m mLr.
have: result - result <= m < result by rewrite subnn leq0n.
move: {mLr}m.
have F : check_less (nat2l result) result = false
  by vm_cast_no_check (refl_equal false).
apply: (Pcheck_less F).
Qed.

End Nombre.


Require Import ssreflect ssrbool.

Goal forall A B, A /\ B -> B /\ A.
Proof.
    by move=> A B [].
Qed.

Goal forall A B, A \/ B -> B \/ A.
Proof.
    by move=> A B [a | b] ; [right | left].
Qed.

Goal forall (A B : bool), A \/ B -> B \/ A.
Proof.
    by move=> [] [] AoB; apply/orP; move/orP : AoB.
Qed.

Goal forall A (a : A) (P : A -> Prop),  (forall P : Prop, P \/ ~P) -> exists x, P x -> forall y, P y.
Proof.
    move=> A a P EP.
    case: (EP (exists u : A, ~P u)) => [[u nPu]| nenPu].
    by exists u.
    exists a => pa m.
    case: (EP (P m)) => // npm.
    by case: nenPu; exists m.
    Show Proof.
Qed.


(* This file is generated by Why3's Coq driver *)
(* Beware! Only edit allowed sections below    *)
Require Import ZArith.
Require Import Rbase.
Parameter pow2: Z -> R.

Axiom Power_0 : ((pow2 0%Z) = 1%R).

Axiom Power_s : forall (n:Z), (0%Z <= n)%Z ->
  ((pow2 (n + 1%Z)%Z) = (2%R * (pow2 n))%R).

Axiom Power_p : forall (n:Z), (n <= 0%Z)%Z ->
  ((pow2 (n - 1%Z)%Z) = ((05 / 10)%R * (pow2 n))%R).

Axiom Power_s_all : forall (n:Z), ((pow2 (n + 1%Z)%Z) = (2%R * (pow2 n))%R).

Axiom Power_p_all : forall (n:Z),
  ((pow2 (n - 1%Z)%Z) = ((05 / 10)%R * (pow2 n))%R).

Axiom Power_1 : ((pow2 1%Z) = 2%R).

Axiom Power_neg1 : ((pow2 (-1%Z)%Z) = (05 / 10)%R).

(* YOU MAY EDIT THE CONTEXT BELOW *)
Open Scope Z_scope.
(* DO NOT EDIT BELOW *)

Theorem Power_non_null : forall (n:Z), ~ ((pow2 n) = 0%R).
(* YOU MAY EDIT THE PROOF BELOW *)

intro n.
assert (h:n>=0 \/ n<=0) by omega.
destruct h.
cut (0 <= n); auto with zarith.
apply Z_lt_induction with
  (P:= fun n =>
       0 <= n -> pow2 n <> 0%R);auto with zarith.
intros x Hind Hxpos.
assert (hx:x = 0 \/ x >0) by omega.
destruct hx.
subst x.
rewrite Power_0;auto with *.
replace (x) with (x-1+1) by omega.
rewrite Power_s;auto with *.

rewrite Rmult_neq_0_reg with (r1:=2%R) (r2:=pow2 (x - 1)).
apply Hind.

Qed.
(* DO NOT EDIT BELOW *)


(**
* Canonical finite types
[FIN.t n] is a type with [n] members,
one member naturally corresponding to each [m < n].
*)

Module FIN.

  (**
  ** Definitions
  *)

  Inductive Singleton (n : nat) := lift : Singleton n.

  (**
  *** Canonical finite types
  [t 0] is the empty type.
  [t (S n)] is the sum of [t n] and the singleton of [n].
  *)

  Fixpoint t (n : nat) : Type :=
  match n with
  | O   => Empty_set
  | S n => sum (t n) (Singleton n)
  end.

  Definition last {n} : t (S n) := inr (lift n).

  (**
  An element of
  [FIN.t n]
  is also an element of
  [FIN.t (S n)].
  More generally,
  it is also an element of
  [FIN.t (k + n)].
  *)

  Definition up {n} (x : t n) : t (S n) := inl x.
  Local Notation "++ x" := (up x) (at level 6, right associativity).

  Fixpoint up' {n k} (x : t n) : t (k + n) :=
  match k with
  | O   => x
  | S k => ++(up' x)
  end.

  (**
  *** To and from naturals
  *)

  Definition ofNat (n k : nat) : t (k + S n) := up' last.

  Fixpoint toNat {n} : t (S n) -> nat :=
  match n with
  | O   => fun _ => 0
  | S n => fun x => match x with
                    | inl x => toNat x
                    | inr x => n
                    end
  end.

  (**
  ** Notations
  *)

  Module NOTATIONS.

    Delimit Scope FIN with FIN.

    Notation "++ x" := ++x (at level 6, right associativity) : FIN.

  End NOTATIONS.

  (**
  ** Theorems
  *)

  Module Type THMS_SIG.

    Axiom t0Empty : t 0 -> False.

    Axiom t1Singleton : forall x : t 1, x = last.

    (**
    One would want to prove
    the correctness of [ofNat] and [toNat],
    i.e. that [ofNat (toNat x) k = up' x] and
    that [toNat (ofNat n k) = n].
    It is however nontrivial to even state these,
    since [(k + S n) = (S _)] does not automatically typecheck.
    For my purposes,
    I do not need to prove their correctness so
    I will skip doing so.
    *)

  End THMS_SIG.

  (**
  ** Proofs
  *)

  Module THMS : THMS_SIG.

    Theorem t0Empty (x : t 0) : False.
    Proof. destruct x. Qed.

    Theorem t1Singleton (x : t 1) : x = last.
    Proof. destruct x as [x | x]; destruct x. reflexivity. Qed.

  End THMS.

End FIN.


Require Import Omega.
Lemma le_lt_S_eq : forall n p:nat, n <= p -> p < S n -> n = p.
Proof.
 intros; omega.
Qed.


Module NatDefs.

Theorem S_n_eq_add : forall n : nat,
    1 + n = S n.
Proof.
  intros.
  simpl.
  reflexivity.
Qed.

Theorem mult_S_1 : forall n m : nat,
    S n = m ->
    m * (1 + n) = m * m.
Proof.
  intros.
  rewrite S_n_eq_add.
  rewrite H.
  reflexivity.
Qed.


(** Generated by coq-of-ocaml *)
Require Import OCaml.OCaml.

Local Set Primitive Projections.
Local Open Scope string_scope.
Local Open Scope Z_scope.
Local Open Scope type_scope.
Import ListNotations.

Unset Positivity Checking.
Unset Guard Checking.

Inductive nat : Set :=
| O : nat
| S : nat -> nat.

Inductive natural : Set :=
| Zero : natural
| Succ : natural -> natural.

Fixpoint plus (plus_arg0 : natural) (plus_arg1 : natural) {struct plus_arg0}
  : natural :=
  match plus_arg0 with
  | Zero => plus_arg1
  | Succ n => Succ (plus n plus_arg1)
  end.

Fixpoint mult (mult_arg0 : natural) (mult_arg1 : natural) {struct mult_arg0}
  : natural :=
  match mult_arg0 with
  | Zero => Zero
  | Succ n => plus (mult n mult_arg1) mult_arg1
  end.

Definition synth (y : natural) (lf1 : natural) (z : natural) (x : natural)
  : natural := plus z (mult z y).


(**EXERCÍCIO x4.11.
Defina usando equações recursivas a seqüência Fibonacci, como uma funcção de Nat para
Nat*)

Fixpoint fib (n : nat) : nat :=
  match n with
  | O => O
  | S O => S O
  | S (S n'' as n') => match n' with
                        | 0 => 1
                        | S n'' => fib n' + fib n''
                       end
  end.
Compute fib 10.

(**EXERCÍCIO x4.12*)

Inductive nat : Type :=
  | O
  | S (n : nat).

Fixpoint plus (x y : nat) : nat :=
match x with
  | O => y
  | S x' => S(plus x' y)
  end.

lemma plus_0(x : nat) plus x 0= x.
Qed.


(* From mathcomp Require Import all_ssreflect. *)
From mathcomp Require Import all_ssreflect.

Set Implicit Arguments.
Unset Strict Implicit.
Unset Printing Implicit Defensive.

Lemma sum_ord_const n m : \sum_(i < n) m = n * m.
Proof. by rewrite sum_nat_const card_ord. Qed.

(** -------------------------------------------- *)

(** #
# *)

(**
  ----
  ** Exercise 1
*)

(**
   Prove the following state by induction and by following Gauss proof.
 *)

Lemma gauss_ex n : (\sum_(i < n) i).*2 = n * n.-1.
Proof.
Qed.

(** Hints

induction *)

Check big_ord_recr.
Check big_ord0.
Check doubleD.
Check muln2.
Check mulnDr.
Check addn2.
Check mulnC.

(** Hints

Gauss (1 + 2 + 3).*2 = (1 + 2 + 3) + (3 + 2 + 1) = ((1 + 3) + (2 + 2) + (3 + 1) *)

Check addnn.
Check reindex_inj.
Check big_split.
Check sum_ord_const.
Print rev_ord.
Check rev_ord_proof.
Check rev_ord_inj.
Check eq_bigr.

(** -------------------------------------------- *)

(** #
# *)

(**
  ----
   ** Exercise 2
*)

Lemma sum_odd1 n : \sum_(i < n) (2 * i + 1) = n ^ 2.
Proof.
Qed.

(** Hints *)

Check big_split.
Check big_distrr.
Check mul2n.
Check mulnDr.
Check addn1.

(** -------------------------------------------- *)

(** #
# *)

(**
  ----
  ** Exercise 3
*)

Lemma sum_exp x n : x ^ n.+1 - 1 = (x - 1) * \sum_(i < n.+1) x ^ i.
Proof.
Qed.

(** Hints *)

Check mulnBl.
Check big_distrr.
Check big_ord_recr.
Check eq_bigr.

(** -------------------------------------------- *)

(** #
# *)

(**
  ----
 ** Exercise 4
*)

(** Prove the following state by induction and by using a similar trick
   as for Gauss noticing that n ^ 3 = n * (n ^ 2) *)

Lemma bound_square n : \sum_(i < n) i ^ 2 <= n ^ 3.
Proof.
Qed.

(** Hints *)

Check big_ind2.
Check leq_add.
Check leq_exp2r.
Check expnS.
Check ltnW.

(** -------------------------------------------- *)

(** #
# *)

(**
  ----
  ** Exercise 5
*)

(**
  building a monoid law
*)

Section cex.

Variable op2 : nat -> nat -> nat.

Hypothesis op2n0 : right_id 0 op2.

Hypothesis op20n : left_id 0 op2.

Hypothesis op2A : associative op2.

Hypothesis op2add : forall x y, op2 x y = x + y.

Canonical Structure op2Mon : Monoid.law 0 :=
  Monoid.Law op2A op20n op2n0.

(** Prove that *)
Lemma ex_op2 : \big[op2/0]_(i < 3) i = 3.
Proof.
Qed.

End cex.

(** -------------------------------------------- *)

(** #
# *)

(**
  ----
  ** Exercise 6
*)

(**
   Try to formalize the following problem
*)

(**
  Given a parking  where the boolean indicates if the slot is occupied or not
*)

Definition parking n := 'I_n -> 'I_n -> bool.

(**
   Number of cars at line i
*)

Definition sumL n (p : parking n) i := \sum_(j < n) p i j.

(**
   Number of cars at column j
*)

Definition sumC n (p : parking n) j := \sum_(i < n) p i j.

(**
   Show that if 0 < n there is always two lines, or two columns, or a column and a line
   that have the same numbers of cars
*)

(**
#

# *)


 (** * Basics: Functional Programming in Coq *)

(* REMINDER:

          #####################################################
          ###  PLEASE DO NOT DISTRIBUTE SOLUTIONS PUBLICLY  ###
          #####################################################

   (See the [Preface] for why.)
*)

(* ################################################################# *)
(** * Introduction *)

(** The functional style of programming is founded on simple, everyday
    mathematical intuition: If a procedure or method has no side
    effects, then (ignoring efficiency) all we need to understand
    about it is how it maps inputs to outputs -- that is, we can think
    of it as just a concrete method for computing a mathematical
    function.  This is one sense of the word "functional" in
    "functional programming."  The direct connection between programs
    and simple mathematical objects supports both formal correctness
    proofs and sound informal reasoning about program behavior.

    The other sense in which functional programming is "functional" is
    that it emphasizes the use of functions as _first-class_ values --
    i.e., values that can be passed as arguments to other functions,
    returned as results, included in data structures, etc.  The
    recognition that functions can be treated as data gives rise to a
    host of useful and powerful programming idioms.

    Other common features of functional languages include _algebraic
    data types_ and _pattern matching_, which make it easy to
    construct and manipulate rich data structures, and _polymorphic
    type systems_ supporting abstraction and code reuse.  Coq offers
    all of these features.

    The first half of this chapter introduces the most essential
    elements of Coq's native functional programming language, called
    _Gallina_.  The second half introduces some basic _tactics_ that
    can be used to prove properties of Gallina programs. *)

(* ################################################################# *)
(** * Data and Functions *)

(* ================================================================= *)
(** ** Enumerated Types *)

(** One notable aspect of Coq is that its set of built-in
    features is _extremely_ small.  For example, instead of providing
    the usual palette of atomic data types (booleans, integers,
    strings, etc.), Coq offers a powerful mechanism for defining new
    data types from scratch, with all these familiar types as
    instances.

    Naturally, the Coq distribution comes with an extensive standard
    library providing definitions of booleans, numbers, and many
    common data structures like lists and hash tables.  But there is
    nothing magic or primitive about these library definitions.  To
    illustrate this, this course we will explicitly recapitulate
    (almost) all the definitions we need, rather than getting them
    from the standard library. *)

(* ================================================================= *)
(** ** Days of the Week *)

(** To see how this definition mechanism works, let's start with
    a very simple example.  The following declaration tells Coq that
    we are defining a set of data values -- a _type_. *)

Inductive day : Set :=
  | monday
  | tuesday
  | wednesday
  | thursday
  | friday
  | saturday
  | sunday.

(** The new type is called [day], and its members are [monday],
    [tuesday], etc.

    Having defined [day], we can write functions that operate on
    days. *)

Definition next_weekday (d : day) : day
  := match d with
       | monday    => tuesday
       | tuesday   => wednesday
       | wednesday => thursday
       | thursday  => friday
       | friday    => monday
       | saturday  => monday
       | sunday    => monday
      end.

 Compute (next_weekday tuesday).

(** One point to note is that the argument and return types of
    this function are explicitly declared.  Like most functional
    programming languages, Coq can often figure out these types for
    itself when they are not given explicitly -- i.e., it can do _type
    inference_ -- but we'll generally include them to make reading
    easier. *)

(** Having defined a function, we should next check that it
    works on some examples.  There are actually three different ways
    to do the examples in Coq.  First, we can use the command
    [Compute] to evaluate a compound expression involving
    [next_weekday]. *)

Compute (next_weekday friday).
(* ==> monday : day *)

Compute (next_weekday (next_weekday saturday)).
(* ==> tuesday : day *)

(** (We show Coq's responses in comments, but, if you have a
    computer handy, this would be an excellent moment to fire up the
    Coq interpreter under your favorite IDE -- either CoqIde or Proof
    General -- and try it for yourself.  Load this file, [Basics.v],
    from the book's Coq sources, find the above example, submit it to
    Coq, and observe the result.) *)

(** Second, we can record what we _expect_ the result to be in the
    form of a Coq example: *)

Theorem test_next_weekday :
  (next_weekday (next_weekday saturday)) = tuesday.

(** This declaration does two things: it makes an
    assertion (that the second weekday after [saturday] is [tuesday]),
    and it gives the assertion a name that can be used to refer to it
    later.  Having made the assertion, we can also ask Coq to verify
    it like this: *)

Proof. simpl. reflexivity.
Qed.

(** The details are not important just now, but essentially this
    can be read as "The assertion we've just made can be proved by
    observing that both sides of the equality evaluate to the same
    thing."

    Third, we can ask Coq to _extract_, from our [Definition], a
    program in another, more conventional, programming
    language (OCaml, Scheme, or Haskell) with a high-performance
    compiler.  This facility is very interesting, since it gives us a
    path from proved-correct algorithms written in Gallina to
    efficient machine code.  (Of course, we are trusting the
    correctness of the OCaml/Haskell/Scheme compiler, and of Coq's
    extraction facility itself, but this is still a big step forward
    from the way most software is developed today.) Indeed, this is
    one of the main uses for which Coq was developed.  We'll come back
    to this topic in later chapters. *)

(* ================================================================= *)
(** ** Homework Submission Guidelines *)

(** If you are using _Software Foundations_ in a course, your
    instructor may use automatic scripts to help grade your homework
    assignments.  In order for these scripts to work correctly (and
    give you that you get full credit for your work!), please be
    careful to follow these rules:
      - The grading scripts work by extracting marked regions of the
        [.v] files that you submit.  It is therefore important that
        you do not alter the "markup" that delimits exercises: the
        Exercise header, the name of the exercise, the "empty square
        bracket" marker at the end, etc.  Please leave this markup
        exactly as you find it.
      - Do not delete exercises.  If you skip an exercise (e.g.,
        because it is marked "optional," or because you can't solve it),
        it is OK to leave a partial proof in your [.v] file; in
        this case, please make sure it ends with [Admitted] (not, for
        example [Abort]).
      - It is fine to use additional definitions (of helper functions,
        useful lemmas, etc.) in your solutions.  You can put these
        between the exercise header and the theorem you are asked to
        prove.
      - If you introduce a helper lemma that you end up being unable
        to prove, hence end it with [Admitted], then make sure to also
        end the main theorem in which you use it with [Admitted], not
        [Qed].  That will help you get partial credit, in case you
        use that main theorem to solve a later exercise.

    You will also notice that each chapter (like [Basics.v]) is
    accompanied by a _test script_ ([BasicsTest.v]) that automatically
    calculates points for the finished homework problems in the
    chapter.  These scripts are mostly for the auto-grading
    tools, but you may also want to use them to double-check
    that your file is well formatted before handing it in.  In a
    terminal window, either type "[make BasicsTest.vo]" or do the
    following:

       coqc -Q . LF Basics.v
       coqc -Q . LF BasicsTest.v

    See the end of this chapter for more information about how to interpret
    the output of test scripts.

    There is no need to hand in [BasicsTest.v] itself (or [Preface.v]).

    If your class is using the Canvas system to hand in assignments...
      - If you submit multiple versions of the assignment, you may
        notice that they are given different names.  This is fine: The
        most recent submission is the one that will be graded.
      - To hand in multiple files at the same time (if more than one
        chapter is assigned in the same week), you need to make a
        single submission with all the files at once using the button
        "Add another file" just above the comment box. *)

(** The [Require Export] statement on the next line tells Coq to use
    the [String] module from the standard library.  We'll use strings
    ourselves in later chapters, but we need to [Require] it here so
    that the grading scripts can use it for internal purposes. *)

From Coq Require Export String.

(* ================================================================= *)
(** ** Booleans *)

(** In a similar way, we can define the standard type [bool] of
    booleans, with members [true] and [false]. *)

Inductive bool : Type :=
  | true
  	| false.

(** Functions over booleans can be defined in the same way as
    above: *)

Definition negb (b:bool) : bool :=
  match b with
  | true => false
  | false => true
  end.

Definition andb_1 (b1 b2 : bool) : bool
  := match b1, b2 with
       | true, true => true
       | true, false => false
       | false, true => false
       | false, false => false
      end.

Definition andb_2 (b1 b2 : bool) : bool
  := match b1, b2 with
       | true, true => true
       | _, _ => false
      end.

Definition andb_3 (b1 b2 : bool) : bool
  := match b1 with
       | true => match b2 with
                        | true => true
                        | false => false
                        end
       | false => false
      end.

Definition andb (b1:bool) (b2:bool) : bool :=
  match b1 with
  | true => b2
  | false => false
  end.

Definition orb (b1:bool) (b2:bool) : bool :=
  match b1 with
  | true => true
  | false => b2
  end.

(** (Although we are rolling our own booleans here for the sake
    of building up everything from scratch, Coq does, of course,
    provide a default implementation of the booleans, together with a
    multitude of useful functions and lemmas.  Whenever possible,
    we'll name our own definitions and theorems so that they exactly
    coincide with the ones in the standard library.) *)

(** The last two of these illustrate Coq's syntax for
    multi-argument function definitions.  The corresponding
    multi-argument application syntax is illustrated by the following
    "unit tests," which constitute a complete specification -- a truth
    table -- for the [orb] function: *)

Example test_orb1:  (orb true  false) = true.
Proof. simpl. reflexivity.  Qed.
Example test_orb2:  (orb false false) = false.
Proof. simpl. reflexivity.  Qed.
Example test_orb3:  (orb false true)  = true.
Proof. simpl. reflexivity.  Qed.
Example test_orb4:  (orb true  true)  = true.
Proof. simpl. reflexivity.  Qed.
Theorem andb12 :
forall b1 b2 : bool, andb_1 b1 b2 = andb_2 b1 b2.
Proof.
intros. destruct b1, b2; simpl; reflexivity.
Qed.
(** We can also introduce some familiar infix syntax for the
    boolean operations we have just defined. The [Notation] command
    defines a new symbolic notation for an existing definition. *)

Notation "x && y" := (andb x y).
Notation "x || y" := (orb x y).

Example test_orb5:  false || false || true = true.
Proof. simpl. reflexivity. Qed.

(** _A note on notation_: In [.v] files, we use square brackets
    to delimit fragments of Coq code within comments; this convention,
    also used by the [coqdoc] documentation tool, keeps them visually
    separate from the surrounding text.  In the HTML version of the
    files, these pieces of text appear in a [different font]. *)

(** These examples are also an opportunity to introduce one more small
    feature of Coq's programming language: conditional expressions... *)

Definition negb' (b:bool) : bool :=
  if b then false
  else true.
Compute (negb' false).
Definition andb' (b1:bool) (b2:bool) : bool :=
  if b1 then b2
  else false.

Definition orb' (b1:bool) (b2:bool) : bool :=
  if b1 then true
  else b2.

(** Coq's conditionals are exactly like those found in any other
    language, with one small generalization.  Since the [bool] type is
    not built in, Coq actually supports conditional expressions over
    _any_ inductively defined type with exactly two clauses in its
    definition.  The guard is considered true if it evaluates to the
    "constructor" of the first clause of the [Inductive]
    definition (which just happens to be called [true] in this case)
    and false if it evaluates to the second. *)

(** **** Exercise: 1 star, standard (nandb)

    The command [Admitted] can be used as a placeholder for an
    incomplete proof.  We use it in exercises to indicate the parts
    that we're leaving for you -- i.e., your job is to replace
    [Admitted]s with real proofs.

    Remove "[Admitted.]" and complete the definition of the following
    function; then make sure that the [Example] assertions below can
    each be verified by Coq.  (I.e., fill in each proof, following the
    model of the [orb] tests above, and make sure Coq accepts it.) The
    function should return [true] if either or both of its inputs are
    [false]. *)

Definition nandb (b1:bool) (b2:bool) : bool :=
match b1, b2 with
  |true, true => false
  |_, _ => true
	end.

Example test_nandb1:               (nandb true false) = true.
Proof. simpl. reflexivity. Qed.
Example test_nandb2:               (nandb false false) = true.
Proof. simpl. reflexivity. Qed.
Example test_nandb3:               (nandb false true) = true.
Proof. simpl. reflexivity. Qed.
Example test_nandb4:               (nandb true true) = false.
Proof. simpl. reflexivity. Qed.
(** [] *)

(** **** Exercise: 1 star, standard (andb3)

    Do the same for the [andb3] function below. This function should
    return [true] when all of its inputs are [true], and [false]
    otherwise. *)

Definition andb3 (b1:bool) (b2:bool) (b3:bool) : bool :=
match b1, b2, b3 with
  |true, true, true => true
  |_, _, _ => false
	end.

Example test_andb31:                 (andb3 true true true) = true.
Proof. simpl. reflexivity. Qed.
Example test_andb32:                 (andb3 false true true) = false.
Proof. simpl. reflexivity. Qed.
Example test_andb33:                 (andb3 true false true) = false.
Proof. simpl. reflexivity. Qed.
Example test_andb34:                 (andb3 true true false) = false.
Proof. simpl. reflexivity. Qed.
(** [] *)

(* ================================================================= *)
(** ** Types *)

(** Every expression in Coq has a type, describing what sort of
    thing it computes. The [Check] command asks Coq to print the type
    of an expression. *)

Check true.
(* ===> true : bool *)

(** If the expression after [Check] is followed by a colon and a type,
    Coq will verify that the type of the expression matches the given
    type and halt with an error if not. *)

Check true
  : bool.
Check (negb true)
  : bool.

(** Functions like [negb] itself are also data values, just like
    [true] and [false].  Their types are called _function types_, and
    they are written with arrows. *)

Check negb
  : bool -> bool.

Check (andb true).

(** The type of [negb], written [bool -> bool] and pronounced
    "[bool] arrow [bool]," can be read, "Given an input of type
    [bool], this function produces an output of type [bool]."
    Similarly, the type of [andb], written [bool -> bool -> bool], can
    be read, "Given two inputs, each of type [bool], this function
    produces an output of type [bool]." *)

(* ================================================================= *)
(** ** New Types from Old *)

(** The types we have defined so far are examples of "enumerated
    types": their definitions explicitly enumerate a finite set of
    elements, called _constructors_.  Here is a more interesting type
    definition, where one of the constructors takes an argument: *)

Inductive rgb : Type :=
  | red
  | green
  | blue.

Inductive color : Type :=
  | black
  | white
  | primary (p : rgb).

(** Let's look at this in a little more detail.

    An [Inductive] definition does two things:

    - It defines a set of new _constructors_. E.g., [red],
      [primary], [true], [false], [monday], etc. are constructors.

    - It groups them into a new named type, like [bool], [rgb], or
      [color].

    _Constructor expressions_ are formed by applying a constructor
    to zero or more other constructors or constructor expressions,
    obeying the declared number and types of the constructor arguments.
    E.g.,
        - [red]
        - [true]
        - [primary red]
        - etc.
    But not
        - [red primary]
        - [true red]
        - [primary (primary red)]
        - etc.
*)

(** In particular, the definitions of [rgb] and [color] say
    which constructor expressions belong to the sets [rgb] and
    [color]:

    - [red], [green], and [blue] belong to the set [rgb];
    - [black] and [white] belong to the set [color];
    - if [p] is a constructor expression belonging to the set [rgb],
      then [primary p] (pronounced "the constructor [primary] applied
      to the argument [p]") is a constructor expression belonging to
      the set [color]; and
    - constructor expressions formed in these ways are the _only_ ones
      belonging to the sets [rgb] and [color]. *)

(** We can define functions on colors using pattern matching just as
    we did for [day] and [bool]. *)

Definition monochrome (c : color) : bool :=
  match c with
  | black => true
  | white => true
  | primary _ => false
  end.

(** Since the [primary] constructor takes an argument, a pattern
    matching [primary] should include either a variable (as above --
    note that we can choose its name freely) or a constant of
    appropriate type (as below). *)

Definition isred (c : color) : bool :=
  match c with
  | black => false
  | white => false
  | primary red => true
  | primary _ => false
  end.

(** The pattern "[primary _]" here is shorthand for "the constructor
    [primary] applied to any [rgb] constructor except [red]."  (The
    wildcard pattern [_] has the same effect as the dummy pattern
    variable [p] in the definition of [monochrome].) *)

(* ================================================================= *)
(** ** Modules *)

(** Coq provides a _module system_ to aid in organizing large
    developments.  We won't need most of its features,
    but one is useful: If we enclose a collection of declarations
    between [Module X] and [End X] markers, then, in the remainder of
    the file after the [End], these definitions are referred to by
    names like [X.foo] instead of just [foo].  We will use this
    feature to limit the scope of definitions, so that we are free to
    reuse names. *)

Module Playground.
  Definition b : rgb := blue.
End Playground.

Definition b : bool := true.

Check Playground.b : rgb.
Check b : bool.

(* ================================================================= *)
(** ** Tuples *)

Module TuplePlayground.

(** A single constructor with multiple parameters can be used
    to create a tuple type. As an example, consider representing
    the four bits in a nybble (half a byte). We first define
    a datatype [bit] that resembles [bool] (using the
    constructors [B0] and [B1] for the two possible bit values)
    and then define the datatype [nybble], which is essentially
    a tuple of four bits. *)

Inductive bit : Set :=
  | B0
  | B1.

Inductive nybble : Set :=
  |  bits (b0 b1 b2 b3 : bit).

Check (bits B1 B0 B1 B0).

(** The [bits] constructor acts as a wrapper for its contents.
    Unwrapping can be done by pattern-matching, as in the [all_zero]
    function which tests a nybble to see if all its bits are [B0].  We
    use underscore (_) as a _wildcard pattern_ to avoid inventing
    variable names that will not be used. *)

Definition all_zero (nb : nybble) : bool :=
  match nb with
  | bits B0 B0 B0 B0 => true
  | _ => false
  end.

Compute (all_zero (bits B1 B0 B1 B0)).
(* ===> false : bool *)
Compute (all_zero (bits B0 B0 B0 B0)).
(* ===> true : bool *)

End TuplePlayground.

(* ================================================================= *)
(** ** Numbers *)

(** We put this section in a module so that our own definition of
    natural numbers does not interfere with the one from the
    standard library.  In the rest of the book, we'll want to use
    the standard library's. *)

Module NatPlayground.

(** All the types we have defined so far -- both "enumerated
    types" such as [day], [bool], and [bit] and tuple types such as
    [nybble] built from them -- are finite.  The natural numbers, on
    the other hand, are an infinite set, so we'll need to use a
    slightly richer form of type declaration to represent them.

    There are many representations of numbers to choose from. We are
    most familiar with decimal notation (base 10), using the digits 0
    through 9, for example, to form the number 123.  You may have
    encountered hexadecimal notation (base 16), in which the same
    number is represented as 7B, or octal (base 8), where it is 173,
    or binary (base 2), where it is 1111011. Using an enumerated type
    to represent digits, we could use any of these as our
    representation natural numbers. Indeed, there are circumstances
    where each of these choices would be useful.

    The binary representation is valuable in computer hardware because
    the digits can be represented with just two distinct voltage
    levels, resulting in simple circuitry. Analogously, we wish here
    to choose a representation that makes _proofs_ simpler.

    In fact, there is a representation of numbers that is even simpler
    than binary, namely unary (base 1), in which only a single digit
    is used (as one might do to count days in prison by scratching on
    the walls). To represent unary numbers with a Coq datatype, we use
    two constructors. The capital-letter [O] constructor represents
    zero.  When the [S] constructor is applied to the representation
    of the natural number n, the result is the representation of
    n+1, where [S] stands for "successor" (or "scratch" if one is in
    prison).  Here is the complete datatype definition. *)

Inductive nat : Set :=
  | O : nat
  | S : nat -> nat.

Check (S (S (S O))).

(** With this definition, 0 is represented by [O], 1 by [S O],
    2 by [S (S O)], and so on. *)

(** Informally, the clauses of the definition can be read:
      - [O] is a natural number (remember this is the letter "[O],"
        not the numeral "[0]").
      - [S] can be put in front of a natural number to yield another
        one -- if [n] is a natural number, then [S n] is too. *)

(** Again, let's look at this in a little more detail.  The definition
    of [nat] says how expressions in the set [nat] can be built:

    - the constructor expression [O] belongs to the set [nat];
    - if [n] is a constructor expression belonging to the set [nat],
      then [S n] is also a constructor expression belonging to the set
      [nat]; and
    - constructor expressions formed in these two ways are the only
      ones belonging to the set [nat]. *)

(** These conditions are the precise force of the [Inductive]
    declaration.  They imply that the constructor expression [O], the
    constructor expression [S O], the constructor expression [S (S
    O)], the constructor expression [S (S (S O))], and so on all
    belong to the set [nat], while other constructor expressions, like
    [true], [andb true false], [S (S false)], and [O (O (O S))] do
    not.

    A critical point here is that what we've done so far is just to
    define a _representation_ of numbers: a way of writing them down.
    The names [O] and [S] are arbitrary, and at this point they have
    no special meaning -- they are just two different marks that we
    can use to write down numbers (together with a rule that says any
    [nat] will be written as some string of [S] marks followed by an
    [O]).  If we like, we can write essentially the same definition
    this way: *)

Inductive nat' : Set :=
  | stop : nat'
  | tick  : nat' -> nat'.

(** The _interpretation_ of these marks comes from how we use them to
    compute. *)

(** We can do this by writing functions that pattern match on
    representations of natural numbers just as we did above with
    booleans and days -- for example, here is the predecessor
    function: *)

Definition pred (n : nat) : nat :=
  match n with
  | O => O
  | S m => m
  end.

(** The second branch can be read: "if [n] has the form [S n']
    for some [n'], then return [n']."  *)

(** The following [End] command closes the current module, so
    [nat] will refer back to the type from the standard library. *)

Check (S (S (S (S O)))).

End NatPlayground.

(** Because natural numbers are such a pervasive form of data,
    Coq provides a tiny bit of built-in magic for parsing and printing
    them: ordinary decimal numerals can be used as an alternative to
    the "unary" notation defined by the constructors [S] and [O].  Coq
    prints numbers in decimal form by default: *)

Check (S (S (S (S O)))).
(* ===> 4 : nat *)

Definition minustwo (n : nat) : nat :=
  match n with
  | O => O
  | S O => O
  | S (S p) => p
  end.

Compute (minustwo 14).
(* ===> 2 : nat *)

(** The constructor [S] has the type [nat -> nat], just like functions
    such as [pred] and [minustwo]: *)

Check S        : nat -> nat.
Check pred     : nat -> nat.
Check minustwo : nat -> nat.

(** These are all things that can be applied to a number to yield a
    number.  However, there is a fundamental difference between [S]
    and the other two: functions like [pred] and [minustwo] are
    defined by giving _computation rules_ -- e.g., the definition of
    [pred] says that [pred 2] can be simplified to [1] -- while the
    definition of [S] has no such behavior attached.  Although it is
    _like_ a function in the sense that it can be applied to an
    argument, it does not _do_ anything at all!  It is just a way of
    writing down numbers.

    (Think about standard decimal numerals: the numeral [1] is not a
    computation; it's a piece of data.  When we write [111] to mean
    the number one hundred and eleven, we are using [1], three times,
    to write down a concrete representation of a number.)

    Now let's go on and define some more functions over numbers.

    For most interesting computations involving numbers, simple
    pattern matching is not enough: we also need recursion.  For
    example, to check that a number [n] is even, we may need to
    recursively check whether [n-2] is even.  Such functions are
    introduced with the keyword [Fixpoint] instead of [Definition]. *)

Fixpoint even (n:nat) : bool :=
  match n with
  | O        => true
  | S O      => false
  | S (S m) => (even m)
  end.

(** We could define [odd] by a similar [Fixpoint] declaration, but
    here is a simpler way: *)

Definition odd (n:nat) : bool :=
  negb (even n).

Example test_odd1:    odd 1 = true.
Proof. simpl. reflexivity.  Qed.
Example test_odd2:    odd 4 = false.
Proof. simpl. reflexivity.  Qed.

(** (You may notice if you step through these proofs that
    [simpl] actually has no effect on the goal -- all of the work is
    done by [reflexivity].  We'll discuss why that is shortly.)

    Naturally, we can also define multi-argument functions by
    recursion.  *)

Module NatPlayground2.

Fixpoint plus (n : nat) (m : nat) : nat :=
  match n with
  | O => m
  | S p => S (plus p m)
  end.

(** Adding three to two now gives us five, as we'd expect. *)

Compute (plus 13 21).
(* ===> 5 : nat *)

Theorem test_1 (n : nat) : O + n = n.
Proof.
  simpl. reflexivity.
Qed.

(** The steps of simplification that Coq performs can be
    visualized as follows: *)

(*      [plus 3 2]
   i.e. [plus (S (S (S O))) (S (S O))]
    ==> [S (plus (S (S O)) (S (S O)))]
          by the second clause of the [match]
    ==> [S (S (plus (S O) (S (S O))))]
          by the second clause of the [match]
    ==> [S (S (S (plus O (S (S O)))))]
          by the second clause of the [match]
    ==> [S (S (S (S (S O))))]
          by the first clause of the [match]
   i.e. [5]  *)

(** As a notational convenience, if two or more arguments have
    the same type, they can be written together.  In the following
    definition, [(n m : nat)] means just the same as if we had written
    [(n : nat) (m : nat)]. *)

Fixpoint mult (n m : nat) : nat :=
  match n with
  | O => O
  | S p => plus m (mult p m)
  end.

Example test_mult1: (mult 3 3) = 9.
Proof. simpl. reflexivity.  Qed.

(** You can match two expressions at once by putting a comma
    between them: *)

Fixpoint minus (n m:nat) : nat :=
  match n, m with
  | O   , _    => O
  | S _ , O    => n
  | S n', S m' => minus n' m'
  end.

End NatPlayground2.

Fixpoint exp (base power : nat) : nat :=
  match power with
  | O => S O
  | S p => mult base (exp base p)
  end.

(** **** Exercise: 1 star, standard (factorial)

    Recall the standard mathematical factorial function:

       factorial(0)  =  1
       factorial(n)  =  n * factorial(n-1)     (if n>0)

    Translate this into Coq. *)

Compute (minus 23 12).

Fixpoint factorial (n:nat) : nat :=
	match n with
	| O => S(O)
	| S(X) => mult (S( X)) (factorial (X))
end.

Example test_factorial1:          (factorial 3) = 6.
Proof. simpl. reflexivity. Qed.
Example test_factorial2:          (factorial 5) = (mult 10 12).
Proof. simpl. reflexivity. Qed.
(** [] *)

(** Again, we can make numerical expressions easier to read and write
    by introducing notations for addition, multiplication, and
    subtraction. *)

Notation "x + y" := (plus x y)
                       (at level 50, left associativity)
                       : nat_scope.
Notation "x - y" := (minus x y)
                       (at level 50, left associativity)
                       : nat_scope.
Notation "x * y" := (mult x y)
                       (at level 40, left associativity)
                       : nat_scope.

Check ((0 + 1) + 1) : nat.

(** (The [level], [associativity], and [nat_scope] annotations
    control how these notations are treated by Coq's parser.  The
    details are not important for present purposes, but interested
    readers can refer to the "More on Notation" section at the end of
    this chapter.)

    Note that these declarations do not change the definitions we've
    already made: they are simply instructions to the Coq parser to
    accept [x + y] in place of [plus x y] and, conversely, to the Coq
    pretty-printer to display [plus x y] as [x + y]. *)

(** When we say that Coq comes with almost nothing built-in, we really
    mean it: even equality testing is a user-defined operation!
    Here is a function [eqb], which tests natural numbers for
    [eq]uality, yielding a [b]oolean.  Note the use of nested
    [match]es (we could also have used a simultaneous match, as we did
    in [minus].) *)

Fixpoint eqb (n m : nat) : bool :=
  match n with
  | O => match m with
         | O => true
         | S m' => false
         end
  | S n' => match m with
            | O => false
            | S m' => eqb n' m'
            end
  end.

Fixpoint eqb1 (n m : nat) : bool :=
  match n, m with
  | O, O => true
  | O, S _ => false
  | S _, O => false
  | S n', S m' => eqb n' m'
  end.

(** Similarly, the [leb] function tests whether its first argument is
    less than or equal to its second argument, yielding a boolean. *)

Fixpoint leb (n m : nat) : bool :=
  match n with
  | O => true
  | S n' => match m with
                 | O => false
                 | S m' => leb n' m'
                 end
  end.

Example test_leb1:                leb 2 2 = true.
Proof. simpl. reflexivity.  Qed.
Example test_leb2:                leb 2 4 = true.
Proof. simpl. reflexivity.  Qed.
Example test_leb3:                leb 4 2 = false.
Proof. simpl. reflexivity.  Qed.

(** We'll be using these (especially [eqb]) a lot, so let's give
    them infix notations. *)

Notation "x =? y" := (eqb x y) (at level 70) : nat_scope.
Notation "x <=? y" := (leb x y) (at level 70) : nat_scope.

Example test_leb3': (4 <=? 2) = false.
Proof. simpl. reflexivity.  Qed.

(** We now have two symbols that look like equality: [=] and
    [=?].  We'll have much more to say about the differences and
    similarities between them later. For now, the main thing to notice
    is that [x = y] is a logical _claim_ -- a "proposition" -- that we
    can try to prove, while [x =? y] is an _expression_ whose
    value (either [true] or [false]) we can compute. *)

(** **** Exercise: 1 star, standard (ltb)

    The [ltb] function tests natural numbers for [l]ess-[t]han,
    yielding a [b]oolean.  Instead of making up a new [Fixpoint] for
    this one, define it in terms of a previously defined
    function.  (It can be done with just one previously defined
    function, but you can use two if you want.) *)

Definition ltb (n m : nat) : bool:=
	if n =? m then false
	else n <=? m.

Definition ltb2 (n m : nat) : bool:=
	match n =? m with
	|true => false
	|false => n <=? m
	end.

Notation "x <? y" := (ltb x y) (at level 70) : nat_scope.

Example test_ltb1:             (ltb 2 2) = false.
Proof. simpl. reflexivity. Qed.
Example test_ltb2:             (ltb 2 4) = true.
Proof. simpl. reflexivity. Qed.
Example test_ltb3:             (ltb 4 2) = false.
Proof. simpl. reflexivity. Qed.
(** [] *)

(* ################################################################# *)
(** * Proof by Simplification *)

(** Now that we've defined a few datatypes and functions, let's
    turn to stating and proving properties of their behavior.
    Actually, we've already started doing this: each [Example] in the
    previous sections makes a precise claim about the behavior of some
    function on some particular inputs.  The proofs of these claims
    were always the same: use [simpl] to simplify both sides of the
    equation, then use [reflexivity] to check that both sides contain
    identical values.

    The same sort of "proof by simplification" can be used to prove
    more interesting properties as well.  For example, the fact that
    [0] is a "neutral element" for [+] on the left can be proved just
    by observing that [0 + n] reduces to [n] no matter what [n] is -- a
    fact that can be read directly off the definition of [plus]. *)

Theorem plus_O_n : forall n : nat, 0 + n = n.
Proof.
  intros n. simpl. reflexivity.
Qed.

(** (You may notice that the above statement looks different in
    the [.v] file in your IDE than it does in the HTML rendition in
    your browser. In [.v] files, we write the universal quantifier
    [forall] using the reserved identifier "forall."  When the [.v]
    files are converted to HTML, this gets transformed into the
    standard upside-down-A symbol.)

    This is a good place to mention that [reflexivity] is a bit more
    powerful than we have acknowledged. In the examples we have seen,
    the calls to [simpl] were actually not needed, because
    [reflexivity] can perform some simplification automatically when
    checking that two sides are equal; [simpl] was just added so that
    we could see the intermediate state -- after simplification but
    before finishing the proof.  Here is a shorter proof of the
    theorem: *)

Theorem plus_O_n' : forall n : nat, 0 + n = n.
Proof.
  intros n. reflexivity. Qed.

(** Moreover, it will be useful to know that [reflexivity] does
    somewhat _more_ simplification than [simpl] does -- for example,
    it tries "unfolding" defined terms, replacing them with their
    right-hand sides.  The reason for this difference is that, if
    reflexivity succeeds, the whole goal is finished and we don't need
    to look at whatever expanded expressions [reflexivity] has created
    by all this simplification and unfolding; by contrast, [simpl] is
    used in situations where we may have to read and understand the
    new goal that it creates, so we would not want it blindly
    expanding definitions and leaving the goal in a messy state.

    The form of the theorem we just stated and its proof are almost
    exactly the same as the simpler examples we saw earlier; there are
    just a few differences.

    First, we've used the keyword [Theorem] instead of [Example].
    This difference is mostly a matter of style; the keywords
    [Example] and [Theorem] (and a few others, including [Lemma],
    [Fact], and [Remark]) mean pretty much the same thing to Coq.

    Second, we've added the quantifier [forall n:nat], so that our
    theorem talks about _all_ natural numbers [n].  Informally, to
    prove theorems of this form, we generally start by saying "Suppose
    [n] is some number..."  Formally, this is achieved in the proof by
    [intros n], which moves [n] from the quantifier in the goal to a
    _context_ of current assumptions. Note that we could have used
    another identifier instead of [n] in the [intros] clause, (though
    of course this might be confusing to human readers of the proof): *)

Theorem plus_O_n'' : forall n : nat, 0 + n = n.
Proof.
  intros m. reflexivity. Qed.

(** The keywords [intros], [simpl], and [reflexivity] are examples of
    _tactics_.  A tactic is a command that is used between [Proof] and
    [Qed] to guide the process of checking some claim we are making.
    We will see several more tactics in the rest of this chapter and
    many more in future chapters. *)

(** Other similar theorems can be proved with the same pattern. *)

Theorem plus_1_l : forall n:nat, 1 + n = S n.
Proof.
  intros n. reflexivity.  Qed.

Theorem mult_0_l : forall n:nat, 0 * n = 0.
Proof.
  intros n. reflexivity.  Qed.

(** The [_l] suffix in the names of these theorems is
    pronounced "on the left." *)

(** It is worth stepping through these proofs to observe how the
    context and the goal change.  You may want to add calls to [simpl]
    before [reflexivity] to see the simplifications that Coq performs
    on the terms before checking that they are equal. *)

(* ################################################################# *)
(** * Proof by Rewriting *)

(** The following theorem is a bit more interesting than the
    ones we've seen: *)

Theorem plus_id_example : forall n m:nat,
  n = m ->
  n + n = m + m.

(** Instead of making a universal claim about all numbers [n] and [m],
    it talks about a more specialized property that only holds when
    [n = m].  The arrow symbol is pronounced "implies."

    As before, we need to be able to reason by assuming we are given such
    numbers [n] and [m].  We also need to assume the hypothesis
    [n = m]. The [intros] tactic will serve to move all three of these
    from the goal into assumptions in the current context.

    Since [n] and [m] are arbitrary numbers, we can't just use
    simplification to prove this theorem.  Instead, we prove it by
    observing that, if we are assuming [n = m], then we can replace
    [n] with [m] in the goal statement and obtain an equality with the
    same expression on both sides.  The tactic that tells Coq to
    perform this replacement is called [rewrite]. *)

Proof.
  (* move both quantifiers into the context: *)
  intros n m.
  (* move the hypothesis into the context: *)
  intros neqm.
  (* rewrite the goal using the hypothesis: *)
  rewrite neqm.
  reflexivity.
Qed.

(** The first line of the proof moves the universally quantified
    variables [n] and [m] into the context.  The second moves the
    hypothesis [n = m] into the context and gives it the name [H].
    The third tells Coq to rewrite the current goal ([n + n = m + m])
    by replacing the left side of the equality hypothesis [H] with the
    right side.

    (The arrow symbol in the [rewrite] has nothing to do with
    implication: it tells Coq to apply the rewrite from left to right.
    In fact, you can omit the arrow, and Coq will default to rewriting
    in this direction.  To rewrite from right to left, you can use
    [rewrite <-].  Try making this change in the above proof and see
    what difference it makes.) *)

(** **** Exercise: 1 star, standard (plus_id_exercise)

    Remove "[Admitted.]" and fill in the proof. *)

(* SOOMER: KK: [plus_id_exercise] contains multiple hypotheses, and at
   least one student was confused about this. Maybe we can talk about
   [->] being right-associative before it. *)

Theorem plus_id_exercise : forall n m o : nat,
  n = m -> m = o -> n + m = m + o.
Proof.
  intros. rewrite <- H at 2. rewrite <- H0. reflexivity.
Qed.

(** [] *)

(** The [Admitted] command tells Coq that we want to skip trying
    to prove this theorem and just accept it as a given.  This can be
    useful for developing longer proofs, since we can state subsidiary
    lemmas that we believe will be useful for making some larger
    argument, use [Admitted] to accept them on faith for the moment,
    and continue working on the main argument until we are sure it
    makes sense; then we can go back and fill in the proofs we
    skipped.  Be careful, though: every time you say [Admitted] you
    are leaving a door open for total nonsense to enter Coq's nice,
    rigorous, formally checked world! *)

(** The [Check] command can also be used to examine the statements of
    previously declared lemmas and theorems.  The two examples below
    are lemmas about multiplication that are proved in the standard
    library.  (We will see how to prove them ourselves in the next
    chapter.) *)

Check mult_n_O.
(* ===> forall n : nat, 0 = n * 0 *)

Check mult_n_Sm.
(* ===> forall n m : nat, n * m + n = n * S m *)

(** We can use the [rewrite] tactic with a previously proved theorem
    instead of a hypothesis from the context. If the statement of the
    previously proved theorem involves quantified variables, as in the
    example below, Coq tries to instantiate them by matching with the
    current goal. *)

Theorem mult_n_0_m_0 : forall p q : nat,
  (p * 0) + (q * 0) = 0.
Proof.
  intros p q.
  rewrite <- (mult_n_O p).
  rewrite <- (mult_n_O q).
  reflexivity. Qed.

(** **** Exercise: 1 star, standard (mult_n_1)

    Use those two lemmas about multiplication that we just checked to
    prove the following theorem.  Hint: recall that [1] is [S O]. *)

Theorem mult_n_1 : forall p : nat,
  p * 1 = p.
Proof.
	intros p.
	rewrite <- (mult_n_Sm p 0).
	rewrite <- (mult_n_O p).
	reflexivity.
	Qed.

(** [] *)

(* ################################################################# *)
(** * Proof by Case Analysis *)

(** Of course, not everything can be proved by simple
    calculation and rewriting: In general, unknown, hypothetical
    values (arbitrary numbers, booleans, lists, etc.) can block
    simplification.  For example, if we try to prove the following
    fact using the [simpl] tactic as above, we get stuck.  (We then
    use the [Abort] command to give up on it for the moment.)*)

Theorem plus_1_neq_0_firsttry : forall n : nat,
  (n + 1) =? 0 = false.
Proof.
  intros n.
  simpl.  (* does nothing! *)
Abort.

(** The reason for this is that the definitions of both [eqb]
    and [+] begin by performing a [match] on their first argument.
    But here, the first argument to [+] is the unknown number [n] and
    the argument to [eqb] is the compound expression [n + 1]; neither
    can be simplified.

    To make progress, we need to consider the possible forms of [n]
    separately.  If [n] is [O], then we can calculate the final result
    of [(n + 1) =? 0] and check that it is, indeed, [false].  And if
    [n = S n'] for some [n'], then, although we don't know exactly
    what number [n + 1] represents, we can calculate that, at least,
    it will begin with one [S], and this is enough to calculate that,
    again, [(n + 1) =? 0] will yield [false].

    The tactic that tells Coq to consider, separately, the cases where
    [n = O] and where [n = S n'] is called [destruct]. *)

Theorem plus_1_neq_0 : forall n : nat,
  (n + 1) =? 0 = false.
Proof.
  intros n.
  destruct n as [| n'] eqn: E.
  { reflexivity. }
  { reflexivity.
  }
 Qed.

(** The [destruct] generates _two_ subgoals, which we must then
    prove, separately, in order to get Coq to accept the theorem.

    The annotation "[as [| n']]" is called an _intro pattern_.  It
    tells Coq what variable names to introduce in each subgoal.  In
    general, what goes between the square brackets is a _list of
    lists_ of names, separated by [|].  In this case, the first
    component is empty, since the [O] constructor is nullary (it
    doesn't have any arguments).  The second component gives a single
    name, [n'], since [S] is a unary constructor.

    In each subgoal, Coq remembers the assumption about [n] that is
    relevant for this subgoal -- either [n = 0] or [n = S n'] for some
    n'.  The [eqn:E] annotation tells [destruct] to give the name [E]
    to this equation.  Leaving off the [eqn:E] annotation causes Coq
    to elide these assumptions in the subgoals.  This slightly
    streamlines proofs where the assumptions are not explicitly used,
    but it is better practice to keep them for the sake of
    documentation, as they can help keep you oriented when working
    with the subgoals.

    The [-] signs on the second and third lines are called _bullets_,
    and they mark the parts of the proof that correspond to the two
    generated subgoals.  The part of the proof script that comes after
    a bullet is the entire proof for the corresponding subgoal.  In
    this example, each of the subgoals is easily proved by a single
    use of [reflexivity], which itself performs some simplification --
    e.g., the second one simplifies [(S n' + 1) =? 0] to [false] by
    first rewriting [(S n' + 1)] to [S (n' + 1)], then unfolding
    [eqb], and then simplifying the [match].

    Marking cases with bullets is optional: if bullets are not
    present, Coq simply asks you to prove each subgoal in sequence,
    one at a time. But it is a good idea to use bullets.  For one
    thing, they make the structure of a proof apparent, improving
    readability. Also, bullets instruct Coq to ensure that a subgoal
    is complete before trying to verify the next one, preventing
    proofs for different subgoals from getting mixed up. These issues
    become especially important in large developments, where fragile
    proofs lead to long debugging sessions.

    There are no hard and fast rules for how proofs should be
    formatted in Coq -- e.g., where lines should be broken and how
    sections of the proof should be indented to indicate their nested
    structure.  However, if the places where multiple subgoals are
    generated are marked with explicit bullets at the beginning of
    lines, then the proof will be readable almost no matter what
    choices are made about other aspects of layout.

    This is also a good place to mention one other piece of somewhat
    obvious advice about line lengths.  Beginning Coq users sometimes
    tend to the extremes, either writing each tactic on its own line
    or writing entire proofs on a single line.  Good style lies
    somewhere in the middle.  One reasonable guideline is to limit
    yourself to 80-character lines.

    The [destruct] tactic can be used with any inductively defined
    datatype.  For example, we use it next to prove that boolean
    negation is involutive -- i.e., that negation is its own
    inverse. *)

Theorem negb_involutive : forall b : bool,
  negb (negb b) = b.
Proof.
  intros b. destruct b eqn:E.
  - reflexivity.
  - reflexivity.  Qed.

(** Note that the [destruct] here has no [as] clause because
    none of the subcases of the [destruct] need to bind any variables,
    so there is no need to specify any names.  In fact, we can omit
    the [as] clause from _any_ [destruct] and Coq will fill in
    variable names automatically.  This is generally considered bad
    style, since Coq often makes confusing choices of names when left
    to its own devices.

    It is sometimes useful to invoke [destruct] inside a subgoal,
    generating yet more proof obligations. In this case, we use
    different kinds of bullets to mark goals on different "levels."
    For example: *)

Theorem andb_commutative : forall b c, andb b c = andb c b.
Proof.
  intros b c. destruct b eqn:Eb.
  - destruct c eqn:Ec.
    + reflexivity.
    + reflexivity.
  - destruct c eqn:Ec.
    + reflexivity.
    + reflexivity.
Qed.

(** Each pair of calls to [reflexivity] corresponds to the
    subgoals that were generated after the execution of the [destruct c]
    line right above it. *)

(** Besides [-] and [+], we can use [*] (asterisk) or any repetition
    of a bullet symbol (e.g. [--] or [***]) as a bullet.  We can also
    enclose sub-proofs in curly braces: *)

Theorem andb_commutative' : forall b c, andb b c = andb c b.
Proof.
  intros b c. destruct b eqn:Eb.
  { destruct c eqn:Ec.
    { reflexivity. }
    { reflexivity. }
  }
  { destruct c eqn:Ec.
    { reflexivity. }
    { reflexivity. }
  }
Qed.

(** Since curly braces mark both the beginning and the end of a proof,
    they can be used for multiple subgoal levels, as this example
    shows. Furthermore, curly braces allow us to reuse the same bullet
    shapes at multiple levels in a proof. The choice of braces,
    bullets, or a combination of the two is purely a matter of
    taste. *)

Theorem andb3_exchange :
  forall b c d, andb (andb b c) d = andb (andb b d) c.
Proof.
  intros b c d. destruct b eqn:Eb.
  { destruct c eqn:Ec.
    { destruct d eqn:Ed; reflexivity. }
    { destruct d eqn:Ed; reflexivity. }
  }
  { destruct c eqn:Ec.
    { destruct d eqn:Ed; reflexivity. }
    { destruct d eqn:Ed; reflexivity. }
  }
Qed.

(** **** Exercise: 2 stars, standard (andb_true_elim2)

    Prove the following claim, marking cases (and subcases) with
    bullets when you use [destruct]. Hint: delay introducing the
    hypothesis until after you have an opportunity to simplify it. *)

Print andb.
Theorem andb_true_elim2 : forall b c : bool,
  andb b c 	= true -> c = true.
Proof.
  intros [|] [|]; simpl; try reflexivity;
  intros H; rewrite H; reflexivity.
Qed.

(** [] *)

(** Before closing the chapter, let's mention one final
    convenience.  As you may have noticed, many proofs perform case
    analysis on a variable right after introducing it:

       intros x y. destruct y as [|y] eqn:E.

    This pattern is so common that Coq provides a shorthand for it: we
    can perform case analysis on a variable when introducing it by
    using an intro pattern instead of a variable name. For instance,
    here is a shorter proof of the [plus_1_neq_0] theorem
    above.  (You'll also note one downside of this shorthand: we lose
    the equation recording the assumption we are making in each
    subgoal, which we previously got from the [eqn:E] annotation.) *)

Theorem plus_1_neq_0' : forall n : nat,
  (n + 1) =? 0 = false.
Proof.
  intros [|n].
  - reflexivity.
  - reflexivity.  Qed.

(** If there are no constructor arguments that need names, we can just
    write [[]] to get the case analysis. *)

Theorem andb_commutative'' :
  forall b c, andb b c = andb c b.
Proof.
  intros [|] [|]; reflexivity.
Qed.

(** **** Exercise: 1 star, standard (zero_nbeq_plus_1) *)
Theorem zero_nbeq_plus_1 : forall n : nat,
  0 =? (n + 1) = false.
Proof.
	intros [|].
	-reflexivity.
	-reflexivity.
Qed.

(** [] *)

(* ================================================================= *)
(** ** More on Notation (Optional) *)

(** (In general, sections marked Optional are not needed to follow the
    rest of the book, except possibly other Optional sections.  On a
    first reading, you might want to skim these sections so that you
    know what's there for future reference.)

    Recall the notation definitions for infix plus and times: *)

Notation "x + y" := (plus x y)
                       (at level 50, left associativity)
                       : nat_scope.
Notation "x * y" := (mult x y)
                       (at level 40, left associativity)
                       : nat_scope.

(** For each notation symbol in Coq, we can specify its _precedence
    level_ and its _associativity_.  The precedence level [n] is
    specified by writing [at level n]; this helps Coq parse compound
    expressions.  The associativity setting helps to disambiguate
    expressions containing multiple occurrences of the same
    symbol. For example, the parameters specified above for [+] and
    [*] say that the expression [1+2*3*4] is shorthand for
    [(1+((2*3)*4))]. Coq uses precedence levels from 0 to 100, and
    _left_, _right_, or _no_ associativity.  We will see more examples
    of this later, e.g., in the [Lists]
    chapter.

    Each notation symbol is also associated with a _notation scope_.
    Coq tries to guess what scope is meant from context, so when it
    sees [S(O*O)] it guesses [nat_scope], but when it sees the product
    type [bool*bool] (which we'll see in later chapters) it guesses
    [type_scope].  Occasionally, it is necessary to help it out with
    percent-notation by writing [(x*y)%nat], and sometimes in what Coq
    prints it will use [%nat] to indicate what scope a notation is in.

    Notation scopes also apply to numeral notation ([3], [4], [5], [42],
    etc.), so you may sometimes see [0%nat], which means [O] (the
    natural number [0] that we're using in this chapter), or [0%Z],
    which means the integer zero (which comes from a different part of
    the standard library).

    Pro tip: Coq's notation mechanism is not especially powerful.
    Don't expect too much from it. *)

(* ================================================================= *)
(** ** Fixpoints and Structural Recursion (Optional) *)

(** Here is a copy of the definition of addition: *)

Fixpoint plus' (n : nat) (m : nat) : nat :=
  match n with
  | O => m
  | S n' => S (plus' n' m)
  end.

(** When Coq checks this definition, it notes that [plus'] is
    "decreasing on 1st argument."  What this means is that we are
    performing a _structural recursion_ over the argument [n] -- i.e.,
    that we make recursive calls only on strictly smaller values of
    [n].  This implies that all calls to [plus'] will eventually
    terminate.  Coq demands that some argument of _every_ [Fixpoint]
    definition is "decreasing."

    This requirement is a fundamental feature of Coq's design: In
    particular, it guarantees that every function that can be defined
    in Coq will terminate on all inputs.  However, because Coq's
    "decreasing analysis" is not very sophisticated, it is sometimes
    necessary to write functions in slightly unnatural ways. *)

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

    To get a concrete sense of this, find a way to write a sensible
    [Fixpoint] definition (of a simple function on numbers, say) that
    _does_ terminate on all inputs, but that Coq will reject because
    of this restriction.  (If you choose to turn in this optional
    exercise as part of a homework assignment, make sure you comment
    out your solution so that it doesn't cause Coq to reject the whole
    file!) *)

Fail Fixpoint plus2 (n m : nat) : nat :=
  match n with
  | O => m
  | S O => S (plus2 O m)
  | S (S n) => S (plus2 (S n) m)
  end.

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

(** **** Exercise: 1 star, standard (identity_fn_applied_twice)

    Use the tactics you have learned so far to prove the following
    theorem about boolean functions. *)

Theorem identity_fn_applied_twice :
  forall (f : bool -> bool),
  (forall (x : bool), f x = x) ->
  forall (b : bool), f (f b) = b.
Proof.
	intros. do 2 rewrite H. reflexivity.
Qed.

(** [] *)

(** **** Exercise: 1 star, standard (negation_fn_applied_twice)

    Now state and prove a theorem [negation_fn_applied_twice] similar
    to the previous one but where the second hypothesis says that the
    function [f] has the property that [f x = negb x]. *)

Print negb.
Theorem negation_fn_applied_twice :
  forall (f : bool -> bool),
  (forall (x : bool), f x = negb x) ->
  forall (b : bool), f (f b) = b.
Proof.
	intros. do 2 rewrite H. destruct b0; reflexivity.
Qed.

(* Do not modify the following line: *)
Definition manual_grade_for_negation_fn_applied_twice : option (nat*string) := None.
(** (The last definition is used by the autograder.)

    [] *)

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

    Prove the following theorem.  (Hint: This one can be a bit tricky,
    depending on how you approach it.  You will probably need both
    [destruct] and [rewrite], but destructing everything in sight is
    not the best way.) *)

Print test_orb1.
Print test_orb2.
Print test_orb3.
Print test_orb4.

Theorem andb_eq_orb :
  forall (b c : bool),
  (andb b c = orb b c) ->
  b = c.
Proof.
	intros []; simpl; intros; rewrite H; reflexivity.
Qed.

(** [] *)

(** **** Exercise: 3 stars, standard (binary)

    We can generalize our unary representation of natural numbers to
    the more efficient binary representation by treating a binary
    number as a sequence of constructors [B0] and [B1] (representing 0s
    and 1s), terminated by a [Z]. For comparison, in the unary
    representation, a number is a sequence of [S] constructors terminated
    by an [O].

    For example:

        decimal               binary                          unary
           0                       Z                              O
           1                    B1 Z                            S O
           2                B0 (B1 Z)                        S (S O)
           3                B1 (B1 Z)                     S (S (S O))
           4            B0 (B0 (B1 Z))                 S (S (S (S O)))
           5            B1 (B0 (B1 Z))              S (S (S (S (S O))))
           6            B0 (B1 (B1 Z))           S (S (S (S (S (S O)))))
           7            B1 (B1 (B1 Z))        S (S (S (S (S (S (S O))))))
           8        B0 (B0 (B0 (B1 Z)))    S (S (S (S (S (S (S (S O)))))))

    Note that the low-order bit is on the left and the high-order bit
    is on the right -- the opposite of the way binary numbers are
    usually written.  This choice makes them easier to manipulate. *)

Inductive bin : Type :=
  | Z
  | B0 (n : bin)
  | B1 (n : bin).

(** Complete the definitions below of an increment function [incr]
    for binary numbers, and a function [bin_to_nat] to convert
    binary numbers to unary numbers. *)

Fixpoint incr (m:bin) : bin :=
	match m with
	|	Z => B1 Z
	|	B0 m => B1 m
	| B1 m => B0 (incr m)
	end.

Fixpoint bin_to_nat (m:bin) : nat :=
	match m with
	| Z => O
	|	B0 m => 2 * (bin_to_nat m)
	| B1 m => 2 * (bin_to_nat m) + 1
	end.

(** The following "unit tests" of your increment and binary-to-unary
    functions should pass after you have defined those functions correctly.
    Of course, unit tests don't fully demonstrate the correctness of
    your functions!  We'll return to that thought at the end of the
    next chapter. *)

Example test_bin_incr1 : (incr (B1 Z)) = B0 (B1 Z).
Proof. reflexivity. Qed.

Example test_bin_incr2 : (incr (B0 (B1 Z))) = B1 (B1 Z).
Proof. reflexivity. Qed.

Example test_bin_incr3 : (incr (B1 (B1 Z))) = B0 (B0 (B1 Z)).
Proof. reflexivity. Qed.

Example test_bin_incr4 : bin_to_nat (B0 (B1 Z)) = 2.
Proof. reflexivity. Qed.

Example test_bin_incr5 :
        bin_to_nat (incr (B1 Z)) = 1 + bin_to_nat (B1 Z).
Proof. reflexivity. Qed.

Example test_bin_incr6 :
        bin_to_nat (incr (incr (B1 Z))) = 2 + bin_to_nat (B1 Z).
Proof. reflexivity. Qed.

(** [] *)

(* ################################################################# *)
(** * Testing Your Solutions *)

(** Each SF chapter comes with a test file containing scripts that
    check whether you have solved the required exercises. If you're
    using SF as part of a course, your instructors will likely be
    running these test files to autograde your solutions. You can also
    use these test files, if you like, to make sure you haven't missed
    anything.

    Important: This step is _optional_: if you've completed all the
    non-optional exercises and Coq accepts your answers, this already
    shows that you are in good shape.

    The test file for this chapter is [BasicsTest.v]. To run it, make
    sure you have saved [Basics.v] to disk.  Then do this:

       coqc -Q . LF Basics.v
       coqc -Q . LF BasicsTest.v

    If you accidentally deleted an exercise or changed its name, then
    [make BasicsTest.vo] will fail with an error that tells you the
    name of the missing exercise.  Otherwise, you will get a lot of
    useful output:

    - First will be all the output produced by [Basics.v] itself.  At
      the end of that you will see [COQC BasicsTest.v].

    - Second, for each required exercise, there is a report that tells
      you its point value (the number of stars or some fraction
      thereof if there are multiple parts to the exercise), whether
      its type is ok, and what assumptions it relies upon.

      If the _type_ is not [ok], it means you proved the wrong thing:
      most likely, you accidentally modified the theorem statement
      while you were proving it.  The autograder won't give you any
      points for that, so make sure to correct the theorem.

      The _assumptions_ are any unproved theorems which your solution
      relies upon.  "Closed under the global context" is a fancy way
      of saying "none": you have solved the exercise. (Hooray!)  On
      the other hand, a list of axioms means you haven't fully solved
      the exercise. (But see below regarding "Allowed Axioms.") If the
      exercise name itself is in the list, that means you haven't
      solved it; probably you have [Admitted] it.

    - Third, you will see the maximum number of points in standard and
      advanced versions of the assignment.  That number is based on
      the number of stars in the non-optional exercises.

    - Fourth, you will see a list of "Allowed Axioms".  These are
      unproved theorems that your solution is permitted to depend
      upon.  You'll probably see something about
      [functional_extensionality] for this chapter; we'll cover what
      that means in a later chapter.

    - Finally, you will see a summary of whether you have solved each
      exercise.  Note that summary does not include the critical
      information of whether the type is ok (that is, whether you
      accidentally changed the theorem statement): you have to look
      above for that information.

    Exercises that are manually graded will also show up in the
    output.  But since they have to be graded by a human, the test
    script won't be able to tell you much about them.  *)

(* 2021-08-11 15:08 *)


(** Generated by coq-of-ocaml *)
Require Import OCaml.OCaml.

Local Set Primitive Projections.
Local Open Scope string_scope.
Local Open Scope Z_scope.
Local Open Scope type_scope.
Import ListNotations.

Unset Positivity Checking.
Unset Guard Checking.

Inductive nat : Set :=
| O : nat
| S : nat -> nat.

Inductive natural : Set :=
| Zero : natural
| Succ : natural -> natural.

Fixpoint plus (plus_arg0 : natural) (plus_arg1 : natural) {struct plus_arg0}
  : natural :=
  match plus_arg0 with
  | Zero => plus_arg1
  | Succ n => Succ (plus n plus_arg1)
  end.

Fixpoint mult (mult_arg0 : natural) (mult_arg1 : natural) {struct mult_arg0}
  : natural :=
  match mult_arg0 with
  | Zero => Zero
  | Succ n => plus (mult n mult_arg1) mult_arg1
  end.

Definition synth (lf2 : natural) (y : natural) (lf1 : natural) : natural :=
  mult (Succ y) lf2.