Logic.v

(** * Logic: Logic in Coq *)

Set Warnings "-notation-overridden,-parsing".
From LF Require Export Tactics.

(** We have seen many examples of factual claims (_propositions_)
    and ways of presenting evidence of their truth (_proofs_).  In
    particular, we have worked extensively with _equality
    propositions_ ([e1 = e2]), implications ([P -> Q]), and quantified
    propositions ([forall x, P]).  In this chapter, we will see how
    Coq can be used to carry out other familiar forms of logical
    reasoning.

    Before diving into details, let's talk a bit about the status of
    mathematical statements in Coq.  Recall that Coq is a _typed_
    language, which means that every sensible expression in its world
    has an associated type.  Logical claims are no exception: any
    statement we might try to prove in Coq has a type, namely [Prop],
    the type of _propositions_.  We can see this with the [Check]
    command: *)

Check 3 = 3 : Prop.

Check forall n m : nat, n + m = m + n : Prop.

(** Note that _all_ syntactically well-formed propositions have type
    [Prop] in Coq, regardless of whether they are true. *)

(** Simply _being_ a proposition is one thing; being _provable_ is
    something else! *)

Check 2 = 2 : Prop.

Check 3 = 2 : Prop.

Check forall n : nat, n = 2 : Prop.

(** Indeed, propositions not only have types: they are
    _first-class_ entities that can be manipulated in all the same
    ways as any of the other things in Coq's world. *)

(** So far, we've seen one primary place that propositions can appear:
    in [Theorem] (and [Lemma] and [Example]) declarations. *)

Theorem plus_2_2_is_4 :
  2 + 2 = 4.
Proof. reflexivity.  Qed.

(** But propositions can be used in many other ways.  For example, we
    can give a name to a proposition using a [Definition], just as we
    have given names to other kinds of expressions. *)

Definition plus_claim : Prop := 2 + 2 = 4.
Check plus_claim : Prop.

(** We can later use this name in any situation where a proposition is
    expected -- for example, as the claim in a [Theorem] declaration. *)

Theorem plus_claim_is_true :
  plus_claim.
Proof. reflexivity.  Qed.

(** We can also write _parameterized_ propositions -- that is,
    functions that take arguments of some type and return a
    proposition. *)

(** For instance, the following function takes a number
    and returns a proposition asserting that this number is equal to
    three: *)

Definition is_three (n : nat) : Prop :=
  n = 3.
Check is_three : nat -> Prop.

(** In Coq, functions that return propositions are said to define
    _properties_ of their arguments. For instance, here's
    a (polymorphic) property defining the familiar notion of an
    _injective function_. *)

Definition injective {A B} (f : A -> B) :=
  forall x y : A, f x = f y -> x = y.

Lemma succ_inj : injective S.
Proof.
  intros n m H. injection H as H1. apply H1.
Qed.

(** The equality operator [=] is also a function that returns a
    [Prop].

    The expression [n = m] is syntactic sugar for [eq n m] (defined in
    Coq's standard library using the [Notation] mechanism). Because
    [eq] can be used with elements of any type, it is also
    polymorphic: *)

Check @eq : forall A : Type, A -> A -> Prop.

(** (Notice that we wrote [@eq] instead of [eq]: The type
    argument [A] to [eq] is declared as implicit, and we need to turn
    off the inference of this implicit argument to see the full type
    of [eq].) *)

(* ################################################################# *)
(** * Logical Connectives *)

(* ================================================================= *)
(** ** Conjunction *)

(** The _conjunction_, or _logical and_, of propositions [A] and [B]
    is written [A /\ B], representing the claim that both [A] and [B]
    are true. *)

Example and_example : 3 + 4 = 7 /\ 2 * 2 = 4.

(** To prove a conjunction, use the [split] tactic.  It will generate
    two subgoals, one for each part of the statement: *)

Proof.
  split.
  - (* 3 + 4 = 7 *) reflexivity.
  - (* 2 * 2 = 4 *) reflexivity.
Qed.

(** For any propositions [A] and [B], if we assume that [A] is true
    and that [B] is true, we can conclude that [A /\ B] is also
    true. *)

Lemma and_intro : forall A B : Prop, A -> B -> A /\ B.
Proof.
  intros A B HA HB. split.
  - apply HA.
  - apply HB.
Qed.

(** Since applying a theorem with hypotheses to some goal has the
    effect of generating as many subgoals as there are hypotheses for
    that theorem, we can apply [and_intro] to achieve the same effect
    as [split]. *)

Example and_example' : 3 + 4 = 7 /\ 2 * 2 = 4.
Proof.
  apply and_intro.
  - (* 3 + 4 = 7 *) reflexivity.
  - (* 2 + 2 = 4 *) reflexivity.
Qed.

(** **** Exercise: 2 stars, standard (and_exercise)  *)
Example and_exercise :
  forall n m : nat, n + m = 0 -> n = 0 /\ m = 0.
Proof.
  intros. inversion H. destruct n.
  - auto.
  - discriminate.
Qed.
(** [] *)

(** So much for proving conjunctive statements.  To go in the other
    direction -- i.e., to _use_ a conjunctive hypothesis to help prove
    something else -- we employ the [destruct] tactic.

    If the proof context contains a hypothesis [H] of the form
    [A /\ B], writing [destruct H as [HA HB]] will remove [H] from the
    context and add two new hypotheses: [HA], stating that [A] is
    true, and [HB], stating that [B] is true.  *)

Lemma and_example2 :
  forall n m : nat, n = 0 /\ m = 0 -> n + m = 0.
Proof.
  (* WORKED IN CLASS *)
  intros n m H.
  destruct H as [Hn Hm].
  rewrite Hn. rewrite Hm.
  reflexivity.
Qed.

(** As usual, we can also destruct [H] right when we introduce it,
    instead of introducing and then destructing it: *)

Lemma and_example2' :
  forall n m : nat, n = 0 /\ m = 0 -> n + m = 0.
Proof.
  intros n m [Hn Hm].
  rewrite Hn. rewrite Hm.
  reflexivity.
Qed.

(** You may wonder why we bothered packing the two hypotheses [n = 0]
    and [m = 0] into a single conjunction, since we could have also
    stated the theorem with two separate premises: *)

Lemma and_example2'' :
  forall n m : nat, n = 0 -> m = 0 -> n + m = 0.
Proof.
  intros n m Hn Hm.
  rewrite Hn. rewrite Hm.
  reflexivity.
Qed.

(** For this specific theorem, both formulations are fine.  But
    it's important to understand how to work with conjunctive
    hypotheses because conjunctions often arise from intermediate
    steps in proofs, especially in larger developments.  Here's a
    simple example: *)

Lemma and_example3 :
  forall n m : nat, n + m = 0 -> n * m = 0.
Proof.
  (* WORKED IN CLASS *)
  intros n m H.
  apply and_exercise in H.
  destruct H as [Hn Hm].
  rewrite Hn. reflexivity.
Qed.

(** Another common situation with conjunctions is that we know
    [A /\ B] but in some context we need just [A] or just [B].
    In such cases we can do a [destruct] (possibly as part of
    an [intros]) and use an underscore pattern [_] to indicate
    that the unneeded conjunct should just be thrown away. *)

Lemma proj1 : forall P Q : Prop,
  P /\ Q -> P.
Proof.
  intros P Q HPQ.
  destruct HPQ as [HP _].
  apply HP.  Qed.

(** **** Exercise: 1 star, standard, optional (proj2)  *)
Lemma proj2 : forall P Q : Prop,
  P /\ Q -> Q.
Proof.
  intros P Q [_ R]. assumption.
Qed.
(** [] *)

(** Finally, we sometimes need to rearrange the order of conjunctions
    and/or the grouping of multi-way conjunctions.  The following
    commutativity and associativity theorems are handy in such
    cases. *)

Theorem and_commut : forall P Q : Prop,
  P /\ Q -> Q /\ P.
Proof.
  intros P Q [HP HQ].
  split.
    - (* left *) apply HQ.
    - (* right *) apply HP.  Qed.

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

    (In the following proof of associativity, notice how the _nested_
    [intros] pattern breaks the hypothesis [H : P /\ (Q /\ R)] down into
    [HP : P], [HQ : Q], and [HR : R].  Finish the proof from
    there.) *)

Theorem and_assoc : forall P Q R : Prop,
  P /\ (Q /\ R) -> (P /\ Q) /\ R.
Proof.
  intros P Q R [HP [HQ HR]].
  split.
  - split; assumption.
  - assumption.
Qed.
(** [] *)

(** By the way, the infix notation [/\] is actually just syntactic
    sugar for [and A B].  That is, [and] is a Coq operator that takes
    two propositions as arguments and yields a proposition. *)

Check and : Prop -> Prop -> Prop.

(* ================================================================= *)
(** ** Disjunction *)

(** Another important connective is the _disjunction_, or _logical or_,
    of two propositions: [A \/ B] is true when either [A] or [B]
    is.  (This infix notation stands for [or A B], where [or : Prop ->
    Prop -> Prop].) *)

(** To use a disjunctive hypothesis in a proof, we proceed by case
    analysis (which, as with other data types like [nat], can be done
    explicitly with [destruct] or implicitly with an [intros]
    pattern): *)

Lemma eq_mult_0 :
  forall n m : nat, n = 0 \/ m = 0 -> n * m = 0.
Proof.
  (* This pattern implicitly does case analysis on
     [n = 0 \/ m = 0] *)
  intros n m [Hn | Hm].
  - (* Here, [n = 0] *)
    rewrite Hn. reflexivity.
  - (* Here, [m = 0] *)
    rewrite Hm. rewrite <- mult_n_O.
    reflexivity.
Qed.

(** Conversely, to show that a disjunction holds, it suffices to show
    that one of its sides holds. This is done via two tactics, [left]
    and [right].  As their names imply, the first one requires proving
    the left side of the disjunction, while the second requires
    proving its right side.  Here is a trivial use... *)

Lemma or_intro_l : forall A B : Prop, A -> A \/ B.
Proof.
  intros A B HA.
  left.
  apply HA.
Qed.

(** ... and here is a slightly more interesting example requiring both
    [left] and [right]: *)

Lemma zero_or_succ :
  forall n : nat, n = 0 \/ n = S (pred n).
Proof.
  (* WORKED IN CLASS *)
  intros [|n'].
  - left. reflexivity.
  - right. reflexivity.
Qed.

(** **** Exercise: 1 star, standard (mult_eq_0)  *)
Lemma mult_eq_0 :
  forall n m, n * m = 0 -> n = 0 \/ m = 0.
Proof.
  intros. destruct n.
  - left. reflexivity.
  - right. simpl in H. destruct m.
    + reflexivity.
    + discriminate.
Qed.
(** [] *)

(** **** Exercise: 1 star, standard (or_commut)  *)
Theorem or_commut : forall P Q : Prop,
  P \/ Q  -> Q \/ P.
Proof.
  intros. destruct H.
  - right. assumption. 
  - left. assumption.
Qed.
(** [] *)

(* ================================================================= *)
(** ** Falsehood and Negation

    So far, we have mostly been concerned with proving that certain
    things are _true_ -- addition is commutative, appending lists is
    associative, etc.  Of course, we may also be interested in
    negative results, demonstrating that some given proposition is
    _not_ true. Such statements are expressed with the logical
    negation operator [~]. *)

(** To see how negation works, recall the _principle of explosion_
    from the [Tactics] chapter, which asserts that, if we assume a
    contradiction, then any other proposition can be derived. 
    Following this intuition, we could define [~ P] ("not [P]") as
    [forall Q, P -> Q].

    Coq actually makes a slightly different (but equivalent) choice,
    defining [~ P] as [P -> False], where [False] is a specific
    contradictory proposition defined in the standard library. *)

Module MyNot.

Definition not (P:Prop) := P -> False.

Notation "~ x" := (not x) : type_scope.

Check not : Prop -> Prop.

End MyNot.

(** Since [False] is a contradictory proposition, the principle of
    explosion also applies to it. If we get [False] into the proof
    context, we can use [destruct] on it to complete any goal: *)

Theorem ex_falso_quodlibet : forall (P:Prop),
  False -> P.
Proof.
  (* WORKED IN CLASS *)
  intros P contra.
  destruct contra.  Qed.

(** The Latin _ex falso quodlibet_ means, literally, "from falsehood
    follows whatever you like"; this is another common name for the
    principle of explosion. *)

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

    Show that Coq's definition of negation implies the intuitive one
    mentioned above: *)

Fact not_implies_our_not : forall (P:Prop),
  ~ P -> (forall (Q:Prop), P -> Q).
Proof.
  intros. unfold not in H. apply H in H0. contradiction.
Qed.
(** [] *)

(** Inequality is a frequent enough example of negated statement
    that there is a special notation for it, [x <> y]:

      Notation "x <> y" := (~(x = y)).
*)

(** We can use [not] to state that [0] and [1] are different elements
    of [nat]: *)

Theorem zero_not_one : 0 <> 1.
Proof.
  (** The proposition [0 <> 1] is exactly the same as
      [~(0 = 1)], that is [not (0 = 1)], which unfolds to
      [(0 = 1) -> False]. (We use [unfold not] explicitly here
      to illustrate that point, but generally it can be omitted.) *)
  unfold not.
  (** To prove an inequality, we may assume the opposite
      equality... *)
  intros contra.
  (** ... and deduce a contradiction from it. Here, the
      equality [O = S O] contradicts the disjointness of
      constructors [O] and [S], so [discriminate] takes care
      of it. *)
  discriminate contra.
Qed.

(** It takes a little practice to get used to working with negation in
    Coq.  Even though you can see perfectly well why a statement
    involving negation is true, it can be a little tricky at first to
    make Coq understand it!  Here are proofs of a few familiar facts
    to get you warmed up. *)

Theorem not_False :
  ~ False.
Proof.
  unfold not. intros H. destruct H. Qed.

Theorem contradiction_implies_anything : forall P Q : Prop,
  (P /\ ~P) -> Q.
Proof.
  (* WORKED IN CLASS *)
  intros P Q [HP HNA]. unfold not in HNA.
  apply HNA in HP. destruct HP.  Qed.

Theorem double_neg : forall P : Prop,
  P -> ~~P.
Proof.
  (* WORKED IN CLASS *)
  intros P H. unfold not. intros G. apply G. apply H.  Qed.

(** **** Exercise: 2 stars, advanced (double_neg_inf) 

    Write an informal proof of [double_neg]:

   _Theorem_: [P] implies [~~P], for any proposition [P]. *)

(* FILL IN HERE *)

(* Do not modify the following line: *)
Definition manual_grade_for_double_neg_inf : option (nat*string) := None.
(** [] *)

(** **** Exercise: 2 stars, standard, especially useful (contrapositive)  *)
Theorem contrapositive : forall (P Q : Prop),
  (P -> Q) -> (~Q -> ~P).
Proof.
  intros. unfold not in H0.
  (* FILL IN HERE *) Admitted.
(** [] *)

(** **** Exercise: 1 star, standard (not_both_true_and_false)  *)
Theorem not_both_true_and_false : forall P : Prop,
  ~ (P /\ ~P).
Proof.
  unfold not. intros. destruct H.
  apply H0 in H. assumption.
Qed.
(** [] *)

(** **** Exercise: 1 star, advanced (informal_not_PNP) 

    Write an informal proof (in English) of the proposition [forall P
    : Prop, ~(P /\ ~P)]. *)

(* FILL IN HERE *)

(* Do not modify the following line: *)
Definition manual_grade_for_informal_not_PNP : option (nat*string) := None.
(** [] *)

(** Since inequality involves a negation, it also requires a little
    practice to be able to work with it fluently.  Here is one useful
    trick.  If you are trying to prove a goal that is
    nonsensical (e.g., the goal state is [false = true]), apply
    [ex_falso_quodlibet] to change the goal to [False].  This makes it
    easier to use assumptions of the form [~P] that may be available
    in the context -- in particular, assumptions of the form
    [x<>y]. *)

Theorem not_true_is_false : forall b : bool,
  b <> true -> b = false.
Proof.
  intros b H.
  destruct b eqn:HE.
  - (* b = true *)
    unfold not in H.
    apply ex_falso_quodlibet.
    apply H. reflexivity.
  - (* b = false *)
    reflexivity.
Qed.

(** Since reasoning with [ex_falso_quodlibet] is quite common, Coq
    provides a built-in tactic, [exfalso], for applying it. *)

Theorem not_true_is_false' : forall b : bool,
  b <> true -> b = false.
Proof.
  intros [] H.          (* note implicit [destruct b] here *)
  - (* b = true *)
    unfold not in H.
    exfalso.                (* <=== *)
    apply H. reflexivity.
  - (* b = false *) reflexivity.
Qed.

(* ================================================================= *)
(** ** Truth *)

(** Besides [False], Coq's standard library also defines [True], a
    proposition that is trivially true. To prove it, we use the
    predefined constant [I : True]: *)

Lemma True_is_true : True.
Proof. apply I. Qed.

(** Unlike [False], which is used extensively, [True] is used quite
    rarely, since it is trivial (and therefore uninteresting) to prove
    as a goal, and it carries no useful information as a hypothesis.

    But it can be quite useful when defining complex [Prop]s using
    conditionals or as a parameter to higher-order [Prop]s.  We will
    see examples later on. *)

(* ================================================================= *)
(** ** Logical Equivalence *)

(** The handy "if and only if" connective, which asserts that two
    propositions have the same truth value, is simply the conjunction
    of two implications. *)

Module MyIff.

Definition iff (P Q : Prop) := (P -> Q) /\ (Q -> P).

Notation "P <-> Q" := (iff P Q)
                      (at level 95, no associativity)
                      : type_scope.

End MyIff.

Theorem iff_sym : forall P Q : Prop,
  (P <-> Q) -> (Q <-> P).
Proof.
  (* WORKED IN CLASS *)
  intros P Q [HAB HBA].
  split.
  - (* -> *) apply HBA.
  - (* <- *) apply HAB.  Qed.

Lemma not_true_iff_false : forall b,
  b <> true <-> b = false.
Proof.
  (* WORKED IN CLASS *)
  intros b. split.
  - (* -> *) apply not_true_is_false.
  - (* <- *)
    intros H. rewrite H. intros H'. discriminate H'.
Qed.

(** **** Exercise: 1 star, standard, optional (iff_properties) 

    Using the above proof that [<->] is symmetric ([iff_sym]) as
    a guide, prove that it is also reflexive and transitive. *)

Theorem iff_refl : forall P : Prop,
  P <-> P.
Proof.
  split; intros; assumption.
Qed.

Theorem iff_trans : forall P Q R : Prop,
  (P <-> Q) -> (Q <-> R) -> (P <-> R).
Proof.
  intros. split; destruct H; destruct H0; auto.
Qed.
(** [] *)

(** **** Exercise: 3 stars, standard (or_distributes_over_and)  *)
Theorem or_distributes_over_and : forall P Q R : Prop,
  P \/ (Q /\ R) <-> (P \/ Q) /\ (P \/ R).
Proof.
  intros. split; intros.
  - destruct H.
    + split; left || right; assumption.
    + destruct H. split; right; assumption.
  - destruct H.
    + destruct H. destruct H0.
      * left. assumption.
      * left. assumption.
      * destruct H0.
        ** left. assumption.
        ** right. split; assumption.
Qed.
(** [] *)

(* ================================================================= *)
(** ** Setoids and Logical Equivalence *)

(** Some of Coq's tactics treat [iff] statements specially, avoiding
    the need for some low-level proof-state manipulation.  In
    particular, [rewrite] and [reflexivity] can be used with [iff]
    statements, not just equalities.  To enable this behavior, we have
    to import the Coq library that supports it: *)

From Coq Require Import Setoids.Setoid.

(** A "setoid" is a set equipped with an equivalence relation,
    that is, a relation that is reflexive, symmetric, and transitive.
    When two elements of a set are equivalent according to the
    relation, [rewrite] can be used to replace one element with the
    other. We've seen that already with the equality relation [=] in
    Coq: when [x = y], we can use [rewrite] to replace [x] with [y],
    or vice-versa.

    Similarly, the logical equivalence relation [<->] is reflexive,
    symmetric, and transitive, so we can use it to replace one part of
    a proposition with another: if [P <-> Q], then we can use
    [rewrite] to replace [P] with [Q], or vice-versa. *)

(** Here is a simple example demonstrating how these tactics work with
    [iff].  First, let's prove a couple of basic iff equivalences. *)

Lemma mult_0 : forall n m, n * m = 0 <-> n = 0 \/ m = 0.
Proof.
  split.
  - apply mult_eq_0.
  - apply eq_mult_0.
Qed.

Theorem or_assoc :
  forall P Q R : Prop, P \/ (Q \/ R) <-> (P \/ Q) \/ R.
Proof.
  intros P Q R. split.
  - intros [H | [H | H]].
    + left. left. apply H.
    + left. right. apply H.
    + right. apply H.
  - intros [[H | H] | H].
    + left. apply H.
    + right. left. apply H.
    + right. right. apply H.
Qed.

(** We can now use these facts with [rewrite] and [reflexivity]
    to give smooth proofs of statements involving equivalences.  For
    example, here is a ternary version of the previous [mult_0]
    result: *)

Lemma mult_0_3 :
  forall n m p, n * m * p = 0 <-> n = 0 \/ m = 0 \/ p = 0.
Proof.
  intros n m p.
  rewrite mult_0. rewrite mult_0. rewrite or_assoc.
  reflexivity.
Qed.

(** The [apply] tactic can also be used with [<->]. When given an
    equivalence as its argument, [apply] tries to guess which direction of
    the equivalence will be useful. *)

Lemma apply_iff_example :
  forall n m : nat, n * m = 0 -> n = 0 \/ m = 0.
Proof.
  intros n m H. apply mult_0. apply H.
Qed.

(* ================================================================= *)
(** ** Existential Quantification *)

(** Another important logical connective is _existential
    quantification_.  To say that there is some [x] of type [T] such
    that some property [P] holds of [x], we write [exists x : T,
    P]. As with [forall], the type annotation [: T] can be omitted if
    Coq is able to infer from the context what the type of [x] should
    be. *)

(** To prove a statement of the form [exists x, P], we must show that
    [P] holds for some specific choice of value for [x], known as the
    _witness_ of the existential.  This is done in two steps: First,
    we explicitly tell Coq which witness [t] we have in mind by
    invoking the tactic [exists t].  Then we prove that [P] holds after
    all occurrences of [x] are replaced by [t]. *)

Definition even x := exists n : nat, x = double n.

Lemma four_is_even : even 4.
Proof.
  unfold even. exists 2. reflexivity.
Qed.

(** Conversely, if we have an existential hypothesis [exists x, P] in
    the context, we can destruct it to obtain a witness [x] and a
    hypothesis stating that [P] holds of [x]. *)

Theorem exists_example_2 : forall n,
  (exists m, n = 4 + m) ->
  (exists o, n = 2 + o).
Proof.
  (* WORKED IN CLASS *)
  intros n [m Hm]. (* note implicit [destruct] here *)
  exists (2 + m).
  apply Hm.  Qed.

(** **** Exercise: 1 star, standard, especially useful (dist_not_exists) 

    Prove that "[P] holds for all [x]" implies "there is no [x] for
    which [P] does not hold."  (Hint: [destruct H as [x E]] works on
    existential assumptions!)  *)

Theorem dist_not_exists : forall (X:Type) (P : X -> Prop),
  (forall x, P x) -> ~ (exists x, ~ P x).
Proof.
  unfold not. intros. destruct H0. apply H0 in H. assumption.
Qed.
(** [] *)

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

    Prove that existential quantification distributes over
    disjunction. *)

Theorem dist_exists_or : forall (X:Type) (P Q : X -> Prop),
  (exists x, P x \/ Q x) <-> (exists x, P x) \/ (exists x, Q x).
Proof.
  split; intros.
    - destruct H. destruct H.
      + left. exists x. assumption.
      + right. exists x. assumption.
    - destruct H. destruct H.
      + exists x. left. assumption.
      + destruct H. exists x. right. assumption.
Qed.
(** [] *)

(* ################################################################# *)
(** * Programming with Propositions *)

(** The logical connectives that we have seen provide a rich
    vocabulary for defining complex propositions from simpler ones.
    To illustrate, let's look at how to express the claim that an
    element [x] occurs in a list [l].  Notice that this property has a
    simple recursive structure:

       - If [l] is the empty list, then [x] cannot occur in it, so the
         property "[x] appears in [l]" is simply false.

       - Otherwise, [l] has the form [x' :: l'].  In this case, [x]
         occurs in [l] if either it is equal to [x'] or it occurs in
         [l']. *)

(** We can translate this directly into a straightforward recursive
    function taking an element and a list and returning a proposition (!): *)

Fixpoint In {A : Type} (x : A) (l : list A) : Prop :=
  match l with
  | [] => False
  | x' :: l' => x' = x \/ In x l'
  end.

(** When [In] is applied to a concrete list, it expands into a
    concrete sequence of nested disjunctions. *)

Example In_example_1 : In 4 [1; 2; 3; 4; 5].
Proof.
  (* WORKED IN CLASS *)
  simpl. right. right. right. left. reflexivity.
Qed.

Example In_example_2 :
  forall n, In n [2; 4] ->
  exists n', n = 2 * n'.
Proof.
  (* WORKED IN CLASS *)
  simpl.
  intros n [H | [H | []]].
  - exists 1. rewrite <- H. reflexivity.
  - exists 2. rewrite <- H. reflexivity.
Qed.
(** (Notice the use of the empty pattern to discharge the last case
    _en passant_.) *)

(** We can also prove more generic, higher-level lemmas about [In].

    Note, in the first, how [In] starts out applied to a variable and
    only gets expanded when we do case analysis on this variable: *)

Theorem In_map :
  forall (A B : Type) (f : A -> B) (l : list A) (x : A),
    In x l ->
    In (f x) (map f l).
Proof.
  intros A B f l x.
  induction l as [|x' l' IHl'].
  - (* l = nil, contradiction *)
    simpl. intros [].
  - (* l = x' :: l' *)
    simpl. intros [H | H].
    + rewrite H. left. reflexivity.
    + right. apply IHl'. apply H.
Qed.

(** This way of defining propositions recursively, though convenient
    in some cases, also has some drawbacks.  In particular, it is
    subject to Coq's usual restrictions regarding the definition of
    recursive functions, e.g., the requirement that they be "obviously
    terminating."  In the next chapter, we will see how to define
    propositions _inductively_, a different technique with its own set
    of strengths and limitations. *)

(** **** Exercise: 3 stars, standard (In_map_iff)  *)
Theorem In_map_iff :
  forall (A B : Type) (f : A -> B) (l : list A) (y : B),
    In y (map f l) <->
    exists x, f x = y /\ In x l.
Proof.
  intros A B f l y. split; intros. 
    - induction l; simpl in H.
      + contradiction.
      + destruct H.
        * exists x. split.
          -- assumption.
          -- simpl. left. auto.
        * apply IHl in H. simpl. destruct H. exists x0. split.
          -- destruct H. assumption.
          -- right. destruct H. assumption.
    - destruct H. destruct H. rewrite <- H. apply In_map. assumption.
Qed.
(** [] *)


(* assert (Z: forall X, forall (a : X), forall (xs : list X) (ys : list X), In a xs \/ In a ys <-> In a (xs ++ ys)). {
  intros. split.
    - intros.
} *)

(** **** Exercise: 2 stars, standard (In_app_iff)  *)
Theorem In_app_iff : forall A l l' (a:A),
  In a (l++l') <-> In a l \/ In a l'.
Proof.
  intros A l. induction l as [|a' l' IH]; intros.
    - simpl. split.
      + intros. auto.
      + intros. destruct H; contradiction || assumption.
    - simpl. split; intros.
      + apply or_assoc. destruct H.
        * repeat left. assumption.
        * apply IH in H. auto.
      + apply or_assoc in H. destruct H.
        * auto.
        * apply IH in H. auto.
Qed.
(** [] *)

(** **** Exercise: 3 stars, standard, especially useful (All) 

    Recall that functions returning propositions can be seen as
    _properties_ of their arguments. For instance, if [P] has type
    [nat -> Prop], then [P n] states that property [P] holds of [n].

    Drawing inspiration from [In], write a recursive function [All]
    stating that some property [P] holds of all elements of a list
    [l]. To make sure your definition is correct, prove the [All_In]
    lemma below.  (Of course, your definition should _not_ just
    restate the left-hand side of [All_In].) *)

Fixpoint All {T : Type} (P : T -> Prop) (l : list T) : Prop := 
  match l with 
  | [] => True
  | h :: t => P h /\ All P t 
  end.

Theorem All_In :
  forall T (P : T -> Prop) (l : list T),
    (forall x, In x l -> P x) <->
    All P l.
Proof.
  split; intros.
    - intros. induction l.
      + reflexivity.
      + simpl in *. split.
        * apply H. auto.
        * apply IHl. auto.
    - induction l.
      + simpl in *. contradiction.
      + simpl in *. destruct H. destruct H0.
        * rewrite <- H0. assumption.
        * apply IHl.
          -- assumption.
          -- assumption.  
Qed.
(** [] *)

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

    Complete the definition of the [combine_odd_even] function below.
    It takes as arguments two properties of numbers, [Podd] and
    [Peven], and it should return a property [P] such that [P n] is
    equivalent to [Podd n] when [n] is odd and equivalent to [Peven n]
    otherwise. *)

Definition combine_odd_even (Podd Peven : nat -> Prop) : nat -> Prop := 
  fun (x : nat) => (oddb x = true -> Podd x) /\ (oddb x = false -> Peven x).

(** To test your definition, prove the following facts: *)

Theorem combine_odd_even_intro :
  forall (Podd Peven : nat -> Prop) (n : nat),
    (oddb n = true -> Podd n) ->
    (oddb n = false -> Peven n) ->
    combine_odd_even Podd Peven n.
Proof.
  intros. unfold combine_odd_even. split.
    - assumption.
    - assumption.
Qed.

Theorem combine_odd_even_elim_odd :
  forall (Podd Peven : nat -> Prop) (n : nat),
    combine_odd_even Podd Peven n ->
    oddb n = true ->
    Podd n.
Proof.
  intros. unfold combine_odd_even in H. destruct H.
  apply H. assumption.
Qed.

Theorem combine_odd_even_elim_even :
  forall (Podd Peven : nat -> Prop) (n : nat),
    combine_odd_even Podd Peven n ->
    oddb n = false ->
    Peven n.
Proof.
  intros. unfold combine_odd_even in H. destruct H.
  apply H1. assumption.
Qed.

(** [] *)

(* ################################################################# *)
(** * Applying Theorems to Arguments *)

(** One feature that distinguishes Coq from some other popular
    proof assistants (e.g., ACL2 and Isabelle) is that it treats
    _proofs_ as first-class objects.

    There is a great deal to be said about this, but it is not
    necessary to understand it all in detail in order to use Coq.
    This section gives just a taste, while a deeper exploration can be
    found in the optional chapters [ProofObjects] and
    [IndPrinciples]. *)

(** We have seen that we can use [Check] to ask Coq to print the type
    of an expression.  We can also use it to ask what theorem a
    particular identifier refers to. *)

Check plus_comm : forall n m : nat, n + m = m + n.


(** Coq checks the _statement_ of the [plus_comm] theorem (or prints
    it for us, if we leave off the part beginning with the colon) in
    the same way that it checks the _type_ of any term that we ask it
    to [Check]. Why? *)

(** The reason is that the identifier [plus_comm] actually refers to a
    _proof object_, which represents a logical derivation establishing
    of the truth of the statement [forall n m : nat, n + m = m + n].  The
    type of this object is the proposition which it is a proof of. *)

(** Intuitively, this makes sense because the statement of a
    theorem tells us what we can use that theorem for, just as the
    type of a "computational" object tells us what we can do with that
    object -- e.g., if we have a term of type [nat -> nat -> nat], we
    can give it two [nat]s as arguments and get a [nat] back. 
    Similarly, if we have an object of type [n = m -> n + n = m + m]
    and we provide it an "argument" of type [n = m], we can derive
    [n + n = m + m]. *)

(** Operationally, this analogy goes even further: by applying a
    theorem as if it were a function, i.e., applying it to hypotheses
    with matching types, we can specialize its result without having
    to resort to intermediate assertions.  For example, suppose we
    wanted to prove the following result: *)

Lemma plus_comm3 :
  forall x y z, x + (y + z) = (z + y) + x.

(** It appears at first sight that we ought to be able to prove this
    by rewriting with [plus_comm] twice to make the two sides match.
    The problem, however, is that the second [rewrite] will undo the
    effect of the first. *)

Proof.
  (* WORKED IN CLASS *)
  intros x y z.
  rewrite plus_comm.
  rewrite plus_comm.
  (* We are back where we started... *)
Abort.

(** We saw similar problems back in Chapter [Induction], and saw one
    way to work around them by using [assert] to derive a specialized
    version of [plus_comm] that can be used to rewrite exactly where
    we want. *)

Lemma plus_comm3_take2 :
  forall x y z, x + (y + z) = (z + y) + x.
Proof.
  intros x y z.
  rewrite plus_comm.
  assert (H : y + z = z + y).
  { rewrite plus_comm. reflexivity. }
  rewrite H.
  reflexivity.
Qed.

(** A more elegant alternative is to apply [plus_comm] directly
    to the arguments we want to instantiate it with, in much the same
    way as we apply a polymorphic function to a type argument. *)

Lemma plus_comm3_take3 :
  forall x y z, x + (y + z) = (z + y) + x.
Proof.
  intros x y z.
  rewrite plus_comm.
  rewrite (plus_comm y z).
  reflexivity.
Qed.

(** Let's see another example of using a theorem like a
    function. The following theorem says: any list [l]
    containing some element must be nonempty. *)

Theorem in_not_nil :
  forall A (x : A) (l : list A), In x l -> l <> [].
Proof.
  intros A x l H. unfold not. intro Hl.
  rewrite Hl in H.
  simpl in H.
  apply H.
Qed.

(** What makes this interesting is that one quantified variable
    ([x]) does not appear in the conclusion ([l <> []]). *)

(** We should be able to use this theorem to prove the special case
    where [x] is [42]. However, naively, the tactic [apply in_not_nil]
    will fail because it cannot infer the value of [x]. *)

Lemma in_not_nil_42 :
  forall l : list nat, In 42 l -> l <> [].
Proof.
  intros l H.
  Fail apply in_not_nil.
Abort.

(** There are several ways to work around this: *)

(** Use [apply ... with ...] *)
Lemma in_not_nil_42_take2 :
  forall l : list nat, In 42 l -> l <> [].
Proof.
  intros l H.
  apply in_not_nil with (x := 42).
  apply H.
Qed.

(** Use [apply ... in ...] *)
Lemma in_not_nil_42_take3 :
  forall l : list nat, In 42 l -> l <> [].
Proof.
  intros l H.
  apply in_not_nil in H.
  apply H.
Qed.

(** Explicitly apply the lemma to the value for [x]. *)
Lemma in_not_nil_42_take4 :
  forall l : list nat, In 42 l -> l <> [].
Proof.
  intros l H.
  apply (in_not_nil nat 42).
  apply H.
Qed.

(** Explicitly apply the lemma to a hypothesis. *)
Lemma in_not_nil_42_take5 :
  forall l : list nat, In 42 l -> l <> [].
Proof.
  intros l H.
  apply (in_not_nil _ _ _ H).
Qed.

(** You can "use theorems as functions" in this way with almost all
    tactics that take a theorem name as an argument.  Note also that
    theorem application uses the same inference mechanisms as function
    application; thus, it is possible, for example, to supply
    wildcards as arguments to be inferred, or to declare some
    hypotheses to a theorem as implicit by default.  These features
    are illustrated in the proof below. (The details of how this proof
    works are not critical -- the goal here is just to illustrate what
    can be done.) *)

Example lemma_application_ex :
  forall {n : nat} {ns : list nat},
    In n (map (fun m => m * 0) ns) ->
    n = 0.
Proof.
  intros n ns H.
  destruct (proj1 _ _ (In_map_iff _ _ _ _ _) H)
           as [m [Hm _]].
  rewrite mult_0_r in Hm. rewrite <- Hm. reflexivity.
Qed.

(** We will see many more examples in later chapters. *)

(* ################################################################# *)
(** * Coq vs. Set Theory *)

(** Coq's logical core, the _Calculus of Inductive
    Constructions_, differs in some important ways from other formal
    systems that are used by mathematicians to write down precise and
    rigorous definitions and proofs.  For example, in the most popular
    foundation for paper-and-pencil mathematics, Zermelo-Fraenkel Set
    Theory (ZFC), a mathematical object can potentially be a member of
    many different sets; a term in Coq's logic, on the other hand, is
    a member of at most one type.  This difference often leads to
    slightly different ways of capturing informal mathematical
    concepts, but these are, by and large, about equally natural and
    easy to work with.  For example, instead of saying that a natural
    number [n] belongs to the set of even numbers, we would say in Coq
    that [even n] holds, where [even : nat -> Prop] is a property
    describing even numbers.

    However, there are some cases where translating standard
    mathematical reasoning into Coq can be cumbersome or sometimes
    even impossible, unless we enrich the core logic with additional
    axioms.

    We conclude this chapter with a brief discussion of some of the
    most significant differences between the two worlds. *)

(* ================================================================= *)
(** ** Functional Extensionality *)

(** The equality assertions that we have seen so far mostly have
    concerned elements of inductive types ([nat], [bool], etc.).  But,
    since Coq's equality operator is polymorphic, we can use it at
    _any_ type -- in particular, we can write propositions claiming
    that two _functions_ are equal to each other: *)

Example function_equality_ex1 :
  (fun x => 3 + x) = (fun x => (pred 4) + x).
Proof. reflexivity. Qed.

(** In common mathematical practice, two functions [f] and [g] are
    considered equal if they produce the same output on every input:

    (forall x, f x = g x) -> f = g

    This is known as the principle of _functional extensionality_. *)

(** Informally, an "extensional property" is one that pertains to an
    object's observable behavior.  Thus, functional extensionality
    simply means that a function's identity is completely determined
    by what we can observe from it -- i.e., the results we obtain
    after applying it. *)

(** However, functional extensionality is not part of Coq's built-in
    logic.  This means that some apparently "obvious" propositions are
    not provable. *)

Example function_equality_ex2 :
  (fun x => plus x 1) = (fun x => plus 1 x).
Proof.
   (* Stuck *)
Abort.

(** However, we can add functional extensionality to Coq's core using
    the [Axiom] command. *)

Axiom functional_extensionality : forall {X Y: Type}
                                    {f g : X -> Y},
  (forall (x:X), f x = g x) -> f = g.

(** Defining something as an [Axiom] has the same effect as stating a
    theorem and skipping its proof using [Admitted], but it alerts the
    reader that this isn't just something we're going to come back and
    fill in later! *)

(** We can now invoke functional extensionality in proofs: *)

Example function_equality_ex2 :
  (fun x => plus x 1) = (fun x => plus 1 x).
Proof.
  apply functional_extensionality. intros x.
  apply plus_comm.
Qed.

(** Naturally, we must be careful when adding new axioms into Coq's
    logic, as this can render it _inconsistent_ -- that is, it may
    become possible to prove every proposition, including [False],
    [2+2=5], etc.!

    Unfortunately, there is no simple way of telling whether an axiom
    is safe to add: hard work by highly trained mathematicians is
    often required to establish the consistency of any particular
    combination of axioms.

    Fortunately, it is known that adding functional extensionality, in
    particular, _is_ consistent. *)

(** To check whether a particular proof relies on any additional
    axioms, use the [Print Assumptions] command.  *)

Print Assumptions function_equality_ex2.
(* ===>
     Axioms:
     functional_extensionality :
         forall (X Y : Type) (f g : X -> Y),
                (forall x : X, f x = g x) -> f = g *)

(** **** Exercise: 4 stars, standard (tr_rev_correct) 

    One problem with the definition of the list-reversing function
    [rev] that we have is that it performs a call to [app] on each
    step; running [app] takes time asymptotically linear in the size
    of the list, which means that [rev] is asymptotically quadratic.
    We can improve this with the following definitions: *)

Fixpoint rev_append {X} (l1 l2 : list X) : list X :=
  match l1 with
  | [] => l2
  | x :: l1' => rev_append l1' (x :: l2)
  end.

Definition tr_rev {X} (l : list X) : list X :=
  rev_append l [].

(** This version of [rev] is said to be _tail-recursive_, because the
    recursive call to the function is the last operation that needs to
    be performed (i.e., we don't have to execute [++] after the
    recursive call); a decent compiler will generate very efficient
    code in this case.

    Prove that the two definitions are indeed equivalent. *)


Lemma rev_append_l2 : forall X, forall xs ys : list X,
  @rev_append X xs ys = @rev X xs ++ ys.
Proof. intros X xs. induction xs.
  - reflexivity.
  - simpl. intros. induction ys.
    + rewrite app_nil_r. apply IHxs.
    + rewrite <- (app_assoc X (rev xs ) [x] (x0 :: ys)). simpl. apply IHxs.
Qed.

Theorem tr_rev_correct : forall X, @tr_rev X = @rev X.
Proof.
  intros. apply functional_extensionality. unfold tr_rev. 
  intros xs. induction xs as [ | x xs' IHxs].
    - reflexivity.
    - simpl. apply rev_append_l2.
Qed. 
(** [] *)

(* ================================================================= *)
(** ** Propositions vs. Booleans *)

(** We've seen two different ways of expressing logical claims in Coq:
    with _booleans_ (of type [bool]), and with _propositions_ (of type
    [Prop]).

    For instance, to claim that a number [n] is even, we can say
    either... *)

(** ... that [evenb n] evaluates to [true]... *)
Example even_42_bool : evenb 42 = true.
Proof. reflexivity. Qed.

(** ... or that there exists some [k] such that [n = double k]. *)
Example even_42_prop : even 42.
Proof. unfold even. exists 21. reflexivity. Qed.

(** Of course, it would be pretty strange if these two
    characterizations of evenness did not describe the same set of
    natural numbers!  Fortunately, we can prove that they do... *)

(** We first need two helper lemmas. *)
Lemma evenb_double : forall k, evenb (double k) = true.
Proof.
  intros k. induction k as [|k' IHk'].
  - reflexivity.
  - simpl. apply IHk'.
Qed.

(** **** Exercise: 3 stars, standard (evenb_double_conv)  *)
Lemma evenb_double_conv : forall n, exists k,
  n = if evenb n then double k else S (double k).
Proof.
  (* Hint: Use the [evenb_S] lemma from [Induction.v]. *)
  (* FILL IN HERE *) Admitted.
(** [] *)

(** Now the main theorem: *)
Theorem even_bool_prop : forall n,
  evenb n = true <-> even n.
Proof.
  intros n. split.
  - intros H. destruct (evenb_double_conv n) as [k Hk].
    rewrite Hk. rewrite H. exists k. reflexivity.
  - intros [k Hk]. rewrite Hk. apply evenb_double.
Qed.

(** In view of this theorem, we say that the boolean computation
    [evenb n] is _reflected_ in the truth of the proposition
    [exists k, n = double k]. *)

(** Similarly, to state that two numbers [n] and [m] are equal, we can
    say either
      - (1) that [n =? m] returns [true], or
      - (2) that [n = m].
    Again, these two notions are equivalent. *)

Theorem eqb_eq : forall n1 n2 : nat,
  n1 =? n2 = true <-> n1 = n2.
Proof.
  intros n1 n2. split.
  - apply eqb_true.
  - intros H. rewrite H. rewrite <- eqb_refl. reflexivity.
Qed.

(** However, even when the boolean and propositional formulations of a
    claim are equivalent from a purely logical perspective, they are
    often not equivalent from the point of view of convenience for
    some specific purpose. *)

(** In the case of even numbers above, when proving the
    backwards direction of [even_bool_prop] (i.e., [evenb_double],
    going from the propositional to the boolean claim), we used a
    simple induction on [k].  On the other hand, the converse (the
    [evenb_double_conv] exercise) required a clever generalization,
    since we can't directly prove [(evenb n = true) -> even n]. *)

(** For these examples, the propositional claims are more useful than
    their boolean counterparts, but this is not always the case.  For
    instance, we cannot test whether a general proposition is true or
    not in a function definition; as a consequence, the following code
    fragment is rejected: *)

Fail
Definition is_even_prime n :=
  if n = 2 then true
  else false.

(** Coq complains that [n = 2] has type [Prop], while it expects
    an element of [bool] (or some other inductive type with two
    elements).  The reason has to do with the _computational_ nature
    of Coq's core language, which is designed so that every function
    it can express is computable and total.  One reason for this is to
    allow the extraction of executable programs from Coq developments.
    As a consequence, [Prop] in Coq does _not_ have a universal case
    analysis operation telling whether any given proposition is true
    or false, since such an operation would allow us to write
    non-computable functions.

    Beyond the fact that non-computable properties are impossible in
    general to phrase as boolean computations, even many _computable_
    properties are easier to express using [Prop] than [bool], since
    recursive function definitions in Coq are subject to significant
    restrictions.  For instance, the next chapter shows how to define
    the property that a regular expression matches a given string
    using [Prop].  Doing the same with [bool] would amount to writing
    a regular expression matching algorithm, which would be more
    complicated, harder to understand, and harder to reason about than
    a simple (non-algorithmic) definition of this property.

    Conversely, an important side benefit of stating facts using
    booleans is enabling some proof automation through computation
    with Coq terms, a technique known as _proof by reflection_.

    Consider the following statement: *)

Example even_1000 : even 1000.

(** The most direct way to prove this is to give the value of [k]
    explicitly. *)

Proof. unfold even. exists 500. reflexivity. Qed.

(** The proof of the corresponding boolean statement is even
    simpler (because we don't have to invent the witness: Coq's
    computation mechanism does it for us!). *)

Example even_1000' : evenb 1000 = true.
Proof. reflexivity. Qed.

(** What is interesting is that, since the two notions are equivalent,
    we can use the boolean formulation to prove the other one without
    mentioning the value 500 explicitly: *)

Example even_1000'' : even 1000.
Proof. apply even_bool_prop. reflexivity. Qed.

(** Although we haven't gained much in terms of proof-script
    size in this case, larger proofs can often be made considerably
    simpler by the use of reflection.  As an extreme example, a famous
    Coq proof of the even more famous _4-color theorem_ uses
    reflection to reduce the analysis of hundreds of different cases
    to a boolean computation. *)

(** Another notable difference is that the negation of a "boolean
    fact" is straightforward to state and prove: simply flip the
    expected boolean result. *)

Example not_even_1001 : evenb 1001 = false.
Proof.
  (* WORKED IN CLASS *)
  reflexivity.
Qed.

(** In contrast, propositional negation can be more difficult
    to work with directly. *)

Example not_even_1001' : ~(even 1001).
Proof.
  (* WORKED IN CLASS *)
  rewrite <- even_bool_prop.
  unfold not.
  simpl.
  intro H.
  discriminate H.
Qed.

(** Equality provides a complementary example, where it is sometimes
    easier to work in the propositional world. Knowing that
    [n =? m = true] is generally of little direct help in the middle of a
    proof involving [n] and [m]; however, if we convert the statement
    to the equivalent form [n = m], we can rewrite with it. *)

Lemma plus_eqb_example : forall n m p : nat,
    n =? m = true -> n + p =? m + p = true.
Proof.
  (* WORKED IN CLASS *)
  intros n m p H.
    rewrite eqb_eq in H.
  rewrite H.
  rewrite eqb_eq.
  reflexivity.
Qed.

(** We won't discuss reflection any further here, but it serves as a
    good example showing the complementary strengths of booleans and
    general propositions, and being able to cross back and forth
    between the boolean and propositional worlds will often be
    convenient in later chapters. *)

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

    The following theorems relate the propositional connectives studied
    in this chapter to the corresponding boolean operations. *)

Theorem andb_true_iff : forall b1 b2:bool,
  b1 && b2 = true <-> b1 = true /\ b2 = true.
Proof.
  (* FILL IN HERE *) Admitted.

Theorem orb_true_iff : forall b1 b2,
  b1 || b2 = true <-> b1 = true \/ b2 = true.
Proof.
  (* FILL IN HERE *) Admitted.
(** [] *)

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

    The following theorem is an alternate "negative" formulation of
    [eqb_eq] that is more convenient in certain situations.  (We'll see
    examples in later chapters.)  Hint: [not_true_iff_false]. *)

Theorem eqb_neq : forall x y : nat,
  x =? y = false <-> x <> y.
Proof.
  (* FILL IN HERE *) Admitted.
(** [] *)

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

    Given a boolean operator [eqb] for testing equality of elements of
    some type [A], we can define a function [eqb_list] for testing
    equality of lists with elements in [A].  Complete the definition
    of the [eqb_list] function below.  To make sure that your
    definition is correct, prove the lemma [eqb_list_true_iff]. *)

Fixpoint eqb_list {A : Type} (eqb : A -> A -> bool)
                  (l1 l2 : list A) : bool
  (* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.

Theorem eqb_list_true_iff :
  forall A (eqb : A -> A -> bool),
    (forall a1 a2, eqb a1 a2 = true <-> a1 = a2) ->
    forall l1 l2, eqb_list eqb l1 l2 = true <-> l1 = l2.
Proof.
(* FILL IN HERE *) Admitted.

(** [] *)

(** **** Exercise: 2 stars, standard, especially useful (All_forallb) 

    Recall the function [forallb], from the exercise
    [forall_exists_challenge] in chapter [Tactics]: *)

Fixpoint forallb {X : Type} (test : X -> bool) (l : list X) : bool :=
  match l with
  | [] => true
  | x :: l' => andb (test x) (forallb test l')
  end.

(** Prove the theorem below, which relates [forallb] to the [All]
    property defined above. *)

Theorem forallb_true_iff : forall X test (l : list X),
   forallb test l = true <-> All (fun x => test x = true) l.
Proof.
  (* FILL IN HERE *) Admitted.

(** (Ungraded thought question) Are there any important properties of
    the function [forallb] which are not captured by this
    specification? *)

(* FILL IN HERE

    [] *)

(* ================================================================= *)
(** ** Classical vs. Constructive Logic *)

(** We have seen that it is not possible to test whether or not a
    proposition [P] holds while defining a Coq function.  You may be
    surprised to learn that a similar restriction applies to _proofs_!
    In other words, the following intuitive reasoning principle is not
    derivable in Coq: *)

Definition excluded_middle := forall P : Prop,
  P \/ ~ P.

(** To understand operationally why this is the case, recall
    that, to prove a statement of the form [P \/ Q], we use the [left]
    and [right] tactics, which effectively require knowing which side
    of the disjunction holds.  But the universally quantified [P] in
    [excluded_middle] is an _arbitrary_ proposition, which we know
    nothing about.  We don't have enough information to choose which
    of [left] or [right] to apply, just as Coq doesn't have enough
    information to mechanically decide whether [P] holds or not inside
    a function. *)

(** However, if we happen to know that [P] is reflected in some
    boolean term [b], then knowing whether it holds or not is trivial:
    we just have to check the value of [b]. *)

Theorem restricted_excluded_middle : forall P b,
  (P <-> b = true) -> P \/ ~ P.
Proof.
  intros P [] H.
  - left. rewrite H. reflexivity.
  - right. rewrite H. intros contra. discriminate contra.
Qed.

(** In particular, the excluded middle is valid for equations [n = m],
    between natural numbers [n] and [m]. *)

Theorem restricted_excluded_middle_eq : forall (n m : nat),
  n = m \/ n <> m.
Proof.
  intros n m.
  apply (restricted_excluded_middle (n = m) (n =? m)).
  symmetry.
  apply eqb_eq.
Qed.

(** It may seem strange that the general excluded middle is not
    available by default in Coq, since it is a standard feature of
    familiar logics like ZFC.  But there is a distinct advantage in
    not assuming the excluded middle: statements in Coq make stronger
    claims than the analogous statements in standard mathematics.
    Notably, when there is a Coq proof of [exists x, P x], it is
    always possible to explicitly exhibit a value of [x] for which we
    can prove [P x] -- in other words, every proof of existence is
    _constructive_. *)

(** Logics like Coq's, which do not assume the excluded middle, are
    referred to as _constructive logics_.

    More conventional logical systems such as ZFC, in which the
    excluded middle does hold for arbitrary propositions, are referred
    to as _classical_. *)

(** The following example illustrates why assuming the excluded middle
    may lead to non-constructive proofs:

    _Claim_: There exist irrational numbers [a] and [b] such that
    [a ^ b] ([a] to the power [b]) is rational.

    _Proof_: It is not difficult to show that [sqrt 2] is irrational.
    If [sqrt 2 ^ sqrt 2] is rational, it suffices to take [a = b =
    sqrt 2] and we are done.  Otherwise, [sqrt 2 ^ sqrt 2] is
    irrational.  In this case, we can take [a = sqrt 2 ^ sqrt 2] and
    [b = sqrt 2], since [a ^ b = sqrt 2 ^ (sqrt 2 * sqrt 2) = sqrt 2 ^
    2 = 2].  []

    Do you see what happened here?  We used the excluded middle to
    consider separately the cases where [sqrt 2 ^ sqrt 2] is rational
    and where it is not, without knowing which one actually holds!
    Because of that, we finish the proof knowing that such [a] and [b]
    exist but we cannot determine what their actual values are (at least,
    not from this line of argument).

    As useful as constructive logic is, it does have its limitations:
    There are many statements that can easily be proven in classical
    logic but that have only much more complicated constructive proofs, and
    there are some that are known to have no constructive proof at
    all!  Fortunately, like functional extensionality, the excluded
    middle is known to be compatible with Coq's logic, allowing us to
    add it safely as an axiom.  However, we will not need to do so
    here: the results that we cover can be developed entirely
    within constructive logic at negligible extra cost.

    It takes some practice to understand which proof techniques must
    be avoided in constructive reasoning, but arguments by
    contradiction, in particular, are infamous for leading to
    non-constructive proofs.  Here's a typical example: suppose that
    we want to show that there exists [x] with some property [P],
    i.e., such that [P x].  We start by assuming that our conclusion
    is false; that is, [~ exists x, P x]. From this premise, it is not
    hard to derive [forall x, ~ P x].  If we manage to show that this
    intermediate fact results in a contradiction, we arrive at an
    existence proof without ever exhibiting a value of [x] for which
    [P x] holds!

    The technical flaw here, from a constructive standpoint, is that
    we claimed to prove [exists x, P x] using a proof of
    [~ ~ (exists x, P x)].  Allowing ourselves to remove double
    negations from arbitrary statements is equivalent to assuming the
    excluded middle, as shown in one of the exercises below.  Thus,
    this line of reasoning cannot be encoded in Coq without assuming
    additional axioms. *)

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

    Proving the consistency of Coq with the general excluded middle
    axiom requires complicated reasoning that cannot be carried out
    within Coq itself.  However, the following theorem implies that it
    is always safe to assume a decidability axiom (i.e., an instance
    of excluded middle) for any _particular_ Prop [P].  Why?  Because
    we cannot prove the negation of such an axiom.  If we could, we
    would have both [~ (P \/ ~P)] and [~ ~ (P \/ ~P)] (since [P]
    implies [~ ~ P], by lemma [double_neg], which we proved above),
    which would be a  contradiction.  But since we can't, it is safe
    to add [P \/ ~P] as an axiom.

    Succinctly: for any proposition P,
       [Coq is consistent ==> (Coq + P \/ ~P) is consistent].

    (Hint: you will need to come up with a clever assertion as the
    next step in the proof.) *)

Theorem excluded_middle_irrefutable: forall (P:Prop),
  ~ ~ (P \/ ~ P).
Proof.
  unfold not. intros P H.
  (* FILL IN HERE *) Admitted.
(** [] *)

(** **** Exercise: 3 stars, advanced (not_exists_dist) 

    It is a theorem of classical logic that the following two
    assertions are equivalent:

    ~ (exists x, ~ P x)
    forall x, P x

    The [dist_not_exists] theorem above proves one side of this
    equivalence. Interestingly, the other direction cannot be proved
    in constructive logic. Your job is to show that it is implied by
    the excluded middle. *)

Theorem not_exists_dist :
  excluded_middle ->
  forall (X:Type) (P : X -> Prop),
    ~ (exists x, ~ P x) -> (forall x, P x).
Proof.
  (* FILL IN HERE *) Admitted.
(** [] *)

(** **** Exercise: 5 stars, standard, optional (classical_axioms) 

    For those who like a challenge, here is an exercise taken from the
    Coq'Art book by Bertot and Casteran (p. 123).  Each of the
    following four statements, together with [excluded_middle], can be
    considered as characterizing classical logic.  We can't prove any
    of them in Coq, but we can consistently add any one of them as an
    axiom if we wish to work in classical logic.

    Prove that all five propositions (these four plus [excluded_middle])
    are equivalent.

    Hint: Rather than considering all pairs of statements pairwise,
    prove a single circular chain of implications that connects them
    all.
*)

Definition peirce := forall P Q: Prop,
  ((P -> Q) -> P) -> P.

Definition double_negation_elimination := forall P: Prop,
  ~~P -> P.

Definition de_morgan_not_and_not := forall P Q: Prop,
  ~(~P /\ ~Q) -> P \/ Q.

Definition implies_to_or := forall P Q: Prop,
  (P -> Q) -> (~P \/ Q).

(* excluded_middle -> peirce: 直接展开即可 *)
(* peirce -> excluded_middle: 排中律可看作 perice 的 Q 为 ~(P \/ ~P) 时的情况 *)
Theorem exc_mid_peirce:
  excluded_middle <-> peirce.
Proof. unfold excluded_middle, peirce. split.
  - intros. destruct (H P).
    + assumption.
    + apply H0. intros. unfold not in H1. apply H1 in H2. contradiction.
  - intros. apply (H _ (~(P \/ ~P))). unfold not. intro. right. intro. 
    apply H0; left; assumption.
Qed.

Theorem exc_mid_double_negation_elimination: 
  excluded_middle <-> double_negation_elimination.
Proof. unfold excluded_middle, double_negation_elimination. split.
  - intros. destruct (H P).
    + assumption.
    + unfold not in *. apply H0 in H1. contradiction.
  - intros. apply (H (P \/ ~P)). unfold not. 
    intros. apply H0. right.
    intros. apply H0. left. assumption. 
Qed.


(* peirce -> double_negation_elimination: double_negation_elimination 可看作 perice 的 Q 为 False 时的情况 *)
Theorem peirce_double_negation_elimination: 
  peirce <-> double_negation_elimination.
Proof.
  unfold peirce, double_negation_elimination.
  split.   
    - intros. apply (H _ False). intros. unfold not in *. apply H0 in H1. contradiction.
    - intros. apply H0.
Admitted.

Theorem double_negation_elimination_to_de_morgan_not_and_not: 
  double_negation_elimination <-> de_morgan_not_and_not.
Proof.
  unfold double_negation_elimination, de_morgan_not_and_not.
  unfold not. intros. 
Admitted.





(* FILL IN HERE

    [] *)

(* 2020-08-18 21:11 *)