From LF Require Import Logic.
Set Default Goal Selector "!".

(** **** Exercise: 1 star, 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.) *)

Theorem and_assoc : forall P Q R : Prop,
  P /\ (Q /\ R) -> (P /\ Q) /\ R.
Proof.
  intros P Q R [HP [HQ HR]].
  (* FILL IN HERE *) Admitted.

(** **** Exercise: 2 stars, standard (mult_is_O) *)
Lemma mult_is_O :
  forall n m, n * m = 0 -> n = 0 \/ m = 0.
Proof.
  (* FILL IN HERE *) Admitted.
(** [] *)

(** **** Exercise: 1 star, standard (or_commut) *)
Theorem or_commut : forall P Q : Prop,
  P \/ Q  -> Q \/ P.
Proof.
  (* FILL IN HERE *) Admitted.
(** [] *)

(** **** Exercise: 1 star, standard, especially useful (contrapositive) *)
Theorem contrapositive : forall (P Q : Prop),
  (P -> Q) -> (~Q -> ~P).
Proof.
  (* FILL IN HERE *) Admitted.
(** [] *)

(** **** 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.
  (* FILL IN HERE *) Admitted.
(** [] *)

(** **** 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.
  (* FILL IN HERE *) Admitted.
(** [] *)

(** **** 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].
  (* FILL IN HERE *) Admitted.
(** [] *)

(** **** Exercise: 2 stars (nth_error_In) *)
Theorem nth_error_In : forall {A} n (x : A) l,
  nth_error l n = Some x -> In x l.
Proof.
  (* FILL IN HERE *) Admitted.
(** [] *)

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

    We noted above that functions returning propositions can be seen as
    _properties_ of their arguments. For instance, if [P] has type
    [nat -> Prop], then [P n] says 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
  (* REPLACE THIS LINE WITH ":= _your_definition_ ." *). Admitted.

Theorem All_In :
  forall T (P : T -> Prop) (l : list T),
    (forall x, In x l -> P x) <->
    All P l.
Proof.
  (* FILL IN HERE *) Admitted.
(** [] *)

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

(** **** Exercise: 3 stars, standard (logic_practice) *)
Lemma logic_test : forall P Q R S, P /\ (exists x : nat, (S /\ Q x) \/ (S /\ R x)) ->
  (exists y, P /\ Q y) \/ (S /\ exists y, R y).
  (* FILL IN HERE *) Admitted.
(** [] *)

(** **** 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.
(** [] *)

(* grad exercise *)
(** **** 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
    the negation of such an axiom leads to a contradiction.  If [~ (P
    \/ ~P)] were provable, then by [de_morgan_not_or] as proved above,
    [P /\ ~P] would be provable, which would be a contradiction. So, it
    is safe to add [P \/ ~P] as an axiom for any particular [P].

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

Theorem excluded_middle_irrefutable: forall (P : Prop),
  ~ ~ (P \/ ~ P).
Proof.
  (* FILL IN HERE *) Admitted.
(** [] *)