(* Copyright (c) 2008, Adam Chlipala
 * 
 * This work is licensed under a
 * Creative Commons Attribution-Noncommercial-No Derivative Works 3.0
 * Unported License.
 * The license text is available at:
 *   http://creativecommons.org/licenses/by-nc-nd/3.0/
 *)

(* begin hide *)
Require Import Eqdep List.

Require Import Tactics.

Set Implicit Arguments.
(* end hide *)


(** %\chapter{Reasoning About Equality Proofs}% *)

(** In traditional mathematics, the concept of equality is usually taken as a given.  On the other hand, in type theory, equality is a very contentious subject.  There are at least three different notions of equality that are important, and researchers are actively investigating new definitions of what it means for two terms to be equal.  Even once we fix a notion of equality, there are inevitably tricky issues that arise in proving properties of programs that manipulate equality proofs explicitly.  In this chapter, we will focus on design patterns for circumventing these tricky issues, and we will introduce the different notions of equality as they are germane. *)


(** * Heterogeneous Lists Revisited *)

(** One of our example dependent data structures from the last chapter was heterogeneous lists and their associated "cursor" type. *)

Section fhlist.
  Variable A : Type.
  Variable B : A -> Type.

  Fixpoint fhlist (ls : list A) : Type :=
    match ls with
      | nil => unit
      | x :: ls' => B x * fhlist ls'
    end%type.

  Variable elm : A.

  Fixpoint fmember (ls : list A) : Type :=
    match ls with
      | nil => Empty_set
      | x :: ls' => (x = elm) + fmember ls'
    end%type.

  Fixpoint fhget (ls : list A) : fhlist ls -> fmember ls -> B elm :=
    match ls return fhlist ls -> fmember ls -> B elm with
      | nil => fun _ idx => match idx with end
      | _ :: ls' => fun mls idx =>
        match idx with
          | inl pf => match pf with
                        | refl_equal => fst mls
                      end
          | inr idx' => fhget ls' (snd mls) idx'
        end
    end.
End fhlist.

Implicit Arguments fhget [A B elm ls].

(** We can define a [map]-like function for [fhlist]s. *)

Section fhlist_map.
  Variables A : Type.
  Variables B C : A -> Type.
  Variable f : forall x, B x -> C x.

  Fixpoint fhmap (ls : list A) : fhlist B ls -> fhlist C ls :=
    match ls return fhlist B ls -> fhlist C ls with
      | nil => fun _ => tt
      | _ :: _ => fun hls => (f (fst hls), fhmap _ (snd hls))
    end.

  Implicit Arguments fhmap [ls].

  (** For the inductive versions of the [ilist] definitions, we proved a lemma about the interaction of [get] and [imap].  It was a strategic choice not to attempt such a proof for the definitions that we just gave, because that sets us on a collision course with the problems that are the subject of this chapter. *)

  Variable elm : A.

  Theorem get_imap : forall ls (mem : fmember elm ls) (hls : fhlist B ls),
    fhget (fhmap hls) mem = f (fhget hls mem).
    induction ls; crush.

    (** Part of our single remaining subgoal is:

       [[
   
  a0 : a = elm
  ============================
   match a0 in (_ = a2) return (C a2) with
   | refl_equal => f a1
   end = f match a0 in (_ = a2) return (B a2) with
           | refl_equal => a1
           end
       ]]

   This seems like a trivial enough obligation.  The equality proof [a0] must be [refl_equal], since that is the only constructor of [eq].  Therefore, both the [match]es reduce to the point where the conclusion follows by reflexivity.

    [[

    destruct a0.

    [[
User error: Cannot solve a second-order unification problem
]]

    This is one of Coq's standard error messages for informing us that its heuristics for attempting an instance of an undecidable problem about dependent typing have failed.  We might try to nudge things in the right direction by stating the lemma that we believe makes the conclusion trivial.

    [[

    assert (a0 = refl_equal _).

    [[
The term "refl_equal ?98" has type "?98 = ?98"
 while it is expected to have type "a = elm"
    ]]

    In retrospect, the problem is not so hard to see.  Reflexivity proofs only show [x = x] for particular values of [x], whereas here we are thinking in terms of a proof of [a = elm], where the two sides of the equality are not equal syntactically.  Thus, the essential lemma we need does not even type-check!

    Is it time to throw in the towel?  Luckily, the answer is "no."  In this chapter, we will see several useful patterns for proving obligations like this.

    For this particular example, the solution is surprisingly straightforward.  [destruct] has a simpler sibling [case] which should behave identically for any inductive type with one constructor of no arguments. *)

    case a0.
    (** [[

  ============================
   f a1 = f a1
   ]]

    It seems that [destruct] was trying to be too smart for its own good. *)

    reflexivity.
  Qed.

  (** It will be helpful to examine the proof terms generated by this sort of strategy.  A simpler example illustrates what is going on. *)

  Lemma lemma1 : forall x (pf : x = elm), O = match pf with refl_equal => O end.
    simple destruct pf; reflexivity.
  Qed.

  (** [simple destruct pf] is a convenient form for applying [case].  It runs [intro] to bring into scope all quantified variables up to its argument. *)

  Print lemma1.

  (** [[

lemma1 = 
fun (x : A) (pf : x = elm) =>
match pf as e in (_ = y) return (0 = match e with
                                     | refl_equal => 0
                                     end) with
| refl_equal => refl_equal 0
end
     : forall (x : A) (pf : x = elm), 0 = match pf with
                                          | refl_equal => 0
                                          end
    ]]

    Using what we know about shorthands for [match] annotations, we can write this proof in shorter form manually. *)

  Definition lemma1' :=
    fun (x : A) (pf : x = elm) =>
      match pf return (0 = match pf with
                             | refl_equal => 0
                           end) with
        | refl_equal => refl_equal 0
      end.

  (** Surprisingly, what seems at first like a %\textit{%#<i>#simpler#</i>#%}% lemma is harder to prove. *)

  Lemma lemma2 : forall (x : A) (pf : x = x), O = match pf with refl_equal => O end.
    (** [[

    simple destruct pf.
    
      [[

User error: Cannot solve a second-order unification problem
      ]] *)
  Abort.

  (** Nonetheless, we can adapt the last manual proof to handle this theorem. *)

  Definition lemma2' :=
    fun (x : A) (pf : x = x) =>
      match pf return (0 = match pf with
                             | refl_equal => 0
                           end) with
        | refl_equal => refl_equal 0
      end.

  (** We can try to prove a lemma that would simplify proofs of many facts like [lemma2]: *)

  Lemma lemma3 : forall (x : A) (pf : x = x), pf = refl_equal x.
    (** [[

    simple destruct pf.
    
      [[

User error: Cannot solve a second-order unification problem
      ]] *)
  Abort.

  (** This time, even our manual attempt fails.

     [[

  Definition lemma3' :=
    fun (x : A) (pf : x = x) =>
      match pf as pf' in (_ = x') return (pf' = refl_equal x') with
        | refl_equal => refl_equal _
      end.

     [[

The term "refl_equal x'" has type "x' = x'" while it is expected to have type
 "x = x'"
     ]]

     The type error comes from our [return] annotation.  In that annotation, the [as]-bound variable [pf'] has type [x = x'], refering to the [in]-bound variable [x'].  To do a dependent [match], we %\textit{%#<i>#must#</i>#%}% choose a fresh name for the second argument of [eq].  We are just as constrained to use the "real" value [x] for the first argument.  Thus, within the [return] clause, the proof we are matching on %\textit{%#<i>#must#</i>#%}% equate two non-matching terms, which makes it impossible to equate that proof with reflexivity.

     Nonetheless, it turns out that, with one catch, we %\textit{%#<i>#can#</i>#%}% prove this lemma. *)

  Lemma lemma3 : forall (x : A) (pf : x = x), pf = refl_equal x.
    intros; apply UIP_refl.
  Qed.

  Check UIP_refl.
  (** [[

UIP_refl
     : forall (U : Type) (x : U) (p : x = x), p = refl_equal x
     ]]

     [UIP_refl] comes from the [Eqdep] module of the standard library.  Do the Coq authors know of some clever trick for building such proofs that we have not seen yet?  If they do, they did not use it for this proof.  Rather, the proof is based on an %\textit{%#<i>#axiom#</i>#%}%. *)

  Print eq_rect_eq.
  (** [[

eq_rect_eq = 
fun U : Type => Eq_rect_eq.eq_rect_eq U
     : forall (U : Type) (p : U) (Q : U -> Type) (x : Q p) (h : p = p),
       x = eq_rect p Q x p h
      ]]

      [eq_rect_eq] states a "fact" that seems like common sense, once the notation is deciphered.  [eq_rect] is the automatically-generated recursion principle for [eq].  Calling [eq_rect] is another way of [match]ing on an equality proof.  The proof we match on is the argument [h], and [x] is the body of the [match].  [eq_rect_eq] just says that [match]es on proofs of [p = p], for any [p], are superfluous and may be removed.

      Perhaps surprisingly, we cannot prove [eq_rect_eq] from within Coq.  This proposition is introduced as an axiom; that is, a proposition asserted as true without proof.  We cannot assert just any statement without proof.  Adding [False] as an axiom would allow us to prove any proposition, for instance, defeating the point of using a proof assistant.  In general, we need to be sure that we never assert %\textit{%#<i>#inconsistent#</i>#%}% sets of axioms.  A set of axioms is inconsistent if its conjunction implies [False].  For the case of [eq_rect_eq], consistency has been verified outside of Coq via "informal" metatheory.

      This axiom is equivalent to another that is more commonly known and mentioned in type theory circles. *)

  Print Streicher_K.
(** [[

Streicher_K = 
fun U : Type => UIP_refl__Streicher_K U (UIP_refl U)
     : forall (U : Type) (x : U) (P : x = x -> Prop),
       P (refl_equal x) -> forall p : x = x, P p
  ]]

  This is the unfortunately-named "Streicher's axiom K," which says that a predicate on properly-typed equality proofs holds of all such proofs if it holds of reflexivity. *)

End fhlist_map.


(** * Type-Casts in Theorem Statements *)

(** Sometimes we need to use tricks with equality just to state the theorems that we care about.  To illustrate, we start by defining a concatenation function for [fhlist]s. *)

Section fhapp.
  Variable A : Type.
  Variable B : A -> Type.

  Fixpoint fhapp (ls1 ls2 : list A) {struct ls1}
    : fhlist B ls1 -> fhlist B ls2 -> fhlist B (ls1 ++ ls2) :=
    match ls1 return fhlist _ ls1 -> _ -> fhlist _ (ls1 ++ ls2) with
      | nil => fun _ hls2 => hls2
      | _ :: _ => fun hls1 hls2 => (fst hls1, fhapp _ _ (snd hls1) hls2)
    end.

  Implicit Arguments fhapp [ls1 ls2].

  (** We might like to prove that [fhapp] is associative.

     [[

  Theorem fhapp_ass : forall ls1 ls2 ls3
    (hls1 : fhlist B ls1) (hls2 : fhlist B ls2) (hls3 : fhlist B ls3),
    fhapp hls1 (fhapp hls2 hls3) = fhapp (fhapp hls1 hls2) hls3.

     [[

The term
 "fhapp (ls1:=ls1 ++ ls2) (ls2:=ls3) (fhapp (ls1:=ls1) (ls2:=ls2) hls1 hls2)
    hls3" has type "fhlist B ((ls1 ++ ls2) ++ ls3)"
 while it is expected to have type "fhlist B (ls1 ++ ls2 ++ ls3)"
     ]]

     This first cut at the theorem statement does not even type-check.  We know that the two [fhlist] types appearing in the error message are always equal, by associativity of normal list append, but this fact is not apparent to the type checker.  This stems from the fact that Coq's equality is %\textit{%#<i>#intensional#</i>#%}%, in the sense that type equality theorems can never be applied after the fact to get a term to type-check.  Instead, we need to make use of equality explicitly in the theorem statement. *)

  Theorem fhapp_ass : forall ls1 ls2 ls3
    (pf : (ls1 ++ ls2) ++ ls3 = ls1 ++ (ls2 ++ ls3))
    (hls1 : fhlist B ls1) (hls2 : fhlist B ls2) (hls3 : fhlist B ls3),
    fhapp hls1 (fhapp hls2 hls3)
    = match pf in (_ = ls) return fhlist _ ls with
        | refl_equal => fhapp (fhapp hls1 hls2) hls3
      end.
    induction ls1; crush.

    (** The first remaining subgoal looks trivial enough:

       [[

  ============================
   fhapp (ls1:=ls2) (ls2:=ls3) hls2 hls3 =
   match pf in (_ = ls) return (fhlist B ls) with
   | refl_equal => fhapp (ls1:=ls2) (ls2:=ls3) hls2 hls3
   end
       ]]

       We can try what worked in previous examples.

       [[
    case pf.

       [[

User error: Cannot solve a second-order unification problem
         ]]

        It seems we have reached another case where it is unclear how to use a dependent [match] to implement case analysis on our proof.  The [UIP_refl] theorem can come to our rescue again. *)

    rewrite (UIP_refl _ _ pf).
    (** [[

  ============================
   fhapp (ls1:=ls2) (ls2:=ls3) hls2 hls3 =
   fhapp (ls1:=ls2) (ls2:=ls3) hls2 hls3
        ]] *)

    reflexivity.

    (** Our second subgoal is trickier.

       [[

  pf : a :: (ls1 ++ ls2) ++ ls3 = a :: ls1 ++ ls2 ++ ls3
  ============================
   (a0,
   fhapp (ls1:=ls1) (ls2:=ls2 ++ ls3) b
     (fhapp (ls1:=ls2) (ls2:=ls3) hls2 hls3)) =
   match pf in (_ = ls) return (fhlist B ls) with
   | refl_equal =>
       (a0,
       fhapp (ls1:=ls1 ++ ls2) (ls2:=ls3)
         (fhapp (ls1:=ls1) (ls2:=ls2) b hls2) hls3)
   end
       ]]


       [[

    rewrite (UIP_refl _ _ pf).

       [[
The term "pf" has type "a :: (ls1 ++ ls2) ++ ls3 = a :: ls1 ++ ls2 ++ ls3"
 while it is expected to have type "?556 = ?556"
       ]]

       We can only apply [UIP_refl] on proofs of equality with syntactically equal operands, which is not the case of [pf] here.  We will need to manipulate the form of this subgoal to get us to a point where we may use [UIP_refl].  A first step is obtaining a proof suitable to use in applying the induction hypothesis.  Inversion on the structure of [pf] is sufficient for that. *)

    injection pf; intro pf'.
    (** [[

  pf : a :: (ls1 ++ ls2) ++ ls3 = a :: ls1 ++ ls2 ++ ls3
  pf' : (ls1 ++ ls2) ++ ls3 = ls1 ++ ls2 ++ ls3
  ============================
   (a0,
   fhapp (ls1:=ls1) (ls2:=ls2 ++ ls3) b
     (fhapp (ls1:=ls2) (ls2:=ls3) hls2 hls3)) =
   match pf in (_ = ls) return (fhlist B ls) with
   | refl_equal =>
       (a0,
       fhapp (ls1:=ls1 ++ ls2) (ls2:=ls3)
         (fhapp (ls1:=ls1) (ls2:=ls2) b hls2) hls3)
   end
   ]]

   Now we can rewrite using the inductive hypothesis. *)

    rewrite (IHls1 _ _ pf').
    (** [[

  ============================
   (a0,
   match pf' in (_ = ls) return (fhlist B ls) with
   | refl_equal =>
       fhapp (ls1:=ls1 ++ ls2) (ls2:=ls3)
         (fhapp (ls1:=ls1) (ls2:=ls2) b hls2) hls3
   end) =
   match pf in (_ = ls) return (fhlist B ls) with
   | refl_equal =>
       (a0,
       fhapp (ls1:=ls1 ++ ls2) (ls2:=ls3)
         (fhapp (ls1:=ls1) (ls2:=ls2) b hls2) hls3)
   end
        ]]

        We have made an important bit of progress, as now only a single call to [fhapp] appears in the conclusion.  Trying case analysis on our proofs still will not work, but there is a move we can make to enable it.  Not only does just one call to [fhapp] matter to us now, but it also %\textit{%#<i>#does not matter what the result of the call is#</i>#%}%.  In other words, the subgoal should remain true if we replace this [fhapp] call with a fresh variable.  The [generalize] tactic helps us do exactly that. *)

    generalize (fhapp (fhapp b hls2) hls3).
    (** [[

   forall f : fhlist B ((ls1 ++ ls2) ++ ls3),
   (a0,
   match pf' in (_ = ls) return (fhlist B ls) with
   | refl_equal => f
   end) =
   match pf in (_ = ls) return (fhlist B ls) with
   | refl_equal => (a0, f)
   end
        ]]

        The conclusion has gotten markedly simpler.  It seems counterintuitive that we can have an easier time of proving a more general theorem, but that is exactly the case here and for many other proofs that use dependent types heavily.  Speaking informally, the reason why this kind of activity helps is that [match] annotations only support variables in certain positions.  By reducing more elements of a goal to variables, built-in tactics can have more success building [match] terms under the hood.

        In this case, it is helpful to generalize over our two proofs as well. *)

    generalize pf pf'.
    (** [[

   forall (pf0 : a :: (ls1 ++ ls2) ++ ls3 = a :: ls1 ++ ls2 ++ ls3)
     (pf'0 : (ls1 ++ ls2) ++ ls3 = ls1 ++ ls2 ++ ls3)
     (f : fhlist B ((ls1 ++ ls2) ++ ls3)),
   (a0,
   match pf'0 in (_ = ls) return (fhlist B ls) with
   | refl_equal => f
   end) =
   match pf0 in (_ = ls) return (fhlist B ls) with
   | refl_equal => (a0, f)
   end
        ]]

        To an experienced dependent types hacker, the appearance of this goal term calls for a celebration.  The formula has a critical property that indicates that our problems are over.  To get our proofs into the right form to apply [UIP_refl], we need to use associativity of list append to rewrite their types.  We could not do that before because other parts of the goal require the proofs to retain their original types.  In particular, the call to [fhapp] that we generalized must have type [(ls1 ++ ls2) ++ ls3], for some values of the list variables.  If we rewrite the type of the proof used to type-cast this value to something like [ls1 ++ ls2 ++ ls3 = ls1 ++ ls2 ++ ls3], then the lefthand side of the equality would no longer match the type of the term we are trying to cast.

        However, now that we have generalized over the [fhapp] call, the type of the term being type-cast appears explicitly in the goal and %\textit{%#<i>#may be rewritten as well#</i>#%}%.  In particular, the final masterstroke is rewriting everywhere in our goal using associativity of list append. *)

    rewrite app_ass.
    (** [[

  ============================
   forall (pf0 : a :: ls1 ++ ls2 ++ ls3 = a :: ls1 ++ ls2 ++ ls3)
     (pf'0 : ls1 ++ ls2 ++ ls3 = ls1 ++ ls2 ++ ls3)
     (f : fhlist B (ls1 ++ ls2 ++ ls3)),
   (a0,
   match pf'0 in (_ = ls) return (fhlist B ls) with
   | refl_equal => f
   end) =
   match pf0 in (_ = ls) return (fhlist B ls) with
   | refl_equal => (a0, f)
   end
        ]]

        We can see that we have achieved the crucial property: the type of each generalized equality proof has syntactically equal operands.  This makes it easy to finish the proof with [UIP_refl]. *)

    intros.
    rewrite (UIP_refl _ _ pf0).
    rewrite (UIP_refl _ _ pf'0).
    reflexivity.
  Qed.
End fhapp.