Last round of feedback from class at Penn
author Adam Chlipala Sun, 06 Jan 2013 16:23:26 -0500 f38a3af9dd17 31258618ef73
rev   line source
adamc@118 10 (* begin hide *)
adamc@120 11 Require Import Eqdep JMeq List.
adamc@118 16 (* end hide *)
adam@456 21 (** 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 in Coq, 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, I will focus on design patterns for circumventing these tricky issues, and I will introduce the different notions of equality as they are germane. *)
adamc@122 24 (** * The Definitional Equality *)
adam@435 26 (** We have seen many examples so far where proof goals follow "by computation." That is, we apply computational reduction rules to reduce the goal to a normal form, at which point it follows trivially. Exactly when this works and when it does not depends on the details of Coq's%\index{definitional equality}% _definitional equality_. This is an untyped binary relation appearing in the formal metatheory of CIC. CIC contains a typing rule allowing the conclusion [E : T] from the premise [E : T'] and a proof that [T] and [T'] are definitionally equal.
adam@364 28 The %\index{tactics!cbv}%[cbv] tactic will help us illustrate the rules of Coq's definitional equality. We redefine the natural number predecessor function in a somewhat convoluted way and construct a manual proof that it returns [0] when applied to [1]. *)
adamc@122 30 Definition pred' (x : nat) :=
adamc@122 32 | O => O
adamc@122 33 | S n' => let y := n' in y
adamc@122 36 Theorem reduce_me : pred' 1 = 0.
adamc@124 38 (* begin thide *)
adam@364 39 (** CIC follows the traditions of lambda calculus in associating reduction rules with Greek letters. Coq can certainly be said to support the familiar alpha reduction rule, which allows capture-avoiding renaming of bound variables, but we never need to apply alpha explicitly, since Coq uses a de Bruijn representation%~\cite{DeBruijn}% that encodes terms canonically.
adam@364 41 The %\index{delta reduction}%delta rule is for unfolding global definitions. We can use it here to unfold the definition of [pred']. We do this with the [cbv] tactic, which takes a list of reduction rules and makes as many call-by-value reduction steps as possible, using only those rules. There is an analogous tactic %\index{tactics!lazy}%[lazy] for call-by-need reduction. *)
adamc@122 46 (fun x : nat => match x with
adamc@122 47 | 0 => 0
adamc@122 48 | S n' => let y := n' in y
adamc@122 49 end) 1 = 0
adam@364 52 At this point, we want to apply the famous %\index{beta reduction}%beta reduction of lambda calculus, to simplify the application of a known function abstraction. *)
adamc@122 58 | 0 => 0
adamc@122 59 | S n' => let y := n' in y
adam@364 63 Next on the list is the %\index{iota reduction}%iota reduction, which simplifies a single [match] term by determining which pattern matches. *)
adamc@122 68 (fun n' : nat => let y := n' in y) 0 = 0
adamc@122 71 Now we need another beta reduction. *)
adamc@122 76 (let y := 0 in y) = 0
adam@364 79 The final reduction rule is %\index{zeta reduction}%zeta, which replaces a [let] expression by its body with the appropriate term substituted. *)
adamc@124 90 (* end thide *)
adam@365 92 (** The [beta] reduction rule applies to recursive functions as well, and its behavior may be surprising in some instances. For instance, we can run some simple tests using the reduction strategy [compute], which applies all applicable rules of the definitional equality. *)
adam@365 94 Definition id (n : nat) := n.
adam@365 96 Eval compute in fun x => id x.
adam@365 98 = fun x : nat => x
adam@365 102 Fixpoint id' (n : nat) := n.
adam@365 104 Eval compute in fun x => id' x.
adam@365 106 = fun x : nat => (fix id' (n : nat) : nat := n) x
adam@479 109 By running [compute], we ask Coq to run reduction steps until no more apply, so why do we see an application of a known function, where clearly no beta reduction has been performed? The answer has to do with ensuring termination of all Gallina programs. One candidate rule would say that we apply recursive definitions wherever possible. However, this would clearly lead to nonterminating reduction sequences, since the function may appear fully applied within its own definition, and we would %\%naive%{}%ly "simplify" such applications immediately. Instead, Coq only applies the beta rule for a recursive function when _the top-level structure of the recursive argument is known_. For [id'] above, we have only one argument [n], so clearly it is the recursive argument, and the top-level structure of [n] is known when the function is applied to [O] or to some [S e] term. The variable [x] is neither, so reduction is blocked.
adam@365 111 What are recursive arguments in general? Every recursive function is compiled by Coq to a %\index{Gallina terms!fix}%[fix] expression, for anonymous definition of recursive functions. Further, every [fix] with multiple arguments has one designated as the recursive argument via a [struct] annotation. The recursive argument is the one that must decrease across recursive calls, to appease Coq's termination checker. Coq will generally infer which argument is recursive, though we may also specify it manually, if we want to tweak reduction behavior. For instance, consider this definition of a function to add two lists of [nat]s elementwise: *)
adam@365 113 Fixpoint addLists (ls1 ls2 : list nat) : list nat :=
adam@365 114 match ls1, ls2 with
adam@365 115 | n1 :: ls1' , n2 :: ls2' => n1 + n2 :: addLists ls1' ls2'
adam@365 116 | _, _ => nil
adam@365 119 (** By default, Coq chooses [ls1] as the recursive argument. We can see that [ls2] would have been another valid choice. The choice has a critical effect on reduction behavior, as these two examples illustrate: *)
adam@365 123 = fun _ : list nat => nil
adam@365 129 = fun ls : list nat =>
adam@365 130 (fix addLists (ls1 ls2 : list nat) : list nat :=
adam@365 132 | nil => nil
adam@365 133 | n1 :: ls1' =>
adam@365 135 | nil => nil
adam@365 136 | n2 :: ls2' =>
adam@365 137 (fix plus (n m : nat) : nat :=
adam@365 139 | 0 => m
adam@365 140 | S p => S (plus p m)
adam@427 146 The outer application of the [fix] expression for [addLists] was only simplified in the first case, because in the second case the recursive argument is [ls], whose top-level structure is not known.
adam@427 148 The opposite behavior pertains to a version of [addLists] with [ls2] marked as recursive. *)
adam@365 150 Fixpoint addLists' (ls1 ls2 : list nat) {struct ls2} : list nat :=
adam@365 151 match ls1, ls2 with
adam@365 152 | n1 :: ls1' , n2 :: ls2' => n1 + n2 :: addLists' ls1' ls2'
adam@365 153 | _, _ => nil
adam@427 156 (* begin hide *)
adam@437 157 (* begin thide *)
adam@437 158 Definition foo := (@eq, plus).
adam@437 159 (* end thide *)
adam@427 160 (* end hide *)
adam@365 164 = fun ls : list nat => match ls with
adam@365 165 | nil => nil
adam@365 166 | _ :: _ => nil
adam@365 170 We see that all use of recursive functions has been eliminated, though the term has not quite simplified to [nil]. We could get it to do so by switching the order of the [match] discriminees in the definition of [addLists'].
adam@365 172 Recall that co-recursive definitions have a dual rule: a co-recursive call only simplifies when it is the discriminee of a [match]. This condition is built into the beta rule for %\index{Gallina terms!cofix}%[cofix], the anonymous form of [CoFixpoint].
adam@407 176 The standard [eq] relation is critically dependent on the definitional equality. The relation [eq] is often called a%\index{propositional equality}% _propositional equality_, because it reifies definitional equality as a proposition that may or may not hold. Standard axiomatizations of an equality predicate in first-order logic define equality in terms of properties it has, like reflexivity, symmetry, and transitivity. In contrast, for [eq] in Coq, those properties are implicit in the properties of the definitional equality, which are built into CIC's metatheory and the implementation of Gallina. We could add new rules to the definitional equality, and [eq] would keep its definition and methods of use.
adam@427 178 This all may make it sound like the choice of [eq]'s definition is unimportant. To the contrary, in this chapter, we will see examples where alternate definitions may simplify proofs. Before that point, I will introduce proof methods for goals that use proofs of the standard propositional equality "as data." *)
adamc@118 181 (** * Heterogeneous Lists Revisited *)
adam@480 183 (** One of our example dependent data structures from the last chapter (code repeated below) was the heterogeneous list and its associated "cursor" type. The recursive version poses some special challenges related to equality proofs, since it uses such proofs in its definition of [fmember] types. *)
adamc@118 186 Variable A : Type.
adamc@118 187 Variable B : A -> Type.
adamc@118 189 Fixpoint fhlist (ls : list A) : Type :=
adamc@118 191 | nil => unit
adamc@118 192 | x :: ls' => B x * fhlist ls'
adamc@118 195 Variable elm : A.
adamc@118 197 Fixpoint fmember (ls : list A) : Type :=
adamc@118 199 | nil => Empty_set
adamc@118 200 | x :: ls' => (x = elm) + fmember ls'
adamc@118 203 Fixpoint fhget (ls : list A) : fhlist ls -> fmember ls -> B elm :=
adamc@118 204 match ls return fhlist ls -> fmember ls -> B elm with
adamc@118 205 | nil => fun _ idx => match idx with end
adamc@118 206 | _ :: ls' => fun mls idx =>
adamc@118 208 | inl pf => match pf with
adam@426 209 | eq_refl => fst mls
adamc@118 211 | inr idx' => fhget ls' (snd mls) idx'
adamc@118 216 Implicit Arguments fhget [A B elm ls].
adam@427 218 (* begin hide *)
adam@437 219 (* begin thide *)
adam@427 220 Definition map := O.
adam@437 221 (* end thide *)
adam@427 222 (* end hide *)
adamc@118 224 (** We can define a [map]-like function for [fhlist]s. *)
adamc@118 227 Variables A : Type.
adamc@118 228 Variables B C : A -> Type.
adamc@118 229 Variable f : forall x, B x -> C x.
adamc@118 231 Fixpoint fhmap (ls : list A) : fhlist B ls -> fhlist C ls :=
adamc@118 232 match ls return fhlist B ls -> fhlist C ls with
adamc@118 233 | nil => fun _ => tt
adamc@118 234 | _ :: _ => fun hls => (f (fst hls), fhmap _ (snd hls))
adamc@118 237 Implicit Arguments fhmap [ls].
adam@427 239 (* begin hide *)
adam@437 240 (* begin thide *)
adam@427 241 Definition ilist := O.
adam@427 242 Definition get := O.
adam@427 243 Definition imap := O.
adam@437 244 (* end thide *)
adam@427 245 (* end hide *)
adamc@118 247 (** 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. *)
adamc@118 249 Variable elm : A.
adam@479 251 Theorem fhget_fhmap : forall ls (mem : fmember elm ls) (hls : fhlist B ls),
adamc@118 252 fhget (fhmap hls) mem = f (fhget hls mem).
adam@298 253 (* begin hide *)
adam@298 254 induction ls; crush; case a0; reflexivity.
adam@298 255 (* end hide *)
adam@364 260 %\vspace{-.15in}%In Coq 8.2, one subgoal remains at this point. Coq 8.3 has added some tactic improvements that enable [crush] to complete all of both inductive cases. To introduce the basics of reasoning about equality, it will be useful to review what was necessary in Coq 8.2.
adam@297 262 Part of our single remaining subgoal is:
adamc@118 264 a0 : a = elm
adamc@118 266 match a0 in (_ = a2) return (C a2) with
adam@426 267 | eq_refl => f a1
adamc@118 268 end = f match a0 in (_ = a2) return (B a2) with
adam@426 269 | eq_refl => a1
adam@426 273 This seems like a trivial enough obligation. The equality proof [a0] must be [eq_refl], since that is the only constructor of [eq]. Therefore, both the [match]es reduce to the point where the conclusion follows by reflexivity.
adam@364 279 User error: Cannot solve a second-order unification problem
adamc@118 282 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.
adam@426 284 assert (a0 = eq_refl _).
adam@426 288 The term "eq_refl ?98" has type "?98 = ?98"
adamc@118 289 while it is expected to have type "a = elm"
adamc@118 292 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!
adam@427 294 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.
adam@407 296 For this particular example, the solution is surprisingly straightforward. The [destruct] tactic has a simpler sibling [case] which should behave identically for any inductive type with one constructor of no arguments.
adamc@118 301 f a1 = f a1
adam@297 304 It seems that [destruct] was trying to be too smart for its own good.
adamc@124 310 (* end thide *)
adamc@118 312 (** It will be helpful to examine the proof terms generated by this sort of strategy. A simpler example illustrates what is going on. *)
adam@426 314 Lemma lemma1 : forall x (pf : x = elm), O = match pf with eq_refl => O end.
adamc@124 315 (* begin thide *)
adamc@118 316 simple destruct pf; reflexivity.
adamc@124 318 (* end thide *)
adam@364 320 (** The tactic %\index{tactics!simple destruct}%[simple destruct pf] is a convenient form for applying [case]. It runs [intro] to bring into scope all quantified variables up to its argument. *)
adamc@118 325 fun (x : A) (pf : x = elm) =>
adamc@118 326 match pf as e in (_ = y) return (0 = match e with
adam@426 327 | eq_refl => 0
adam@426 329 | eq_refl => eq_refl 0
adamc@118 331 : forall (x : A) (pf : x = elm), 0 = match pf with
adam@426 332 | eq_refl => 0
adamc@118 337 Using what we know about shorthands for [match] annotations, we can write this proof in shorter form manually. *)
adamc@124 339 (* begin thide *)
adam@456 340 Definition lemma1' (x : A) (pf : x = elm) :=
adam@456 341 match pf return (0 = match pf with
adam@456 342 | eq_refl => 0
adam@456 344 | eq_refl => eq_refl 0
adamc@124 346 (* end thide *)
adam@398 348 (** Surprisingly, what seems at first like a _simpler_ lemma is harder to prove. *)
adam@426 350 Lemma lemma2 : forall (x : A) (pf : x = x), O = match pf with eq_refl => O end.
adamc@124 351 (* begin thide *)
adamc@118 357 User error: Cannot solve a second-order unification problem
adamc@118 362 (** Nonetheless, we can adapt the last manual proof to handle this theorem. *)
adamc@124 364 (* begin thide *)
adamc@118 366 fun (x : A) (pf : x = x) =>
adamc@118 367 match pf return (0 = match pf with
adam@426 368 | eq_refl => 0
adam@426 370 | eq_refl => eq_refl 0
adamc@124 372 (* end thide *)
adamc@118 374 (** We can try to prove a lemma that would simplify proofs of many facts like [lemma2]: *)
adam@427 376 (* begin hide *)
adam@437 377 (* begin thide *)
adam@427 378 Definition lemma3' := O.
adam@437 379 (* end thide *)
adam@427 380 (* end hide *)
adam@426 382 Lemma lemma3 : forall (x : A) (pf : x = x), pf = eq_refl x.
adamc@124 383 (* begin thide *)
adamc@118 389 User error: Cannot solve a second-order unification problem
adamc@118 395 (** This time, even our manual attempt fails.
adamc@118 398 fun (x : A) (pf : x = x) =>
adam@426 399 match pf as pf' in (_ = x') return (pf' = eq_refl x') with
adam@426 400 | eq_refl => eq_refl _
adam@426 405 The term "eq_refl x'" has type "x' = x'" while it is expected to have type
adam@427 409 The type error comes from our [return] annotation. In that annotation, the [as]-bound variable [pf'] has type [x = x'], referring to the [in]-bound variable [x']. To do a dependent [match], we _must_ 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 _must_ equate two non-matching terms, which makes it impossible to equate that proof with reflexivity.
adam@398 411 Nonetheless, it turns out that, with one catch, we _can_ prove this lemma. *)
adam@426 413 Lemma lemma3 : forall (x : A) (pf : x = x), pf = eq_refl x.
adam@426 420 : forall (U : Type) (x : U) (p : x = x), p = eq_refl x
adam@480 423 The theorem %\index{Gallina terms!UIP\_refl}%[UIP_refl] comes from the [Eqdep] module of the standard library. (Its name uses the acronym "UIP" for "unicity of identity proofs.") 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 _axiom_, the term [eq_rect_eq] below. *)
adam@436 425 (* begin hide *)
adam@436 427 (* end hide *)
adam@436 430 *** [ eq_rect_eq :
adam@436 431 forall (U : Type) (p : U) (Q : U -> Type) (x : Q p) (h : p = p),
adam@436 432 x = eq_rect p Q x p h ]
adam@427 435 The axiom %\index{Gallina terms!eq\_rect\_eq}%[eq_rect_eq] states a "fact" that seems like common sense, once the notation is deciphered. The term [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]. The statement of [eq_rect_eq] just says that [match]es on proofs of [p = p], for any [p], are superfluous and may be removed. We can see this intuition better in code by asking Coq to simplify the theorem statement with the [compute] reduction strategy (which, by the way, applies all applicable rules of the definitional equality presented in this chapter's first section). *)
adam@427 437 (* begin hide *)
adam@437 438 (* begin thide *)
adam@427 439 Definition False' := False.
adam@437 440 (* end thide *)
adam@427 441 (* end hide *)
adam@364 443 Eval compute in (forall (U : Type) (p : U) (Q : U -> Type) (x : Q p) (h : p = p),
adam@364 444 x = eq_rect p Q x p h).
adam@364 446 = forall (U : Type) (p : U) (Q : U -> Type) (x : Q p) (h : p = p),
adam@364 447 x = match h in (_ = y) return (Q y) with
adam@364 448 | eq_refl => x
adam@427 452 Perhaps surprisingly, we cannot prove [eq_rect_eq] from within Coq. This proposition is introduced as an %\index{axioms}%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 _inconsistent_ 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%~\cite{AxiomK}%, in a study that also established unprovability of the axiom in CIC.
adamc@118 454 This axiom is equivalent to another that is more commonly known and mentioned in type theory circles. *)
adam@427 456 (* begin hide *)
adam@437 457 (* begin thide *)
adam@437 458 Definition Streicher_K' := UIP_refl__Streicher_K.
adam@437 459 (* end thide *)
adam@427 460 (* end hide *)
adamc@124 463 (* end thide *)
adamc@118 466 : forall (U : Type) (x : U) (P : x = x -> Prop),
adam@480 467 P eq_refl -> forall p : x = x, P p
adam@480 470 This is the opaquely named %\index{axiom K}%"Streicher's axiom K," which says that a predicate on properly typed equality proofs holds of all such proofs if it holds of reflexivity. *)
adam@436 474 (* begin hide *)
adam@437 475 (* begin thide *)
adam@437 477 (* end thide *)
adam@436 478 (* end hide *)
adam@364 480 (** It is worth remarking that it is possible to avoid axioms altogether for equalities on types with decidable equality. The [Eqdep_dec] module of the standard library contains a parametric proof of [UIP_refl] for such cases. To simplify presentation, we will stick with the axiom version in the rest of this chapter. *)
adamc@119 483 (** * Type-Casts in Theorem Statements *)
adamc@119 485 (** 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. *)
adamc@119 488 Variable A : Type.
adamc@119 489 Variable B : A -> Type.
adamc@218 491 Fixpoint fhapp (ls1 ls2 : list A)
adamc@119 492 : fhlist B ls1 -> fhlist B ls2 -> fhlist B (ls1 ++ ls2) :=
adamc@119 494 | nil => fun _ hls2 => hls2
adamc@119 495 | _ :: _ => fun hls1 hls2 => (fst hls1, fhapp _ _ (snd hls1) hls2)
adamc@119 498 Implicit Arguments fhapp [ls1 ls2].
adamc@124 500 (* EX: Prove that fhapp is associative. *)
adamc@124 501 (* begin thide *)
adamc@119 503 (** We might like to prove that [fhapp] is associative.
adam@479 505 Theorem fhapp_assoc : forall ls1 ls2 ls3
adamc@119 506 (hls1 : fhlist B ls1) (hls2 : fhlist B ls2) (hls3 : fhlist B ls3),
adamc@119 507 fhapp hls1 (fhapp hls2 hls3) = fhapp (fhapp hls1 hls2) hls3.
adamc@119 512 "fhapp (ls1:=ls1 ++ ls2) (ls2:=ls3) (fhapp (ls1:=ls1) (ls2:=ls2) hls1 hls2)
adamc@119 513 hls3" has type "fhlist B ((ls1 ++ ls2) ++ ls3)"
adamc@119 514 while it is expected to have type "fhlist B (ls1 ++ ls2 ++ ls3)"
adam@407 517 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%\index{intensional type theory}% _intensional_, 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. *)
adam@479 519 Theorem fhapp_assoc : forall ls1 ls2 ls3
adamc@119 520 (pf : (ls1 ++ ls2) ++ ls3 = ls1 ++ (ls2 ++ ls3))
adamc@119 521 (hls1 : fhlist B ls1) (hls2 : fhlist B ls2) (hls3 : fhlist B ls3),
adamc@119 522 fhapp hls1 (fhapp hls2 hls3)
adamc@119 523 = match pf in (_ = ls) return fhlist _ ls with
adam@426 524 | eq_refl => fhapp (fhapp hls1 hls2) hls3
adamc@119 528 (** The first remaining subgoal looks trivial enough:
adamc@119 531 fhapp (ls1:=ls2) (ls2:=ls3) hls2 hls3 =
adamc@119 532 match pf in (_ = ls) return (fhlist B ls) with
adam@426 533 | eq_refl => fhapp (ls1:=ls2) (ls2:=ls3) hls2 hls3
adamc@119 537 We can try what worked in previous examples.
adamc@119 543 User error: Cannot solve a second-order unification problem
adamc@119 546 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. *)
adamc@119 548 rewrite (UIP_refl _ _ pf).
adamc@119 551 fhapp (ls1:=ls2) (ls2:=ls3) hls2 hls3 =
adamc@119 552 fhapp (ls1:=ls2) (ls2:=ls3) hls2 hls3
adamc@119 558 (** Our second subgoal is trickier.
adamc@119 560 pf : a :: (ls1 ++ ls2) ++ ls3 = a :: ls1 ++ ls2 ++ ls3
adamc@119 563 fhapp (ls1:=ls1) (ls2:=ls2 ++ ls3) b
adamc@119 564 (fhapp (ls1:=ls2) (ls2:=ls3) hls2 hls3)) =
adamc@119 565 match pf in (_ = ls) return (fhlist B ls) with
adamc@119 568 fhapp (ls1:=ls1 ++ ls2) (ls2:=ls3)
adamc@119 569 (fhapp (ls1:=ls1) (ls2:=ls2) b hls2) hls3)
adamc@119 572 rewrite (UIP_refl _ _ pf).
adamc@119 576 The term "pf" has type "a :: (ls1 ++ ls2) ++ ls3 = a :: ls1 ++ ls2 ++ ls3"
adamc@119 577 while it is expected to have type "?556 = ?556"
adamc@119 580 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. *)
adamc@119 582 injection pf; intro pf'.
adamc@119 584 pf : a :: (ls1 ++ ls2) ++ ls3 = a :: ls1 ++ ls2 ++ ls3
adamc@119 585 pf' : (ls1 ++ ls2) ++ ls3 = ls1 ++ ls2 ++ ls3
adamc@119 588 fhapp (ls1:=ls1) (ls2:=ls2 ++ ls3) b
adamc@119 589 (fhapp (ls1:=ls2) (ls2:=ls3) hls2 hls3)) =
adamc@119 590 match pf in (_ = ls) return (fhlist B ls) with
adamc@119 593 fhapp (ls1:=ls1 ++ ls2) (ls2:=ls3)
adamc@119 594 (fhapp (ls1:=ls1) (ls2:=ls2) b hls2) hls3)
adamc@119 598 Now we can rewrite using the inductive hypothesis. *)
adamc@119 600 rewrite (IHls1 _ _ pf').
adamc@119 604 match pf' in (_ = ls) return (fhlist B ls) with
adamc@119 606 fhapp (ls1:=ls1 ++ ls2) (ls2:=ls3)
adamc@119 607 (fhapp (ls1:=ls1) (ls2:=ls2) b hls2) hls3
adamc@119 609 match pf in (_ = ls) return (fhlist B ls) with
adamc@119 612 fhapp (ls1:=ls1 ++ ls2) (ls2:=ls3)
adamc@119 613 (fhapp (ls1:=ls1) (ls2:=ls2) b hls2) hls3)
adam@398 617 We have made an important bit of progress, as now only a single call to [fhapp] appears in the conclusion, repeated twice. 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 _does not matter what the result of the call is_. In other words, the subgoal should remain true if we replace this [fhapp] call with a fresh variable. The %\index{tactics!generalize}%[generalize] tactic helps us do exactly that. *)
adamc@119 619 generalize (fhapp (fhapp b hls2) hls3).
adamc@119 621 forall f : fhlist B ((ls1 ++ ls2) ++ ls3),
adamc@119 623 match pf' in (_ = ls) return (fhlist B ls) with
adam@426 624 | eq_refl => f
adamc@119 626 match pf in (_ = ls) return (fhlist B ls) with
adam@426 627 | eq_refl => (a0, f)
adamc@119 631 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.
adamc@119 633 In this case, it is helpful to generalize over our two proofs as well. *)
adamc@119 637 forall (pf0 : a :: (ls1 ++ ls2) ++ ls3 = a :: ls1 ++ ls2 ++ ls3)
adamc@119 638 (pf'0 : (ls1 ++ ls2) ++ ls3 = ls1 ++ ls2 ++ ls3)
adamc@119 639 (f : fhlist B ((ls1 ++ ls2) ++ ls3)),
adamc@119 641 match pf'0 in (_ = ls) return (fhlist B ls) with
adam@426 642 | eq_refl => f
adamc@119 644 match pf0 in (_ = ls) return (fhlist B ls) with
adam@426 645 | eq_refl => (a0, f)
adamc@119 649 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.
adam@398 651 However, now that we have generalized over the [fhapp] call, the type of the term being type-cast appears explicitly in the goal and _may be rewritten as well_. In particular, the final masterstroke is rewriting everywhere in our goal using associativity of list append. *)
adamc@119 656 forall (pf0 : a :: ls1 ++ ls2 ++ ls3 = a :: ls1 ++ ls2 ++ ls3)
adamc@119 657 (pf'0 : ls1 ++ ls2 ++ ls3 = ls1 ++ ls2 ++ ls3)
adamc@119 658 (f : fhlist B (ls1 ++ ls2 ++ ls3)),
adamc@119 660 match pf'0 in (_ = ls) return (fhlist B ls) with
adam@426 661 | eq_refl => f
adamc@119 663 match pf0 in (_ = ls) return (fhlist B ls) with
adam@426 664 | eq_refl => (a0, f)
adamc@119 668 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]. *)
adamc@119 671 rewrite (UIP_refl _ _ pf0).
adamc@119 672 rewrite (UIP_refl _ _ pf'0).
adamc@124 675 (* end thide *)
adamc@120 678 Implicit Arguments fhapp [A B ls1 ls2].
adam@364 680 (** This proof strategy was cumbersome and unorthodox, from the perspective of mainstream mathematics. The next section explores an alternative that leads to simpler developments in some cases. *)
adamc@120 683 (** * Heterogeneous Equality *)
adam@407 685 (** There is another equality predicate, defined in the %\index{Gallina terms!JMeq}%[JMeq] module of the standard library, implementing%\index{heterogeneous equality}% _heterogeneous equality_. *)
adamc@120 689 Inductive JMeq (A : Type) (x : A) : forall B : Type, B -> Prop :=
adamc@120 690 JMeq_refl : JMeq x x
adam@480 693 The identifier [JMeq] stands for %\index{John Major equality}%"John Major equality," a name coined by Conor McBride%~\cite{JMeq}% as an inside joke about British politics. The definition [JMeq] starts out looking a lot like the definition of [eq]. The crucial difference is that we may use [JMeq] _on arguments of different types_. For instance, a lemma that we failed to establish before is trivial with [JMeq]. It makes for prettier theorem statements to define some syntactic shorthand first. *)
adamc@120 695 Infix "==" := JMeq (at level 70, no associativity).
adam@426 697 (* EX: Prove UIP_refl' : forall (A : Type) (x : A) (pf : x = x), pf == eq_refl x *)
adamc@124 698 (* begin thide *)
adam@426 699 Definition UIP_refl' (A : Type) (x : A) (pf : x = x) : pf == eq_refl x :=
adam@426 700 match pf return (pf == eq_refl _) with
adam@426 701 | eq_refl => JMeq_refl _
adamc@124 703 (* end thide *)
adamc@120 705 (** There is no quick way to write such a proof by tactics, but the underlying proof term that we want is trivial.
adamc@271 707 Suppose that we want to use [UIP_refl'] to establish another lemma of the kind we have run into several times so far. *)
adamc@120 709 Lemma lemma4 : forall (A : Type) (x : A) (pf : x = x),
adam@426 710 O = match pf with eq_refl => O end.
adamc@124 711 (* begin thide *)
adamc@121 712 intros; rewrite (UIP_refl' pf); reflexivity.
adamc@124 714 (* end thide *)
adamc@120 716 (** All in all, refreshingly straightforward, but there really is no such thing as a free lunch. The use of [rewrite] is implemented in terms of an axiom: *)
adamc@120 721 : forall (A : Type) (x y : A), x == y -> x = y
adam@480 724 It may be surprising that we cannot prove that heterogeneous equality implies normal equality. The difficulties are the same kind we have seen so far, based on limitations of [match] annotations. The [JMeq_eq] axiom has been proved on paper to be consistent, but asserting it may still be considered to complicate the logic we work in, so there is some motivation for avoiding it.
adamc@120 726 We can redo our [fhapp] associativity proof based around [JMeq]. *)
adamc@120 729 Variable A : Type.
adamc@120 730 Variable B : A -> Type.
adam@484 732 (** This time, the %\%naive%{}% theorem statement type-checks. *)
adamc@124 734 (* EX: Prove [fhapp] associativity using [JMeq]. *)
adamc@124 736 (* begin thide *)
adam@479 737 Theorem fhapp_assoc' : forall ls1 ls2 ls3 (hls1 : fhlist B ls1) (hls2 : fhlist B ls2)
adam@364 738 (hls3 : fhlist B ls3),
adamc@120 739 fhapp hls1 (fhapp hls2 hls3) == fhapp (fhapp hls1 hls2) hls3.
adamc@120 742 (** Even better, [crush] discharges the first subgoal automatically. The second subgoal is:
adam@297 745 (a0, fhapp b (fhapp hls2 hls3)) == (a0, fhapp (fhapp b hls2) hls3)
adam@297 748 It looks like one rewrite with the inductive hypothesis should be enough to make the goal trivial. Here is what happens when we try that in Coq 8.2:
adamc@120 754 Error: Impossible to unify "fhlist B ((ls1 ++ ?1572) ++ ?1573)" with
adamc@120 755 "fhlist B (ls1 ++ ?1572 ++ ?1573)"
adam@407 758 Coq 8.4 currently gives an error message about an uncaught exception. Perhaps that will be fixed soon. In any case, it is educational to consider a more explicit approach.
adamc@120 760 We see that [JMeq] is not a silver bullet. We can use it to simplify the statements of equality facts, but the Coq type-checker uses non-trivial heterogeneous equality facts no more readily than it uses standard equality facts. Here, the problem is that the form [(e1, e2)] is syntactic sugar for an explicit application of a constructor of an inductive type. That application mentions the type of each tuple element explicitly, and our [rewrite] tries to change one of those elements without updating the corresponding type argument.
adamc@120 762 We can get around this problem by another multiple use of [generalize]. We want to bring into the goal the proper instance of the inductive hypothesis, and we also want to generalize the two relevant uses of [fhapp]. *)
adamc@120 764 generalize (fhapp b (fhapp hls2 hls3))
adamc@120 765 (fhapp (fhapp b hls2) hls3)
adamc@120 766 (IHls1 _ _ b hls2 hls3).
adamc@120 769 forall (f : fhlist B (ls1 ++ ls2 ++ ls3))
adamc@120 770 (f0 : fhlist B ((ls1 ++ ls2) ++ ls3)), f == f0 -> (a0, f) == (a0, f0)
adamc@120 773 Now we can rewrite with append associativity, as before. *)
adamc@120 778 forall f f0 : fhlist B (ls1 ++ ls2 ++ ls3), f == f0 -> (a0, f) == (a0, f0)
adamc@120 781 From this point, the goal is trivial. *)
adamc@120 783 intros f f0 H; rewrite H; reflexivity.
adamc@124 785 (* end thide *)
adam@385 789 (** This example illustrates a general pattern: heterogeneous equality often simplifies theorem statements, but we still need to do some work to line up some dependent pattern matches that tactics will generate for us.
adam@385 791 The proof we have found relies on the [JMeq_eq] axiom, which we can verify with a command%\index{Vernacular commands!Print Assumptions}% that we will discuss more in two chapters. *)
adam@385 796 JMeq_eq : forall (A : Type) (x y : A), x == y -> x = y
adam@385 799 It was the [rewrite H] tactic that implicitly appealed to the axiom. By restructuring the proof, we can avoid axiom dependence. A general lemma about pairs provides the key element. (Our use of [generalize] above can be thought of as reducing the proof to another, more complex and specialized lemma.) *)
adam@385 801 Lemma pair_cong : forall A1 A2 B1 B2 (x1 : A1) (x2 : A2) (y1 : B1) (y2 : B2),
adam@385 803 -> y1 == y2
adam@385 804 -> (x1, y1) == (x2, y2).
adam@385 805 intros until y2; intros Hx Hy; rewrite Hx; rewrite Hy; reflexivity.
adam@385 811 Variable A : Type.
adam@385 812 Variable B : A -> Type.
adam@479 814 Theorem fhapp_assoc'' : forall ls1 ls2 ls3 (hls1 : fhlist B ls1) (hls2 : fhlist B ls2)
adam@385 815 (hls3 : fhlist B ls3),
adam@385 816 fhapp hls1 (fhapp hls2 hls3) == fhapp (fhapp hls1 hls2) hls3.
adam@385 823 Closed under the global context
adam@479 826 One might wonder exactly which elements of a proof involving [JMeq] imply that [JMeq_eq] must be used. For instance, above we noticed that [rewrite] had brought [JMeq_eq] into the proof of [fhapp_assoc'], yet here we have also used [rewrite] with [JMeq] hypotheses while avoiding axioms! One illuminating exercise is comparing the types of the lemmas that [rewrite] uses under the hood to implement the rewrites. Here is the normal lemma for [eq] rewriting:%\index{Gallina terms!eq\_ind\_r}% *)
adam@385 831 : forall (A : Type) (x : A) (P : A -> Prop),
adam@385 832 P x -> forall y : A, y = x -> P y
adam@398 835 The corresponding lemma used for [JMeq] in the proof of [pair_cong] is %\index{Gallina terms!internal\_JMeq\_rew\_r}%[internal_JMeq_rew_r], which, confusingly, is defined by [rewrite] as needed, so it is not available for checking until after we apply it. *)
adam@385 840 : forall (A : Type) (x : A) (B : Type) (b : B)
adam@385 841 (P : forall B0 : Type, B0 -> Type), P B b -> x == b -> P A x
adam@479 844 The key difference is that, where the [eq] lemma is parameterized on a predicate of type [A -> Prop], the [JMeq] lemma is parameterized on a predicate of type more like [forall A : Type, A -> Prop]. To apply [eq_ind_r] with a proof of [x = y], it is only necessary to rearrange the goal into an application of a [fun] abstraction to [y]. In contrast, to apply [internal_JMeq_rew_r], it is necessary to rearrange the goal to an application of a [fun] abstraction to both [y] and _its type_. In other words, the predicate must be _polymorphic_ in [y]'s type; any type must make sense, from a type-checking standpoint. There may be cases where the former rearrangement is easy to do in a type-correct way, but the second rearrangement done %\%naive%{}%ly leads to a type error.
adam@398 846 When [rewrite] cannot figure out how to apply [internal_JMeq_rew_r] for [x == y] where [x] and [y] have the same type, the tactic can instead use an alternate theorem, which is easy to prove as a composition of [eq_ind_r] and [JMeq_eq]. *)
adam@385 851 : forall (A : Type) (x : A) (P : A -> Prop),
adam@385 852 P x -> forall y : A, y == x -> P y
adam@479 855 Ironically, where in the proof of [fhapp_assoc'] we used [rewrite app_assoc] to make it clear that a use of [JMeq] was actually homogeneously typed, we created a situation where [rewrite] applied the axiom-based [JMeq_ind_r] instead of the axiom-free [internal_JMeq_rew_r]!
adam@385 857 For another simple example, consider this theorem that applies a heterogeneous equality to prove a congruence fact. *)
adam@385 859 Theorem out_of_luck : forall n m : nat,
adam@385 861 -> S n == S m.
adam@385 862 intros n m H.
adam@480 864 (** Applying [JMeq_ind_r] is easy, as the %\index{tactics!pattern}%[pattern] tactic will transform the goal into an application of an appropriate [fun] to a term that we want to abstract. (In general, [pattern] abstracts over a term by introducing a new anonymous function taking that term as argument.) *)
adam@385 870 H : n == m
adam@385 872 (fun n0 : nat => S n0 == S m) n
adam@385 875 apply JMeq_ind_r with (x := m); auto.
adam@398 877 (** However, we run into trouble trying to get the goal into a form compatible with [internal_JMeq_rew_r.] *)
adam@385 884 Error: The abstracted term "fun (P : Set) (n0 : P) => S n0 == S m"
adam@385 885 is not well typed.
adam@385 886 Illegal application (Type Error):
adam@385 887 The term "S" of type "nat -> nat"
adam@385 888 cannot be applied to the term
adam@385 890 This term has type "P" which should be coercible to
adam@385 894 In other words, the successor function [S] is insufficiently polymorphic. If we try to generalize over the type of [n], we find that [S] is no longer legal to apply to [n]. *)
adam@479 898 (** Why did we not run into this problem in our proof of [fhapp_assoc'']? The reason is that the pair constructor is polymorphic in the types of the pair components, while functions like [S] are not polymorphic at all. Use of such non-polymorphic functions with [JMeq] tends to push toward use of axioms. The example with [nat] here is a bit unrealistic; more likely cases would involve functions that have _some_ polymorphism, but not enough to allow abstractions of the sort we attempted above with [pattern]. For instance, we might have an equality between two lists, where the goal only type-checks when the terms involved really are lists, though everything is polymorphic in the types of list data elements. The {{http://www.mpi-sws.org/~gil/Heq/}Heq} library builds up a slightly different foundation to help avoid such problems. *)
adamc@121 901 (** * Equivalence of Equality Axioms *)
adamc@124 903 (* EX: Show that the approaches based on K and JMeq are equivalent logically. *)
adamc@124 905 (* begin thide *)
adamc@272 906 (** Assuming axioms (like axiom K and [JMeq_eq]) is a hazardous business. The due diligence associated with it is necessarily global in scope, since two axioms may be consistent alone but inconsistent together. It turns out that all of the major axioms proposed for reasoning about equality in Coq are logically equivalent, so that we only need to pick one to assert without proof. In this section, we demonstrate this by showing how each of the previous two sections' approaches reduces to the other logically.
adamc@121 908 To show that [JMeq] and its axiom let us prove [UIP_refl], we start from the lemma [UIP_refl'] from the previous section. The rest of the proof is trivial. *)
adam@426 910 Lemma UIP_refl'' : forall (A : Type) (x : A) (pf : x = x), pf = eq_refl x.
adamc@121 911 intros; rewrite (UIP_refl' pf); reflexivity.
adamc@121 914 (** The other direction is perhaps more interesting. Assume that we only have the axiom of the [Eqdep] module available. We can define [JMeq] in a way that satisfies the same interface as the combination of the [JMeq] module's inductive definition and axiom. *)
adamc@121 916 Definition JMeq' (A : Type) (x : A) (B : Type) (y : B) : Prop :=
adam@426 917 exists pf : B = A, x = match pf with eq_refl => y end.
adamc@121 919 Infix "===" := JMeq' (at level 70, no associativity).
adam@427 921 (** remove printing exists *)
adamc@121 923 (** We say that, by definition, [x] and [y] are equal if and only if there exists a proof [pf] that their types are equal, such that [x] equals the result of casting [y] with [pf]. This statement can look strange from the standpoint of classical math, where we almost never mention proofs explicitly with quantifiers in formulas, but it is perfectly legal Coq code.
adamc@121 925 We can easily prove a theorem with the same type as that of the [JMeq_refl] constructor of [JMeq]. *)
adamc@121 927 Theorem JMeq_refl' : forall (A : Type) (x : A), x === x.
adam@426 928 intros; unfold JMeq'; exists (eq_refl A); reflexivity.
adamc@121 931 (** printing exists $\exists$ *)
adamc@121 933 (** The proof of an analogue to [JMeq_eq] is a little more interesting, but most of the action is in appealing to [UIP_refl]. *)
adamc@121 935 Theorem JMeq_eq' : forall (A : Type) (x y : A),
adamc@121 936 x === y -> x = y.
adamc@121 939 H : exists pf : A = A,
adamc@121 940 x = match pf in (_ = T) return T with
adam@426 941 | eq_refl => y
adamc@121 950 x0 : A = A
adamc@121 951 H : x = match x0 in (_ = T) return T with
adam@426 952 | eq_refl => y
adamc@121 961 x0 : A = A
adamc@121 963 match x0 in (_ = T) return T with
adam@426 964 | eq_refl => y
adamc@121 969 rewrite (UIP_refl _ _ x0); reflexivity.
adam@427 972 (** We see that, in a very formal sense, we are free to switch back and forth between the two styles of proofs about equality proofs. One style may be more convenient than the other for some proofs, but we can always interconvert between our results. The style that does not use heterogeneous equality may be preferable in cases where many results do not require the tricks of this chapter, since then the use of axioms is avoided altogether for the simple cases, and a wider audience will be able to follow those "simple" proofs. On the other hand, heterogeneous equality often makes for shorter and more readable theorem statements. *)
adamc@124 974 (* end thide *)
adamc@123 977 (** * Equality of Functions *)
adam@444 979 (** The following seems like a reasonable theorem to want to hold, and it does hold in set theory.
adam@456 981 Theorem two_funs : (fun n => n) = (fun n => n + 0).
adam@444 983 %\vspace{-.15in}%Unfortunately, this theorem is not provable in CIC without additional axioms. None of the definitional equality rules force function equality to be%\index{extensionality of function equality}% _extensional_. That is, the fact that two functions return equal results on equal inputs does not imply that the functions are equal. We _can_ assert function extensionality as an axiom, and indeed the standard library already contains that axiom. *)
adam@407 989 forall (A B : Type) (f g : A -> B), (forall x : A, f x = g x) -> f = g
adam@456 993 (** This axiom has been verified metatheoretically to be consistent with CIC and the two equality axioms we considered previously. With it, the proof of [two_funs] is trivial. *)
adam@456 995 Theorem two_funs : (fun n => n) = (fun n => n + 0).
adamc@124 996 (* begin thide *)
adamc@124 999 (* end thide *)
adam@427 1001 (** The same axiom can help us prove equality of types, where we need to "reason under quantifiers." *)
adamc@123 1003 Theorem forall_eq : (forall x : nat, match x with
adamc@123 1004 | O => True
adamc@123 1005 | S _ => True
adamc@123 1007 = (forall _ : nat, True).
adam@456 1009 (** There are no immediate opportunities to apply [functional_extensionality], but we can use %\index{tactics!change}%[change] to fix that problem. *)
adamc@124 1011 (* begin thide *)
adamc@123 1012 change ((forall x : nat, (fun x => match x with
adamc@123 1013 | 0 => True
adamc@123 1014 | S _ => True
adamc@123 1015 end) x) = (nat -> True)).
adam@407 1016 rewrite (functional_extensionality (fun x => match x with
adam@407 1017 | 0 => True
adam@407 1018 | S _ => True
adam@407 1019 end) (fun _ => True)).
adamc@123 1024 (nat -> True) = (nat -> True)
adamc@123 1027 forall x : nat, match x with
adamc@123 1028 | 0 => True
adamc@123 1029 | S _ => True