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Reflection exercise
author Adam Chlipala <adamc@hcoop.net>
date Wed, 29 Oct 2008 17:41:38 -0400
parents bc1f7d3687e7
children cadeb49dc1ef
rev   line source
adamc@118 1 (* Copyright (c) 2008, Adam Chlipala
adamc@118 2 *
adamc@118 3 * This work is licensed under a
adamc@118 4 * Creative Commons Attribution-Noncommercial-No Derivative Works 3.0
adamc@118 5 * Unported License.
adamc@118 6 * The license text is available at:
adamc@118 7 * http://creativecommons.org/licenses/by-nc-nd/3.0/
adamc@118 8 *)
adamc@118 9
adamc@118 10 (* begin hide *)
adamc@120 11 Require Import Eqdep JMeq List.
adamc@118 12
adamc@132 13 Require Import Tactics.
adamc@118 14
adamc@118 15 Set Implicit Arguments.
adamc@118 16 (* end hide *)
adamc@118 17
adamc@118 18
adamc@118 19 (** %\chapter{Reasoning About Equality Proofs}% *)
adamc@118 20
adamc@118 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, 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. *)
adamc@118 22
adamc@118 23
adamc@122 24 (** * The Definitional Equality *)
adamc@122 25
adamc@122 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 %\textit{%#<i>#definitional equality#</i>#%}%. 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.
adamc@122 27
adamc@122 28 The [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 [1] when applied to [0]. *)
adamc@122 29
adamc@122 30 Definition pred' (x : nat) :=
adamc@122 31 match x with
adamc@122 32 | O => O
adamc@122 33 | S n' => let y := n' in y
adamc@122 34 end.
adamc@122 35
adamc@122 36 Theorem reduce_me : pred' 1 = 0.
adamc@124 37 (* begin thide *)
adamc@122 38 (** 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 that encodes terms canonically.
adamc@122 39
adamc@131 40 The 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 [lazy] for call-by-need reduction. *)
adamc@122 41
adamc@122 42 cbv delta.
adamc@122 43 (** [[
adamc@122 44
adamc@122 45 ============================
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
adamc@122 50 ]]
adamc@122 51
adamc@122 52 At this point, we want to apply the famous beta reduction of lambda calculus, to simplify the application of a known function abstraction. *)
adamc@122 53
adamc@122 54 cbv beta.
adamc@122 55 (** [[
adamc@122 56
adamc@122 57 ============================
adamc@122 58 match 1 with
adamc@122 59 | 0 => 0
adamc@122 60 | S n' => let y := n' in y
adamc@122 61 end = 0
adamc@122 62 ]]
adamc@122 63
adamc@122 64 Next on the list is the iota reduction, which simplifies a single [match] term by determining which pattern matches. *)
adamc@122 65
adamc@122 66 cbv iota.
adamc@122 67 (** [[
adamc@122 68
adamc@122 69 ============================
adamc@122 70 (fun n' : nat => let y := n' in y) 0 = 0
adamc@122 71 ]]
adamc@122 72
adamc@122 73 Now we need another beta reduction. *)
adamc@122 74
adamc@122 75 cbv beta.
adamc@122 76 (** [[
adamc@122 77
adamc@122 78 ============================
adamc@122 79 (let y := 0 in y) = 0
adamc@122 80 ]]
adamc@122 81
adamc@122 82 The final reduction rule is zeta, which replaces a [let] expression by its body with the appropriate term subsituted. *)
adamc@122 83
adamc@122 84 cbv zeta.
adamc@122 85 (** [[
adamc@122 86
adamc@122 87 ============================
adamc@122 88 0 = 0
adamc@122 89 ]] *)
adamc@122 90
adamc@122 91 reflexivity.
adamc@122 92 Qed.
adamc@124 93 (* end thide *)
adamc@122 94
adamc@122 95 (** The standard [eq] relation is critically dependent on the definitional equality. [eq] is often called a %\textit{%#<i>#propositional equality#</i>#%}%, 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.
adamc@122 96
adamc@122 97 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, we will introduce effective proof methods for goals that use proofs of the standard propositional equality "as data." *)
adamc@122 98
adamc@122 99
adamc@118 100 (** * Heterogeneous Lists Revisited *)
adamc@118 101
adamc@118 102 (** One of our example dependent data structures from the last chapter was heterogeneous lists and their associated "cursor" type. *)
adamc@118 103
adamc@118 104 Section fhlist.
adamc@118 105 Variable A : Type.
adamc@118 106 Variable B : A -> Type.
adamc@118 107
adamc@118 108 Fixpoint fhlist (ls : list A) : Type :=
adamc@118 109 match ls with
adamc@118 110 | nil => unit
adamc@118 111 | x :: ls' => B x * fhlist ls'
adamc@118 112 end%type.
adamc@118 113
adamc@118 114 Variable elm : A.
adamc@118 115
adamc@118 116 Fixpoint fmember (ls : list A) : Type :=
adamc@118 117 match ls with
adamc@118 118 | nil => Empty_set
adamc@118 119 | x :: ls' => (x = elm) + fmember ls'
adamc@118 120 end%type.
adamc@118 121
adamc@118 122 Fixpoint fhget (ls : list A) : fhlist ls -> fmember ls -> B elm :=
adamc@118 123 match ls return fhlist ls -> fmember ls -> B elm with
adamc@118 124 | nil => fun _ idx => match idx with end
adamc@118 125 | _ :: ls' => fun mls idx =>
adamc@118 126 match idx with
adamc@118 127 | inl pf => match pf with
adamc@118 128 | refl_equal => fst mls
adamc@118 129 end
adamc@118 130 | inr idx' => fhget ls' (snd mls) idx'
adamc@118 131 end
adamc@118 132 end.
adamc@118 133 End fhlist.
adamc@118 134
adamc@118 135 Implicit Arguments fhget [A B elm ls].
adamc@118 136
adamc@118 137 (** We can define a [map]-like function for [fhlist]s. *)
adamc@118 138
adamc@118 139 Section fhlist_map.
adamc@118 140 Variables A : Type.
adamc@118 141 Variables B C : A -> Type.
adamc@118 142 Variable f : forall x, B x -> C x.
adamc@118 143
adamc@118 144 Fixpoint fhmap (ls : list A) : fhlist B ls -> fhlist C ls :=
adamc@118 145 match ls return fhlist B ls -> fhlist C ls with
adamc@118 146 | nil => fun _ => tt
adamc@118 147 | _ :: _ => fun hls => (f (fst hls), fhmap _ (snd hls))
adamc@118 148 end.
adamc@118 149
adamc@118 150 Implicit Arguments fhmap [ls].
adamc@118 151
adamc@118 152 (** 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@118 153
adamc@118 154 Variable elm : A.
adamc@118 155
adamc@118 156 Theorem get_imap : forall ls (mem : fmember elm ls) (hls : fhlist B ls),
adamc@118 157 fhget (fhmap hls) mem = f (fhget hls mem).
adamc@124 158 (* begin thide *)
adamc@118 159 induction ls; crush.
adamc@118 160
adamc@118 161 (** Part of our single remaining subgoal is:
adamc@118 162
adamc@118 163 [[
adamc@118 164
adamc@118 165 a0 : a = elm
adamc@118 166 ============================
adamc@118 167 match a0 in (_ = a2) return (C a2) with
adamc@118 168 | refl_equal => f a1
adamc@118 169 end = f match a0 in (_ = a2) return (B a2) with
adamc@118 170 | refl_equal => a1
adamc@118 171 end
adamc@118 172 ]]
adamc@118 173
adamc@118 174 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.
adamc@118 175
adamc@118 176 [[
adamc@118 177
adamc@118 178 destruct a0.
adamc@118 179
adamc@118 180 [[
adamc@118 181 User error: Cannot solve a second-order unification problem
adamc@118 182 ]]
adamc@118 183
adamc@118 184 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.
adamc@118 185
adamc@118 186 [[
adamc@118 187
adamc@118 188 assert (a0 = refl_equal _).
adamc@118 189
adamc@118 190 [[
adamc@118 191 The term "refl_equal ?98" has type "?98 = ?98"
adamc@118 192 while it is expected to have type "a = elm"
adamc@118 193 ]]
adamc@118 194
adamc@118 195 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!
adamc@118 196
adamc@118 197 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.
adamc@118 198
adamc@118 199 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. *)
adamc@118 200
adamc@118 201 case a0.
adamc@118 202 (** [[
adamc@118 203
adamc@118 204 ============================
adamc@118 205 f a1 = f a1
adamc@118 206 ]]
adamc@118 207
adamc@118 208 It seems that [destruct] was trying to be too smart for its own good. *)
adamc@118 209
adamc@118 210 reflexivity.
adamc@118 211 Qed.
adamc@124 212 (* end thide *)
adamc@118 213
adamc@118 214 (** It will be helpful to examine the proof terms generated by this sort of strategy. A simpler example illustrates what is going on. *)
adamc@118 215
adamc@118 216 Lemma lemma1 : forall x (pf : x = elm), O = match pf with refl_equal => O end.
adamc@124 217 (* begin thide *)
adamc@118 218 simple destruct pf; reflexivity.
adamc@118 219 Qed.
adamc@124 220 (* end thide *)
adamc@118 221
adamc@118 222 (** [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@118 223
adamc@118 224 Print lemma1.
adamc@118 225
adamc@118 226 (** [[
adamc@118 227
adamc@118 228 lemma1 =
adamc@118 229 fun (x : A) (pf : x = elm) =>
adamc@118 230 match pf as e in (_ = y) return (0 = match e with
adamc@118 231 | refl_equal => 0
adamc@118 232 end) with
adamc@118 233 | refl_equal => refl_equal 0
adamc@118 234 end
adamc@118 235 : forall (x : A) (pf : x = elm), 0 = match pf with
adamc@118 236 | refl_equal => 0
adamc@118 237 end
adamc@118 238 ]]
adamc@118 239
adamc@118 240 Using what we know about shorthands for [match] annotations, we can write this proof in shorter form manually. *)
adamc@118 241
adamc@124 242 (* begin thide *)
adamc@118 243 Definition lemma1' :=
adamc@118 244 fun (x : A) (pf : x = elm) =>
adamc@118 245 match pf return (0 = match pf with
adamc@118 246 | refl_equal => 0
adamc@118 247 end) with
adamc@118 248 | refl_equal => refl_equal 0
adamc@118 249 end.
adamc@124 250 (* end thide *)
adamc@118 251
adamc@118 252 (** Surprisingly, what seems at first like a %\textit{%#<i>#simpler#</i>#%}% lemma is harder to prove. *)
adamc@118 253
adamc@118 254 Lemma lemma2 : forall (x : A) (pf : x = x), O = match pf with refl_equal => O end.
adamc@124 255 (* begin thide *)
adamc@118 256 (** [[
adamc@118 257
adamc@118 258 simple destruct pf.
adamc@118 259
adamc@118 260 [[
adamc@118 261
adamc@118 262 User error: Cannot solve a second-order unification problem
adamc@118 263 ]] *)
adamc@118 264 Abort.
adamc@118 265
adamc@118 266 (** Nonetheless, we can adapt the last manual proof to handle this theorem. *)
adamc@118 267
adamc@124 268 (* begin thide *)
adamc@124 269 Definition lemma2 :=
adamc@118 270 fun (x : A) (pf : x = x) =>
adamc@118 271 match pf return (0 = match pf with
adamc@118 272 | refl_equal => 0
adamc@118 273 end) with
adamc@118 274 | refl_equal => refl_equal 0
adamc@118 275 end.
adamc@124 276 (* end thide *)
adamc@118 277
adamc@118 278 (** We can try to prove a lemma that would simplify proofs of many facts like [lemma2]: *)
adamc@118 279
adamc@118 280 Lemma lemma3 : forall (x : A) (pf : x = x), pf = refl_equal x.
adamc@124 281 (* begin thide *)
adamc@118 282 (** [[
adamc@118 283
adamc@118 284 simple destruct pf.
adamc@118 285
adamc@118 286 [[
adamc@118 287
adamc@118 288 User error: Cannot solve a second-order unification problem
adamc@118 289 ]] *)
adamc@118 290 Abort.
adamc@118 291
adamc@118 292 (** This time, even our manual attempt fails.
adamc@118 293
adamc@118 294 [[
adamc@118 295
adamc@118 296 Definition lemma3' :=
adamc@118 297 fun (x : A) (pf : x = x) =>
adamc@118 298 match pf as pf' in (_ = x') return (pf' = refl_equal x') with
adamc@118 299 | refl_equal => refl_equal _
adamc@118 300 end.
adamc@118 301
adamc@118 302 [[
adamc@118 303
adamc@118 304 The term "refl_equal x'" has type "x' = x'" while it is expected to have type
adamc@118 305 "x = x'"
adamc@118 306 ]]
adamc@118 307
adamc@118 308 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.
adamc@118 309
adamc@118 310 Nonetheless, it turns out that, with one catch, we %\textit{%#<i>#can#</i>#%}% prove this lemma. *)
adamc@118 311
adamc@118 312 Lemma lemma3 : forall (x : A) (pf : x = x), pf = refl_equal x.
adamc@118 313 intros; apply UIP_refl.
adamc@118 314 Qed.
adamc@118 315
adamc@118 316 Check UIP_refl.
adamc@118 317 (** [[
adamc@118 318
adamc@118 319 UIP_refl
adamc@118 320 : forall (U : Type) (x : U) (p : x = x), p = refl_equal x
adamc@118 321 ]]
adamc@118 322
adamc@118 323 [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>#%}%. *)
adamc@118 324
adamc@118 325 Print eq_rect_eq.
adamc@118 326 (** [[
adamc@118 327
adamc@118 328 eq_rect_eq =
adamc@118 329 fun U : Type => Eq_rect_eq.eq_rect_eq U
adamc@118 330 : forall (U : Type) (p : U) (Q : U -> Type) (x : Q p) (h : p = p),
adamc@118 331 x = eq_rect p Q x p h
adamc@118 332 ]]
adamc@118 333
adamc@118 334 [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.
adamc@118 335
adamc@118 336 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.
adamc@118 337
adamc@118 338 This axiom is equivalent to another that is more commonly known and mentioned in type theory circles. *)
adamc@118 339
adamc@118 340 Print Streicher_K.
adamc@124 341 (* end thide *)
adamc@118 342 (** [[
adamc@118 343
adamc@118 344 Streicher_K =
adamc@118 345 fun U : Type => UIP_refl__Streicher_K U (UIP_refl U)
adamc@118 346 : forall (U : Type) (x : U) (P : x = x -> Prop),
adamc@118 347 P (refl_equal x) -> forall p : x = x, P p
adamc@118 348 ]]
adamc@118 349
adamc@118 350 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. *)
adamc@118 351
adamc@118 352 End fhlist_map.
adamc@118 353
adamc@119 354
adamc@119 355 (** * Type-Casts in Theorem Statements *)
adamc@119 356
adamc@119 357 (** 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@119 358
adamc@119 359 Section fhapp.
adamc@119 360 Variable A : Type.
adamc@119 361 Variable B : A -> Type.
adamc@119 362
adamc@119 363 Fixpoint fhapp (ls1 ls2 : list A) {struct ls1}
adamc@119 364 : fhlist B ls1 -> fhlist B ls2 -> fhlist B (ls1 ++ ls2) :=
adamc@119 365 match ls1 return fhlist _ ls1 -> _ -> fhlist _ (ls1 ++ ls2) with
adamc@119 366 | nil => fun _ hls2 => hls2
adamc@119 367 | _ :: _ => fun hls1 hls2 => (fst hls1, fhapp _ _ (snd hls1) hls2)
adamc@119 368 end.
adamc@119 369
adamc@119 370 Implicit Arguments fhapp [ls1 ls2].
adamc@119 371
adamc@124 372 (* EX: Prove that fhapp is associative. *)
adamc@124 373 (* begin thide *)
adamc@124 374
adamc@119 375 (** We might like to prove that [fhapp] is associative.
adamc@119 376
adamc@119 377 [[
adamc@119 378
adamc@119 379 Theorem fhapp_ass : forall ls1 ls2 ls3
adamc@119 380 (hls1 : fhlist B ls1) (hls2 : fhlist B ls2) (hls3 : fhlist B ls3),
adamc@119 381 fhapp hls1 (fhapp hls2 hls3) = fhapp (fhapp hls1 hls2) hls3.
adamc@119 382
adamc@119 383 [[
adamc@119 384
adamc@119 385 The term
adamc@119 386 "fhapp (ls1:=ls1 ++ ls2) (ls2:=ls3) (fhapp (ls1:=ls1) (ls2:=ls2) hls1 hls2)
adamc@119 387 hls3" has type "fhlist B ((ls1 ++ ls2) ++ ls3)"
adamc@119 388 while it is expected to have type "fhlist B (ls1 ++ ls2 ++ ls3)"
adamc@119 389 ]]
adamc@119 390
adamc@119 391 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. *)
adamc@119 392
adamc@119 393 Theorem fhapp_ass : forall ls1 ls2 ls3
adamc@119 394 (pf : (ls1 ++ ls2) ++ ls3 = ls1 ++ (ls2 ++ ls3))
adamc@119 395 (hls1 : fhlist B ls1) (hls2 : fhlist B ls2) (hls3 : fhlist B ls3),
adamc@119 396 fhapp hls1 (fhapp hls2 hls3)
adamc@119 397 = match pf in (_ = ls) return fhlist _ ls with
adamc@119 398 | refl_equal => fhapp (fhapp hls1 hls2) hls3
adamc@119 399 end.
adamc@119 400 induction ls1; crush.
adamc@119 401
adamc@119 402 (** The first remaining subgoal looks trivial enough:
adamc@119 403
adamc@119 404 [[
adamc@119 405
adamc@119 406 ============================
adamc@119 407 fhapp (ls1:=ls2) (ls2:=ls3) hls2 hls3 =
adamc@119 408 match pf in (_ = ls) return (fhlist B ls) with
adamc@119 409 | refl_equal => fhapp (ls1:=ls2) (ls2:=ls3) hls2 hls3
adamc@119 410 end
adamc@119 411 ]]
adamc@119 412
adamc@119 413 We can try what worked in previous examples.
adamc@119 414
adamc@119 415 [[
adamc@119 416 case pf.
adamc@119 417
adamc@119 418 [[
adamc@119 419
adamc@119 420 User error: Cannot solve a second-order unification problem
adamc@119 421 ]]
adamc@119 422
adamc@119 423 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@119 424
adamc@119 425 rewrite (UIP_refl _ _ pf).
adamc@119 426 (** [[
adamc@119 427
adamc@119 428 ============================
adamc@119 429 fhapp (ls1:=ls2) (ls2:=ls3) hls2 hls3 =
adamc@119 430 fhapp (ls1:=ls2) (ls2:=ls3) hls2 hls3
adamc@119 431 ]] *)
adamc@119 432
adamc@119 433 reflexivity.
adamc@119 434
adamc@119 435 (** Our second subgoal is trickier.
adamc@119 436
adamc@119 437 [[
adamc@119 438
adamc@119 439 pf : a :: (ls1 ++ ls2) ++ ls3 = a :: ls1 ++ ls2 ++ ls3
adamc@119 440 ============================
adamc@119 441 (a0,
adamc@119 442 fhapp (ls1:=ls1) (ls2:=ls2 ++ ls3) b
adamc@119 443 (fhapp (ls1:=ls2) (ls2:=ls3) hls2 hls3)) =
adamc@119 444 match pf in (_ = ls) return (fhlist B ls) with
adamc@119 445 | refl_equal =>
adamc@119 446 (a0,
adamc@119 447 fhapp (ls1:=ls1 ++ ls2) (ls2:=ls3)
adamc@119 448 (fhapp (ls1:=ls1) (ls2:=ls2) b hls2) hls3)
adamc@119 449 end
adamc@119 450 ]]
adamc@119 451
adamc@119 452
adamc@119 453 [[
adamc@119 454
adamc@119 455 rewrite (UIP_refl _ _ pf).
adamc@119 456
adamc@119 457 [[
adamc@119 458 The term "pf" has type "a :: (ls1 ++ ls2) ++ ls3 = a :: ls1 ++ ls2 ++ ls3"
adamc@119 459 while it is expected to have type "?556 = ?556"
adamc@119 460 ]]
adamc@119 461
adamc@119 462 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@119 463
adamc@119 464 injection pf; intro pf'.
adamc@119 465 (** [[
adamc@119 466
adamc@119 467 pf : a :: (ls1 ++ ls2) ++ ls3 = a :: ls1 ++ ls2 ++ ls3
adamc@119 468 pf' : (ls1 ++ ls2) ++ ls3 = ls1 ++ ls2 ++ ls3
adamc@119 469 ============================
adamc@119 470 (a0,
adamc@119 471 fhapp (ls1:=ls1) (ls2:=ls2 ++ ls3) b
adamc@119 472 (fhapp (ls1:=ls2) (ls2:=ls3) hls2 hls3)) =
adamc@119 473 match pf in (_ = ls) return (fhlist B ls) with
adamc@119 474 | refl_equal =>
adamc@119 475 (a0,
adamc@119 476 fhapp (ls1:=ls1 ++ ls2) (ls2:=ls3)
adamc@119 477 (fhapp (ls1:=ls1) (ls2:=ls2) b hls2) hls3)
adamc@119 478 end
adamc@119 479 ]]
adamc@119 480
adamc@119 481 Now we can rewrite using the inductive hypothesis. *)
adamc@119 482
adamc@119 483 rewrite (IHls1 _ _ pf').
adamc@119 484 (** [[
adamc@119 485
adamc@119 486 ============================
adamc@119 487 (a0,
adamc@119 488 match pf' in (_ = ls) return (fhlist B ls) with
adamc@119 489 | refl_equal =>
adamc@119 490 fhapp (ls1:=ls1 ++ ls2) (ls2:=ls3)
adamc@119 491 (fhapp (ls1:=ls1) (ls2:=ls2) b hls2) hls3
adamc@119 492 end) =
adamc@119 493 match pf in (_ = ls) return (fhlist B ls) with
adamc@119 494 | refl_equal =>
adamc@119 495 (a0,
adamc@119 496 fhapp (ls1:=ls1 ++ ls2) (ls2:=ls3)
adamc@119 497 (fhapp (ls1:=ls1) (ls2:=ls2) b hls2) hls3)
adamc@119 498 end
adamc@119 499 ]]
adamc@119 500
adamc@119 501 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. *)
adamc@119 502
adamc@119 503 generalize (fhapp (fhapp b hls2) hls3).
adamc@119 504 (** [[
adamc@119 505
adamc@119 506 forall f : fhlist B ((ls1 ++ ls2) ++ ls3),
adamc@119 507 (a0,
adamc@119 508 match pf' in (_ = ls) return (fhlist B ls) with
adamc@119 509 | refl_equal => f
adamc@119 510 end) =
adamc@119 511 match pf in (_ = ls) return (fhlist B ls) with
adamc@119 512 | refl_equal => (a0, f)
adamc@119 513 end
adamc@119 514 ]]
adamc@119 515
adamc@119 516 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@119 517
adamc@119 518 In this case, it is helpful to generalize over our two proofs as well. *)
adamc@119 519
adamc@119 520 generalize pf pf'.
adamc@119 521 (** [[
adamc@119 522
adamc@119 523 forall (pf0 : a :: (ls1 ++ ls2) ++ ls3 = a :: ls1 ++ ls2 ++ ls3)
adamc@119 524 (pf'0 : (ls1 ++ ls2) ++ ls3 = ls1 ++ ls2 ++ ls3)
adamc@119 525 (f : fhlist B ((ls1 ++ ls2) ++ ls3)),
adamc@119 526 (a0,
adamc@119 527 match pf'0 in (_ = ls) return (fhlist B ls) with
adamc@119 528 | refl_equal => f
adamc@119 529 end) =
adamc@119 530 match pf0 in (_ = ls) return (fhlist B ls) with
adamc@119 531 | refl_equal => (a0, f)
adamc@119 532 end
adamc@119 533 ]]
adamc@119 534
adamc@119 535 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.
adamc@119 536
adamc@119 537 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. *)
adamc@119 538
adamc@119 539 rewrite app_ass.
adamc@119 540 (** [[
adamc@119 541
adamc@119 542 ============================
adamc@119 543 forall (pf0 : a :: ls1 ++ ls2 ++ ls3 = a :: ls1 ++ ls2 ++ ls3)
adamc@119 544 (pf'0 : ls1 ++ ls2 ++ ls3 = ls1 ++ ls2 ++ ls3)
adamc@119 545 (f : fhlist B (ls1 ++ ls2 ++ ls3)),
adamc@119 546 (a0,
adamc@119 547 match pf'0 in (_ = ls) return (fhlist B ls) with
adamc@119 548 | refl_equal => f
adamc@119 549 end) =
adamc@119 550 match pf0 in (_ = ls) return (fhlist B ls) with
adamc@119 551 | refl_equal => (a0, f)
adamc@119 552 end
adamc@119 553 ]]
adamc@119 554
adamc@119 555 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@119 556
adamc@119 557 intros.
adamc@119 558 rewrite (UIP_refl _ _ pf0).
adamc@119 559 rewrite (UIP_refl _ _ pf'0).
adamc@119 560 reflexivity.
adamc@119 561 Qed.
adamc@124 562 (* end thide *)
adamc@119 563 End fhapp.
adamc@120 564
adamc@120 565 Implicit Arguments fhapp [A B ls1 ls2].
adamc@120 566
adamc@120 567
adamc@120 568 (** * Heterogeneous Equality *)
adamc@120 569
adamc@120 570 (** There is another equality predicate, defined in the [JMeq] module of the standard library, implementing %\textit{%#<i>#heterogeneous equality#</i>#%}%. *)
adamc@120 571
adamc@120 572 Print JMeq.
adamc@120 573 (** [[
adamc@120 574
adamc@120 575 Inductive JMeq (A : Type) (x : A) : forall B : Type, B -> Prop :=
adamc@120 576 JMeq_refl : JMeq x x
adamc@120 577 ]]
adamc@120 578
adamc@120 579 [JMeq] stands for "John Major equality," a name coined by Conor McBride as a sort of pun about British politics. [JMeq] starts out looking a lot like [eq]. The crucial difference is that we may use [JMeq] %\textit{%#<i>#on arguments of different types#</i>#%}%. 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@120 580
adamc@120 581 Infix "==" := JMeq (at level 70, no associativity).
adamc@120 582
adamc@124 583 (* EX: Prove UIP_refl' : forall (A : Type) (x : A) (pf : x = x), pf == refl_equal x *)
adamc@124 584 (* begin thide *)
adamc@121 585 Definition UIP_refl' (A : Type) (x : A) (pf : x = x) : pf == refl_equal x :=
adamc@120 586 match pf return (pf == refl_equal _) with
adamc@120 587 | refl_equal => JMeq_refl _
adamc@120 588 end.
adamc@124 589 (* end thide *)
adamc@120 590
adamc@120 591 (** There is no quick way to write such a proof by tactics, but the underlying proof term that we want is trivial.
adamc@120 592
adamc@121 593 Suppose that we want to use [UIP_refl'] to establish another lemma of the kind of we have run into several times so far. *)
adamc@120 594
adamc@120 595 Lemma lemma4 : forall (A : Type) (x : A) (pf : x = x),
adamc@120 596 O = match pf with refl_equal => O end.
adamc@124 597 (* begin thide *)
adamc@121 598 intros; rewrite (UIP_refl' pf); reflexivity.
adamc@120 599 Qed.
adamc@124 600 (* end thide *)
adamc@120 601
adamc@120 602 (** 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@120 603
adamc@120 604 Check JMeq_eq.
adamc@120 605 (** [[
adamc@120 606
adamc@120 607 JMeq_eq
adamc@120 608 : forall (A : Type) (x y : A), x == y -> x = y
adamc@120 609 ]] *)
adamc@120 610
adamc@120 611 (** 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.
adamc@120 612
adamc@120 613 We can redo our [fhapp] associativity proof based around [JMeq]. *)
adamc@120 614
adamc@120 615 Section fhapp'.
adamc@120 616 Variable A : Type.
adamc@120 617 Variable B : A -> Type.
adamc@120 618
adamc@120 619 (** This time, the naive theorem statement type-checks. *)
adamc@120 620
adamc@124 621 (* EX: Prove [fhapp] associativity using [JMeq]. *)
adamc@124 622
adamc@124 623 (* begin thide *)
adamc@120 624 Theorem fhapp_ass' : forall ls1 ls2 ls3
adamc@120 625 (hls1 : fhlist B ls1) (hls2 : fhlist B ls2) (hls3 : fhlist B ls3),
adamc@120 626 fhapp hls1 (fhapp hls2 hls3) == fhapp (fhapp hls1 hls2) hls3.
adamc@120 627 induction ls1; crush.
adamc@120 628
adamc@120 629 (** Even better, [crush] discharges the first subgoal automatically. The second subgoal is:
adamc@120 630
adamc@120 631 [[
adamc@120 632
adamc@120 633 ============================
adamc@120 634 (a0,
adamc@120 635 fhapp (B:=B) (ls1:=ls1) (ls2:=ls2 ++ ls3) b
adamc@120 636 (fhapp (B:=B) (ls1:=ls2) (ls2:=ls3) hls2 hls3)) ==
adamc@120 637 (a0,
adamc@120 638 fhapp (B:=B) (ls1:=ls1 ++ ls2) (ls2:=ls3)
adamc@120 639 (fhapp (B:=B) (ls1:=ls1) (ls2:=ls2) b hls2) hls3)
adamc@120 640 ]]
adamc@120 641
adamc@120 642 It looks like one rewrite with the inductive hypothesis should be enough to make the goal trivial.
adamc@120 643
adamc@120 644 [[
adamc@120 645
adamc@120 646 rewrite IHls1.
adamc@120 647
adamc@120 648 [[
adamc@120 649
adamc@120 650 Error: Impossible to unify "fhlist B ((ls1 ++ ?1572) ++ ?1573)" with
adamc@120 651 "fhlist B (ls1 ++ ?1572 ++ ?1573)"
adamc@120 652 ]]
adamc@120 653
adamc@120 654 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@120 655
adamc@120 656 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@120 657
adamc@120 658 generalize (fhapp b (fhapp hls2 hls3))
adamc@120 659 (fhapp (fhapp b hls2) hls3)
adamc@120 660 (IHls1 _ _ b hls2 hls3).
adamc@120 661 (** [[
adamc@120 662
adamc@120 663 ============================
adamc@120 664 forall (f : fhlist B (ls1 ++ ls2 ++ ls3))
adamc@120 665 (f0 : fhlist B ((ls1 ++ ls2) ++ ls3)), f == f0 -> (a0, f) == (a0, f0)
adamc@120 666 ]]
adamc@120 667
adamc@120 668 Now we can rewrite with append associativity, as before. *)
adamc@120 669
adamc@120 670 rewrite app_ass.
adamc@120 671 (** [[
adamc@120 672
adamc@120 673 ============================
adamc@120 674 forall f f0 : fhlist B (ls1 ++ ls2 ++ ls3), f == f0 -> (a0, f) == (a0, f0)
adamc@120 675 ]]
adamc@120 676
adamc@120 677 From this point, the goal is trivial. *)
adamc@120 678
adamc@120 679 intros f f0 H; rewrite H; reflexivity.
adamc@120 680 Qed.
adamc@124 681 (* end thide *)
adamc@120 682 End fhapp'.
adamc@121 683
adamc@121 684
adamc@121 685 (** * Equivalence of Equality Axioms *)
adamc@121 686
adamc@124 687 (* EX: Show that the approaches based on K and JMeq are equivalent logically. *)
adamc@124 688
adamc@124 689 (* begin thide *)
adamc@121 690 (** 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 the previous two sections' approaches reduces to the other logically.
adamc@121 691
adamc@121 692 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. *)
adamc@121 693
adamc@121 694 Lemma UIP_refl'' : forall (A : Type) (x : A) (pf : x = x), pf = refl_equal x.
adamc@121 695 intros; rewrite (UIP_refl' pf); reflexivity.
adamc@121 696 Qed.
adamc@121 697
adamc@121 698 (** 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@121 699
adamc@121 700 Definition JMeq' (A : Type) (x : A) (B : Type) (y : B) : Prop :=
adamc@121 701 exists pf : B = A, x = match pf with refl_equal => y end.
adamc@121 702
adamc@121 703 Infix "===" := JMeq' (at level 70, no associativity).
adamc@121 704
adamc@121 705 (** 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@121 706
adamc@121 707 We can easily prove a theorem with the same type as that of the [JMeq_refl] constructor of [JMeq]. *)
adamc@121 708
adamc@121 709 (** remove printing exists *)
adamc@121 710 Theorem JMeq_refl' : forall (A : Type) (x : A), x === x.
adamc@121 711 intros; unfold JMeq'; exists (refl_equal A); reflexivity.
adamc@121 712 Qed.
adamc@121 713
adamc@121 714 (** printing exists $\exists$ *)
adamc@121 715
adamc@121 716 (** 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@121 717
adamc@121 718 Theorem JMeq_eq' : forall (A : Type) (x y : A),
adamc@121 719 x === y -> x = y.
adamc@121 720 unfold JMeq'; intros.
adamc@121 721 (** [[
adamc@121 722
adamc@121 723 H : exists pf : A = A,
adamc@121 724 x = match pf in (_ = T) return T with
adamc@121 725 | refl_equal => y
adamc@121 726 end
adamc@121 727 ============================
adamc@121 728 x = y
adamc@121 729 ]] *)
adamc@121 730
adamc@121 731 destruct H.
adamc@121 732 (** [[
adamc@121 733
adamc@121 734 x0 : A = A
adamc@121 735 H : x = match x0 in (_ = T) return T with
adamc@121 736 | refl_equal => y
adamc@121 737 end
adamc@121 738 ============================
adamc@121 739 x = y
adamc@121 740 ]] *)
adamc@121 741
adamc@121 742 rewrite H.
adamc@121 743 (** [[
adamc@121 744
adamc@121 745 x0 : A = A
adamc@121 746 ============================
adamc@121 747 match x0 in (_ = T) return T with
adamc@121 748 | refl_equal => y
adamc@121 749 end = y
adamc@121 750 ]] *)
adamc@121 751
adamc@121 752 rewrite (UIP_refl _ _ x0); reflexivity.
adamc@121 753 Qed.
adamc@121 754
adamc@123 755 (** 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 intercovert 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@123 756
adamc@123 757 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. *)
adamc@124 758 (* end thide *)
adamc@123 759
adamc@123 760
adamc@123 761 (** * Equality of Functions *)
adamc@123 762
adamc@123 763 (** The following seems like a reasonable theorem to want to hold, and it does hold in set theory. [[
adamc@123 764
adamc@123 765 Theorem S_eta : S = (fun n => S n).
adamc@123 766
adamc@123 767 Unfortunately, this theorem is not provable in CIC without additional axioms. None of the definitional equality rules force function equality to be %\textit{%#<i>#extensional#</i>#%}%. That is, the fact that two functions return equal results on equal inputs does not imply that the functions are equal. We %\textit{%#<i>#can#</i>#%}% assert function extensionality as an axiom. *)
adamc@123 768
adamc@124 769 (* begin thide *)
adamc@123 770 Axiom ext_eq : forall A B (f g : A -> B),
adamc@123 771 (forall x, f x = g x)
adamc@123 772 -> f = g.
adamc@124 773 (* end thide *)
adamc@123 774
adamc@123 775 (** This axiom has been verified metatheoretically to be consistent with CIC and the two equality axioms we considered previously. With it, the proof of [S_eta] is trivial. *)
adamc@123 776
adamc@123 777 Theorem S_eta : S = (fun n => S n).
adamc@124 778 (* begin thide *)
adamc@123 779 apply ext_eq; reflexivity.
adamc@123 780 Qed.
adamc@124 781 (* end thide *)
adamc@123 782
adamc@123 783 (** The same axiom can help us prove equality of types, where we need to "reason under quantifiers." *)
adamc@123 784
adamc@123 785 Theorem forall_eq : (forall x : nat, match x with
adamc@123 786 | O => True
adamc@123 787 | S _ => True
adamc@123 788 end)
adamc@123 789 = (forall _ : nat, True).
adamc@123 790
adamc@123 791 (** There are no immediate opportunities to apply [ext_eq], but we can use [change] to fix that. *)
adamc@123 792
adamc@124 793 (* begin thide *)
adamc@123 794 change ((forall x : nat, (fun x => match x with
adamc@123 795 | 0 => True
adamc@123 796 | S _ => True
adamc@123 797 end) x) = (nat -> True)).
adamc@123 798 rewrite (ext_eq (fun x => match x with
adamc@123 799 | 0 => True
adamc@123 800 | S _ => True
adamc@123 801 end) (fun _ => True)).
adamc@123 802 (** [[
adamc@123 803
adamc@123 804 2 subgoals
adamc@123 805
adamc@123 806 ============================
adamc@123 807 (nat -> True) = (nat -> True)
adamc@123 808
adamc@123 809 subgoal 2 is:
adamc@123 810 forall x : nat, match x with
adamc@123 811 | 0 => True
adamc@123 812 | S _ => True
adamc@123 813 end = True
adamc@123 814 ]] *)
adamc@123 815
adamc@123 816
adamc@123 817 reflexivity.
adamc@123 818
adamc@123 819 destruct x; constructor.
adamc@123 820 Qed.
adamc@124 821 (* end thide *)
adamc@127 822
adamc@127 823
adamc@127 824 (** * Exercises *)
adamc@127 825
adamc@127 826 (** %\begin{enumerate}%#<ol>#
adamc@127 827
adamc@127 828 %\item%#<li># Implement and prove correct a substitution function for simply-typed lambda calculus. In particular:
adamc@127 829 %\begin{enumerate}%#<ol>#
adamc@127 830 %\item%#<li># Define a datatype [type] of lambda types, including just booleans and function types.#</li>#
adamc@127 831 %\item%#<li># Define a type family [exp : list type -> type -> Type] of lambda expressions, including boolean constants, variables, and function application and abstraction.#</li>#
adamc@127 832 %\item%#<li># Implement a definitional interpreter for [exp]s, by way of a recursive function over expressions and substitutions for free variables, like in the related example from the last chapter.#</li>#
adamc@127 833 %\item%#<li># Implement a function [subst : forall t' ts t, exp (t' :: ts) t -> exp ts t' -> exp ts t]. The type of the first expression indicates that its most recently bound free variable has type [t']. The second expression also has type [t'], and the job of [subst] is to substitute the second expression for every occurrence of the "first" variable of the first expression.#</li>#
adamc@127 834 %\item%#<li># Prove that [subst] preserves program meanings. That is, prove
adamc@127 835 [[
adamc@127 836 forall t' ts t (e : exp (t' :: ts) t) (e' : exp ts t') (s : hlist typeDenote ts),
adamc@127 837 expDenote (subst e e') s = expDenote e (expDenote e' s ::: s)
adamc@127 838 ]]
adamc@127 839 where [:::] is an infix operator for heterogeneous "cons" that is defined in the book's [DepList] module.#</li>#
adamc@127 840 #</ol>#%\end{enumerate}%
adamc@127 841 The material presented up to this point should be sufficient to enable a good solution of this exercise, with enough ingenuity. If you get stuck, it may be helpful to use the following structure. None of these elements need to appear in your solution, but we can at least guarantee that there is a reasonable solution based on them.
adamc@127 842 %\begin{enumerate}%#<ol>#
adamc@127 843 %\item%#<li># The [DepList] module will be useful. You can get the standard dependent list definitions there, instead of copying-and-pasting from the last chapter. It is worth reading the source for that module over, since it defines some new helpful functions and notations that we did not use last chapter.#</li>#
adamc@127 844 %\item%#<li># Define a recursive function [liftVar : forall ts1 ts2 t t', member t (ts1 ++ ts2) -> member t (ts1 ++ t' :: ts2)]. This function should "lift" a de Bruijn variable so that its type refers to a new variable inserted somewhere in the index list.#</li>#
adamc@127 845 %\item%#<li># Define a recursive function [lift' : forall ts t (e : exp ts t) ts1 ts2 t', ts = ts1 ++ ts2 -> exp (ts1 ++ t' :: ts2) t] which performs a similar lifting on an [exp]. The convoluted type is to get around restrictions on [match] annotations. We delay "realizing" that the first index of [e] is built with list concatenation until after a dependent [match], and the new explicit proof argument must be used to cast some terms that come up in the [match] body.#</li>#
adamc@127 846 %\item%#<li># Define a function [lift : forall ts t t', exp ts t -> exp (t' :: ts) t], which handles simpler top-level lifts. This should be an easy one-liner based on [lift'].#</li>#
adamc@127 847 %\item%#<li># Define a recursive function [substVar : forall ts1 ts2 t t', member t (ts1 ++ t' :: ts2) -> (t' = t) + member t (ts1 ++ ts2)]. This function is the workhorse behind substitution applied to a variable. It returns [inl] to indicate that the variable we pass to it is the variable that we are substituting for, and it returns [inr] to indicate that the variable we are examining is %\textit{%#<i>#not#</i>#%}% the one we are substituting for. In the first case, we get a proof that the necessary typing relationship holds, and, in the second case, we get the original variable modified to reflect the removal of the substitutee from the typing context.#</li>#
adamc@127 848 %\item%#<li># Define a recursive function [subst' : forall ts t (e : exp ts t) ts1 t' ts2, ts = ts1 ++ t' :: ts2 -> exp (ts1 ++ ts2) t' -> exp (ts1 ++ ts2) t]. This is the workhorse of substitution in expressions, employing the same proof-passing trick as for [lift']. You will probably want to use [lift] somewhere in the definition of [subst'].#</li>#
adamc@127 849 %\item%#<li># Now [subst] should be a one-liner, defined in terms of [subst'].#</li>#
adamc@127 850 %\item%#<li># Prove a correctness theorem for each auxiliary function, leading up to the proof of [subst] correctness.#</li>#
adamc@127 851 %\item%#<li># All of the reasoning about equality proofs in these theorems follows a regular pattern. If you have an equality proof that you want to replace with [refl_equal] somehow, run [generalize] on that proof variable. Your goal is to get to the point where you can [rewrite] with the original proof to change the type of the generalized version. To avoid type errors (the infamous "second-order unification" failure messages), it will be helpful to run [generalize] on other pieces of the proof context that mention the equality's lefthand side. You might also want to use [generalize dependent], which generalizes not just one variable but also all variables whose types depend on it. [generalize dependent] has the sometimes-helpful property of removing from the context all variables that it generalizes. Once you do manage the mind-bending trick of using the equality proof to rewrite its own type, you will be able to rewrite with [UIP_refl].#</li>#
adamc@127 852 %\item%#<li># A variant of the [ext_eq] axiom from the end of this chapter is available in the book module [Axioms], and you will probably want to use it in the [lift'] and [subst'] correctness proofs.#</li>#
adamc@127 853 %\item%#<li># The [change] tactic should come in handy in the proofs about [lift] and [subst], where you want to introduce "extraneous" list concatenations with [nil] to match the forms of earlier theorems.#</li>#
adamc@127 854 %\item%#<li># Be careful about [destruct]ing a term "too early." You can use [generalize] on proof terms to bring into the proof context any important propositions about the term. Then, when you [destruct] the term, it is updated in the extra propositions, too. The [case_eq] tactic is another alternative to this approach, based on saving an equality between the original term and its new form.#</li>#
adamc@127 855 #</ol>#%\end{enumerate}%
adamc@127 856 #</li>#
adamc@127 857
adamc@127 858 #</ol>#%\end{enumerate}% *)