Mercurial > cpdt > repo

annotate src/Equality.v @ 297:b441010125d4

Find changesets by keywords (author, files, the commit message), revision number or hash, or revset expression.
Everything compiles in Coq 8.3pl1
author Adam Chlipala <adam@chlipala.net>
date Fri, 14 Jan 2011 14:39:12 -0500
parents 559ec7328410
children 123f466faedc
rev   line source
adam@297 1 (* Copyright (c) 2008-2011, 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
adam@294 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, 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@118 22
adamc@118 23
adamc@122 24 (** * The Definitional Equality *)
adamc@122 25
adam@292 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@199 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 [0] when applied to [1]. *)
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@218 37
adamc@124 38 (* begin thide *)
adamc@122 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 that encodes terms canonically.
adamc@122 40
adamc@131 41 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 42
adamc@122 43 cbv delta.
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@218 50
adamc@122 51 ]]
adamc@122 52
adamc@122 53 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 54
adamc@122 55 cbv beta.
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@218 62
adamc@122 63 ]]
adamc@122 64
adamc@122 65 Next on the list is the iota reduction, which simplifies a single [match] term by determining which pattern matches. *)
adamc@122 66
adamc@122 67 cbv iota.
adamc@122 68 (** [[
adamc@122 69 ============================
adamc@122 70 (fun n' : nat => let y := n' in y) 0 = 0
adamc@218 71
adamc@122 72 ]]
adamc@122 73
adamc@122 74 Now we need another beta reduction. *)
adamc@122 75
adamc@122 76 cbv beta.
adamc@122 77 (** [[
adamc@122 78 ============================
adamc@122 79 (let y := 0 in y) = 0
adamc@218 80
adamc@122 81 ]]
adamc@122 82
adam@296 83 The final reduction rule is zeta, which replaces a [let] expression by its body with the appropriate term substituted. *)
adamc@122 84
adamc@122 85 cbv zeta.
adamc@122 86 (** [[
adamc@122 87 ============================
adamc@122 88 0 = 0
adamc@218 89
adamc@122 90 ]] *)
adamc@122 91
adamc@122 92 reflexivity.
adamc@122 93 Qed.
adamc@124 94 (* end thide *)
adamc@122 95
adamc@122 96 (** 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 97
adam@294 98 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@122 99
adamc@122 100
adamc@118 101 (** * Heterogeneous Lists Revisited *)
adamc@118 102
adam@292 103 (** One of our example dependent data structures from the last chapter was heterogeneous lists and their associated %``%#"#cursor#"#%''% type. The recursive version poses some special challenges related to equality proofs, since it uses such proofs in its definition of [member] types. *)
adamc@118 104
adamc@118 105 Section fhlist.
adamc@118 106 Variable A : Type.
adamc@118 107 Variable B : A -> Type.
adamc@118 108
adamc@118 109 Fixpoint fhlist (ls : list A) : Type :=
adamc@118 110 match ls with
adamc@118 111 | nil => unit
adamc@118 112 | x :: ls' => B x * fhlist ls'
adamc@118 113 end%type.
adamc@118 114
adamc@118 115 Variable elm : A.
adamc@118 116
adamc@118 117 Fixpoint fmember (ls : list A) : Type :=
adamc@118 118 match ls with
adamc@118 119 | nil => Empty_set
adamc@118 120 | x :: ls' => (x = elm) + fmember ls'
adamc@118 121 end%type.
adamc@118 122
adamc@118 123 Fixpoint fhget (ls : list A) : fhlist ls -> fmember ls -> B elm :=
adamc@118 124 match ls return fhlist ls -> fmember ls -> B elm with
adamc@118 125 | nil => fun _ idx => match idx with end
adamc@118 126 | _ :: ls' => fun mls idx =>
adamc@118 127 match idx with
adamc@118 128 | inl pf => match pf with
adamc@118 129 | refl_equal => fst mls
adamc@118 130 end
adamc@118 131 | inr idx' => fhget ls' (snd mls) idx'
adamc@118 132 end
adamc@118 133 end.
adamc@118 134 End fhlist.
adamc@118 135
adamc@118 136 Implicit Arguments fhget [A B elm ls].
adamc@118 137
adamc@118 138 (** We can define a [map]-like function for [fhlist]s. *)
adamc@118 139
adamc@118 140 Section fhlist_map.
adamc@118 141 Variables A : Type.
adamc@118 142 Variables B C : A -> Type.
adamc@118 143 Variable f : forall x, B x -> C x.
adamc@118 144
adamc@118 145 Fixpoint fhmap (ls : list A) : fhlist B ls -> fhlist C ls :=
adamc@118 146 match ls return fhlist B ls -> fhlist C ls with
adamc@118 147 | nil => fun _ => tt
adamc@118 148 | _ :: _ => fun hls => (f (fst hls), fhmap _ (snd hls))
adamc@118 149 end.
adamc@118 150
adamc@118 151 Implicit Arguments fhmap [ls].
adamc@118 152
adamc@118 153 (** 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@118 154
adamc@118 155 Variable elm : A.
adamc@118 156
adamc@118 157 Theorem get_imap : forall ls (mem : fmember elm ls) (hls : fhlist B ls),
adamc@118 158 fhget (fhmap hls) mem = f (fhget hls mem).
adamc@124 159 (* begin thide *)
adamc@118 160 induction ls; crush.
adamc@118 161
adam@297 162 (** 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@297 163
adam@297 164 Part of our single remaining subgoal is:
adamc@118 165
adamc@118 166 [[
adamc@118 167 a0 : a = elm
adamc@118 168 ============================
adamc@118 169 match a0 in (_ = a2) return (C a2) with
adamc@118 170 | refl_equal => f a1
adamc@118 171 end = f match a0 in (_ = a2) return (B a2) with
adamc@118 172 | refl_equal => a1
adamc@118 173 end
adamc@218 174
adamc@118 175 ]]
adamc@118 176
adamc@118 177 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@118 178
adamc@118 179 [[
adamc@118 180 destruct a0.
adamc@118 181
adamc@118 182 User error: Cannot solve a second-order unification problem
adamc@218 183
adamc@118 184 ]]
adamc@118 185
adamc@118 186 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@118 187
adamc@118 188 [[
adamc@118 189 assert (a0 = refl_equal _).
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@218 193
adamc@118 194 ]]
adamc@118 195
adamc@118 196 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@118 197
adam@292 198 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@118 199
adam@297 200 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@118 201
adam@297 202 [[
adamc@118 203 case a0.
adam@297 204
adamc@118 205 ============================
adamc@118 206 f a1 = f a1
adamc@218 207
adamc@118 208 ]]
adamc@118 209
adam@297 210 It seems that [destruct] was trying to be too smart for its own good.
adamc@118 211
adam@297 212 [[
adamc@118 213 reflexivity.
adam@297 214
adam@297 215 ]] *)
adamc@118 216 Qed.
adamc@124 217 (* end thide *)
adamc@118 218
adamc@118 219 (** It will be helpful to examine the proof terms generated by this sort of strategy. A simpler example illustrates what is going on. *)
adamc@118 220
adamc@118 221 Lemma lemma1 : forall x (pf : x = elm), O = match pf with refl_equal => O end.
adamc@124 222 (* begin thide *)
adamc@118 223 simple destruct pf; reflexivity.
adamc@118 224 Qed.
adamc@124 225 (* end thide *)
adamc@118 226
adamc@118 227 (** [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@118 228
adamc@118 229 Print lemma1.
adamc@218 230 (** %\vspace{-.15in}% [[
adamc@118 231 lemma1 =
adamc@118 232 fun (x : A) (pf : x = elm) =>
adamc@118 233 match pf as e in (_ = y) return (0 = match e with
adamc@118 234 | refl_equal => 0
adamc@118 235 end) with
adamc@118 236 | refl_equal => refl_equal 0
adamc@118 237 end
adamc@118 238 : forall (x : A) (pf : x = elm), 0 = match pf with
adamc@118 239 | refl_equal => 0
adamc@118 240 end
adamc@218 241
adamc@118 242 ]]
adamc@118 243
adamc@118 244 Using what we know about shorthands for [match] annotations, we can write this proof in shorter form manually. *)
adamc@118 245
adamc@124 246 (* begin thide *)
adamc@118 247 Definition lemma1' :=
adamc@118 248 fun (x : A) (pf : x = elm) =>
adamc@118 249 match pf return (0 = match pf with
adamc@118 250 | refl_equal => 0
adamc@118 251 end) with
adamc@118 252 | refl_equal => refl_equal 0
adamc@118 253 end.
adamc@124 254 (* end thide *)
adamc@118 255
adamc@118 256 (** Surprisingly, what seems at first like a %\textit{%#<i>#simpler#</i>#%}% lemma is harder to prove. *)
adamc@118 257
adamc@118 258 Lemma lemma2 : forall (x : A) (pf : x = x), O = match pf with refl_equal => O end.
adamc@124 259 (* begin thide *)
adamc@118 260 (** [[
adamc@118 261 simple destruct pf.
adamc@205 262
adamc@118 263 User error: Cannot solve a second-order unification problem
adamc@218 264
adamc@118 265 ]] *)
adamc@118 266 Abort.
adamc@118 267
adamc@118 268 (** Nonetheless, we can adapt the last manual proof to handle this theorem. *)
adamc@118 269
adamc@124 270 (* begin thide *)
adamc@124 271 Definition lemma2 :=
adamc@118 272 fun (x : A) (pf : x = x) =>
adamc@118 273 match pf return (0 = match pf with
adamc@118 274 | refl_equal => 0
adamc@118 275 end) with
adamc@118 276 | refl_equal => refl_equal 0
adamc@118 277 end.
adamc@124 278 (* end thide *)
adamc@118 279
adamc@118 280 (** We can try to prove a lemma that would simplify proofs of many facts like [lemma2]: *)
adamc@118 281
adamc@118 282 Lemma lemma3 : forall (x : A) (pf : x = x), pf = refl_equal x.
adamc@124 283 (* begin thide *)
adamc@118 284 (** [[
adamc@118 285 simple destruct pf.
adamc@205 286
adamc@118 287 User error: Cannot solve a second-order unification problem
adamc@118 288 ]] *)
adamc@218 289
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 Definition lemma3' :=
adamc@118 296 fun (x : A) (pf : x = x) =>
adamc@118 297 match pf as pf' in (_ = x') return (pf' = refl_equal x') with
adamc@118 298 | refl_equal => refl_equal _
adamc@118 299 end.
adamc@118 300
adamc@118 301 The term "refl_equal x'" has type "x' = x'" while it is expected to have type
adamc@118 302 "x = x'"
adamc@218 303
adamc@118 304 ]]
adamc@118 305
adam@296 306 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 %\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@118 307
adamc@118 308 Nonetheless, it turns out that, with one catch, we %\textit{%#<i>#can#</i>#%}% prove this lemma. *)
adamc@118 309
adamc@118 310 Lemma lemma3 : forall (x : A) (pf : x = x), pf = refl_equal x.
adamc@118 311 intros; apply UIP_refl.
adamc@118 312 Qed.
adamc@118 313
adamc@118 314 Check UIP_refl.
adamc@218 315 (** %\vspace{-.15in}% [[
adamc@118 316 UIP_refl
adamc@118 317 : forall (U : Type) (x : U) (p : x = x), p = refl_equal x
adamc@218 318
adamc@118 319 ]]
adamc@118 320
adamc@118 321 [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@118 322
adamc@118 323 Print eq_rect_eq.
adamc@218 324 (** %\vspace{-.15in}% [[
adamc@118 325 eq_rect_eq =
adamc@118 326 fun U : Type => Eq_rect_eq.eq_rect_eq U
adamc@118 327 : forall (U : Type) (p : U) (Q : U -> Type) (x : Q p) (h : p = p),
adamc@118 328 x = eq_rect p Q x p h
adamc@218 329
adamc@118 330 ]]
adamc@118 331
adam@292 332 [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@118 333
adam@292 334 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@118 335
adamc@118 336 This axiom is equivalent to another that is more commonly known and mentioned in type theory circles. *)
adamc@118 337
adamc@118 338 Print Streicher_K.
adamc@124 339 (* end thide *)
adamc@218 340 (** %\vspace{-.15in}% [[
adamc@118 341 Streicher_K =
adamc@118 342 fun U : Type => UIP_refl__Streicher_K U (UIP_refl U)
adamc@118 343 : forall (U : Type) (x : U) (P : x = x -> Prop),
adamc@118 344 P (refl_equal x) -> forall p : x = x, P p
adamc@218 345
adamc@118 346 ]]
adamc@118 347
adam@292 348 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@118 349
adamc@118 350 End fhlist_map.
adamc@118 351
adamc@119 352
adamc@119 353 (** * Type-Casts in Theorem Statements *)
adamc@119 354
adamc@119 355 (** 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@119 356
adamc@119 357 Section fhapp.
adamc@119 358 Variable A : Type.
adamc@119 359 Variable B : A -> Type.
adamc@119 360
adamc@218 361 Fixpoint fhapp (ls1 ls2 : list A)
adamc@119 362 : fhlist B ls1 -> fhlist B ls2 -> fhlist B (ls1 ++ ls2) :=
adamc@218 363 match ls1 with
adamc@119 364 | nil => fun _ hls2 => hls2
adamc@119 365 | _ :: _ => fun hls1 hls2 => (fst hls1, fhapp _ _ (snd hls1) hls2)
adamc@119 366 end.
adamc@119 367
adamc@119 368 Implicit Arguments fhapp [ls1 ls2].
adamc@119 369
adamc@124 370 (* EX: Prove that fhapp is associative. *)
adamc@124 371 (* begin thide *)
adamc@124 372
adamc@119 373 (** We might like to prove that [fhapp] is associative.
adamc@119 374
adamc@119 375 [[
adamc@119 376 Theorem fhapp_ass : forall ls1 ls2 ls3
adamc@119 377 (hls1 : fhlist B ls1) (hls2 : fhlist B ls2) (hls3 : fhlist B ls3),
adamc@119 378 fhapp hls1 (fhapp hls2 hls3) = fhapp (fhapp hls1 hls2) hls3.
adamc@119 379
adamc@119 380 The term
adamc@119 381 "fhapp (ls1:=ls1 ++ ls2) (ls2:=ls3) (fhapp (ls1:=ls1) (ls2:=ls2) hls1 hls2)
adamc@119 382 hls3" has type "fhlist B ((ls1 ++ ls2) ++ ls3)"
adamc@119 383 while it is expected to have type "fhlist B (ls1 ++ ls2 ++ ls3)"
adamc@218 384
adamc@119 385 ]]
adamc@119 386
adamc@119 387 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@119 388
adamc@119 389 Theorem fhapp_ass : forall ls1 ls2 ls3
adamc@119 390 (pf : (ls1 ++ ls2) ++ ls3 = ls1 ++ (ls2 ++ ls3))
adamc@119 391 (hls1 : fhlist B ls1) (hls2 : fhlist B ls2) (hls3 : fhlist B ls3),
adamc@119 392 fhapp hls1 (fhapp hls2 hls3)
adamc@119 393 = match pf in (_ = ls) return fhlist _ ls with
adamc@119 394 | refl_equal => fhapp (fhapp hls1 hls2) hls3
adamc@119 395 end.
adamc@119 396 induction ls1; crush.
adamc@119 397
adamc@119 398 (** The first remaining subgoal looks trivial enough:
adamc@119 399
adamc@119 400 [[
adamc@119 401 ============================
adamc@119 402 fhapp (ls1:=ls2) (ls2:=ls3) hls2 hls3 =
adamc@119 403 match pf in (_ = ls) return (fhlist B ls) with
adamc@119 404 | refl_equal => fhapp (ls1:=ls2) (ls2:=ls3) hls2 hls3
adamc@119 405 end
adamc@218 406
adamc@119 407 ]]
adamc@119 408
adamc@119 409 We can try what worked in previous examples.
adamc@119 410
adamc@119 411 [[
adamc@119 412 case pf.
adamc@119 413
adamc@119 414 User error: Cannot solve a second-order unification problem
adamc@218 415
adamc@119 416 ]]
adamc@119 417
adamc@119 418 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@119 419
adamc@119 420 rewrite (UIP_refl _ _ pf).
adamc@119 421 (** [[
adamc@119 422 ============================
adamc@119 423 fhapp (ls1:=ls2) (ls2:=ls3) hls2 hls3 =
adamc@119 424 fhapp (ls1:=ls2) (ls2:=ls3) hls2 hls3
adamc@218 425
adamc@119 426 ]] *)
adamc@119 427
adamc@119 428 reflexivity.
adamc@119 429
adamc@119 430 (** Our second subgoal is trickier.
adamc@119 431
adamc@119 432 [[
adamc@119 433 pf : a :: (ls1 ++ ls2) ++ ls3 = a :: ls1 ++ ls2 ++ ls3
adamc@119 434 ============================
adamc@119 435 (a0,
adamc@119 436 fhapp (ls1:=ls1) (ls2:=ls2 ++ ls3) b
adamc@119 437 (fhapp (ls1:=ls2) (ls2:=ls3) hls2 hls3)) =
adamc@119 438 match pf in (_ = ls) return (fhlist B ls) with
adamc@119 439 | refl_equal =>
adamc@119 440 (a0,
adamc@119 441 fhapp (ls1:=ls1 ++ ls2) (ls2:=ls3)
adamc@119 442 (fhapp (ls1:=ls1) (ls2:=ls2) b hls2) hls3)
adamc@119 443 end
adamc@119 444
adamc@119 445 rewrite (UIP_refl _ _ pf).
adamc@119 446
adamc@119 447 The term "pf" has type "a :: (ls1 ++ ls2) ++ ls3 = a :: ls1 ++ ls2 ++ ls3"
adamc@119 448 while it is expected to have type "?556 = ?556"
adamc@218 449
adamc@119 450 ]]
adamc@119 451
adamc@119 452 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@119 453
adamc@119 454 injection pf; intro pf'.
adamc@119 455 (** [[
adamc@119 456 pf : a :: (ls1 ++ ls2) ++ ls3 = a :: ls1 ++ ls2 ++ ls3
adamc@119 457 pf' : (ls1 ++ ls2) ++ ls3 = ls1 ++ ls2 ++ ls3
adamc@119 458 ============================
adamc@119 459 (a0,
adamc@119 460 fhapp (ls1:=ls1) (ls2:=ls2 ++ ls3) b
adamc@119 461 (fhapp (ls1:=ls2) (ls2:=ls3) hls2 hls3)) =
adamc@119 462 match pf in (_ = ls) return (fhlist B ls) with
adamc@119 463 | refl_equal =>
adamc@119 464 (a0,
adamc@119 465 fhapp (ls1:=ls1 ++ ls2) (ls2:=ls3)
adamc@119 466 (fhapp (ls1:=ls1) (ls2:=ls2) b hls2) hls3)
adamc@119 467 end
adamc@218 468
adamc@119 469 ]]
adamc@119 470
adamc@119 471 Now we can rewrite using the inductive hypothesis. *)
adamc@119 472
adamc@119 473 rewrite (IHls1 _ _ pf').
adamc@119 474 (** [[
adamc@119 475 ============================
adamc@119 476 (a0,
adamc@119 477 match pf' in (_ = ls) return (fhlist B ls) with
adamc@119 478 | refl_equal =>
adamc@119 479 fhapp (ls1:=ls1 ++ ls2) (ls2:=ls3)
adamc@119 480 (fhapp (ls1:=ls1) (ls2:=ls2) b hls2) hls3
adamc@119 481 end) =
adamc@119 482 match pf in (_ = ls) return (fhlist B ls) with
adamc@119 483 | refl_equal =>
adamc@119 484 (a0,
adamc@119 485 fhapp (ls1:=ls1 ++ ls2) (ls2:=ls3)
adamc@119 486 (fhapp (ls1:=ls1) (ls2:=ls2) b hls2) hls3)
adamc@119 487 end
adamc@218 488
adamc@119 489 ]]
adamc@119 490
adam@294 491 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 %\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@119 492
adamc@119 493 generalize (fhapp (fhapp b hls2) hls3).
adamc@119 494 (** [[
adamc@119 495 forall f : fhlist B ((ls1 ++ ls2) ++ ls3),
adamc@119 496 (a0,
adamc@119 497 match pf' in (_ = ls) return (fhlist B ls) with
adamc@119 498 | refl_equal => f
adamc@119 499 end) =
adamc@119 500 match pf in (_ = ls) return (fhlist B ls) with
adamc@119 501 | refl_equal => (a0, f)
adamc@119 502 end
adamc@218 503
adamc@119 504 ]]
adamc@119 505
adamc@119 506 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@119 507
adamc@119 508 In this case, it is helpful to generalize over our two proofs as well. *)
adamc@119 509
adamc@119 510 generalize pf pf'.
adamc@119 511 (** [[
adamc@119 512 forall (pf0 : a :: (ls1 ++ ls2) ++ ls3 = a :: ls1 ++ ls2 ++ ls3)
adamc@119 513 (pf'0 : (ls1 ++ ls2) ++ ls3 = ls1 ++ ls2 ++ ls3)
adamc@119 514 (f : fhlist B ((ls1 ++ ls2) ++ ls3)),
adamc@119 515 (a0,
adamc@119 516 match pf'0 in (_ = ls) return (fhlist B ls) with
adamc@119 517 | refl_equal => f
adamc@119 518 end) =
adamc@119 519 match pf0 in (_ = ls) return (fhlist B ls) with
adamc@119 520 | refl_equal => (a0, f)
adamc@119 521 end
adamc@218 522
adamc@119 523 ]]
adamc@119 524
adamc@119 525 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@119 526
adamc@119 527 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@119 528
adamc@119 529 rewrite app_ass.
adamc@119 530 (** [[
adamc@119 531 ============================
adamc@119 532 forall (pf0 : a :: ls1 ++ ls2 ++ ls3 = a :: ls1 ++ ls2 ++ ls3)
adamc@119 533 (pf'0 : ls1 ++ ls2 ++ ls3 = ls1 ++ ls2 ++ ls3)
adamc@119 534 (f : fhlist B (ls1 ++ ls2 ++ ls3)),
adamc@119 535 (a0,
adamc@119 536 match pf'0 in (_ = ls) return (fhlist B ls) with
adamc@119 537 | refl_equal => f
adamc@119 538 end) =
adamc@119 539 match pf0 in (_ = ls) return (fhlist B ls) with
adamc@119 540 | refl_equal => (a0, f)
adamc@119 541 end
adamc@218 542
adamc@119 543 ]]
adamc@119 544
adamc@119 545 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@119 546
adamc@119 547 intros.
adamc@119 548 rewrite (UIP_refl _ _ pf0).
adamc@119 549 rewrite (UIP_refl _ _ pf'0).
adamc@119 550 reflexivity.
adamc@119 551 Qed.
adamc@124 552 (* end thide *)
adamc@119 553 End fhapp.
adamc@120 554
adamc@120 555 Implicit Arguments fhapp [A B ls1 ls2].
adamc@120 556
adamc@120 557
adamc@120 558 (** * Heterogeneous Equality *)
adamc@120 559
adamc@120 560 (** There is another equality predicate, defined in the [JMeq] module of the standard library, implementing %\textit{%#<i>#heterogeneous equality#</i>#%}%. *)
adamc@120 561
adamc@120 562 Print JMeq.
adamc@218 563 (** %\vspace{-.15in}% [[
adamc@120 564 Inductive JMeq (A : Type) (x : A) : forall B : Type, B -> Prop :=
adamc@120 565 JMeq_refl : JMeq x x
adamc@218 566
adamc@120 567 ]]
adamc@120 568
adam@292 569 [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@120 570
adamc@120 571 Infix "==" := JMeq (at level 70, no associativity).
adamc@120 572
adamc@124 573 (* EX: Prove UIP_refl' : forall (A : Type) (x : A) (pf : x = x), pf == refl_equal x *)
adamc@124 574 (* begin thide *)
adamc@121 575 Definition UIP_refl' (A : Type) (x : A) (pf : x = x) : pf == refl_equal x :=
adamc@120 576 match pf return (pf == refl_equal _) with
adamc@120 577 | refl_equal => JMeq_refl _
adamc@120 578 end.
adamc@124 579 (* end thide *)
adamc@120 580
adamc@120 581 (** There is no quick way to write such a proof by tactics, but the underlying proof term that we want is trivial.
adamc@120 582
adamc@271 583 Suppose that we want to use [UIP_refl'] to establish another lemma of the kind we have run into several times so far. *)
adamc@120 584
adamc@120 585 Lemma lemma4 : forall (A : Type) (x : A) (pf : x = x),
adamc@120 586 O = match pf with refl_equal => O end.
adamc@124 587 (* begin thide *)
adamc@121 588 intros; rewrite (UIP_refl' pf); reflexivity.
adamc@120 589 Qed.
adamc@124 590 (* end thide *)
adamc@120 591
adamc@120 592 (** 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@120 593
adamc@120 594 Check JMeq_eq.
adamc@218 595 (** %\vspace{-.15in}% [[
adamc@120 596 JMeq_eq
adamc@120 597 : forall (A : Type) (x y : A), x == y -> x = y
adamc@218 598
adamc@218 599 ]]
adamc@120 600
adamc@218 601 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@120 602
adamc@120 603 We can redo our [fhapp] associativity proof based around [JMeq]. *)
adamc@120 604
adamc@120 605 Section fhapp'.
adamc@120 606 Variable A : Type.
adamc@120 607 Variable B : A -> Type.
adamc@120 608
adamc@120 609 (** This time, the naive theorem statement type-checks. *)
adamc@120 610
adamc@124 611 (* EX: Prove [fhapp] associativity using [JMeq]. *)
adamc@124 612
adamc@124 613 (* begin thide *)
adam@297 614 Theorem fhapp_ass' : forall ls1 ls2 ls3 (hls1 : fhlist B ls1) (hls2 : fhlist B ls2) (hls3 : fhlist B ls3),
adamc@120 615 fhapp hls1 (fhapp hls2 hls3) == fhapp (fhapp hls1 hls2) hls3.
adamc@120 616 induction ls1; crush.
adamc@120 617
adamc@120 618 (** Even better, [crush] discharges the first subgoal automatically. The second subgoal is:
adamc@120 619
adamc@120 620 [[
adamc@120 621 ============================
adam@297 622 (a0, fhapp b (fhapp hls2 hls3)) == (a0, fhapp (fhapp b hls2) hls3)
adamc@218 623
adamc@120 624 ]]
adamc@120 625
adam@297 626 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@120 627
adamc@120 628 [[
adamc@120 629 rewrite IHls1.
adamc@120 630
adamc@120 631 Error: Impossible to unify "fhlist B ((ls1 ++ ?1572) ++ ?1573)" with
adamc@120 632 "fhlist B (ls1 ++ ?1572 ++ ?1573)"
adamc@218 633
adamc@120 634 ]]
adamc@120 635
adam@297 636 Coq 8.3 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.
adam@297 637
adamc@120 638 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@120 639
adamc@120 640 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@120 641
adamc@120 642 generalize (fhapp b (fhapp hls2 hls3))
adamc@120 643 (fhapp (fhapp b hls2) hls3)
adamc@120 644 (IHls1 _ _ b hls2 hls3).
adamc@120 645 (** [[
adamc@120 646 ============================
adamc@120 647 forall (f : fhlist B (ls1 ++ ls2 ++ ls3))
adamc@120 648 (f0 : fhlist B ((ls1 ++ ls2) ++ ls3)), f == f0 -> (a0, f) == (a0, f0)
adamc@218 649
adamc@120 650 ]]
adamc@120 651
adamc@120 652 Now we can rewrite with append associativity, as before. *)
adamc@120 653
adamc@120 654 rewrite app_ass.
adamc@120 655 (** [[
adamc@120 656 ============================
adamc@120 657 forall f f0 : fhlist B (ls1 ++ ls2 ++ ls3), f == f0 -> (a0, f) == (a0, f0)
adamc@218 658
adamc@120 659 ]]
adamc@120 660
adamc@120 661 From this point, the goal is trivial. *)
adamc@120 662
adamc@120 663 intros f f0 H; rewrite H; reflexivity.
adamc@120 664 Qed.
adamc@124 665 (* end thide *)
adamc@120 666 End fhapp'.
adamc@121 667
adamc@121 668
adamc@121 669 (** * Equivalence of Equality Axioms *)
adamc@121 670
adamc@124 671 (* EX: Show that the approaches based on K and JMeq are equivalent logically. *)
adamc@124 672
adamc@124 673 (* begin thide *)
adamc@272 674 (** 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@121 675
adamc@121 676 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@121 677
adamc@121 678 Lemma UIP_refl'' : forall (A : Type) (x : A) (pf : x = x), pf = refl_equal x.
adamc@121 679 intros; rewrite (UIP_refl' pf); reflexivity.
adamc@121 680 Qed.
adamc@121 681
adamc@121 682 (** 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@121 683
adamc@121 684 Definition JMeq' (A : Type) (x : A) (B : Type) (y : B) : Prop :=
adamc@121 685 exists pf : B = A, x = match pf with refl_equal => y end.
adamc@121 686
adamc@121 687 Infix "===" := JMeq' (at level 70, no associativity).
adamc@121 688
adamc@121 689 (** 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@121 690
adamc@121 691 We can easily prove a theorem with the same type as that of the [JMeq_refl] constructor of [JMeq]. *)
adamc@121 692
adamc@121 693 (** remove printing exists *)
adamc@121 694 Theorem JMeq_refl' : forall (A : Type) (x : A), x === x.
adamc@121 695 intros; unfold JMeq'; exists (refl_equal A); reflexivity.
adamc@121 696 Qed.
adamc@121 697
adamc@121 698 (** printing exists $\exists$ *)
adamc@121 699
adamc@121 700 (** 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@121 701
adamc@121 702 Theorem JMeq_eq' : forall (A : Type) (x y : A),
adamc@121 703 x === y -> x = y.
adamc@121 704 unfold JMeq'; intros.
adamc@121 705 (** [[
adamc@121 706 H : exists pf : A = A,
adamc@121 707 x = match pf in (_ = T) return T with
adamc@121 708 | refl_equal => y
adamc@121 709 end
adamc@121 710 ============================
adamc@121 711 x = y
adamc@218 712
adamc@121 713 ]] *)
adamc@121 714
adamc@121 715 destruct H.
adamc@121 716 (** [[
adamc@121 717 x0 : A = A
adamc@121 718 H : x = match x0 in (_ = T) return T with
adamc@121 719 | refl_equal => y
adamc@121 720 end
adamc@121 721 ============================
adamc@121 722 x = y
adamc@218 723
adamc@121 724 ]] *)
adamc@121 725
adamc@121 726 rewrite H.
adamc@121 727 (** [[
adamc@121 728 x0 : A = A
adamc@121 729 ============================
adamc@121 730 match x0 in (_ = T) return T with
adamc@121 731 | refl_equal => y
adamc@121 732 end = y
adamc@218 733
adamc@121 734 ]] *)
adamc@121 735
adamc@121 736 rewrite (UIP_refl _ _ x0); reflexivity.
adamc@121 737 Qed.
adamc@121 738
adam@292 739 (** 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@123 740
adamc@123 741 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@124 742 (* end thide *)
adamc@123 743
adamc@123 744
adamc@123 745 (** * Equality of Functions *)
adamc@123 746
adamc@123 747 (** The following seems like a reasonable theorem to want to hold, and it does hold in set theory. [[
adamc@123 748 Theorem S_eta : S = (fun n => S n).
adamc@218 749
adamc@205 750 ]]
adamc@205 751
adamc@123 752 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@123 753
adamc@124 754 (* begin thide *)
adamc@123 755 Axiom ext_eq : forall A B (f g : A -> B),
adamc@123 756 (forall x, f x = g x)
adamc@123 757 -> f = g.
adamc@124 758 (* end thide *)
adamc@123 759
adamc@123 760 (** 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@123 761
adamc@123 762 Theorem S_eta : S = (fun n => S n).
adamc@124 763 (* begin thide *)
adamc@123 764 apply ext_eq; reflexivity.
adamc@123 765 Qed.
adamc@124 766 (* end thide *)
adamc@123 767
adam@292 768 (** The same axiom can help us prove equality of types, where we need to %``%#"#reason under quantifiers.#"#%''% *)
adamc@123 769
adamc@123 770 Theorem forall_eq : (forall x : nat, match x with
adamc@123 771 | O => True
adamc@123 772 | S _ => True
adamc@123 773 end)
adamc@123 774 = (forall _ : nat, True).
adamc@123 775
adamc@123 776 (** There are no immediate opportunities to apply [ext_eq], but we can use [change] to fix that. *)
adamc@123 777
adamc@124 778 (* begin thide *)
adamc@123 779 change ((forall x : nat, (fun x => match x with
adamc@123 780 | 0 => True
adamc@123 781 | S _ => True
adamc@123 782 end) x) = (nat -> True)).
adamc@123 783 rewrite (ext_eq (fun x => match x with
adamc@123 784 | 0 => True
adamc@123 785 | S _ => True
adamc@123 786 end) (fun _ => True)).
adamc@123 787 (** [[
adamc@123 788 2 subgoals
adamc@123 789
adamc@123 790 ============================
adamc@123 791 (nat -> True) = (nat -> True)
adamc@123 792
adamc@123 793 subgoal 2 is:
adamc@123 794 forall x : nat, match x with
adamc@123 795 | 0 => True
adamc@123 796 | S _ => True
adamc@123 797 end = True
adamc@123 798 ]] *)
adamc@123 799
adamc@123 800 reflexivity.
adamc@123 801
adamc@123 802 destruct x; constructor.
adamc@123 803 Qed.
adamc@124 804 (* end thide *)
adamc@127 805
adamc@127 806
adamc@127 807 (** * Exercises *)
adamc@127 808
adamc@127 809 (** %\begin{enumerate}%#<ol>#
adamc@127 810
adamc@127 811 %\item%#<li># Implement and prove correct a substitution function for simply-typed lambda calculus. In particular:
adamc@127 812 %\begin{enumerate}%#<ol>#
adamc@127 813 %\item%#<li># Define a datatype [type] of lambda types, including just booleans and function types.#</li>#
adamc@127 814 %\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@127 815 %\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>#
adam@292 816 %\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@127 817 %\item%#<li># Prove that [subst] preserves program meanings. That is, prove
adamc@127 818 [[
adamc@127 819 forall t' ts t (e : exp (t' :: ts) t) (e' : exp ts t') (s : hlist typeDenote ts),
adamc@127 820 expDenote (subst e e') s = expDenote e (expDenote e' s ::: s)
adamc@218 821
adamc@127 822 ]]
adam@292 823 where [:::] is an infix operator for heterogeneous %``%#"#cons#"#%''% that is defined in the book's [DepList] module.#</li>#
adamc@127 824 #</ol>#%\end{enumerate}%
adamc@127 825 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@127 826 %\begin{enumerate}%#<ol>#
adamc@127 827 %\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>#
adam@292 828 %\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>#
adam@292 829 %\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@127 830 %\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@127 831 %\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@127 832 %\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@127 833 %\item%#<li># Now [subst] should be a one-liner, defined in terms of [subst'].#</li>#
adamc@127 834 %\item%#<li># Prove a correctness theorem for each auxiliary function, leading up to the proof of [subst] correctness.#</li>#
adam@292 835 %\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@127 836 %\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>#
adam@292 837 %\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>#
adam@292 838 %\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@127 839 #</ol>#%\end{enumerate}%
adamc@127 840 #</li>#
adamc@127 841
adamc@127 842 #</ol>#%\end{enumerate}% *)