Mercurial > cpdt > repo

annotate src/Equality.v @ 298:123f466faedc

Find changesets by keywords (author, files, the commit message), revision number or hash, or revset expression.
Small tweak to keep things working in 8.2
author Adam Chlipala <adam@chlipala.net>
date Fri, 14 Jan 2011 14:55:32 -0500
parents b441010125d4
children 7b38729be069
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).
adam@298 159 (* begin hide *)
adam@298 160 induction ls; crush; case a0; reflexivity.
adam@298 161 (* end hide *)
adam@298 162 (** [[
adamc@118 163 induction ls; crush.
adam@298 164
adam@298 165 ]]
adamc@118 166
adam@298 167 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@297 168
adam@297 169 Part of our single remaining subgoal is:
adamc@118 170
adamc@118 171 [[
adamc@118 172 a0 : a = elm
adamc@118 173 ============================
adamc@118 174 match a0 in (_ = a2) return (C a2) with
adamc@118 175 | refl_equal => f a1
adamc@118 176 end = f match a0 in (_ = a2) return (B a2) with
adamc@118 177 | refl_equal => a1
adamc@118 178 end
adamc@218 179
adamc@118 180 ]]
adamc@118 181
adamc@118 182 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@118 183
adamc@118 184 [[
adamc@118 185 destruct a0.
adamc@118 186
adamc@118 187 User error: Cannot solve a second-order unification problem
adamc@218 188
adamc@118 189 ]]
adamc@118 190
adamc@118 191 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@118 192
adamc@118 193 [[
adamc@118 194 assert (a0 = refl_equal _).
adamc@118 195
adamc@118 196 The term "refl_equal ?98" has type "?98 = ?98"
adamc@118 197 while it is expected to have type "a = elm"
adamc@218 198
adamc@118 199 ]]
adamc@118 200
adamc@118 201 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@118 202
adam@292 203 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@118 204
adam@297 205 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@118 206
adam@297 207 [[
adamc@118 208 case a0.
adam@297 209
adamc@118 210 ============================
adamc@118 211 f a1 = f a1
adamc@218 212
adamc@118 213 ]]
adamc@118 214
adam@297 215 It seems that [destruct] was trying to be too smart for its own good.
adamc@118 216
adam@297 217 [[
adamc@118 218 reflexivity.
adam@297 219
adam@297 220 ]] *)
adam@298 221
adamc@118 222 Qed.
adamc@124 223 (* end thide *)
adamc@118 224
adamc@118 225 (** It will be helpful to examine the proof terms generated by this sort of strategy. A simpler example illustrates what is going on. *)
adamc@118 226
adamc@118 227 Lemma lemma1 : forall x (pf : x = elm), O = match pf with refl_equal => O end.
adamc@124 228 (* begin thide *)
adamc@118 229 simple destruct pf; reflexivity.
adamc@118 230 Qed.
adamc@124 231 (* end thide *)
adamc@118 232
adamc@118 233 (** [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@118 234
adamc@118 235 Print lemma1.
adamc@218 236 (** %\vspace{-.15in}% [[
adamc@118 237 lemma1 =
adamc@118 238 fun (x : A) (pf : x = elm) =>
adamc@118 239 match pf as e in (_ = y) return (0 = match e with
adamc@118 240 | refl_equal => 0
adamc@118 241 end) with
adamc@118 242 | refl_equal => refl_equal 0
adamc@118 243 end
adamc@118 244 : forall (x : A) (pf : x = elm), 0 = match pf with
adamc@118 245 | refl_equal => 0
adamc@118 246 end
adamc@218 247
adamc@118 248 ]]
adamc@118 249
adamc@118 250 Using what we know about shorthands for [match] annotations, we can write this proof in shorter form manually. *)
adamc@118 251
adamc@124 252 (* begin thide *)
adamc@118 253 Definition lemma1' :=
adamc@118 254 fun (x : A) (pf : x = elm) =>
adamc@118 255 match pf return (0 = match pf with
adamc@118 256 | refl_equal => 0
adamc@118 257 end) with
adamc@118 258 | refl_equal => refl_equal 0
adamc@118 259 end.
adamc@124 260 (* end thide *)
adamc@118 261
adamc@118 262 (** Surprisingly, what seems at first like a %\textit{%#<i>#simpler#</i>#%}% lemma is harder to prove. *)
adamc@118 263
adamc@118 264 Lemma lemma2 : forall (x : A) (pf : x = x), O = match pf with refl_equal => O end.
adamc@124 265 (* begin thide *)
adamc@118 266 (** [[
adamc@118 267 simple destruct pf.
adamc@205 268
adamc@118 269 User error: Cannot solve a second-order unification problem
adamc@218 270
adamc@118 271 ]] *)
adamc@118 272 Abort.
adamc@118 273
adamc@118 274 (** Nonetheless, we can adapt the last manual proof to handle this theorem. *)
adamc@118 275
adamc@124 276 (* begin thide *)
adamc@124 277 Definition lemma2 :=
adamc@118 278 fun (x : A) (pf : x = x) =>
adamc@118 279 match pf return (0 = match pf with
adamc@118 280 | refl_equal => 0
adamc@118 281 end) with
adamc@118 282 | refl_equal => refl_equal 0
adamc@118 283 end.
adamc@124 284 (* end thide *)
adamc@118 285
adamc@118 286 (** We can try to prove a lemma that would simplify proofs of many facts like [lemma2]: *)
adamc@118 287
adamc@118 288 Lemma lemma3 : forall (x : A) (pf : x = x), pf = refl_equal x.
adamc@124 289 (* begin thide *)
adamc@118 290 (** [[
adamc@118 291 simple destruct pf.
adamc@205 292
adamc@118 293 User error: Cannot solve a second-order unification problem
adamc@118 294 ]] *)
adamc@218 295
adamc@118 296 Abort.
adamc@118 297
adamc@118 298 (** This time, even our manual attempt fails.
adamc@118 299
adamc@118 300 [[
adamc@118 301 Definition lemma3' :=
adamc@118 302 fun (x : A) (pf : x = x) =>
adamc@118 303 match pf as pf' in (_ = x') return (pf' = refl_equal x') with
adamc@118 304 | refl_equal => refl_equal _
adamc@118 305 end.
adamc@118 306
adamc@118 307 The term "refl_equal x'" has type "x' = x'" while it is expected to have type
adamc@118 308 "x = x'"
adamc@218 309
adamc@118 310 ]]
adamc@118 311
adam@296 312 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@118 313
adamc@118 314 Nonetheless, it turns out that, with one catch, we %\textit{%#<i>#can#</i>#%}% prove this lemma. *)
adamc@118 315
adamc@118 316 Lemma lemma3 : forall (x : A) (pf : x = x), pf = refl_equal x.
adamc@118 317 intros; apply UIP_refl.
adamc@118 318 Qed.
adamc@118 319
adamc@118 320 Check UIP_refl.
adamc@218 321 (** %\vspace{-.15in}% [[
adamc@118 322 UIP_refl
adamc@118 323 : forall (U : Type) (x : U) (p : x = x), p = refl_equal x
adamc@218 324
adamc@118 325 ]]
adamc@118 326
adamc@118 327 [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@118 328
adamc@118 329 Print eq_rect_eq.
adamc@218 330 (** %\vspace{-.15in}% [[
adamc@118 331 eq_rect_eq =
adamc@118 332 fun U : Type => Eq_rect_eq.eq_rect_eq U
adamc@118 333 : forall (U : Type) (p : U) (Q : U -> Type) (x : Q p) (h : p = p),
adamc@118 334 x = eq_rect p Q x p h
adamc@218 335
adamc@118 336 ]]
adamc@118 337
adam@292 338 [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@118 339
adam@292 340 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@118 341
adamc@118 342 This axiom is equivalent to another that is more commonly known and mentioned in type theory circles. *)
adamc@118 343
adamc@118 344 Print Streicher_K.
adamc@124 345 (* end thide *)
adamc@218 346 (** %\vspace{-.15in}% [[
adamc@118 347 Streicher_K =
adamc@118 348 fun U : Type => UIP_refl__Streicher_K U (UIP_refl U)
adamc@118 349 : forall (U : Type) (x : U) (P : x = x -> Prop),
adamc@118 350 P (refl_equal x) -> forall p : x = x, P p
adamc@218 351
adamc@118 352 ]]
adamc@118 353
adam@292 354 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@118 355
adamc@118 356 End fhlist_map.
adamc@118 357
adamc@119 358
adamc@119 359 (** * Type-Casts in Theorem Statements *)
adamc@119 360
adamc@119 361 (** 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@119 362
adamc@119 363 Section fhapp.
adamc@119 364 Variable A : Type.
adamc@119 365 Variable B : A -> Type.
adamc@119 366
adamc@218 367 Fixpoint fhapp (ls1 ls2 : list A)
adamc@119 368 : fhlist B ls1 -> fhlist B ls2 -> fhlist B (ls1 ++ ls2) :=
adamc@218 369 match ls1 with
adamc@119 370 | nil => fun _ hls2 => hls2
adamc@119 371 | _ :: _ => fun hls1 hls2 => (fst hls1, fhapp _ _ (snd hls1) hls2)
adamc@119 372 end.
adamc@119 373
adamc@119 374 Implicit Arguments fhapp [ls1 ls2].
adamc@119 375
adamc@124 376 (* EX: Prove that fhapp is associative. *)
adamc@124 377 (* begin thide *)
adamc@124 378
adamc@119 379 (** We might like to prove that [fhapp] is associative.
adamc@119 380
adamc@119 381 [[
adamc@119 382 Theorem fhapp_ass : forall ls1 ls2 ls3
adamc@119 383 (hls1 : fhlist B ls1) (hls2 : fhlist B ls2) (hls3 : fhlist B ls3),
adamc@119 384 fhapp hls1 (fhapp hls2 hls3) = fhapp (fhapp hls1 hls2) hls3.
adamc@119 385
adamc@119 386 The term
adamc@119 387 "fhapp (ls1:=ls1 ++ ls2) (ls2:=ls3) (fhapp (ls1:=ls1) (ls2:=ls2) hls1 hls2)
adamc@119 388 hls3" has type "fhlist B ((ls1 ++ ls2) ++ ls3)"
adamc@119 389 while it is expected to have type "fhlist B (ls1 ++ ls2 ++ ls3)"
adamc@218 390
adamc@119 391 ]]
adamc@119 392
adamc@119 393 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@119 394
adamc@119 395 Theorem fhapp_ass : forall ls1 ls2 ls3
adamc@119 396 (pf : (ls1 ++ ls2) ++ ls3 = ls1 ++ (ls2 ++ ls3))
adamc@119 397 (hls1 : fhlist B ls1) (hls2 : fhlist B ls2) (hls3 : fhlist B ls3),
adamc@119 398 fhapp hls1 (fhapp hls2 hls3)
adamc@119 399 = match pf in (_ = ls) return fhlist _ ls with
adamc@119 400 | refl_equal => fhapp (fhapp hls1 hls2) hls3
adamc@119 401 end.
adamc@119 402 induction ls1; crush.
adamc@119 403
adamc@119 404 (** The first remaining subgoal looks trivial enough:
adamc@119 405
adamc@119 406 [[
adamc@119 407 ============================
adamc@119 408 fhapp (ls1:=ls2) (ls2:=ls3) hls2 hls3 =
adamc@119 409 match pf in (_ = ls) return (fhlist B ls) with
adamc@119 410 | refl_equal => fhapp (ls1:=ls2) (ls2:=ls3) hls2 hls3
adamc@119 411 end
adamc@218 412
adamc@119 413 ]]
adamc@119 414
adamc@119 415 We can try what worked in previous examples.
adamc@119 416
adamc@119 417 [[
adamc@119 418 case pf.
adamc@119 419
adamc@119 420 User error: Cannot solve a second-order unification problem
adamc@218 421
adamc@119 422 ]]
adamc@119 423
adamc@119 424 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@119 425
adamc@119 426 rewrite (UIP_refl _ _ pf).
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@218 431
adamc@119 432 ]] *)
adamc@119 433
adamc@119 434 reflexivity.
adamc@119 435
adamc@119 436 (** Our second subgoal is trickier.
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 rewrite (UIP_refl _ _ pf).
adamc@119 452
adamc@119 453 The term "pf" has type "a :: (ls1 ++ ls2) ++ ls3 = a :: ls1 ++ ls2 ++ ls3"
adamc@119 454 while it is expected to have type "?556 = ?556"
adamc@218 455
adamc@119 456 ]]
adamc@119 457
adamc@119 458 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@119 459
adamc@119 460 injection pf; intro pf'.
adamc@119 461 (** [[
adamc@119 462 pf : a :: (ls1 ++ ls2) ++ ls3 = a :: ls1 ++ ls2 ++ ls3
adamc@119 463 pf' : (ls1 ++ ls2) ++ ls3 = ls1 ++ ls2 ++ ls3
adamc@119 464 ============================
adamc@119 465 (a0,
adamc@119 466 fhapp (ls1:=ls1) (ls2:=ls2 ++ ls3) b
adamc@119 467 (fhapp (ls1:=ls2) (ls2:=ls3) hls2 hls3)) =
adamc@119 468 match pf in (_ = ls) return (fhlist B ls) with
adamc@119 469 | refl_equal =>
adamc@119 470 (a0,
adamc@119 471 fhapp (ls1:=ls1 ++ ls2) (ls2:=ls3)
adamc@119 472 (fhapp (ls1:=ls1) (ls2:=ls2) b hls2) hls3)
adamc@119 473 end
adamc@218 474
adamc@119 475 ]]
adamc@119 476
adamc@119 477 Now we can rewrite using the inductive hypothesis. *)
adamc@119 478
adamc@119 479 rewrite (IHls1 _ _ pf').
adamc@119 480 (** [[
adamc@119 481 ============================
adamc@119 482 (a0,
adamc@119 483 match pf' in (_ = ls) return (fhlist B ls) with
adamc@119 484 | refl_equal =>
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@119 488 match pf in (_ = ls) return (fhlist B ls) with
adamc@119 489 | refl_equal =>
adamc@119 490 (a0,
adamc@119 491 fhapp (ls1:=ls1 ++ ls2) (ls2:=ls3)
adamc@119 492 (fhapp (ls1:=ls1) (ls2:=ls2) b hls2) hls3)
adamc@119 493 end
adamc@218 494
adamc@119 495 ]]
adamc@119 496
adam@294 497 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@119 498
adamc@119 499 generalize (fhapp (fhapp b hls2) hls3).
adamc@119 500 (** [[
adamc@119 501 forall f : fhlist B ((ls1 ++ ls2) ++ ls3),
adamc@119 502 (a0,
adamc@119 503 match pf' in (_ = ls) return (fhlist B ls) with
adamc@119 504 | refl_equal => f
adamc@119 505 end) =
adamc@119 506 match pf in (_ = ls) return (fhlist B ls) with
adamc@119 507 | refl_equal => (a0, f)
adamc@119 508 end
adamc@218 509
adamc@119 510 ]]
adamc@119 511
adamc@119 512 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@119 513
adamc@119 514 In this case, it is helpful to generalize over our two proofs as well. *)
adamc@119 515
adamc@119 516 generalize pf pf'.
adamc@119 517 (** [[
adamc@119 518 forall (pf0 : a :: (ls1 ++ ls2) ++ ls3 = a :: ls1 ++ ls2 ++ ls3)
adamc@119 519 (pf'0 : (ls1 ++ ls2) ++ ls3 = ls1 ++ ls2 ++ ls3)
adamc@119 520 (f : fhlist B ((ls1 ++ ls2) ++ ls3)),
adamc@119 521 (a0,
adamc@119 522 match pf'0 in (_ = ls) return (fhlist B ls) with
adamc@119 523 | refl_equal => f
adamc@119 524 end) =
adamc@119 525 match pf0 in (_ = ls) return (fhlist B ls) with
adamc@119 526 | refl_equal => (a0, f)
adamc@119 527 end
adamc@218 528
adamc@119 529 ]]
adamc@119 530
adamc@119 531 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@119 532
adamc@119 533 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@119 534
adamc@119 535 rewrite app_ass.
adamc@119 536 (** [[
adamc@119 537 ============================
adamc@119 538 forall (pf0 : a :: ls1 ++ ls2 ++ ls3 = a :: ls1 ++ ls2 ++ ls3)
adamc@119 539 (pf'0 : ls1 ++ ls2 ++ ls3 = ls1 ++ ls2 ++ ls3)
adamc@119 540 (f : fhlist B (ls1 ++ ls2 ++ ls3)),
adamc@119 541 (a0,
adamc@119 542 match pf'0 in (_ = ls) return (fhlist B ls) with
adamc@119 543 | refl_equal => f
adamc@119 544 end) =
adamc@119 545 match pf0 in (_ = ls) return (fhlist B ls) with
adamc@119 546 | refl_equal => (a0, f)
adamc@119 547 end
adamc@218 548
adamc@119 549 ]]
adamc@119 550
adamc@119 551 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@119 552
adamc@119 553 intros.
adamc@119 554 rewrite (UIP_refl _ _ pf0).
adamc@119 555 rewrite (UIP_refl _ _ pf'0).
adamc@119 556 reflexivity.
adamc@119 557 Qed.
adamc@124 558 (* end thide *)
adamc@119 559 End fhapp.
adamc@120 560
adamc@120 561 Implicit Arguments fhapp [A B ls1 ls2].
adamc@120 562
adamc@120 563
adamc@120 564 (** * Heterogeneous Equality *)
adamc@120 565
adamc@120 566 (** There is another equality predicate, defined in the [JMeq] module of the standard library, implementing %\textit{%#<i>#heterogeneous equality#</i>#%}%. *)
adamc@120 567
adamc@120 568 Print JMeq.
adamc@218 569 (** %\vspace{-.15in}% [[
adamc@120 570 Inductive JMeq (A : Type) (x : A) : forall B : Type, B -> Prop :=
adamc@120 571 JMeq_refl : JMeq x x
adamc@218 572
adamc@120 573 ]]
adamc@120 574
adam@292 575 [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@120 576
adamc@120 577 Infix "==" := JMeq (at level 70, no associativity).
adamc@120 578
adamc@124 579 (* EX: Prove UIP_refl' : forall (A : Type) (x : A) (pf : x = x), pf == refl_equal x *)
adamc@124 580 (* begin thide *)
adamc@121 581 Definition UIP_refl' (A : Type) (x : A) (pf : x = x) : pf == refl_equal x :=
adamc@120 582 match pf return (pf == refl_equal _) with
adamc@120 583 | refl_equal => JMeq_refl _
adamc@120 584 end.
adamc@124 585 (* end thide *)
adamc@120 586
adamc@120 587 (** There is no quick way to write such a proof by tactics, but the underlying proof term that we want is trivial.
adamc@120 588
adamc@271 589 Suppose that we want to use [UIP_refl'] to establish another lemma of the kind we have run into several times so far. *)
adamc@120 590
adamc@120 591 Lemma lemma4 : forall (A : Type) (x : A) (pf : x = x),
adamc@120 592 O = match pf with refl_equal => O end.
adamc@124 593 (* begin thide *)
adamc@121 594 intros; rewrite (UIP_refl' pf); reflexivity.
adamc@120 595 Qed.
adamc@124 596 (* end thide *)
adamc@120 597
adamc@120 598 (** 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@120 599
adamc@120 600 Check JMeq_eq.
adamc@218 601 (** %\vspace{-.15in}% [[
adamc@120 602 JMeq_eq
adamc@120 603 : forall (A : Type) (x y : A), x == y -> x = y
adamc@218 604
adamc@218 605 ]]
adamc@120 606
adamc@218 607 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@120 608
adamc@120 609 We can redo our [fhapp] associativity proof based around [JMeq]. *)
adamc@120 610
adamc@120 611 Section fhapp'.
adamc@120 612 Variable A : Type.
adamc@120 613 Variable B : A -> Type.
adamc@120 614
adamc@120 615 (** This time, the naive theorem statement type-checks. *)
adamc@120 616
adamc@124 617 (* EX: Prove [fhapp] associativity using [JMeq]. *)
adamc@124 618
adamc@124 619 (* begin thide *)
adam@297 620 Theorem fhapp_ass' : forall ls1 ls2 ls3 (hls1 : fhlist B ls1) (hls2 : fhlist B ls2) (hls3 : fhlist B ls3),
adamc@120 621 fhapp hls1 (fhapp hls2 hls3) == fhapp (fhapp hls1 hls2) hls3.
adamc@120 622 induction ls1; crush.
adamc@120 623
adamc@120 624 (** Even better, [crush] discharges the first subgoal automatically. The second subgoal is:
adamc@120 625
adamc@120 626 [[
adamc@120 627 ============================
adam@297 628 (a0, fhapp b (fhapp hls2 hls3)) == (a0, fhapp (fhapp b hls2) hls3)
adamc@218 629
adamc@120 630 ]]
adamc@120 631
adam@297 632 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@120 633
adamc@120 634 [[
adamc@120 635 rewrite IHls1.
adamc@120 636
adamc@120 637 Error: Impossible to unify "fhlist B ((ls1 ++ ?1572) ++ ?1573)" with
adamc@120 638 "fhlist B (ls1 ++ ?1572 ++ ?1573)"
adamc@218 639
adamc@120 640 ]]
adamc@120 641
adam@297 642 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@297 643
adamc@120 644 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@120 645
adamc@120 646 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@120 647
adamc@120 648 generalize (fhapp b (fhapp hls2 hls3))
adamc@120 649 (fhapp (fhapp b hls2) hls3)
adamc@120 650 (IHls1 _ _ b hls2 hls3).
adamc@120 651 (** [[
adamc@120 652 ============================
adamc@120 653 forall (f : fhlist B (ls1 ++ ls2 ++ ls3))
adamc@120 654 (f0 : fhlist B ((ls1 ++ ls2) ++ ls3)), f == f0 -> (a0, f) == (a0, f0)
adamc@218 655
adamc@120 656 ]]
adamc@120 657
adamc@120 658 Now we can rewrite with append associativity, as before. *)
adamc@120 659
adamc@120 660 rewrite app_ass.
adamc@120 661 (** [[
adamc@120 662 ============================
adamc@120 663 forall f f0 : fhlist B (ls1 ++ ls2 ++ ls3), f == f0 -> (a0, f) == (a0, f0)
adamc@218 664
adamc@120 665 ]]
adamc@120 666
adamc@120 667 From this point, the goal is trivial. *)
adamc@120 668
adamc@120 669 intros f f0 H; rewrite H; reflexivity.
adamc@120 670 Qed.
adamc@124 671 (* end thide *)
adamc@120 672 End fhapp'.
adamc@121 673
adamc@121 674
adamc@121 675 (** * Equivalence of Equality Axioms *)
adamc@121 676
adamc@124 677 (* EX: Show that the approaches based on K and JMeq are equivalent logically. *)
adamc@124 678
adamc@124 679 (* begin thide *)
adamc@272 680 (** 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@121 681
adamc@121 682 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@121 683
adamc@121 684 Lemma UIP_refl'' : forall (A : Type) (x : A) (pf : x = x), pf = refl_equal x.
adamc@121 685 intros; rewrite (UIP_refl' pf); reflexivity.
adamc@121 686 Qed.
adamc@121 687
adamc@121 688 (** 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@121 689
adamc@121 690 Definition JMeq' (A : Type) (x : A) (B : Type) (y : B) : Prop :=
adamc@121 691 exists pf : B = A, x = match pf with refl_equal => y end.
adamc@121 692
adamc@121 693 Infix "===" := JMeq' (at level 70, no associativity).
adamc@121 694
adamc@121 695 (** 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@121 696
adamc@121 697 We can easily prove a theorem with the same type as that of the [JMeq_refl] constructor of [JMeq]. *)
adamc@121 698
adamc@121 699 (** remove printing exists *)
adamc@121 700 Theorem JMeq_refl' : forall (A : Type) (x : A), x === x.
adamc@121 701 intros; unfold JMeq'; exists (refl_equal A); reflexivity.
adamc@121 702 Qed.
adamc@121 703
adamc@121 704 (** printing exists $\exists$ *)
adamc@121 705
adamc@121 706 (** 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@121 707
adamc@121 708 Theorem JMeq_eq' : forall (A : Type) (x y : A),
adamc@121 709 x === y -> x = y.
adamc@121 710 unfold JMeq'; intros.
adamc@121 711 (** [[
adamc@121 712 H : exists pf : A = A,
adamc@121 713 x = match pf in (_ = T) return T with
adamc@121 714 | refl_equal => y
adamc@121 715 end
adamc@121 716 ============================
adamc@121 717 x = y
adamc@218 718
adamc@121 719 ]] *)
adamc@121 720
adamc@121 721 destruct H.
adamc@121 722 (** [[
adamc@121 723 x0 : A = A
adamc@121 724 H : x = match x0 in (_ = T) return T with
adamc@121 725 | refl_equal => y
adamc@121 726 end
adamc@121 727 ============================
adamc@121 728 x = y
adamc@218 729
adamc@121 730 ]] *)
adamc@121 731
adamc@121 732 rewrite H.
adamc@121 733 (** [[
adamc@121 734 x0 : A = A
adamc@121 735 ============================
adamc@121 736 match x0 in (_ = T) return T with
adamc@121 737 | refl_equal => y
adamc@121 738 end = y
adamc@218 739
adamc@121 740 ]] *)
adamc@121 741
adamc@121 742 rewrite (UIP_refl _ _ x0); reflexivity.
adamc@121 743 Qed.
adamc@121 744
adam@292 745 (** 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@123 746
adamc@123 747 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@124 748 (* end thide *)
adamc@123 749
adamc@123 750
adamc@123 751 (** * Equality of Functions *)
adamc@123 752
adamc@123 753 (** The following seems like a reasonable theorem to want to hold, and it does hold in set theory. [[
adamc@123 754 Theorem S_eta : S = (fun n => S n).
adamc@218 755
adamc@205 756 ]]
adamc@205 757
adamc@123 758 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@123 759
adamc@124 760 (* begin thide *)
adamc@123 761 Axiom ext_eq : forall A B (f g : A -> B),
adamc@123 762 (forall x, f x = g x)
adamc@123 763 -> f = g.
adamc@124 764 (* end thide *)
adamc@123 765
adamc@123 766 (** 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@123 767
adamc@123 768 Theorem S_eta : S = (fun n => S n).
adamc@124 769 (* begin thide *)
adamc@123 770 apply ext_eq; reflexivity.
adamc@123 771 Qed.
adamc@124 772 (* end thide *)
adamc@123 773
adam@292 774 (** The same axiom can help us prove equality of types, where we need to %``%#"#reason under quantifiers.#"#%''% *)
adamc@123 775
adamc@123 776 Theorem forall_eq : (forall x : nat, match x with
adamc@123 777 | O => True
adamc@123 778 | S _ => True
adamc@123 779 end)
adamc@123 780 = (forall _ : nat, True).
adamc@123 781
adamc@123 782 (** There are no immediate opportunities to apply [ext_eq], but we can use [change] to fix that. *)
adamc@123 783
adamc@124 784 (* begin thide *)
adamc@123 785 change ((forall x : nat, (fun x => match x with
adamc@123 786 | 0 => True
adamc@123 787 | S _ => True
adamc@123 788 end) x) = (nat -> True)).
adamc@123 789 rewrite (ext_eq (fun x => match x with
adamc@123 790 | 0 => True
adamc@123 791 | S _ => True
adamc@123 792 end) (fun _ => True)).
adamc@123 793 (** [[
adamc@123 794 2 subgoals
adamc@123 795
adamc@123 796 ============================
adamc@123 797 (nat -> True) = (nat -> True)
adamc@123 798
adamc@123 799 subgoal 2 is:
adamc@123 800 forall x : nat, match x with
adamc@123 801 | 0 => True
adamc@123 802 | S _ => True
adamc@123 803 end = True
adamc@123 804 ]] *)
adamc@123 805
adamc@123 806 reflexivity.
adamc@123 807
adamc@123 808 destruct x; constructor.
adamc@123 809 Qed.
adamc@124 810 (* end thide *)
adamc@127 811
adamc@127 812
adamc@127 813 (** * Exercises *)
adamc@127 814
adamc@127 815 (** %\begin{enumerate}%#<ol>#
adamc@127 816
adamc@127 817 %\item%#<li># Implement and prove correct a substitution function for simply-typed lambda calculus. In particular:
adamc@127 818 %\begin{enumerate}%#<ol>#
adamc@127 819 %\item%#<li># Define a datatype [type] of lambda types, including just booleans and function types.#</li>#
adamc@127 820 %\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@127 821 %\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@292 822 %\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@127 823 %\item%#<li># Prove that [subst] preserves program meanings. That is, prove
adamc@127 824 [[
adamc@127 825 forall t' ts t (e : exp (t' :: ts) t) (e' : exp ts t') (s : hlist typeDenote ts),
adamc@127 826 expDenote (subst e e') s = expDenote e (expDenote e' s ::: s)
adamc@218 827
adamc@127 828 ]]
adam@292 829 where [:::] is an infix operator for heterogeneous %``%#"#cons#"#%''% that is defined in the book's [DepList] module.#</li>#
adamc@127 830 #</ol>#%\end{enumerate}%
adamc@127 831 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@127 832 %\begin{enumerate}%#<ol>#
adamc@127 833 %\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@292 834 %\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@292 835 %\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@127 836 %\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@127 837 %\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@127 838 %\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@127 839 %\item%#<li># Now [subst] should be a one-liner, defined in terms of [subst'].#</li>#
adamc@127 840 %\item%#<li># Prove a correctness theorem for each auxiliary function, leading up to the proof of [subst] correctness.#</li>#
adam@292 841 %\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@127 842 %\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@292 843 %\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@292 844 %\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@127 845 #</ol>#%\end{enumerate}%
adamc@127 846 #</li>#
adamc@127 847
adamc@127 848 #</ol>#%\end{enumerate}% *)