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Easy direction of Intensional
author Adam Chlipala <adamc@hcoop.net>
date Wed, 16 Dec 2009 15:54:50 -0500
parents c8508d277a00
children aa3c054afce0
rev   line source
adamc@118 1 (* Copyright (c) 2008, Adam Chlipala
adamc@118 2 *
adamc@118 3 * This work is licensed under a
adamc@118 4 * Creative Commons Attribution-Noncommercial-No Derivative Works 3.0
adamc@118 5 * Unported License.
adamc@118 6 * The license text is available at:
adamc@118 7 * http://creativecommons.org/licenses/by-nc-nd/3.0/
adamc@118 8 *)
adamc@118 9
adamc@118 10 (* begin hide *)
adamc@120 11 Require Import Eqdep JMeq List.
adamc@118 12
adamc@132 13 Require Import Tactics.
adamc@118 14
adamc@118 15 Set Implicit Arguments.
adamc@118 16 (* end hide *)
adamc@118 17
adamc@118 18
adamc@118 19 (** %\chapter{Reasoning About Equality Proofs}% *)
adamc@118 20
adamc@118 21 (** In traditional mathematics, the concept of equality is usually taken as a given. On the other hand, in type theory, equality is a very contentious subject. There are at least three different notions of equality that are important, and researchers are actively investigating new definitions of what it means for two terms to be equal. Even once we fix a notion of equality, there are inevitably tricky issues that arise in proving properties of programs that manipulate equality proofs explicitly. In this chapter, we will focus on design patterns for circumventing these tricky issues, and we will introduce the different notions of equality as they are germane. *)
adamc@118 22
adamc@118 23
adamc@122 24 (** * The Definitional Equality *)
adamc@122 25
adamc@122 26 (** We have seen many examples so far where proof goals follow "by computation." That is, we apply computational reduction rules to reduce the goal to a normal form, at which point it follows trivially. Exactly when this works and when it does not depends on the details of Coq's %\textit{%#<i>#definitional equality#</i>#%}%. This is an untyped binary relation appearing in the formal metatheory of CIC. CIC contains a typing rule allowing the conclusion $E : T$ from the premise $E : T'$ and a proof that $T$ and $T'$ are definitionally equal.
adamc@122 27
adamc@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
adamc@122 83 The final reduction rule is zeta, which replaces a [let] expression by its body with the appropriate term subsituted. *)
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
adamc@122 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, we will introduce effective 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
adamc@218 103 (** 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
adamc@118 162 (** Part of our single remaining subgoal is:
adamc@118 163
adamc@118 164 [[
adamc@118 165 a0 : a = elm
adamc@118 166 ============================
adamc@118 167 match a0 in (_ = a2) return (C a2) with
adamc@118 168 | refl_equal => f a1
adamc@118 169 end = f match a0 in (_ = a2) return (B a2) with
adamc@118 170 | refl_equal => a1
adamc@118 171 end
adamc@218 172
adamc@118 173 ]]
adamc@118 174
adamc@118 175 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@118 176
adamc@118 177 [[
adamc@118 178 destruct a0.
adamc@118 179
adamc@118 180 User error: Cannot solve a second-order unification problem
adamc@218 181
adamc@118 182 ]]
adamc@118 183
adamc@118 184 This is one of Coq's standard error messages for informing us that its heuristics for attempting an instance of an undecidable problem about dependent typing have failed. We might try to nudge things in the right direction by stating the lemma that we believe makes the conclusion trivial.
adamc@118 185
adamc@118 186 [[
adamc@118 187 assert (a0 = refl_equal _).
adamc@118 188
adamc@118 189 The term "refl_equal ?98" has type "?98 = ?98"
adamc@118 190 while it is expected to have type "a = elm"
adamc@218 191
adamc@118 192 ]]
adamc@118 193
adamc@118 194 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@118 195
adamc@118 196 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@118 197
adamc@118 198 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@118 199
adamc@118 200 case a0.
adamc@118 201 (** [[
adamc@118 202 ============================
adamc@118 203 f a1 = f a1
adamc@218 204
adamc@118 205 ]]
adamc@118 206
adamc@118 207 It seems that [destruct] was trying to be too smart for its own good. *)
adamc@118 208
adamc@118 209 reflexivity.
adamc@118 210 Qed.
adamc@124 211 (* end thide *)
adamc@118 212
adamc@118 213 (** It will be helpful to examine the proof terms generated by this sort of strategy. A simpler example illustrates what is going on. *)
adamc@118 214
adamc@118 215 Lemma lemma1 : forall x (pf : x = elm), O = match pf with refl_equal => O end.
adamc@124 216 (* begin thide *)
adamc@118 217 simple destruct pf; reflexivity.
adamc@118 218 Qed.
adamc@124 219 (* end thide *)
adamc@118 220
adamc@118 221 (** [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@118 222
adamc@118 223 Print lemma1.
adamc@218 224 (** %\vspace{-.15in}% [[
adamc@118 225 lemma1 =
adamc@118 226 fun (x : A) (pf : x = elm) =>
adamc@118 227 match pf as e in (_ = y) return (0 = match e with
adamc@118 228 | refl_equal => 0
adamc@118 229 end) with
adamc@118 230 | refl_equal => refl_equal 0
adamc@118 231 end
adamc@118 232 : forall (x : A) (pf : x = elm), 0 = match pf with
adamc@118 233 | refl_equal => 0
adamc@118 234 end
adamc@218 235
adamc@118 236 ]]
adamc@118 237
adamc@118 238 Using what we know about shorthands for [match] annotations, we can write this proof in shorter form manually. *)
adamc@118 239
adamc@124 240 (* begin thide *)
adamc@118 241 Definition lemma1' :=
adamc@118 242 fun (x : A) (pf : x = elm) =>
adamc@118 243 match pf return (0 = match pf with
adamc@118 244 | refl_equal => 0
adamc@118 245 end) with
adamc@118 246 | refl_equal => refl_equal 0
adamc@118 247 end.
adamc@124 248 (* end thide *)
adamc@118 249
adamc@118 250 (** Surprisingly, what seems at first like a %\textit{%#<i>#simpler#</i>#%}% lemma is harder to prove. *)
adamc@118 251
adamc@118 252 Lemma lemma2 : forall (x : A) (pf : x = x), O = match pf with refl_equal => O end.
adamc@124 253 (* begin thide *)
adamc@118 254 (** [[
adamc@118 255 simple destruct pf.
adamc@205 256
adamc@118 257 User error: Cannot solve a second-order unification problem
adamc@218 258
adamc@118 259 ]] *)
adamc@118 260 Abort.
adamc@118 261
adamc@118 262 (** Nonetheless, we can adapt the last manual proof to handle this theorem. *)
adamc@118 263
adamc@124 264 (* begin thide *)
adamc@124 265 Definition lemma2 :=
adamc@118 266 fun (x : A) (pf : x = x) =>
adamc@118 267 match pf return (0 = match pf with
adamc@118 268 | refl_equal => 0
adamc@118 269 end) with
adamc@118 270 | refl_equal => refl_equal 0
adamc@118 271 end.
adamc@124 272 (* end thide *)
adamc@118 273
adamc@118 274 (** We can try to prove a lemma that would simplify proofs of many facts like [lemma2]: *)
adamc@118 275
adamc@118 276 Lemma lemma3 : forall (x : A) (pf : x = x), pf = refl_equal x.
adamc@124 277 (* begin thide *)
adamc@118 278 (** [[
adamc@118 279 simple destruct pf.
adamc@205 280
adamc@118 281 User error: Cannot solve a second-order unification problem
adamc@118 282 ]] *)
adamc@218 283
adamc@118 284 Abort.
adamc@118 285
adamc@118 286 (** This time, even our manual attempt fails.
adamc@118 287
adamc@118 288 [[
adamc@118 289 Definition lemma3' :=
adamc@118 290 fun (x : A) (pf : x = x) =>
adamc@118 291 match pf as pf' in (_ = x') return (pf' = refl_equal x') with
adamc@118 292 | refl_equal => refl_equal _
adamc@118 293 end.
adamc@118 294
adamc@118 295 The term "refl_equal x'" has type "x' = x'" while it is expected to have type
adamc@118 296 "x = x'"
adamc@218 297
adamc@118 298 ]]
adamc@118 299
adamc@118 300 The type error comes from our [return] annotation. In that annotation, the [as]-bound variable [pf'] has type [x = x'], refering to the [in]-bound variable [x']. To do a dependent [match], we %\textit{%#<i>#must#</i>#%}% choose a fresh name for the second argument of [eq]. We are just as constrained to use the "real" value [x] for the first argument. Thus, within the [return] clause, the proof we are matching on %\textit{%#<i>#must#</i>#%}% equate two non-matching terms, which makes it impossible to equate that proof with reflexivity.
adamc@118 301
adamc@118 302 Nonetheless, it turns out that, with one catch, we %\textit{%#<i>#can#</i>#%}% prove this lemma. *)
adamc@118 303
adamc@118 304 Lemma lemma3 : forall (x : A) (pf : x = x), pf = refl_equal x.
adamc@118 305 intros; apply UIP_refl.
adamc@118 306 Qed.
adamc@118 307
adamc@118 308 Check UIP_refl.
adamc@218 309 (** %\vspace{-.15in}% [[
adamc@118 310 UIP_refl
adamc@118 311 : forall (U : Type) (x : U) (p : x = x), p = refl_equal x
adamc@218 312
adamc@118 313 ]]
adamc@118 314
adamc@118 315 [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@118 316
adamc@118 317 Print eq_rect_eq.
adamc@218 318 (** %\vspace{-.15in}% [[
adamc@118 319 eq_rect_eq =
adamc@118 320 fun U : Type => Eq_rect_eq.eq_rect_eq U
adamc@118 321 : forall (U : Type) (p : U) (Q : U -> Type) (x : Q p) (h : p = p),
adamc@118 322 x = eq_rect p Q x p h
adamc@218 323
adamc@118 324 ]]
adamc@118 325
adamc@118 326 [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@118 327
adamc@118 328 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@118 329
adamc@118 330 This axiom is equivalent to another that is more commonly known and mentioned in type theory circles. *)
adamc@118 331
adamc@118 332 Print Streicher_K.
adamc@124 333 (* end thide *)
adamc@218 334 (** %\vspace{-.15in}% [[
adamc@118 335 Streicher_K =
adamc@118 336 fun U : Type => UIP_refl__Streicher_K U (UIP_refl U)
adamc@118 337 : forall (U : Type) (x : U) (P : x = x -> Prop),
adamc@118 338 P (refl_equal x) -> forall p : x = x, P p
adamc@218 339
adamc@118 340 ]]
adamc@118 341
adamc@118 342 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@118 343
adamc@118 344 End fhlist_map.
adamc@118 345
adamc@119 346
adamc@119 347 (** * Type-Casts in Theorem Statements *)
adamc@119 348
adamc@119 349 (** 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@119 350
adamc@119 351 Section fhapp.
adamc@119 352 Variable A : Type.
adamc@119 353 Variable B : A -> Type.
adamc@119 354
adamc@218 355 Fixpoint fhapp (ls1 ls2 : list A)
adamc@119 356 : fhlist B ls1 -> fhlist B ls2 -> fhlist B (ls1 ++ ls2) :=
adamc@218 357 match ls1 with
adamc@119 358 | nil => fun _ hls2 => hls2
adamc@119 359 | _ :: _ => fun hls1 hls2 => (fst hls1, fhapp _ _ (snd hls1) hls2)
adamc@119 360 end.
adamc@119 361
adamc@119 362 Implicit Arguments fhapp [ls1 ls2].
adamc@119 363
adamc@124 364 (* EX: Prove that fhapp is associative. *)
adamc@124 365 (* begin thide *)
adamc@124 366
adamc@119 367 (** We might like to prove that [fhapp] is associative.
adamc@119 368
adamc@119 369 [[
adamc@119 370 Theorem fhapp_ass : forall ls1 ls2 ls3
adamc@119 371 (hls1 : fhlist B ls1) (hls2 : fhlist B ls2) (hls3 : fhlist B ls3),
adamc@119 372 fhapp hls1 (fhapp hls2 hls3) = fhapp (fhapp hls1 hls2) hls3.
adamc@119 373
adamc@119 374 The term
adamc@119 375 "fhapp (ls1:=ls1 ++ ls2) (ls2:=ls3) (fhapp (ls1:=ls1) (ls2:=ls2) hls1 hls2)
adamc@119 376 hls3" has type "fhlist B ((ls1 ++ ls2) ++ ls3)"
adamc@119 377 while it is expected to have type "fhlist B (ls1 ++ ls2 ++ ls3)"
adamc@218 378
adamc@119 379 ]]
adamc@119 380
adamc@119 381 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@119 382
adamc@119 383 Theorem fhapp_ass : forall ls1 ls2 ls3
adamc@119 384 (pf : (ls1 ++ ls2) ++ ls3 = ls1 ++ (ls2 ++ ls3))
adamc@119 385 (hls1 : fhlist B ls1) (hls2 : fhlist B ls2) (hls3 : fhlist B ls3),
adamc@119 386 fhapp hls1 (fhapp hls2 hls3)
adamc@119 387 = match pf in (_ = ls) return fhlist _ ls with
adamc@119 388 | refl_equal => fhapp (fhapp hls1 hls2) hls3
adamc@119 389 end.
adamc@119 390 induction ls1; crush.
adamc@119 391
adamc@119 392 (** The first remaining subgoal looks trivial enough:
adamc@119 393
adamc@119 394 [[
adamc@119 395 ============================
adamc@119 396 fhapp (ls1:=ls2) (ls2:=ls3) hls2 hls3 =
adamc@119 397 match pf in (_ = ls) return (fhlist B ls) with
adamc@119 398 | refl_equal => fhapp (ls1:=ls2) (ls2:=ls3) hls2 hls3
adamc@119 399 end
adamc@218 400
adamc@119 401 ]]
adamc@119 402
adamc@119 403 We can try what worked in previous examples.
adamc@119 404
adamc@119 405 [[
adamc@119 406 case pf.
adamc@119 407
adamc@119 408 User error: Cannot solve a second-order unification problem
adamc@218 409
adamc@119 410 ]]
adamc@119 411
adamc@119 412 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@119 413
adamc@119 414 rewrite (UIP_refl _ _ pf).
adamc@119 415 (** [[
adamc@119 416 ============================
adamc@119 417 fhapp (ls1:=ls2) (ls2:=ls3) hls2 hls3 =
adamc@119 418 fhapp (ls1:=ls2) (ls2:=ls3) hls2 hls3
adamc@218 419
adamc@119 420 ]] *)
adamc@119 421
adamc@119 422 reflexivity.
adamc@119 423
adamc@119 424 (** Our second subgoal is trickier.
adamc@119 425
adamc@119 426 [[
adamc@119 427 pf : a :: (ls1 ++ ls2) ++ ls3 = a :: ls1 ++ ls2 ++ ls3
adamc@119 428 ============================
adamc@119 429 (a0,
adamc@119 430 fhapp (ls1:=ls1) (ls2:=ls2 ++ ls3) b
adamc@119 431 (fhapp (ls1:=ls2) (ls2:=ls3) hls2 hls3)) =
adamc@119 432 match pf in (_ = ls) return (fhlist B ls) with
adamc@119 433 | refl_equal =>
adamc@119 434 (a0,
adamc@119 435 fhapp (ls1:=ls1 ++ ls2) (ls2:=ls3)
adamc@119 436 (fhapp (ls1:=ls1) (ls2:=ls2) b hls2) hls3)
adamc@119 437 end
adamc@119 438
adamc@119 439 rewrite (UIP_refl _ _ pf).
adamc@119 440
adamc@119 441 The term "pf" has type "a :: (ls1 ++ ls2) ++ ls3 = a :: ls1 ++ ls2 ++ ls3"
adamc@119 442 while it is expected to have type "?556 = ?556"
adamc@218 443
adamc@119 444 ]]
adamc@119 445
adamc@119 446 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@119 447
adamc@119 448 injection pf; intro pf'.
adamc@119 449 (** [[
adamc@119 450 pf : a :: (ls1 ++ ls2) ++ ls3 = a :: ls1 ++ ls2 ++ ls3
adamc@119 451 pf' : (ls1 ++ ls2) ++ ls3 = ls1 ++ ls2 ++ ls3
adamc@119 452 ============================
adamc@119 453 (a0,
adamc@119 454 fhapp (ls1:=ls1) (ls2:=ls2 ++ ls3) b
adamc@119 455 (fhapp (ls1:=ls2) (ls2:=ls3) hls2 hls3)) =
adamc@119 456 match pf in (_ = ls) return (fhlist B ls) with
adamc@119 457 | refl_equal =>
adamc@119 458 (a0,
adamc@119 459 fhapp (ls1:=ls1 ++ ls2) (ls2:=ls3)
adamc@119 460 (fhapp (ls1:=ls1) (ls2:=ls2) b hls2) hls3)
adamc@119 461 end
adamc@218 462
adamc@119 463 ]]
adamc@119 464
adamc@119 465 Now we can rewrite using the inductive hypothesis. *)
adamc@119 466
adamc@119 467 rewrite (IHls1 _ _ pf').
adamc@119 468 (** [[
adamc@119 469 ============================
adamc@119 470 (a0,
adamc@119 471 match pf' in (_ = ls) return (fhlist B ls) with
adamc@119 472 | refl_equal =>
adamc@119 473 fhapp (ls1:=ls1 ++ ls2) (ls2:=ls3)
adamc@119 474 (fhapp (ls1:=ls1) (ls2:=ls2) b hls2) hls3
adamc@119 475 end) =
adamc@119 476 match pf in (_ = ls) return (fhlist B ls) with
adamc@119 477 | refl_equal =>
adamc@119 478 (a0,
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@218 482
adamc@119 483 ]]
adamc@119 484
adamc@119 485 We have made an important bit of progress, as now only a single call to [fhapp] appears in the conclusion. Trying case analysis on our proofs still will not work, but there is a move we can make to enable it. Not only does just one call to [fhapp] matter to us now, but it also %\textit{%#<i>#does not matter what the result of the call is#</i>#%}%. In other words, the subgoal should remain true if we replace this [fhapp] call with a fresh variable. The [generalize] tactic helps us do exactly that. *)
adamc@119 486
adamc@119 487 generalize (fhapp (fhapp b hls2) hls3).
adamc@119 488 (** [[
adamc@119 489 forall f : fhlist B ((ls1 ++ ls2) ++ ls3),
adamc@119 490 (a0,
adamc@119 491 match pf' in (_ = ls) return (fhlist B ls) with
adamc@119 492 | refl_equal => f
adamc@119 493 end) =
adamc@119 494 match pf in (_ = ls) return (fhlist B ls) with
adamc@119 495 | refl_equal => (a0, f)
adamc@119 496 end
adamc@218 497
adamc@119 498 ]]
adamc@119 499
adamc@119 500 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@119 501
adamc@119 502 In this case, it is helpful to generalize over our two proofs as well. *)
adamc@119 503
adamc@119 504 generalize pf pf'.
adamc@119 505 (** [[
adamc@119 506 forall (pf0 : a :: (ls1 ++ ls2) ++ ls3 = a :: ls1 ++ ls2 ++ ls3)
adamc@119 507 (pf'0 : (ls1 ++ ls2) ++ ls3 = ls1 ++ ls2 ++ ls3)
adamc@119 508 (f : fhlist B ((ls1 ++ ls2) ++ ls3)),
adamc@119 509 (a0,
adamc@119 510 match pf'0 in (_ = ls) return (fhlist B ls) with
adamc@119 511 | refl_equal => f
adamc@119 512 end) =
adamc@119 513 match pf0 in (_ = ls) return (fhlist B ls) with
adamc@119 514 | refl_equal => (a0, f)
adamc@119 515 end
adamc@218 516
adamc@119 517 ]]
adamc@119 518
adamc@119 519 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@119 520
adamc@119 521 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@119 522
adamc@119 523 rewrite app_ass.
adamc@119 524 (** [[
adamc@119 525 ============================
adamc@119 526 forall (pf0 : a :: ls1 ++ ls2 ++ ls3 = a :: ls1 ++ ls2 ++ ls3)
adamc@119 527 (pf'0 : ls1 ++ ls2 ++ ls3 = ls1 ++ ls2 ++ ls3)
adamc@119 528 (f : fhlist B (ls1 ++ ls2 ++ ls3)),
adamc@119 529 (a0,
adamc@119 530 match pf'0 in (_ = ls) return (fhlist B ls) with
adamc@119 531 | refl_equal => f
adamc@119 532 end) =
adamc@119 533 match pf0 in (_ = ls) return (fhlist B ls) with
adamc@119 534 | refl_equal => (a0, f)
adamc@119 535 end
adamc@218 536
adamc@119 537 ]]
adamc@119 538
adamc@119 539 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@119 540
adamc@119 541 intros.
adamc@119 542 rewrite (UIP_refl _ _ pf0).
adamc@119 543 rewrite (UIP_refl _ _ pf'0).
adamc@119 544 reflexivity.
adamc@119 545 Qed.
adamc@124 546 (* end thide *)
adamc@119 547 End fhapp.
adamc@120 548
adamc@120 549 Implicit Arguments fhapp [A B ls1 ls2].
adamc@120 550
adamc@120 551
adamc@120 552 (** * Heterogeneous Equality *)
adamc@120 553
adamc@120 554 (** There is another equality predicate, defined in the [JMeq] module of the standard library, implementing %\textit{%#<i>#heterogeneous equality#</i>#%}%. *)
adamc@120 555
adamc@120 556 Print JMeq.
adamc@218 557 (** %\vspace{-.15in}% [[
adamc@120 558 Inductive JMeq (A : Type) (x : A) : forall B : Type, B -> Prop :=
adamc@120 559 JMeq_refl : JMeq x x
adamc@218 560
adamc@120 561 ]]
adamc@120 562
adamc@120 563 [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@120 564
adamc@120 565 Infix "==" := JMeq (at level 70, no associativity).
adamc@120 566
adamc@124 567 (* EX: Prove UIP_refl' : forall (A : Type) (x : A) (pf : x = x), pf == refl_equal x *)
adamc@124 568 (* begin thide *)
adamc@121 569 Definition UIP_refl' (A : Type) (x : A) (pf : x = x) : pf == refl_equal x :=
adamc@120 570 match pf return (pf == refl_equal _) with
adamc@120 571 | refl_equal => JMeq_refl _
adamc@120 572 end.
adamc@124 573 (* end thide *)
adamc@120 574
adamc@120 575 (** There is no quick way to write such a proof by tactics, but the underlying proof term that we want is trivial.
adamc@120 576
adamc@121 577 Suppose that we want to use [UIP_refl'] to establish another lemma of the kind of we have run into several times so far. *)
adamc@120 578
adamc@120 579 Lemma lemma4 : forall (A : Type) (x : A) (pf : x = x),
adamc@120 580 O = match pf with refl_equal => O end.
adamc@124 581 (* begin thide *)
adamc@121 582 intros; rewrite (UIP_refl' pf); reflexivity.
adamc@120 583 Qed.
adamc@124 584 (* end thide *)
adamc@120 585
adamc@120 586 (** 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@120 587
adamc@120 588 Check JMeq_eq.
adamc@218 589 (** %\vspace{-.15in}% [[
adamc@120 590 JMeq_eq
adamc@120 591 : forall (A : Type) (x y : A), x == y -> x = y
adamc@218 592
adamc@218 593 ]]
adamc@120 594
adamc@218 595 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@120 596
adamc@120 597 We can redo our [fhapp] associativity proof based around [JMeq]. *)
adamc@120 598
adamc@120 599 Section fhapp'.
adamc@120 600 Variable A : Type.
adamc@120 601 Variable B : A -> Type.
adamc@120 602
adamc@120 603 (** This time, the naive theorem statement type-checks. *)
adamc@120 604
adamc@124 605 (* EX: Prove [fhapp] associativity using [JMeq]. *)
adamc@124 606
adamc@124 607 (* begin thide *)
adamc@120 608 Theorem fhapp_ass' : forall ls1 ls2 ls3
adamc@120 609 (hls1 : fhlist B ls1) (hls2 : fhlist B ls2) (hls3 : fhlist B ls3),
adamc@120 610 fhapp hls1 (fhapp hls2 hls3) == fhapp (fhapp hls1 hls2) hls3.
adamc@120 611 induction ls1; crush.
adamc@120 612
adamc@120 613 (** Even better, [crush] discharges the first subgoal automatically. The second subgoal is:
adamc@120 614
adamc@120 615 [[
adamc@120 616 ============================
adamc@120 617 (a0,
adamc@120 618 fhapp (B:=B) (ls1:=ls1) (ls2:=ls2 ++ ls3) b
adamc@120 619 (fhapp (B:=B) (ls1:=ls2) (ls2:=ls3) hls2 hls3)) ==
adamc@120 620 (a0,
adamc@120 621 fhapp (B:=B) (ls1:=ls1 ++ ls2) (ls2:=ls3)
adamc@120 622 (fhapp (B:=B) (ls1:=ls1) (ls2:=ls2) b hls2) hls3)
adamc@218 623
adamc@120 624 ]]
adamc@120 625
adamc@120 626 It looks like one rewrite with the inductive hypothesis should be enough to make the goal trivial.
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
adamc@120 636 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@120 637
adamc@120 638 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@120 639
adamc@120 640 generalize (fhapp b (fhapp hls2 hls3))
adamc@120 641 (fhapp (fhapp b hls2) hls3)
adamc@120 642 (IHls1 _ _ b hls2 hls3).
adamc@120 643 (** [[
adamc@120 644 ============================
adamc@120 645 forall (f : fhlist B (ls1 ++ ls2 ++ ls3))
adamc@120 646 (f0 : fhlist B ((ls1 ++ ls2) ++ ls3)), f == f0 -> (a0, f) == (a0, f0)
adamc@218 647
adamc@120 648 ]]
adamc@120 649
adamc@120 650 Now we can rewrite with append associativity, as before. *)
adamc@120 651
adamc@120 652 rewrite app_ass.
adamc@120 653 (** [[
adamc@120 654 ============================
adamc@120 655 forall f f0 : fhlist B (ls1 ++ ls2 ++ ls3), f == f0 -> (a0, f) == (a0, f0)
adamc@218 656
adamc@120 657 ]]
adamc@120 658
adamc@120 659 From this point, the goal is trivial. *)
adamc@120 660
adamc@120 661 intros f f0 H; rewrite H; reflexivity.
adamc@120 662 Qed.
adamc@124 663 (* end thide *)
adamc@120 664 End fhapp'.
adamc@121 665
adamc@121 666
adamc@121 667 (** * Equivalence of Equality Axioms *)
adamc@121 668
adamc@124 669 (* EX: Show that the approaches based on K and JMeq are equivalent logically. *)
adamc@124 670
adamc@124 671 (* begin thide *)
adamc@121 672 (** Assuming axioms (like axiom K and [JMeq_eq]) is a hazardous business. The due diligence associated with it is necessarily global in scope, since two axioms may be consistent alone but inconsistent together. It turns out that all of the major axioms proposed for reasoning about equality in Coq are logically equivalent, so that we only need to pick one to assert without proof. In this section, we demonstrate this by showing how each the previous two sections' approaches reduces to the other logically.
adamc@121 673
adamc@121 674 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@121 675
adamc@121 676 Lemma UIP_refl'' : forall (A : Type) (x : A) (pf : x = x), pf = refl_equal x.
adamc@121 677 intros; rewrite (UIP_refl' pf); reflexivity.
adamc@121 678 Qed.
adamc@121 679
adamc@121 680 (** 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@121 681
adamc@121 682 Definition JMeq' (A : Type) (x : A) (B : Type) (y : B) : Prop :=
adamc@121 683 exists pf : B = A, x = match pf with refl_equal => y end.
adamc@121 684
adamc@121 685 Infix "===" := JMeq' (at level 70, no associativity).
adamc@121 686
adamc@121 687 (** 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@121 688
adamc@121 689 We can easily prove a theorem with the same type as that of the [JMeq_refl] constructor of [JMeq]. *)
adamc@121 690
adamc@121 691 (** remove printing exists *)
adamc@121 692 Theorem JMeq_refl' : forall (A : Type) (x : A), x === x.
adamc@121 693 intros; unfold JMeq'; exists (refl_equal A); reflexivity.
adamc@121 694 Qed.
adamc@121 695
adamc@121 696 (** printing exists $\exists$ *)
adamc@121 697
adamc@121 698 (** 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@121 699
adamc@121 700 Theorem JMeq_eq' : forall (A : Type) (x y : A),
adamc@121 701 x === y -> x = y.
adamc@121 702 unfold JMeq'; intros.
adamc@121 703 (** [[
adamc@121 704 H : exists pf : A = A,
adamc@121 705 x = match pf in (_ = T) return T with
adamc@121 706 | refl_equal => y
adamc@121 707 end
adamc@121 708 ============================
adamc@121 709 x = y
adamc@218 710
adamc@121 711 ]] *)
adamc@121 712
adamc@121 713 destruct H.
adamc@121 714 (** [[
adamc@121 715 x0 : A = A
adamc@121 716 H : x = match x0 in (_ = T) return T with
adamc@121 717 | refl_equal => y
adamc@121 718 end
adamc@121 719 ============================
adamc@121 720 x = y
adamc@218 721
adamc@121 722 ]] *)
adamc@121 723
adamc@121 724 rewrite H.
adamc@121 725 (** [[
adamc@121 726 x0 : A = A
adamc@121 727 ============================
adamc@121 728 match x0 in (_ = T) return T with
adamc@121 729 | refl_equal => y
adamc@121 730 end = y
adamc@218 731
adamc@121 732 ]] *)
adamc@121 733
adamc@121 734 rewrite (UIP_refl _ _ x0); reflexivity.
adamc@121 735 Qed.
adamc@121 736
adamc@123 737 (** We see that, in a very formal sense, we are free to switch back and forth between the two styles of proofs about equality proofs. One style may be more convenient than the other for some proofs, but we can always intercovert between our results. The style that does not use heterogeneous equality may be preferable in cases where many results do not require the tricks of this chapter, since then the use of axioms is avoided altogether for the simple cases, and a wider audience will be able to follow those "simple" proofs. On the other hand, heterogeneous equality often makes for shorter and more readable theorem statements.
adamc@123 738
adamc@123 739 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@124 740 (* end thide *)
adamc@123 741
adamc@123 742
adamc@123 743 (** * Equality of Functions *)
adamc@123 744
adamc@123 745 (** The following seems like a reasonable theorem to want to hold, and it does hold in set theory. [[
adamc@123 746 Theorem S_eta : S = (fun n => S n).
adamc@218 747
adamc@205 748 ]]
adamc@205 749
adamc@123 750 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@123 751
adamc@124 752 (* begin thide *)
adamc@123 753 Axiom ext_eq : forall A B (f g : A -> B),
adamc@123 754 (forall x, f x = g x)
adamc@123 755 -> f = g.
adamc@124 756 (* end thide *)
adamc@123 757
adamc@123 758 (** 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@123 759
adamc@123 760 Theorem S_eta : S = (fun n => S n).
adamc@124 761 (* begin thide *)
adamc@123 762 apply ext_eq; reflexivity.
adamc@123 763 Qed.
adamc@124 764 (* end thide *)
adamc@123 765
adamc@123 766 (** The same axiom can help us prove equality of types, where we need to "reason under quantifiers." *)
adamc@123 767
adamc@123 768 Theorem forall_eq : (forall x : nat, match x with
adamc@123 769 | O => True
adamc@123 770 | S _ => True
adamc@123 771 end)
adamc@123 772 = (forall _ : nat, True).
adamc@123 773
adamc@123 774 (** There are no immediate opportunities to apply [ext_eq], but we can use [change] to fix that. *)
adamc@123 775
adamc@124 776 (* begin thide *)
adamc@123 777 change ((forall x : nat, (fun x => match x with
adamc@123 778 | 0 => True
adamc@123 779 | S _ => True
adamc@123 780 end) x) = (nat -> True)).
adamc@123 781 rewrite (ext_eq (fun x => match x with
adamc@123 782 | 0 => True
adamc@123 783 | S _ => True
adamc@123 784 end) (fun _ => True)).
adamc@123 785 (** [[
adamc@123 786 2 subgoals
adamc@123 787
adamc@123 788 ============================
adamc@123 789 (nat -> True) = (nat -> True)
adamc@123 790
adamc@123 791 subgoal 2 is:
adamc@123 792 forall x : nat, match x with
adamc@123 793 | 0 => True
adamc@123 794 | S _ => True
adamc@123 795 end = True
adamc@123 796 ]] *)
adamc@123 797
adamc@123 798 reflexivity.
adamc@123 799
adamc@123 800 destruct x; constructor.
adamc@123 801 Qed.
adamc@124 802 (* end thide *)
adamc@127 803
adamc@127 804
adamc@127 805 (** * Exercises *)
adamc@127 806
adamc@127 807 (** %\begin{enumerate}%#<ol>#
adamc@127 808
adamc@127 809 %\item%#<li># Implement and prove correct a substitution function for simply-typed lambda calculus. In particular:
adamc@127 810 %\begin{enumerate}%#<ol>#
adamc@127 811 %\item%#<li># Define a datatype [type] of lambda types, including just booleans and function types.#</li>#
adamc@127 812 %\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@127 813 %\item%#<li># Implement a definitional interpreter for [exp]s, by way of a recursive function over expressions and substitutions for free variables, like in the related example from the last chapter.#</li>#
adamc@127 814 %\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@127 815 %\item%#<li># Prove that [subst] preserves program meanings. That is, prove
adamc@127 816 [[
adamc@127 817 forall t' ts t (e : exp (t' :: ts) t) (e' : exp ts t') (s : hlist typeDenote ts),
adamc@127 818 expDenote (subst e e') s = expDenote e (expDenote e' s ::: s)
adamc@218 819
adamc@127 820 ]]
adamc@127 821 where [:::] is an infix operator for heterogeneous "cons" that is defined in the book's [DepList] module.#</li>#
adamc@127 822 #</ol>#%\end{enumerate}%
adamc@127 823 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@127 824 %\begin{enumerate}%#<ol>#
adamc@127 825 %\item%#<li># The [DepList] module will be useful. You can get the standard dependent list definitions there, instead of copying-and-pasting from the last chapter. It is worth reading the source for that module over, since it defines some new helpful functions and notations that we did not use last chapter.#</li>#
adamc@127 826 %\item%#<li># Define a recursive function [liftVar : forall ts1 ts2 t t', member t (ts1 ++ ts2) -> member t (ts1 ++ t' :: ts2)]. This function should "lift" a de Bruijn variable so that its type refers to a new variable inserted somewhere in the index list.#</li>#
adamc@127 827 %\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@127 828 %\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@127 829 %\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@127 830 %\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@127 831 %\item%#<li># Now [subst] should be a one-liner, defined in terms of [subst'].#</li>#
adamc@127 832 %\item%#<li># Prove a correctness theorem for each auxiliary function, leading up to the proof of [subst] correctness.#</li>#
adamc@127 833 %\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@127 834 %\item%#<li># A variant of the [ext_eq] axiom from the end of this chapter is available in the book module [Axioms], and you will probably want to use it in the [lift'] and [subst'] correctness proofs.#</li>#
adamc@127 835 %\item%#<li># The [change] tactic should come in handy in the proofs about [lift] and [subst], where you want to introduce "extraneous" list concatenations with [nil] to match the forms of earlier theorems.#</li>#
adamc@127 836 %\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@127 837 #</ol>#%\end{enumerate}%
adamc@127 838 #</li>#
adamc@127 839
adamc@127 840 #</ol>#%\end{enumerate}% *)