annotate src/Predicates.v @ 310:cf67b7d6ceac

Fix .hgignore and check in .bib
author Adam Chlipala <adam@chlipala.net>
date Mon, 29 Aug 2011 09:47:39 -0400
parents 7b38729be069
children d5787b70cf48
rev   line source
adam@281 1 (* Copyright (c) 2008-2010, Adam Chlipala
adamc@45 2 *
adamc@45 3 * This work is licensed under a
adamc@45 4 * Creative Commons Attribution-Noncommercial-No Derivative Works 3.0
adamc@45 5 * Unported License.
adamc@45 6 * The license text is available at:
adamc@45 7 * http://creativecommons.org/licenses/by-nc-nd/3.0/
adamc@45 8 *)
adamc@45 9
adamc@45 10 (* begin hide *)
adamc@45 11 Require Import List.
adamc@45 12
adamc@45 13 Require Import Tactics.
adamc@45 14
adamc@45 15 Set Implicit Arguments.
adamc@45 16 (* end hide *)
adamc@45 17
adamc@45 18
adamc@45 19 (** %\chapter{Inductive Predicates}% *)
adamc@45 20
adam@292 21 (** The so-called %``%#"#Curry-Howard Correspondence#"#%''% states a formal connection between functional programs and mathematical proofs. In the last chapter, we snuck in a first introduction to this subject in Coq. Witness the close similarity between the types [unit] and [True] from the standard library: *)
adamc@45 22
adamc@45 23 Print unit.
adamc@209 24 (** %\vspace{-.15in}% [[
adamc@209 25 Inductive unit : Set := tt : unit
adam@302 26 ]]
adam@302 27 *)
adamc@45 28
adamc@45 29 Print True.
adamc@209 30 (** %\vspace{-.15in}% [[
adamc@209 31 Inductive True : Prop := I : True
adam@302 32 ]]
adam@302 33 *)
adamc@45 34
adamc@45 35 (** Recall that [unit] is the type with only one value, and [True] is the proposition that always holds. Despite this superficial difference between the two concepts, in both cases we can use the same inductive definition mechanism. The connection goes further than this. We see that we arrive at the definition of [True] by replacing [unit] by [True], [tt] by [I], and [Set] by [Prop]. The first two of these differences are superficial changes of names, while the third difference is the crucial one for separating programs from proofs. A term [T] of type [Set] is a type of programs, and a term of type [T] is a program. A term [T] of type [Prop] is a logical proposition, and its proofs are of type [T].
adamc@45 36
adamc@45 37 [unit] has one value, [tt]. [True] has one proof, [I]. Why distinguish between these two types? Many people who have read about Curry-Howard in an abstract context and not put it to use in proof engineering answer that the two types in fact %\textit{%#<i>#should not#</i>#%}% be distinguished. There is a certain aesthetic appeal to this point of view, but I want to argue that it is best to treat Curry-Howard very loosely in practical proving. There are Coq-specific reasons for preferring the distinction, involving efficient compilation and avoidance of paradoxes in the presence of classical math, but I will argue that there is a more general principle that should lead us to avoid conflating programming and proving.
adamc@45 38
adamc@45 39 The essence of the argument is roughly this: to an engineer, not all functions of type [A -> B] are created equal, but all proofs of a proposition [P -> Q] are. This idea is known as %\textit{%#<i>#proof irrelevance#</i>#%}%, and its formalizations in logics prevent us from distinguishing between alternate proofs of the same proposition. Proof irrelevance is compatible with, but not derivable in, Gallina. Apart from this theoretical concern, I will argue that it is most effective to do engineering with Coq by employing different techniques for programs versus proofs. Most of this book is organized around that distinction, describing how to program, by applying standard functional programming techniques in the presence of dependent types; and how to prove, by writing custom Ltac decision procedures.
adamc@45 40
adam@292 41 With that perspective in mind, this chapter is sort of a mirror image of the last chapter, introducing how to define predicates with inductive definitions. We will point out similarities in places, but much of the effective Coq user's bag of tricks is disjoint for predicates versus %``%#"#datatypes.#"#%''% This chapter is also a covert introduction to dependent types, which are the foundation on which interesting inductive predicates are built, though we will rely on tactics to build dependently-typed proof terms for us for now. A future chapter introduces more manual application of dependent types. *)
adamc@45 42
adamc@45 43
adamc@48 44 (** * Propositional Logic *)
adamc@45 45
adamc@45 46 (** Let us begin with a brief tour through the definitions of the connectives for propositional logic. We will work within a Coq section that provides us with a set of propositional variables. In Coq parlance, these are just terms of type [Prop.] *)
adamc@45 47
adamc@45 48 Section Propositional.
adamc@46 49 Variables P Q R : Prop.
adamc@45 50
adamc@45 51 (** In Coq, the most basic propositional connective is implication, written [->], which we have already used in almost every proof. Rather than being defined inductively, implication is built into Coq as the function type constructor.
adamc@45 52
adamc@45 53 We have also already seen the definition of [True]. For a demonstration of a lower-level way of establishing proofs of inductive predicates, we turn to this trivial theorem. *)
adamc@45 54
adamc@45 55 Theorem obvious : True.
adamc@55 56 (* begin thide *)
adamc@45 57 apply I.
adamc@55 58 (* end thide *)
adamc@45 59 Qed.
adamc@45 60
adamc@45 61 (** We may always use the [apply] tactic to take a proof step based on applying a particular constructor of the inductive predicate that we are trying to establish. Sometimes there is only one constructor that could possibly apply, in which case a shortcut is available: *)
adamc@45 62
adamc@55 63 (* begin thide *)
adamc@45 64 Theorem obvious' : True.
adamc@45 65 constructor.
adamc@45 66 Qed.
adamc@45 67
adamc@55 68 (* end thide *)
adamc@55 69
adamc@45 70 (** There is also a predicate [False], which is the Curry-Howard mirror image of [Empty_set] from the last chapter. *)
adamc@45 71
adamc@45 72 Print False.
adamc@209 73 (** %\vspace{-.15in}% [[
adamc@209 74 Inductive False : Prop :=
adamc@209 75
adamc@209 76 ]]
adamc@45 77
adamc@209 78 We can conclude anything from [False], doing case analysis on a proof of [False] in the same way we might do case analysis on, say, a natural number. Since there are no cases to consider, any such case analysis succeeds immediately in proving the goal. *)
adamc@45 79
adamc@45 80 Theorem False_imp : False -> 2 + 2 = 5.
adamc@55 81 (* begin thide *)
adamc@45 82 destruct 1.
adamc@55 83 (* end thide *)
adamc@45 84 Qed.
adamc@45 85
adamc@45 86 (** In a consistent context, we can never build a proof of [False]. In inconsistent contexts that appear in the courses of proofs, it is usually easiest to proceed by demonstrating that inconsistency with an explicit proof of [False]. *)
adamc@45 87
adamc@45 88 Theorem arith_neq : 2 + 2 = 5 -> 9 + 9 = 835.
adamc@55 89 (* begin thide *)
adamc@45 90 intro.
adamc@45 91
adamc@45 92 (** At this point, we have an inconsistent hypothesis [2 + 2 = 5], so the specific conclusion is not important. We use the [elimtype] tactic to state a proposition, telling Coq that we wish to construct a proof of the new proposition and then prove the original goal by case analysis on the structure of the new auxiliary proof. Since [False] has no constructors, [elimtype False] simply leaves us with the obligation to prove [False]. *)
adamc@45 93
adamc@45 94 elimtype False.
adamc@45 95 (** [[
adamc@45 96 H : 2 + 2 = 5
adamc@45 97 ============================
adamc@45 98 False
adamc@209 99
adamc@209 100 ]]
adamc@45 101
adamc@209 102 For now, we will leave the details of this proof about arithmetic to [crush]. *)
adamc@45 103
adamc@45 104 crush.
adamc@55 105 (* end thide *)
adamc@45 106 Qed.
adamc@45 107
adamc@45 108 (** A related notion to [False] is logical negation. *)
adamc@45 109
adamc@45 110 Print not.
adamc@209 111 (** %\vspace{-.15in}% [[
adamc@209 112 not = fun A : Prop => A -> False
adamc@209 113 : Prop -> Prop
adamc@209 114
adamc@209 115 ]]
adamc@45 116
adam@280 117 We see that [not] is just shorthand for implication of [False]. We can use that fact explicitly in proofs. The syntax [~ P] expands to [not P]. *)
adamc@45 118
adamc@45 119 Theorem arith_neq' : ~ (2 + 2 = 5).
adamc@55 120 (* begin thide *)
adamc@45 121 unfold not.
adamc@45 122 (** [[
adamc@45 123 ============================
adamc@45 124 2 + 2 = 5 -> False
adam@302 125 ]]
adam@302 126 *)
adamc@45 127
adamc@45 128 crush.
adamc@55 129 (* end thide *)
adamc@45 130 Qed.
adamc@45 131
adamc@45 132 (** We also have conjunction, which we introduced in the last chapter. *)
adamc@45 133
adamc@45 134 Print and.
adamc@209 135 (** %\vspace{-.15in}% [[
adamc@209 136 Inductive and (A : Prop) (B : Prop) : Prop := conj : A -> B -> A /\ B
adamc@209 137
adamc@209 138 ]]
adamc@209 139
adamc@210 140 The interested reader can check that [and] has a Curry-Howard doppelganger called [prod], the type of pairs. However, it is generally most convenient to reason about conjunction using tactics. An explicit proof of commutativity of [and] illustrates the usual suspects for such tasks. [/\] is an infix shorthand for [and]. *)
adamc@45 141
adamc@45 142 Theorem and_comm : P /\ Q -> Q /\ P.
adamc@209 143
adamc@55 144 (* begin thide *)
adamc@45 145 (** We start by case analysis on the proof of [P /\ Q]. *)
adamc@45 146
adamc@45 147 destruct 1.
adamc@45 148 (** [[
adamc@45 149 H : P
adamc@45 150 H0 : Q
adamc@45 151 ============================
adamc@45 152 Q /\ P
adamc@209 153
adamc@209 154 ]]
adamc@45 155
adamc@209 156 Every proof of a conjunction provides proofs for both conjuncts, so we get a single subgoal reflecting that. We can proceed by splitting this subgoal into a case for each conjunct of [Q /\ P]. *)
adamc@45 157
adamc@45 158 split.
adamc@45 159 (** [[
adamc@45 160 2 subgoals
adamc@45 161
adamc@45 162 H : P
adamc@45 163 H0 : Q
adamc@45 164 ============================
adamc@45 165 Q
adamc@45 166
adamc@45 167 subgoal 2 is:
adamc@45 168 P
adamc@209 169
adamc@209 170 ]]
adamc@45 171
adamc@209 172 In each case, the conclusion is among our hypotheses, so the [assumption] tactic finishes the process. *)
adamc@45 173
adamc@45 174 assumption.
adamc@45 175 assumption.
adamc@55 176 (* end thide *)
adamc@45 177 Qed.
adamc@45 178
adamc@45 179 (** Coq disjunction is called [or] and abbreviated with the infix operator [\/]. *)
adamc@45 180
adamc@45 181 Print or.
adamc@209 182 (** %\vspace{-.15in}% [[
adamc@209 183 Inductive or (A : Prop) (B : Prop) : Prop :=
adamc@209 184 or_introl : A -> A \/ B | or_intror : B -> A \/ B
adamc@209 185
adamc@209 186 ]]
adamc@45 187
adamc@209 188 We see that there are two ways to prove a disjunction: prove the first disjunct or prove the second. The Curry-Howard analogue of this is the Coq [sum] type. We can demonstrate the main tactics here with another proof of commutativity. *)
adamc@45 189
adamc@45 190 Theorem or_comm : P \/ Q -> Q \/ P.
adamc@55 191
adamc@55 192 (* begin thide *)
adamc@45 193 (** As in the proof for [and], we begin with case analysis, though this time we are met by two cases instead of one. *)
adamc@209 194
adamc@45 195 destruct 1.
adamc@45 196 (** [[
adamc@45 197 2 subgoals
adamc@45 198
adamc@45 199 H : P
adamc@45 200 ============================
adamc@45 201 Q \/ P
adamc@45 202
adamc@45 203 subgoal 2 is:
adamc@45 204 Q \/ P
adamc@209 205
adamc@209 206 ]]
adamc@45 207
adamc@209 208 We can see that, in the first subgoal, we want to prove the disjunction by proving its second disjunct. The [right] tactic telegraphs this intent. *)
adamc@209 209
adamc@45 210 right; assumption.
adamc@45 211
adamc@45 212 (** The second subgoal has a symmetric proof.
adamc@45 213
adamc@45 214 [[
adamc@45 215 1 subgoal
adamc@45 216
adamc@45 217 H : Q
adamc@45 218 ============================
adamc@45 219 Q \/ P
adam@302 220 ]]
adam@302 221 *)
adamc@45 222
adamc@45 223 left; assumption.
adamc@55 224 (* end thide *)
adamc@45 225 Qed.
adamc@45 226
adamc@46 227
adamc@46 228 (* begin hide *)
adamc@46 229 (* In-class exercises *)
adamc@46 230
adamc@46 231 Theorem contra : P -> ~P -> R.
adamc@52 232 (* begin thide *)
adamc@52 233 unfold not.
adamc@52 234 intros.
adamc@52 235 elimtype False.
adamc@52 236 apply H0.
adamc@52 237 assumption.
adamc@52 238 (* end thide *)
adamc@46 239 Admitted.
adamc@46 240
adamc@46 241 Theorem and_assoc : (P /\ Q) /\ R -> P /\ (Q /\ R).
adamc@52 242 (* begin thide *)
adamc@52 243 intros.
adamc@52 244 destruct H.
adamc@52 245 destruct H.
adamc@52 246 split.
adamc@52 247 assumption.
adamc@52 248 split.
adamc@52 249 assumption.
adamc@52 250 assumption.
adamc@52 251 (* end thide *)
adamc@46 252 Admitted.
adamc@46 253
adamc@46 254 Theorem or_assoc : (P \/ Q) \/ R -> P \/ (Q \/ R).
adamc@52 255 (* begin thide *)
adamc@52 256 intros.
adamc@52 257 destruct H.
adamc@52 258 destruct H.
adamc@52 259 left.
adamc@52 260 assumption.
adamc@52 261 right.
adamc@52 262 left.
adamc@52 263 assumption.
adamc@52 264 right.
adamc@52 265 right.
adamc@52 266 assumption.
adamc@52 267 (* end thide *)
adamc@46 268 Admitted.
adamc@46 269
adamc@46 270 (* end hide *)
adamc@46 271
adamc@46 272
adam@292 273 (** It would be a shame to have to plod manually through all proofs about propositional logic. Luckily, there is no need. One of the most basic Coq automation tactics is [tauto], which is a complete decision procedure for constructive propositional logic. (More on what %``%#"#constructive#"#%''% means in the next section.) We can use [tauto] to dispatch all of the purely propositional theorems we have proved so far. *)
adamc@46 274
adamc@46 275 Theorem or_comm' : P \/ Q -> Q \/ P.
adamc@55 276 (* begin thide *)
adamc@46 277 tauto.
adamc@55 278 (* end thide *)
adamc@46 279 Qed.
adamc@46 280
adamc@46 281 (** Sometimes propositional reasoning forms important plumbing for the proof of a theorem, but we still need to apply some other smarts about, say, arithmetic. [intuition] is a generalization of [tauto] that proves everything it can using propositional reasoning. When some goals remain, it uses propositional laws to simplify them as far as possible. Consider this example, which uses the list concatenation operator [++] from the standard library. *)
adamc@46 282
adamc@46 283 Theorem arith_comm : forall ls1 ls2 : list nat,
adamc@46 284 length ls1 = length ls2 \/ length ls1 + length ls2 = 6
adamc@46 285 -> length (ls1 ++ ls2) = 6 \/ length ls1 = length ls2.
adamc@55 286 (* begin thide *)
adamc@46 287 intuition.
adamc@46 288
adamc@46 289 (** A lot of the proof structure has been generated for us by [intuition], but the final proof depends on a fact about lists. The remaining subgoal hints at what cleverness we need to inject. *)
adamc@46 290
adamc@46 291 (** [[
adamc@46 292 ls1 : list nat
adamc@46 293 ls2 : list nat
adamc@46 294 H0 : length ls1 + length ls2 = 6
adamc@46 295 ============================
adamc@46 296 length (ls1 ++ ls2) = 6 \/ length ls1 = length ls2
adamc@209 297
adamc@209 298 ]]
adamc@46 299
adamc@209 300 We can see that we need a theorem about lengths of concatenated lists, which we proved last chapter and is also in the standard library. *)
adamc@46 301
adamc@46 302 rewrite app_length.
adamc@46 303 (** [[
adamc@46 304 ls1 : list nat
adamc@46 305 ls2 : list nat
adamc@46 306 H0 : length ls1 + length ls2 = 6
adamc@46 307 ============================
adamc@46 308 length ls1 + length ls2 = 6 \/ length ls1 = length ls2
adamc@209 309
adamc@209 310 ]]
adamc@46 311
adamc@209 312 Now the subgoal follows by purely propositional reasoning. That is, we could replace [length ls1 + length ls2 = 6] with [P] and [length ls1 = length ls2] with [Q] and arrive at a tautology of propositional logic. *)
adamc@46 313
adamc@46 314 tauto.
adamc@55 315 (* end thide *)
adamc@46 316 Qed.
adamc@46 317
adamc@46 318 (** [intuition] is one of the main bits of glue in the implementation of [crush], so, with a little help, we can get a short automated proof of the theorem. *)
adamc@46 319
adamc@55 320 (* begin thide *)
adamc@46 321 Theorem arith_comm' : forall ls1 ls2 : list nat,
adamc@46 322 length ls1 = length ls2 \/ length ls1 + length ls2 = 6
adamc@46 323 -> length (ls1 ++ ls2) = 6 \/ length ls1 = length ls2.
adamc@46 324 Hint Rewrite app_length : cpdt.
adamc@46 325
adamc@46 326 crush.
adamc@46 327 Qed.
adamc@55 328 (* end thide *)
adamc@46 329
adamc@45 330 End Propositional.
adamc@45 331
adamc@46 332
adamc@47 333 (** * What Does It Mean to Be Constructive? *)
adamc@46 334
adamc@47 335 (** One potential point of confusion in the presentation so far is the distinction between [bool] and [Prop]. [bool] is a datatype whose two values are [true] and [false], while [Prop] is a more primitive type that includes among its members [True] and [False]. Why not collapse these two concepts into one, and why must there be more than two states of mathematical truth?
adamc@46 336
adam@292 337 The answer comes from the fact that Coq implements %\textit{%#<i>#constructive#</i>#%}% or %\textit{%#<i>#intuitionistic#</i>#%}% logic, in contrast to the %\textit{%#<i>#classical#</i>#%}% logic that you may be more familiar with. In constructive logic, classical tautologies like [~ ~ P -> P] and [P \/ ~ P] do not always hold. In general, we can only prove these tautologies when [P] is %\textit{%#<i>#decidable#</i>#%}%, in the sense of computability theory. The Curry-Howard encoding that Coq uses for [or] allows us to extract either a proof of [P] or a proof of [~ P] from any proof of [P \/ ~ P]. Since our proofs are just functional programs which we can run, this would give us a decision procedure for the halting problem, where the instantiations of [P] would be formulas like %``%#"#this particular Turing machine halts.#"#%''%
adamc@47 338
adam@292 339 Hence the distinction between [bool] and [Prop]. Programs of type [bool] are computational by construction; we can always run them to determine their results. Many [Prop]s are undecidable, and so we can write more expressive formulas with [Prop]s than with [bool]s, but the inevitable consequence is that we cannot simply %``%#"#run a [Prop] to determine its truth.#"#%''%
adamc@47 340
adamc@47 341 Constructive logic lets us define all of the logical connectives in an aesthetically-appealing way, with orthogonal inductive definitions. That is, each connective is defined independently using a simple, shared mechanism. Constructivity also enables a trick called %\textit{%#<i>#program extraction#</i>#%}%, where we write programs by phrasing them as theorems to be proved. Since our proofs are just functional programs, we can extract executable programs from our final proofs, which we could not do as naturally with classical proofs.
adamc@47 342
adamc@47 343 We will see more about Coq's program extraction facility in a later chapter. However, I think it is worth interjecting another warning at this point, following up on the prior warning about taking the Curry-Howard correspondence too literally. It is possible to write programs by theorem-proving methods in Coq, but hardly anyone does it. It is almost always most useful to maintain the distinction between programs and proofs. If you write a program by proving a theorem, you are likely to run into algorithmic inefficiencies that you introduced in your proof to make it easier to prove. It is a shame to have to worry about such situations while proving tricky theorems, and it is a happy state of affairs that you almost certainly will not need to, with the ideal of extracting programs from proofs being confined mostly to theoretical studies. *)
adamc@48 344
adamc@48 345
adamc@48 346 (** * First-Order Logic *)
adamc@48 347
adam@292 348 (** The [forall] connective of first-order logic, which we have seen in many examples so far, is built into Coq. Getting ahead of ourselves a bit, we can see it as the dependent function type constructor. In fact, implication and universal quantification are just different syntactic shorthands for the same Coq mechanism. A formula [P -> Q] is equivalent to [forall x : P, Q], where [x] does not appear in [Q]. That is, the %``%#"#real#"#%''% type of the implication says %``%#"#for every proof of [P], there exists a proof of [Q].#"#%''%
adamc@48 349
adamc@48 350 Existential quantification is defined in the standard library. *)
adamc@48 351
adamc@48 352 Print ex.
adamc@209 353 (** %\vspace{-.15in}% [[
adamc@209 354 Inductive ex (A : Type) (P : A -> Prop) : Prop :=
adamc@209 355 ex_intro : forall x : A, P x -> ex P
adamc@209 356
adamc@209 357 ]]
adamc@48 358
adamc@209 359 [ex] is parameterized by the type [A] that we quantify over, and by a predicate [P] over [A]s. We prove an existential by exhibiting some [x] of type [A], along with a proof of [P x]. As usual, there are tactics that save us from worrying about the low-level details most of the time. We use the equality operator [=], which, depending on the settings in which they learned logic, different people will say either is or is not part of first-order logic. For our purposes, it is. *)
adamc@48 360
adamc@48 361 Theorem exist1 : exists x : nat, x + 1 = 2.
adamc@55 362 (* begin thide *)
adamc@67 363 (** remove printing exists *)
adam@292 364 (** We can start this proof with a tactic [exists], which should not be confused with the formula constructor shorthand of the same name. (In the PDF version of this document, the reverse 'E' appears instead of the text %``%#"#exists#"#%''% in formulas.) *)
adamc@209 365
adamc@48 366 exists 1.
adamc@48 367
adamc@209 368 (** The conclusion is replaced with a version using the existential witness that we announced.
adamc@48 369
adamc@209 370 [[
adamc@48 371 ============================
adamc@48 372 1 + 1 = 2
adam@302 373 ]]
adam@302 374 *)
adamc@48 375
adamc@48 376 reflexivity.
adamc@55 377 (* end thide *)
adamc@48 378 Qed.
adamc@48 379
adamc@48 380 (** printing exists $\exists$ *)
adamc@48 381
adamc@48 382 (** We can also use tactics to reason about existential hypotheses. *)
adamc@48 383
adamc@48 384 Theorem exist2 : forall n m : nat, (exists x : nat, n + x = m) -> n <= m.
adamc@55 385 (* begin thide *)
adamc@48 386 (** We start by case analysis on the proof of the existential fact. *)
adamc@209 387
adamc@48 388 destruct 1.
adamc@48 389 (** [[
adamc@48 390 n : nat
adamc@48 391 m : nat
adamc@48 392 x : nat
adamc@48 393 H : n + x = m
adamc@48 394 ============================
adamc@48 395 n <= m
adamc@209 396
adamc@209 397 ]]
adamc@48 398
adamc@209 399 The goal has been replaced by a form where there is a new free variable [x], and where we have a new hypothesis that the body of the existential holds with [x] substituted for the old bound variable. From here, the proof is just about arithmetic and is easy to automate. *)
adamc@48 400
adamc@48 401 crush.
adamc@55 402 (* end thide *)
adamc@48 403 Qed.
adamc@48 404
adamc@48 405
adamc@48 406 (* begin hide *)
adamc@48 407 (* In-class exercises *)
adamc@48 408
adamc@48 409 Theorem forall_exists_commute : forall (A B : Type) (P : A -> B -> Prop),
adamc@48 410 (exists x : A, forall y : B, P x y) -> (forall y : B, exists x : A, P x y).
adamc@52 411 (* begin thide *)
adamc@52 412 intros.
adamc@52 413 destruct H.
adamc@52 414 exists x.
adamc@52 415 apply H.
adamc@52 416 (* end thide *)
adamc@48 417 Admitted.
adamc@48 418
adamc@48 419 (* end hide *)
adamc@48 420
adamc@48 421
adamc@48 422 (** The tactic [intuition] has a first-order cousin called [firstorder]. [firstorder] proves many formulas when only first-order reasoning is needed, and it tries to perform first-order simplifications in any case. First-order reasoning is much harder than propositional reasoning, so [firstorder] is much more likely than [intuition] to get stuck in a way that makes it run for long enough to be useless. *)
adamc@49 423
adamc@49 424
adamc@49 425 (** * Predicates with Implicit Equality *)
adamc@49 426
adamc@49 427 (** We start our exploration of a more complicated class of predicates with a simple example: an alternative way of characterizing when a natural number is zero. *)
adamc@49 428
adamc@49 429 Inductive isZero : nat -> Prop :=
adamc@49 430 | IsZero : isZero 0.
adamc@49 431
adamc@49 432 Theorem isZero_zero : isZero 0.
adamc@55 433 (* begin thide *)
adamc@49 434 constructor.
adamc@55 435 (* end thide *)
adamc@49 436 Qed.
adamc@49 437
adamc@49 438 (** We can call [isZero] a %\textit{%#<i>#judgment#</i>#%}%, in the sense often used in the semantics of programming languages. Judgments are typically defined in the style of %\textit{%#<i>#natural deduction#</i>#%}%, where we write a number of %\textit{%#<i>#inference rules#</i>#%}% with premises appearing above a solid line and a conclusion appearing below the line. In this example, the sole constructor [IsZero] of [isZero] can be thought of as the single inference rule for deducing [isZero], with nothing above the line and [isZero 0] below it. The proof of [isZero_zero] demonstrates how we can apply an inference rule.
adamc@49 439
adamc@49 440 The definition of [isZero] differs in an important way from all of the other inductive definitions that we have seen in this and the previous chapter. Instead of writing just [Set] or [Prop] after the colon, here we write [nat -> Prop]. We saw examples of parameterized types like [list], but there the parameters appeared with names %\textit{%#<i>#before#</i>#%}% the colon. Every constructor of a parameterized inductive type must have a range type that uses the same parameter, whereas the form we use here enables us to use different arguments to the type for different constructors.
adamc@49 441
adamc@49 442 For instance, [isZero] forces its argument to be [0]. We can see that the concept of equality is somehow implicit in the inductive definition mechanism. The way this is accomplished is similar to the way that logic variables are used in Prolog, and it is a very powerful mechanism that forms a foundation for formalizing all of mathematics. In fact, though it is natural to think of inductive types as folding in the functionality of equality, in Coq, the true situation is reversed, with equality defined as just another inductive type! *)
adamc@49 443
adamc@49 444 Print eq.
adamc@209 445 (** %\vspace{-.15in}% [[
adamc@209 446 Inductive eq (A : Type) (x : A) : A -> Prop := refl_equal : x = x
adamc@209 447
adamc@209 448 ]]
adamc@49 449
adamc@209 450 [eq] is the type we get behind the scenes when uses of infix [=] are expanded. We see that [eq] has both a parameter [x] that is fixed and an extra unnamed argument of the same type. The type of [eq] allows us to state any equalities, even those that are provably false. However, examining the type of equality's sole constructor [refl_equal], we see that we can only %\textit{%#<i>#prove#</i>#%}% equality when its two arguments are syntactically equal. This definition turns out to capture all of the basic properties of equality, and the equality-manipulating tactics that we have seen so far, like [reflexivity] and [rewrite], are implemented treating [eq] as just another inductive type with a well-chosen definition.
adamc@49 451
adamc@49 452 Returning to the example of [isZero], we can see how to make use of hypotheses that use this predicate. *)
adamc@49 453
adamc@49 454 Theorem isZero_plus : forall n m : nat, isZero m -> n + m = n.
adamc@55 455 (* begin thide *)
adamc@49 456 (** We want to proceed by cases on the proof of the assumption about [isZero]. *)
adamc@209 457
adamc@49 458 destruct 1.
adamc@49 459 (** [[
adamc@49 460 n : nat
adamc@49 461 ============================
adamc@49 462 n + 0 = n
adamc@209 463
adamc@209 464 ]]
adamc@49 465
adamc@209 466 Since [isZero] has only one constructor, we are presented with only one subgoal. The argument [m] to [isZero] is replaced with that type's argument from the single constructor [IsZero]. From this point, the proof is trivial. *)
adamc@49 467
adamc@49 468 crush.
adamc@55 469 (* end thide *)
adamc@49 470 Qed.
adamc@49 471
adamc@49 472 (** Another example seems at first like it should admit an analogous proof, but in fact provides a demonstration of one of the most basic gotchas of Coq proving. *)
adamc@49 473
adamc@49 474 Theorem isZero_contra : isZero 1 -> False.
adamc@55 475 (* begin thide *)
adamc@49 476 (** Let us try a proof by cases on the assumption, as in the last proof. *)
adamc@209 477
adamc@49 478 destruct 1.
adamc@49 479 (** [[
adamc@49 480 ============================
adamc@49 481 False
adamc@209 482
adamc@209 483 ]]
adamc@49 484
adamc@209 485 It seems that case analysis has not helped us much at all! Our sole hypothesis disappears, leaving us, if anything, worse off than we were before. What went wrong? We have met an important restriction in tactics like [destruct] and [induction] when applied to types with arguments. If the arguments are not already free variables, they will be replaced by new free variables internally before doing the case analysis or induction. Since the argument [1] to [isZero] is replaced by a fresh variable, we lose the crucial fact that it is not equal to [0].
adamc@49 486
adam@292 487 Why does Coq use this restriction? We will discuss the issue in detail in a future chapter, when we see the dependently-typed programming techniques that would allow us to write this proof term manually. For now, we just say that the algorithmic problem of %``%#"#logically complete case analysis#"#%''% is undecidable when phrased in Coq's logic. A few tactics and design patterns that we will present in this chapter suffice in almost all cases. For the current example, what we want is a tactic called [inversion], which corresponds to the concept of inversion that is frequently used with natural deduction proof systems. *)
adamc@49 488
adamc@49 489 Undo.
adamc@49 490 inversion 1.
adamc@55 491 (* end thide *)
adamc@49 492 Qed.
adamc@49 493
adamc@49 494 (** What does [inversion] do? Think of it as a version of [destruct] that does its best to take advantage of the structure of arguments to inductive types. In this case, [inversion] completed the proof immediately, because it was able to detect that we were using [isZero] with an impossible argument.
adamc@49 495
adamc@49 496 Sometimes using [destruct] when you should have used [inversion] can lead to confusing results. To illustrate, consider an alternate proof attempt for the last theorem. *)
adamc@49 497
adamc@49 498 Theorem isZero_contra' : isZero 1 -> 2 + 2 = 5.
adamc@49 499 destruct 1.
adamc@49 500 (** [[
adamc@49 501 ============================
adamc@49 502 1 + 1 = 4
adamc@209 503
adamc@209 504 ]]
adamc@49 505
adam@280 506 What on earth happened here? Internally, [destruct] replaced [1] with a fresh variable, and, trying to be helpful, it also replaced the occurrence of [1] within the unary representation of each number in the goal. This has the net effect of decrementing each of these numbers. *)
adamc@209 507
adamc@49 508 Abort.
adamc@49 509
adam@280 510 (** To see more clearly what is happening, we can consider the type of [isZero]'s induction principle. *)
adam@280 511
adam@280 512 Check isZero_ind.
adam@280 513 (** %\vspace{-.15in}% [[
adam@280 514 isZero_ind
adam@280 515 : forall P : nat -> Prop, P 0 -> forall n : nat, isZero n -> P n
adam@280 516
adam@280 517 ]]
adam@280 518
adam@280 519 In our last proof script, [destruct] chose to instantiate [P] as [fun n => S n + S n = S (S (S (S n)))]. You can verify for yourself that this specialization of the principle applies to the goal and that the hypothesis [P 0] then matches the subgoal we saw generated. If you are doing a proof and encounter a strange transmutation like this, there is a good chance that you should go back and replace a use of [destruct] with [inversion]. *)
adam@280 520
adamc@49 521
adamc@49 522 (* begin hide *)
adamc@49 523 (* In-class exercises *)
adamc@49 524
adamc@49 525 (* EX: Define an inductive type capturing when a list has exactly two elements. Prove that your predicate does not hold of the empty list, and prove that, whenever it holds of a list, the length of that list is two. *)
adamc@49 526
adamc@52 527 (* begin thide *)
adamc@52 528 Section twoEls.
adamc@52 529 Variable A : Type.
adamc@52 530
adamc@52 531 Inductive twoEls : list A -> Prop :=
adamc@52 532 | TwoEls : forall x y, twoEls (x :: y :: nil).
adamc@52 533
adamc@52 534 Theorem twoEls_nil : twoEls nil -> False.
adamc@52 535 inversion 1.
adamc@52 536 Qed.
adamc@52 537
adamc@52 538 Theorem twoEls_two : forall ls, twoEls ls -> length ls = 2.
adamc@52 539 inversion 1.
adamc@52 540 reflexivity.
adamc@52 541 Qed.
adamc@52 542 End twoEls.
adamc@52 543 (* end thide *)
adamc@52 544
adamc@49 545 (* end hide *)
adamc@49 546
adamc@50 547
adamc@50 548 (** * Recursive Predicates *)
adamc@50 549
adamc@50 550 (** We have already seen all of the ingredients we need to build interesting recursive predicates, like this predicate capturing even-ness. *)
adamc@50 551
adamc@50 552 Inductive even : nat -> Prop :=
adamc@50 553 | EvenO : even O
adamc@50 554 | EvenSS : forall n, even n -> even (S (S n)).
adamc@50 555
adamc@50 556 (** Think of [even] as another judgment defined by natural deduction rules. [EvenO] is a rule with nothing above the line and [even O] below the line, and [EvenSS] is a rule with [even n] above the line and [even (S (S n))] below.
adamc@50 557
adamc@50 558 The proof techniques of the last section are easily adapted. *)
adamc@50 559
adamc@50 560 Theorem even_0 : even 0.
adamc@55 561 (* begin thide *)
adamc@50 562 constructor.
adamc@55 563 (* end thide *)
adamc@50 564 Qed.
adamc@50 565
adamc@50 566 Theorem even_4 : even 4.
adamc@55 567 (* begin thide *)
adamc@50 568 constructor; constructor; constructor.
adamc@55 569 (* end thide *)
adamc@50 570 Qed.
adamc@50 571
adamc@50 572 (** It is not hard to see that sequences of constructor applications like the above can get tedious. We can avoid them using Coq's hint facility. *)
adamc@50 573
adamc@55 574 (* begin thide *)
adamc@50 575 Hint Constructors even.
adamc@50 576
adamc@50 577 Theorem even_4' : even 4.
adamc@50 578 auto.
adamc@50 579 Qed.
adamc@50 580
adamc@55 581 (* end thide *)
adamc@55 582
adamc@50 583 Theorem even_1_contra : even 1 -> False.
adamc@55 584 (* begin thide *)
adamc@50 585 inversion 1.
adamc@55 586 (* end thide *)
adamc@50 587 Qed.
adamc@50 588
adamc@50 589 Theorem even_3_contra : even 3 -> False.
adamc@55 590 (* begin thide *)
adamc@50 591 inversion 1.
adamc@50 592 (** [[
adamc@50 593 H : even 3
adamc@50 594 n : nat
adamc@50 595 H1 : even 1
adamc@50 596 H0 : n = 1
adamc@50 597 ============================
adamc@50 598 False
adamc@209 599
adamc@209 600 ]]
adamc@50 601
adamc@209 602 [inversion] can be a little overzealous at times, as we can see here with the introduction of the unused variable [n] and an equality hypothesis about it. For more complicated predicates, though, adding such assumptions is critical to dealing with the undecidability of general inversion. *)
adamc@50 603
adamc@50 604 inversion H1.
adamc@55 605 (* end thide *)
adamc@50 606 Qed.
adamc@50 607
adamc@50 608 (** We can also do inductive proofs about [even]. *)
adamc@50 609
adamc@50 610 Theorem even_plus : forall n m, even n -> even m -> even (n + m).
adamc@55 611 (* begin thide *)
adamc@50 612 (** It seems a reasonable first choice to proceed by induction on [n]. *)
adamc@209 613
adamc@50 614 induction n; crush.
adamc@50 615 (** [[
adamc@50 616 n : nat
adamc@50 617 IHn : forall m : nat, even n -> even m -> even (n + m)
adamc@50 618 m : nat
adamc@50 619 H : even (S n)
adamc@50 620 H0 : even m
adamc@50 621 ============================
adamc@50 622 even (S (n + m))
adamc@209 623
adamc@209 624 ]]
adamc@50 625
adamc@209 626 We will need to use the hypotheses [H] and [H0] somehow. The most natural choice is to invert [H]. *)
adamc@50 627
adamc@50 628 inversion H.
adamc@50 629 (** [[
adamc@50 630 n : nat
adamc@50 631 IHn : forall m : nat, even n -> even m -> even (n + m)
adamc@50 632 m : nat
adamc@50 633 H : even (S n)
adamc@50 634 H0 : even m
adamc@50 635 n0 : nat
adamc@50 636 H2 : even n0
adamc@50 637 H1 : S n0 = n
adamc@50 638 ============================
adamc@50 639 even (S (S n0 + m))
adamc@209 640
adamc@209 641 ]]
adamc@50 642
adamc@209 643 Simplifying the conclusion brings us to a point where we can apply a constructor. *)
adamc@209 644
adamc@50 645 simpl.
adamc@50 646 (** [[
adamc@50 647 ============================
adamc@50 648 even (S (S (n0 + m)))
adam@302 649 ]]
adam@302 650 *)
adamc@50 651
adamc@50 652 constructor.
adamc@50 653 (** [[
adamc@50 654 ============================
adamc@50 655 even (n0 + m)
adamc@209 656
adamc@209 657 ]]
adamc@50 658
adamc@209 659 At this point, we would like to apply the inductive hypothesis, which is:
adamc@209 660
adamc@209 661 [[
adamc@50 662
adamc@50 663 IHn : forall m : nat, even n -> even m -> even (n + m)
adamc@209 664
adamc@209 665 ]]
adamc@50 666
adam@292 667 Unfortunately, the goal mentions [n0] where it would need to mention [n] to match [IHn]. We could keep looking for a way to finish this proof from here, but it turns out that we can make our lives much easier by changing our basic strategy. Instead of inducting on the structure of [n], we should induct %\textit{%#<i>#on the structure of one of the [even] proofs#</i>#%}%. This technique is commonly called %\textit{%#<i>#rule induction#</i>#%}% in programming language semantics. In the setting of Coq, we have already seen how predicates are defined using the same inductive type mechanism as datatypes, so the fundamental unity of rule induction with %``%#"#normal#"#%''% induction is apparent. *)
adamc@50 668
adamc@50 669 Restart.
adamc@50 670
adamc@50 671 induction 1.
adamc@50 672 (** [[
adamc@50 673 m : nat
adamc@50 674 ============================
adamc@50 675 even m -> even (0 + m)
adamc@50 676
adamc@50 677 subgoal 2 is:
adamc@50 678 even m -> even (S (S n) + m)
adamc@209 679
adamc@209 680 ]]
adamc@50 681
adamc@209 682 The first case is easily discharged by [crush], based on the hint we added earlier to try the constructors of [even]. *)
adamc@50 683
adamc@50 684 crush.
adamc@50 685
adamc@50 686 (** Now we focus on the second case: *)
adamc@209 687
adamc@50 688 intro.
adamc@50 689
adamc@50 690 (** [[
adamc@50 691 m : nat
adamc@50 692 n : nat
adamc@50 693 H : even n
adamc@50 694 IHeven : even m -> even (n + m)
adamc@50 695 H0 : even m
adamc@50 696 ============================
adamc@50 697 even (S (S n) + m)
adamc@209 698
adamc@209 699 ]]
adamc@50 700
adamc@209 701 We simplify and apply a constructor, as in our last proof attempt. *)
adamc@50 702
adamc@50 703 simpl; constructor.
adamc@50 704 (** [[
adamc@50 705 ============================
adamc@50 706 even (n + m)
adamc@209 707
adamc@209 708 ]]
adamc@50 709
adamc@209 710 Now we have an exact match with our inductive hypothesis, and the remainder of the proof is trivial. *)
adamc@50 711
adamc@50 712 apply IHeven; assumption.
adamc@50 713
adamc@50 714 (** In fact, [crush] can handle all of the details of the proof once we declare the induction strategy. *)
adamc@50 715
adamc@50 716 Restart.
adamc@50 717 induction 1; crush.
adamc@55 718 (* end thide *)
adamc@50 719 Qed.
adamc@50 720
adamc@50 721 (** Induction on recursive predicates has similar pitfalls to those we encountered with inversion in the last section. *)
adamc@50 722
adamc@50 723 Theorem even_contra : forall n, even (S (n + n)) -> False.
adamc@55 724 (* begin thide *)
adamc@50 725 induction 1.
adamc@50 726 (** [[
adamc@50 727 n : nat
adamc@50 728 ============================
adamc@50 729 False
adamc@50 730
adamc@50 731 subgoal 2 is:
adamc@50 732 False
adamc@209 733
adamc@209 734 ]]
adamc@50 735
adam@280 736 We are already sunk trying to prove the first subgoal, since the argument to [even] was replaced by a fresh variable internally. This time, we find it easier to prove this theorem by way of a lemma. Instead of trusting [induction] to replace expressions with fresh variables, we do it ourselves, explicitly adding the appropriate equalities as new assumptions. *)
adamc@209 737
adamc@50 738 Abort.
adamc@50 739
adamc@50 740 Lemma even_contra' : forall n', even n' -> forall n, n' = S (n + n) -> False.
adamc@50 741 induction 1; crush.
adamc@50 742
adamc@54 743 (** At this point, it is useful to consider all cases of [n] and [n0] being zero or nonzero. Only one of these cases has any trickiness to it. *)
adamc@209 744
adamc@50 745 destruct n; destruct n0; crush.
adamc@50 746
adamc@50 747 (** [[
adamc@50 748 n : nat
adamc@50 749 H : even (S n)
adamc@50 750 IHeven : forall n0 : nat, S n = S (n0 + n0) -> False
adamc@50 751 n0 : nat
adamc@50 752 H0 : S n = n0 + S n0
adamc@50 753 ============================
adamc@50 754 False
adamc@209 755
adamc@209 756 ]]
adamc@50 757
adam@280 758 At this point it is useful to use a theorem from the standard library, which we also proved with a different name in the last chapter. We can search for a theorem that allows us to rewrite terms of the form [x + S y]. *)
adamc@209 759
adam@280 760 SearchRewrite (_ + S _).
adamc@209 761 (** %\vspace{-.15in}% [[
adam@280 762 plus_n_Sm : forall n m : nat, S (n + m) = n + S m
adam@302 763 ]]
adam@302 764 *)
adamc@50 765
adamc@50 766 rewrite <- plus_n_Sm in H0.
adamc@50 767
adamc@50 768 (** The induction hypothesis lets us complete the proof. *)
adamc@209 769
adamc@50 770 apply IHeven with n0; assumption.
adamc@50 771
adamc@202 772 (** As usual, we can rewrite the proof to avoid referencing any locally-generated names, which makes our proof script more readable and more robust to changes in the theorem statement. We use the notation [<-] to request a hint that does right-to-left rewriting, just like we can with the [rewrite] tactic. *)
adamc@209 773
adamc@209 774 Restart.
adamc@50 775 Hint Rewrite <- plus_n_Sm : cpdt.
adamc@50 776
adamc@50 777 induction 1; crush;
adamc@50 778 match goal with
adamc@50 779 | [ H : S ?N = ?N0 + ?N0 |- _ ] => destruct N; destruct N0
adamc@50 780 end; crush; eauto.
adamc@50 781 Qed.
adamc@50 782
adamc@50 783 (** We write the proof in a way that avoids the use of local variable or hypothesis names, using the [match] tactic form to do pattern-matching on the goal. We use unification variables prefixed by question marks in the pattern, and we take advantage of the possibility to mention a unification variable twice in one pattern, to enforce equality between occurrences. The hint to rewrite with [plus_n_Sm] in a particular direction saves us from having to figure out the right place to apply that theorem, and we also take critical advantage of a new tactic, [eauto].
adamc@50 784
adam@292 785 [crush] uses the tactic [intuition], which, when it runs out of tricks to try using only propositional logic, by default tries the tactic [auto], which we saw in an earlier example. [auto] attempts Prolog-style logic programming, searching through all proof trees up to a certain depth that are built only out of hints that have been registered with [Hint] commands. Compared to Prolog, [auto] places an important restriction: it never introduces new unification variables during search. That is, every time a rule is applied during proof search, all of its arguments must be deducible by studying the form of the goal. [eauto] relaxes this restriction, at the cost of possibly exponentially greater running time. In this particular case, we know that [eauto] has only a small space of proofs to search, so it makes sense to run it. It is common in effectively-automated Coq proofs to see a bag of standard tactics applied to pick off the %``%#"#easy#"#%''% subgoals, finishing with [eauto] to handle the tricky parts that can benefit from ad-hoc exhaustive search.
adamc@50 786
adamc@50 787 The original theorem now follows trivially from our lemma. *)
adamc@50 788
adamc@50 789 Theorem even_contra : forall n, even (S (n + n)) -> False.
adamc@52 790 intros; eapply even_contra'; eauto.
adamc@50 791 Qed.
adamc@52 792
adamc@52 793 (** We use a variant [eapply] of [apply] which has the same relationship to [apply] as [eauto] has to [auto]. [apply] only succeeds if all arguments to the rule being used can be determined from the form of the goal, whereas [eapply] will introduce unification variables for undetermined arguments. [eauto] is able to determine the right values for those unification variables.
adamc@52 794
adamc@52 795 By considering an alternate attempt at proving the lemma, we can see another common pitfall of inductive proofs in Coq. Imagine that we had tried to prove [even_contra'] with all of the [forall] quantifiers moved to the front of the lemma statement. *)
adamc@52 796
adamc@52 797 Lemma even_contra'' : forall n' n, even n' -> n' = S (n + n) -> False.
adamc@52 798 induction 1; crush;
adamc@52 799 match goal with
adamc@52 800 | [ H : S ?N = ?N0 + ?N0 |- _ ] => destruct N; destruct N0
adamc@52 801 end; crush; eauto.
adamc@52 802
adamc@209 803 (** One subgoal remains:
adamc@52 804
adamc@209 805 [[
adamc@52 806 n : nat
adamc@52 807 H : even (S (n + n))
adamc@52 808 IHeven : S (n + n) = S (S (S (n + n))) -> False
adamc@52 809 ============================
adamc@52 810 False
adamc@209 811
adamc@209 812 ]]
adamc@52 813
adamc@209 814 We are out of luck here. The inductive hypothesis is trivially true, since its assumption is false. In the version of this proof that succeeded, [IHeven] had an explicit quantification over [n]. This is because the quantification of [n] %\textit{%#<i>#appeared after the thing we are inducting on#</i>#%}% in the theorem statement. In general, quantified variables and hypotheses that appear before the induction object in the theorem statement stay fixed throughout the inductive proof. Variables and hypotheses that are quantified after the induction object may be varied explicitly in uses of inductive hypotheses.
adamc@52 815
adamc@52 816 Why should Coq implement [induction] this way? One answer is that it avoids burdening this basic tactic with additional heuristic smarts, but that is not the whole picture. Imagine that [induction] analyzed dependencies among variables and reordered quantifiers to preserve as much freedom as possible in later uses of inductive hypotheses. This could make the inductive hypotheses more complex, which could in turn cause particular automation machinery to fail when it would have succeeded before. In general, we want to avoid quantifiers in our proofs whenever we can, and that goal is furthered by the refactoring that the [induction] tactic forces us to do. *)
adamc@55 817 (* end thide *)
adamc@209 818
adamc@51 819 Abort.
adamc@51 820
adamc@52 821
adamc@52 822 (* begin hide *)
adamc@52 823 (* In-class exercises *)
adamc@52 824
adam@292 825 (* EX: Define a type [prop] of simple boolean formulas made up only of truth, falsehood, binary conjunction, and binary disjunction. Define an inductive predicate [holds] that captures when [prop]s are valid, and define a predicate [falseFree] that captures when a [prop] does not contain the %``%#"#false#"#%''% formula. Prove that every false-free [prop] is valid. *)
adamc@52 826
adamc@52 827 (* begin thide *)
adamc@52 828 Inductive prop : Set :=
adamc@52 829 | Tru : prop
adamc@52 830 | Fals : prop
adamc@52 831 | And : prop -> prop -> prop
adamc@52 832 | Or : prop -> prop -> prop.
adamc@52 833
adamc@52 834 Inductive holds : prop -> Prop :=
adamc@52 835 | HTru : holds Tru
adamc@52 836 | HAnd : forall p1 p2, holds p1 -> holds p2 -> holds (And p1 p2)
adamc@52 837 | HOr1 : forall p1 p2, holds p1 -> holds (Or p1 p2)
adamc@52 838 | HOr2 : forall p1 p2, holds p2 -> holds (Or p1 p2).
adamc@52 839
adamc@52 840 Inductive falseFree : prop -> Prop :=
adamc@52 841 | FFTru : falseFree Tru
adamc@52 842 | FFAnd : forall p1 p2, falseFree p1 -> falseFree p2 -> falseFree (And p1 p2)
adamc@52 843 | FFNot : forall p1 p2, falseFree p1 -> falseFree p2 -> falseFree (Or p1 p2).
adamc@52 844
adamc@52 845 Hint Constructors holds.
adamc@52 846
adamc@52 847 Theorem falseFree_holds : forall p, falseFree p -> holds p.
adamc@52 848 induction 1; crush.
adamc@52 849 Qed.
adamc@52 850 (* end thide *)
adamc@52 851
adamc@52 852
adamc@52 853 (* EX: Define an inductive type [prop'] that is the same as [prop] but omits the possibility for falsehood. Define a proposition [holds'] for [prop'] that is analogous to [holds]. Define a function [propify] for translating [prop']s to [prop]s. Prove that, for any [prop'] [p], if [propify p] is valid, then so is [p]. *)
adamc@52 854
adamc@52 855 (* begin thide *)
adamc@52 856 Inductive prop' : Set :=
adamc@52 857 | Tru' : prop'
adamc@52 858 | And' : prop' -> prop' -> prop'
adamc@52 859 | Or' : prop' -> prop' -> prop'.
adamc@52 860
adamc@52 861 Inductive holds' : prop' -> Prop :=
adamc@52 862 | HTru' : holds' Tru'
adamc@52 863 | HAnd' : forall p1 p2, holds' p1 -> holds' p2 -> holds' (And' p1 p2)
adamc@52 864 | HOr1' : forall p1 p2, holds' p1 -> holds' (Or' p1 p2)
adamc@52 865 | HOr2' : forall p1 p2, holds' p2 -> holds' (Or' p1 p2).
adamc@52 866
adamc@52 867 Fixpoint propify (p : prop') : prop :=
adamc@52 868 match p with
adamc@52 869 | Tru' => Tru
adamc@52 870 | And' p1 p2 => And (propify p1) (propify p2)
adamc@52 871 | Or' p1 p2 => Or (propify p1) (propify p2)
adamc@52 872 end.
adamc@52 873
adamc@52 874 Hint Constructors holds'.
adamc@52 875
adamc@52 876 Lemma propify_holds' : forall p', holds p' -> forall p, p' = propify p -> holds' p.
adamc@52 877 induction 1; crush; destruct p; crush.
adamc@52 878 Qed.
adamc@52 879
adamc@52 880 Theorem propify_holds : forall p, holds (propify p) -> holds' p.
adamc@52 881 intros; eapply propify_holds'; eauto.
adamc@52 882 Qed.
adamc@52 883 (* end thide *)
adamc@52 884
adamc@52 885 (* end hide *)
adamc@58 886
adamc@58 887
adamc@58 888 (** * Exercises *)
adamc@58 889
adamc@58 890 (** %\begin{enumerate}%#<ol>#
adamc@58 891
adamc@58 892 %\item%#<li># Prove these tautologies of propositional logic, using only the tactics [apply], [assumption], [constructor], [destruct], [intro], [intros], [left], [right], [split], and [unfold].
adamc@58 893 %\begin{enumerate}%#<ol>#
adamc@58 894 %\item%#<li># [(True \/ False) /\ (False \/ True)]#</li>#
adamc@209 895 %\item%#<li># [P -> ~ ~ P]#</li>#
adamc@58 896 %\item%#<li># [P /\ (Q \/ R) -> (P /\ Q) \/ (P /\ R)]#</li>#
adamc@61 897 #</ol> </li>#%\end{enumerate}%
adamc@58 898
adam@292 899 %\item%#<li># Prove the following tautology of first-order logic, using only the tactics [apply], [assert], [assumption], [destruct], [eapply], [eassumption], and %\textit{%#<tt>#exists#</tt>#%}%. You will probably find [assert] useful for stating and proving an intermediate lemma, enabling a kind of %``%#"#forward reasoning,#"#%''% in contrast to the %``%#"#backward reasoning#"#%''% that is the default for Coq tactics. [eassumption] is a version of [assumption] that will do matching of unification variables. Let some variable [T] of type [Set] be the set of individuals. [x] is a constant symbol, [p] is a unary predicate symbol, [q] is a binary predicate symbol, and [f] is a unary function symbol.
adamc@61 900 %\begin{enumerate}%#<ol>#
adamc@58 901 %\item%#<li># [p x -> (forall x, p x -> exists y, q x y) -> (forall x y, q x y -> q y (f y)) -> exists z, q z (f z)]#</li>#
adamc@58 902 #</ol> </li>#%\end{enumerate}%
adamc@58 903
adam@292 904 %\item%#<li># Define an inductive predicate capturing when a natural number is an integer multiple of either 6 or 10. Prove that 13 does not satisfy your predicate, and prove that any number satisfying the predicate is not odd. It is probably easiest to prove the second theorem by indicating %``%#"#odd-ness#"#%''% as equality to [2 * n + 1] for some [n].#</li>#
adamc@59 905
adamc@60 906 %\item%#<li># Define a simple programming language, its semantics, and its typing rules, and then prove that well-typed programs cannot go wrong. Specifically:
adamc@60 907 %\begin{enumerate}%#<ol>#
adamc@60 908 %\item%#<li># Define [var] as a synonym for the natural numbers.#</li>#
adamc@60 909 %\item%#<li># Define an inductive type [exp] of expressions, containing natural number constants, natural number addition, pairing of two other expressions, extraction of the first component of a pair, extraction of the second component of a pair, and variables (based on the [var] type you defined).#</li>#
adamc@60 910 %\item%#<li># Define an inductive type [cmd] of commands, containing expressions and variable assignments. A variable assignment node should contain the variable being assigned, the expression being assigned to it, and the command to run afterward.#</li>#
adamc@60 911 %\item%#<li># Define an inductive type [val] of values, containing natural number constants and pairings of values.#</li>#
adamc@60 912 %\item%#<li># Define a type of variable assignments, which assign a value to each variable.#</li>#
adam@292 913 %\item%#<li># Define a big-step evaluation relation [eval], capturing what it means for an expression to evaluate to a value under a particular variable assignment. %``%#"#Big step#"#%''% means that the evaluation of every expression should be proved with a single instance of the inductive predicate you will define. For instance, %``%#"#[1 + 1] evaluates to [2] under assignment [va]#"#%''% should be derivable for any assignment [va].#</li>#
adamc@60 914 %\item%#<li># Define a big-step evaluation relation [run], capturing what it means for a command to run to a value under a particular variable assignment. The value of a command is the result of evaluating its final expression.#</li>#
adamc@60 915 %\item%#<li># Define a type of variable typings, which are like variable assignments, but map variables to types instead of values. You might use polymorphism to share some code with your variable assignments.#</li>#
adamc@60 916 %\item%#<li># Define typing judgments for expressions, values, and commands. The expression and command cases will be in terms of a typing assignment.#</li>#
adamc@60 917 %\item%#<li># Define a predicate [varsType] to express when a variable assignment and a variable typing agree on the types of variables.#</li>#
adamc@60 918 %\item%#<li># Prove that any expression that has type [t] under variable typing [vt] evaluates under variable assignment [va] to some value that also has type [t] in [vt], as long as [va] and [vt] agree.#</li>#
adamc@60 919 %\item%#<li># Prove that any command that has type [t] under variable typing [vt] evaluates under variable assignment [va] to some value that also has type [t] in [vt], as long as [va] and [vt] agree.#</li>#
adamc@60 920 #</ol> </li>#%\end{enumerate}%
adamc@60 921 A few hints that may be helpful:
adamc@60 922 %\begin{enumerate}%#<ol>#
adamc@60 923 %\item%#<li># One easy way of defining variable assignments and typings is to define both as instances of a polymorphic map type. The map type at parameter [T] can be defined to be the type of arbitrary functions from variables to [T]. A helpful function for implementing insertion into such a functional map is [eq_nat_dec], which you can make available with [Require Import Arith.]. [eq_nat_dec] has a dependent type that tells you that it makes accurate decisions on whether two natural numbers are equal, but you can use it as if it returned a boolean, e.g., [if eq_nat_dec n m then E1 else E2].#</li>#
adamc@60 924 %\item%#<li># If you follow the last hint, you may find yourself writing a proof that involves an expression with [eq_nat_dec] that you would like to simplify. Running [destruct] on the particular call to [eq_nat_dec] should do the trick. You can automate this advice with a piece of Ltac: [[
adamc@60 925
adamc@60 926 match goal with
adamc@60 927 | [ |- context[eq_nat_dec ?X ?Y] ] => destruct (eq_nat_dec X Y)
adamc@60 928 end
adam@302 929 ]]
adam@302 930 #</li>#
adamc@60 931 %\item%#<li># You probably do not want to use an inductive definition for compatibility of variable assignments and typings.#</li>#
adam@292 932 %\item%#<li># The [Tactics] module from this book contains a variant [crush'] of [crush]. [crush'] takes two arguments. The first argument is a list of lemmas and other functions to be tried automatically in %``%#"#forward reasoning#"#%''% style, where we add new facts without being sure yet that they link into a proof of the conclusion. The second argument is a list of predicates on which inversion should be attempted automatically. For instance, running [crush' (lemma1, lemma2) pred] will search for chances to apply [lemma1] and [lemma2] to hypotheses that are already available, adding the new concluded fact if suitable hypotheses can be found. Inversion will be attempted on any hypothesis using [pred], but only those inversions that narrow the field of possibilities to one possible rule will be kept. The format of the list arguments to [crush'] is that you can pass an empty list as [tt], a singleton list as the unadorned single element, and a multiple-element list as a tuple of the elements.#</li>#
adamc@60 933 %\item%#<li># If you want [crush'] to apply polymorphic lemmas, you may have to do a little extra work, if the type parameter is not a free variable of your proof context (so that [crush'] does not know to try it). For instance, if you define a polymorphic map insert function [assign] of some type [forall T : Set, ...], and you want particular applications of [assign] added automatically with type parameter [U], you would need to include [assign] in the lemma list as [assign U] (if you have implicit arguments off) or [assign (T := U)] or [@assign U] (if you have implicit arguments on).#</li>#
adamc@60 934 #</ol> </li>#%\end{enumerate}%
adamc@60 935
adamc@60 936 #</li>#
adamc@60 937
adamc@58 938 #</ol>#%\end{enumerate}% *)